Modern

MODERN ALGEBRA
Prof. Manindra Agrawal

Computer Science andEngineering
IIT Kanpur
INDEX
S.No. Topic Page No.

WEEK- 1
1 Groups: Introduction to abstraction 01
2 Groups: Subgroups and homomorphism 15
WEEK- 2
3 Groups: Isomorphism 30
4 Groups: Quotienting 44
WEEK- 3
5 Groups: Structure Theorem 57
6 Groups: Applications 69
WEEK- 4
7 Rings: Introduction 84
8 Rings: Failure of Unique Factorization 96
9 Rings: Birth of Ideals 106
WEEK- 5
10 Rings: Ideal Arithmetic 118

11 Rings: Special Ideals 130
WEEK- 6
12 Rings: Dedekind Domains 146

13 Rings: Quotient Rings 157
WEEK- 7
14 Fields 170
15 Cauchy sequences and real numbers 181
16 Properties of Fields 195
WEEK- 8
17 Finite Fields 209

18 Application of Fields 224
Modern Algebra
Department of Computer Science and Engineering
Indian Institute of Technology, Kanpur
Lecture - 01
Groups: Introduction to abstraction
(Refer Slide Time: 00:20)
So, this is a course on algebra, the number is 203B. And in this course, we will essentially
introduce the algebra again to you. I am sure you have all done algebra in some form of other,
used algebra in many forms actually. But, what I am going to talk about is the modern
version of algebra, the way it is used to understand and study different mathematical
concepts. And that is something that perhaps many of you have not seen, may be partly seen,
but not seen as a development from the basics to the higher level and that is what we will do
in this course.
The focus will be on abstractions that how do we look at various concepts, abstract out the
properties of those concepts, and then study properties in the abstract level. What would be
the advantage? Let me just write down. The first advantage of abstraction is that it allows us
to study concepts without unnecessary details. It gets rid of the specific details that are not
relevant, and you can just look at it only at the core aspects.
The second one is it allows us to study common properties of different concepts. You have
multiple concepts and you can abstract their properties out. Some of them will be common.
Once you study the common properties, you can apply it across the concepts.
We will see examples of these later. I am sure you have come across examples earlier as well.
And third one is that it allows us to take out the properties that you have studied in abstract,
and apply it into a completely different context which is another way of saying that it allows
for generalizations. So, there is a specific concept and we abstracted its properties out, study
those properties and now this is the abstract world, so we can apply it anywhere that we wish
to.
So, in a sense what we study would be a much more general version of what we started with.
These are the three very big advantages of abstraction. Of course, there is down side of it
also, which is that it becomes perhaps a little boring or rather too abstract, too mathematical,
i.e the intuition gets lost many times and that you see lot of symbols and manipulation of
those symbols and it is not clear why exactly are we doing it. So, it is therefore very
important to keep in mind the concrete concepts from which we have abstracted out, so that
we never lose sight of what is the effect of our abstract manipulations on those concrete
concepts.
Let us see some examples of abstraction and these are something that you have already, come
across. What are the two most fundamental objects in mathematics? Numbers, yes numbers
are one, clearly. Operators - are an abstraction of when you want to study certain operations
and then you use those, but I am saying not going into the abstract world in the concrete
setting in mathematics, what are the two most fundamental objects it will we like to study.
One is numbers. Second?
Student: Properties.
Properties of various things sure but objects, for concrete objects that you want to study. You
have studied it for long in school.
Student: Variables.
Variables is again an abstraction, use variables to study some concrete things. Well,
geometric objects, curves, planes and so various surfaces that’s also come from the real life,
just like numbers in the sense come from real life. You want to counts certain things. So, you
introduce and use new numbers. Similarly, in real life the whole world is a whole collection
of geometric objects and you would like to study how these geometric objects behave, what
are their properties.
So, with this in mind, that is how the mathematics starts in the ancient times. The study was
done for numbers or about numbers and about the geometric objects. And already the
abstraction particularly in the case of geometric objects was found extremely useful. Take for
example, a circle - that’s a geometric object. Now if I want to study properties of a circle -
one way is to draw the circle and see now geometrically various properties by drawing other
curves. Do you know another way? Something that you have been using in the school very.
Yes, so use compass, that’s geometric way of studying circle and you draw circles, lines and
then see how they interact. Learning geometry, yes, that’s what you did in the school. All
these basic curves you studied using coordinate geometry and what coordinate geometry give
is, for a circle instead of drawing a circle, we write this equation x2 + y 2 = r2 and this
represents a circle. So now, we introduce coordinates here which is the center is at point
(0, 0) and there are any point of the circle is at coordinates are (x, y ) . And, the radius of the
circle is r and then this circle is represented by this equation x2 + y 2 = r2 .
And then suddenly we can study the circle by using this equation. You can differentiate it, get
the tangent, equation of the tangent at various points, you can look at two circles and see
whether they intersect. Whatever you do by drawing, you can do it algebraically. Now this is
already an abstraction, because on its own, this equation is just a collection of symbols. There
is an x , there is a 2 on top of x , then there is a + sign, then there is y , two on top of y and
then etc. What we do is we assign meaning to this and the meaning we assign is that the x
represents the x coordinate value of a point on circle, the y represents the corresponding y
coordinate of that point and the x with 2 on top is again a representation of multiplying x
with x itself and so on.
So, what we did was that we use this abstraction to represent the circle with an equation and
in such a way that all the properties of that circle are inherent in this equation and then we
study this equation. And then you as you would recall that coordinate geometry is extremely
useful in deriving properties of geometric objects. Things that you would not be able to prove
using just the pictures, you can prove using this abstraction. Can you remember or give me an
example of a non-trivial property that you could prove using coordinate geometry. Tangent
to a hyperbola does not intersect what.
Student: The curve
The curve ok, I will believe you, I do not recall this property, but he saying, so that tangent to
hyperbola does not intersect the curve at any other point in it.
Student:
Yeah, there are two parts of the hyperbola, and they it doesn’t intersect. Okay fine. That you
can prove using this. But if you were to prove it using geometry, how would you do that?
You could draw a the hyperbola and draw a tangent at one point and observe that it does not
intersect, but that doesn’t mean that for no hyperbola for any point on the hyperbola the
tangent will not intersect the curve. So, not only you get an easier way of studying property
you get a more powerful way of studying properties by applying this abstraction. So, that’s
the observation that we can prove. So, that is very important observation and that already
demonstrates this power of abstraction.
Now, later on in this course, I will show you further abstraction of the same. There is a circle,
we abstracted it out as equation and then I will go one more step. This equation and the study
of its properties comes under coordinate geometry. I will go one more step and abstract data
out even more and that that field is called algebraic geometry. So, we will use more algebra
to abstract out the circle and similarly other curves and use that representation to study
properties and it turns out that that representation is even more powerful than the coordinate
geometry abstraction. And we can prove even more properties using that abstraction than
with coordinate geometry.
Okay, let’s stay on this picture for little more while. Here you see that equation x2 + y 2 = r2
. This seems like an equation over numbers. Now, there is addition and multiplication in this
equation and addition and multiplication is what we do for numbers. So, that is a curious
things that exists here. Firstly, there is this geometric object that is been seemingly translated
into numbers. In fact that’s what the name coordinate geometry represents that you assign
coordinates and coordinates are collection of numbers to every point and therefore represent a
geometric object as a collection of numbers and then you see properties or relationship
between those numbers that is given by this equation.
Second curious thing about this is that not everything in this is a number. In fact, almost
nothing is a number. x, y, r are not numbers. These are symbols. So, how can we treat them
as numbers. We are seen to be treating them as numbers here. We are multiply x to itself,
adding it to y 2 . So, how can we treat them as a number? Is something not right with this or
are we taking too many liberties with our notations?
Student: There is some abstraction.
There is an abstraction that is happening here is yes, but we need to be very careful when we
abstract by precisely defining the meaning of every symbols that we use in the abstraction.
So, when we say that x, y , r here, they are going to be thought of as numbers, is that
justified?
Student: Set of numbers.
When we are doing abstraction we are at liberty to assign a meaning to any symbol that we
introduce. What we need to ensure is that that meaning is consistent with the operations we
carry out. So, if you say x represents say set of numbers, collection of numbers, then what
does this x2 represents? In what way the collection of numbers of x2 is related to collection
of number of x .
And we say that this collection is multiply every number from one collection to every number
in other collection and is that what we say? That is not what we intent to mean anyways.
Remember our intention is (x, y ) represents, specifically one coordinate point. It represents
many coordinate point but take one coordinate point then x2 is the square of that number in x
coordinate and y 2 takes the square over y coordinate.
The thing is that x and y don’t represent a single number. They don’t stand for a single
number, they stand for a collection number. But that collection of number is also not any
collection. x represents the numbers that correspond to x coordinates of points lying on the
circle. So, what is the meaning that we assign to x here? Do we say that, either we can say
just draw the circle look at the x coordinates of all the points and those represent those are
represented by x here? But that’s really a kind of messy. Firstly somebody has to draw the
circle then assign the meaning or atleast imagine a circle in a mind. Instead what we can say
here is it x represents any number, y represents any number such that collectively they
satisfy this equation, that x2 + y 2 = r2 . Does that make sense?
So, x is some number here, y is some other number here. They need to satisfy this
relationship x2 + y 2 = r2 , r is a fixed number. r is not does not change. So, we can take r
to be 1. Then we say this equation corresponds to collection of all pairs of numbers, which
satisfy this - for x If you substitute the first of the pair, for y substitute second of the pair,
then this equation must be true. So that’s the interpretation we assign to this abstraction. But
again it is not yet very precise, because I said x can be in any number and y can be any
number and such that they together satisfy this equation.
But when I say number what do I mean? Do I mean integers? Do I mean rational numbers?
Do I mean real numbers or do I mean complex numbers? There are different types of
numbers also. What would you suggest we should treat x as? What kind of number? Real
numbers? Yeah, that’s a natural way of thinking, because this geometric object this circle we
are thinking over a real plane. So, we can therefore, think of x and y as a real numbers.
That’s good. But suppose I want to study the rational numbers lying on the circle, that is
points with have rational coordinates, which lie on the circle.
Then we can say that the it is all the rational numbers x and y , which satisfy this equation
are the points of interest to us, which mean that we are saying that now we no longer think of
x and y as real numbers, but think of them as a rational numbers. The equation remains the
same just that we are assigning a different meaning to x and y and we can change the
meaning may be some other time we want to study all complex number satisfying this
equation, so we change the meaning to complex numbers or something else. So, that
abstraction remains the same, but the meaning we assign to it changes which is very good
thing because then if we study the properties of this in abstract domain, then no matter what
meaning we assign to this eventually, those properties will continue to be true. So that that is
an important observation.

This is not hundred percent true because each meaning will have its own quirks also. For at
least the common properties we can handle simultaneously. So, its means we would like
ideally to study this equation without assigning meaning to x and y . I have at least hopefully
managed to convince you that that’s a good idea. Let’s not think of x is a rational number, let
us not think of it as real number. That we will do later when we require, as and when
required. To begin with in a real abstract sense I will not assign any meaning to x and y . It is
a dangerous statement that I making because if you do not assign a meaning to x and y ,
what is it meaning of x times x , how can you add x and y ?
Student: (Refer Time: 26:49).
That’s one way of doing it, but then you have to do it at least three examples that I give
rational, reals and complex. For each of them you have to then study it separately.
So, each addition may have slightly different properties. So, then you will have to worry
about those peculiarities of each meaning. Ideally in abstraction we would like to get rid of
the specific quirks and you just like abstract it out so that we can study objects, which are
applicable across multiple meanings, which in this case we would be able to do if we don’t
assign meaning to x and y . The only hitch there is that how do we define addition and
multiplication for x , for y for a combination of these.
Student: Let’s assume some properties of addition.
About.
Student: Addition.
Addition, yes of course that is a solution, yes, you have given it.
So, let’s attack in a more direct fashion, the concept of numbers itself. What is a number?
Okay, who can tell what is a number is?
Student: Any object that is countable.
Any object that is countable that is a number, but there are uncountable reals which are
uncountable, so that is not a good enough definition.
(Refer Time: 29:24).
By countable?
Either power set of count countable (Refer Time: 29:36).
Okay so, let’s may give another argument against this. Let’s pick up a collection of countable
objects, which is may be collection of all stars in the galaxy. Are they numbers? Yes
countable, surely. Are they numbers? Can you add two stars and get a third star? How about
multiplying two stars? That’s not a good definition. Let’s try to cover with a better definition
of a number.
Student: It is an ordered set with addition and multiplication.
That is a good start. See what we really want to do with numbers is arithmetic on it. We have
four fundamental arithmetic operations addition, subtraction, multiplication and division. We
would like to do these operations on numbers. Now of course for certain collection of
numbers not all four of these operations are possible. For example, for integers, division is
not possible. Addition, subtraction, multiplication is possible. But some others for example,
rational numbers division is also possible. So, leaving a side this slight distinction we would
like to do arithmetic with numbers.
So, let us define numbers as the collection of objects on which we can do arithmetic. But then
we have to now define what does it mean to do arithmetic and you pointed already one
important property of doing arithmetic - the closure; that is if I add two objects I should get a
third object from my collection. Now, we are dealing with numbers as just object, we are not
assigning any other meaning to them. So, we just have to say what it means to do arithmetic
on those objects. So, let us try to write down the properties we want of those objects.

So, this is a formal definition of the closure property, that capital A is the collection of
objects that we are going to think of as numbers. The first requirement is that this collection
of objects must admit the addition operation, which I am going to represent with the symbol
+. This satisfies the following properties - the first one is the closure property that for any a
and b in the collection a + b must be defined and must be a another object in the collection.
Okay? What other property should addition have?
Student: Associativity.
Associativity, excellent, that for every a, b, c you want to add all three of them. So, it does
not matter in which order you add them. Anything else? Commutativity yes, a + b is same
as b + a . There are more properties that are required that hold for typical numbers that we
have in mind. So, we should abstract those properties out as well. One very important of
these properties is this very special number 0. It has a property that if you add 0 to any other
number, the number does not change and the 0 is called the identity of addition.
So, when you add 0 to a , you get a itself and finally, that is negative numbers or this
property allows you to define subtraction. Just addition is not the sufficient to operate a
numbers, you will also like to do subtraction. Subtraction is simply when you say a − b we
are adding a with − b and what is − b ? − b is that number, which you when add to b you
get 0. So, the fact that − b exist is guaranteed by 5th property.
So, we need this property in order to define abstraction in order to define subtraction. And in
terms of abstraction, we call this property the inverse property - the existence of inverse of
every element. So, these are the 5 properties we require in order to define addition and
subtraction. So happens that this is the comprehensive set of properties that the addition
operation satisfies for numbers.
And now with this written down, we can define numbers to be any collection with an addition
operation - subtraction comes for free which satisfies all these 5 properties. Of course, this is
not quite there because a numbers also have multiplication and division and I am not using
that. So, we should also have division and multiplication definition. But for now let us just
stay on these properties and I will introduce the multiplication, division slightly later to have
the numbers. And the reason why I am stopping here for now is that this properties for a
collection is already a very interesting abstraction.
This collection with the addition operation defined as we just did is called a commutative
group. The name has its own the history. Let’s not get into why is it called a group, but the
key thing or important thing is that groups are an abstraction, which are extremely useful. We
started with numbers, in order to abstract that we to came to groups. But like I initially had
suggested that once you do an abstraction, you can apply to completely different domains and
that is true with groups as well.
Let me give you an example of course, the obvious examples of groups you already know.
Integers with addition operation is a group. That’s how we started actually. Not just integers,
you take rational with addition, reals with addition, complex numbers with addition. These
are all groups for obvious reasons commutative group that is. By the way I should have said
that if you drop property three which is the commutativity property ( In case of numbers we
don’t need to drop it that this is definitely satisfied by numbers, but since we have would like
to generalize this notion and apply it in other domains also there are situation where this
property doesn’t exist) and the remaining property are satisfied by a collection, in that case
the collection is called simply a group. So now, come back to some examples. So, these are
the obvious examples coming from numbers. Let me give you an examples for groups, which
do not come from numbers. Okay, any of you have a suggestion. Have you come across
groups some time earlier?
Student: Matrices.
Matrices, the excellent example; yes, so let’s say n × n matrices under addition - that is a
group - the identity element is the all 0 matrix. You can add. It is a commutative group. Also
the other properties are easy to see that they are true. Okay. More examples.
Student: Permutation.
Permutation, yes, collection of the permutations. Let’s say look at the permutations on
numbers 1 to n . Each permutation by definition is a mapping from 1 to n to itself, which is
a one-one onto mapping. So, what is the addition operation for the permutations? How do
you add two permutations?
Student: Compose.
Composition of two permutations is a permutation. So, let’s just quickly go through all the
property. Closure is true. Associativity is true again it is very easy to see. It is not
commutative. If we take two permutations, and it matters in which order you apply those
permutation. ϕ1 ° ϕ2 is not necessarily equal to ϕ2 ° ϕ1 . ϕ1 may map 1 to 2 and ϕ2 may
map 2 to 3. So, if you apply first ϕ1 and then ϕ2 then you will map 1 to 3.
On the other hand, if you apply ϕ2 first, ϕ2 may map 1 to 7; and ϕ1 may map 7 to 20, if
you apply first ϕ2 and then ϕ1 , then 1 is will goes to 20. So, it’s definitely not commutative,
but identity exist. The identity maps 1 to 1, 2 to 2, 3 to 3, because we compose it with any
other permutation you get back that same permutation. Inverse? Inverse also exists. Inverse
of a permutation is just an inverse map - if we compose an inverse map you get the identity
permutation.
More examples. Q* which I use to denote all nonzero rational numbers under multiplication
operation. That is a group. Why is that a group? Closure holds, associative holds,
commutative holds. Identity? 1 is the identity. Inverse? 1 by a . So, if a is a rational number,
1 by a is also a rational number and that is the inverse.
So, this is actually a multiplication operation, which is different than addition operation for
numbers. But if we view it in that abstract fashion, it is same as the addition operation. This is
already a remarkable fact which was not at all evidence and this becomes evident only when
we abstract out the properties of addition, abstract out properties of multiplication see they
are the same.
Student: Multiplication is repeated addition.
Multiplication is repeated addition true, but the fact that properties would remain the same is
not at all clear. For example, if you start with integers. There also multiplication is repeated
addition, but integers under multiplication do not form a group, because the inverse does not
exist, but over rational they do form a group. So, this is a good time to close, because we
have defined groups, we have given some examples of groups, and already thrown up certain
an unexpected fact.
So, what I would like you to do is go back, think about it. Tomorrow we will meet again at
12, and will continue the discussion.
Modern Algebra
Lecture – 02
Groups: Subgroups and homomorphism
We looked at the definition of a commutative group and also a group, which I have not
written down here.
But I assume that it’s very clear that except for the property number three here, which is
commutativity, all other properties that are satisfied, which is the case in for example three
here. All other examples that we looked at yesterday were commutative groups. Now, in this
course, we will primarily focus on commutative groups. There will be occasions when I talk
about non commutative groups or general groups. But mostly, we will stick to commutative
groups and that is why I have given the definition of commutative group first. Now, of the
two examples given yesterday, the first one, which is numbers under addition, is a natural
obvious example of a commutative group, that’s how we actually derive the properties of the
group as well.
But if you look at the fourth one, that’s already a bit of a surprise, because it is a group of
nonzero numbers under multiplication. It’s a commutative group because it satisfies all the
properties as we had listed down. But the nature of these two operations - addition and
multiplication seems very different. And what we have learned from this abstraction is that
while on the face of it these two operations look very different, at the certain abstraction level
they are the same and that’s a first important learning we have. And already we have seen the
benefits of abstraction, now we can actually look at both the operation with the same
perspective.
Now that we have defined what a group is, there are two directions in which we can proceed.
First one is to continue with our abstraction process, which we started with numbers. Right
now, we are only at a halfway stage, because we have just abstracted out the addition
operation property (or multiplication operation property as well). But we have not talked
about the entire arithmetic. Remember these numbers admit both addition and multiplication
operation, and we don’t yet have a definition of a structure in an abstract way, which admits
two operations, addition and multiplication. The other direction would be the way to just start
focusing on groups and see what are the properties of groups. So, what I am going to do is
very quickly I will do the full arithmetic abstraction. Then we will shift our attention to
groups, and then later on once we are done with groups, we will move on to the full
arithmetic abstraction.
Let me first abstract it out and give a definition. So, for a full arithmetic to happen over a set
of numbers, we have to define two operations - addition, multiplication and then of course,
addition comes in conjunction within subtraction, and similarly multiplication comes in
conjunction with division.
Now with division, we have already observed in certain set of numbers division is not
possible. For example, in integers division is not possible. So, with division, there is always
this bit of, let us say carefulness that is needed to distinguish cases where division is possible
and where division is not possible. So, let’s first define a set up or a structure, which may or
may not have division, which is more general types of numbers which you know span
everything. So, what are the other properties that we need, just like we identified for groups,
we had these five properties for addition, surely when we are talking of two operations and
addition should still retain those properties.
And now let’s define a structure. So, I will have two operations: addition and multiplication
such that A is a commutative group under addition that is again borrowing the property from
numbers. Now, let us bring in multiplication. We should not say that A is commutative group
under multiplication although there are certain types of number where it is true, but we do not
want to say that yet. So, let us look at integers and what are the properties that multiplication
satisfy for integers. If we just go backward to the group properties, we will see that A is
closed under multiplication.
Then the associativity holds for multiplication, multiplication satisfies commutativity

property. Then, there is the identity also present. Again I am simply picking up properties
from the definition of groups. The identity is 1 for multiplication. The only one that I have
left out is the inverse, because inverse again may or may not exist. So, let’s leave out inverse,
because we want to capture numbers in general. Is there anything else?
Student: Distributive(Refer Time: 08:19)
Yes, because multiplication and addition interact in a certain way and that we have not yet
captured. And that’s captured by a very simple property called distributivity, which is that;
for any three elements a, b, c in A , a * (b + c) is same as (a * b) + (a * c) . That’s again very
obvious property that multiplication and addition satisfy and that we want to carry over or
abstract out. This is called distributivity property. And that’s it. That is the structure with of
two operations, which mimics the arithmetic operations over numbers.
So, structures like this which has now addition and multiplication is called a commutative
ring. Again the commutative prefix is because multiplication satisfies commutativity
property. By definition in a ring, addition is always commutative. it’s only multiplication
that may or may not be commutative and if it is commutative, the ring is called a
commutative ring. If it is not commutative it’s just called a ring. And just like with groups we
will almost always be concerned with commutative rings in this course, so that’s why we are
defining it first. Any questions on this?
Student: If you are (Refer Time: 11:00) then it is valid.
Well, in certain, you are right. In certain more general definitions of a ring the multiplicative
identity may not exist, but we will not define it in that general sense. So, we will always
assume that multiplicative identity that is 1 exists. So, this abstraction tells us that set of
integers under addition and multiplicative operations are a commutative ring.
So, is the set of rational numbers, real numbers, complex numbers etc. They all are
commutative rings. Some of them even admit division. So, that’s a next step. Our next
definition that I want to define is this structure, which is a commutative ring also having an
inverse for every nonzero element - inverse under multiplication operation, which is another
way of saying is that for every a in this collection which is nonzero, there exists b , such that
a * b = 1 . This defines division because 1/a is b .
In that case, the structure is called a field. Examples of this - whole set of integers under
addition; multiplication is a ring is a commutative ring. If you look at rational numbers under
addition multiplication, that’s a field. So are real numbers under addition multiplication and
complex numbers under addition multiplication. We are going back to yesterday’s example
set of n × n matrices under matrix addition and matrix multiplication.
What structure is that? We saw that under addition it’s a commutative group? How about
multiplication. Does it, let’s see. Closure property is it there? Associativity? There we have
seen that a * (b + c) = (a * b) + (a * c) or whichever way you multiply, you get the same
result. Commutativity? Not necessarily true. Identity? Yes the identity matrix is the identity
element.
Then finally, the existence of inverse is not necessarily true. A matrix may or may not be
invertible. So, this is a ring. Not a commutative ring, it is a ring. By the way, why have I not
included inverse of 0? When I defined a field, I said that for every nonzero element a has an
inverse. What about 0? Well for numbers certainly 0 does not have an inverse, there is 1 by 0
is infinity, which is not a number. But that does not rule out in general when we have
abstracted it out. So, could it be possible that there exists a set of different set of numbers
with arithmetic on them such that there is a multiplicative inverse of 0.
Student: (Refer Time: 16:43) What if we are able to prove that 0 × a = 0 , if that is true
then.
0 × a = 0 , that is another interesting observation. Is it true it is true for numbers, but is it

true for a field or a commutative ring.
So, let us prove a theorem - our first theorem. We can prove both the properties. Let’s try to
prove the first one. a × 0 = 0 for every a . How do you prove? This is a good exercise,
because while we can try to take the intuition from numbers we cannot be working on
numbers here, we are just working with symbols with a certain properties. So, we have to
keep that in mind. How do we prove a × 0 = 0 . It is very simple. Let’s say a × 0 = b ,
then b = a × 0 .
By definition of 0, 0 = a − a . Now use distributivity property of multiplication over

addition. See although I have written it here a − a it is really a + (− a) . So, I can distribute
this multiplication. So, that is a * a + a * (− a) .
Now, what is a * (− a) ? Is it − (a * a) ? For numbers of course, yes. Now these are not
numbers of the kind that we are familiar with. So, we will have to prove this. Can I write it
this way? Is − a = (− 1) * a ? Should be, but is it. Or you see it’s obvious. Is it obvious?
Nothing is obvious. We have to prove everything. It is like learning arithmetic in class 1. We
got to redo that learning because although these structures are similar to number they are not
really numbers of the kind we know of. So, let’s therefore, look at (− 1) * a and I want to
prove that it is − a . So, add a to it, this is (− 1) * a + 1 * a that’s because 1 is the identity of
multiplication.
Now by distributivity property this is (− 1 + 1) * a and that is 0 * a and what is 0 * a that is
what we are trying to prove. So, we have cycled back to that original question. So, we are still
stuck - if we can prove one of them we will prove the other one, but this way we cannot
prove any of them, because one property is dependent on the other property. So, we need to
come up with an alternative way of proving this. So, this is an excellent assignment problem.
So, instead of proving it here, it will be very useful if you prove it yourself - just play around
with these symbols. You have all the properties at your disposal and try to prove this.
But we will take the statement of the theorem as given to us and which means that the
a × 0 = 0 . So, which means that once you prove one, two is pretty straightforward second
property that a * 0 =/ 1 . So, 0 has no multiplicative inverse in the collection. Which is why
in the definition you know I have to explicitly write that multiplicative inverse exists only for
nonzero elements. So that’s commutative ring and field. We will now not talk about rings and
fields until some more time. We will pick this thread up once we are finished with groups,
because in order to understand rings and a fields we first have to understand groups, because
under addition there is a group all sitting in both of that. So, let us try to understand groups
first and their properties.
Now, for the study of groups let’s try to or let’s adopt certain notation. So, the structure that
we will represent as groups typically we use the capital G to denote the set of elements or
objects and the group operation, which earlier even I gave the definition I used “+”. Now, I
am going to use “ · ” as the representative group operation. This has certain logic to this.
Because generally the addition operation in whatever context we use it is commutative, but
multiplication operation depending on context may be commutative may not be commutative.
We will see an examples of it - composition which is also denoted as multiplication or matrix
multiplication is not commutative.
So, when we talk of a general group which could be commutative or not commutative, it’s
better to use a multiplication symbol as its operation and that’s why we have we write groups
multiplicatively more often than we write them additively. So, let (G,·) be a general group.
Now when we want to study a structure what exactly should be we looking for? One very
important aspect is the structure of the group. How really does it look like?
Once if we have understand the structure of the group then we can derive the properties of the
group a lot more easily. But what does it mean to say the structure of a group? Well, looking
at the structure of integers you can visualize them as lying on a number line equally spaced
and there is a each next point can be achieved by adding 1 to the previous point. So, that
basically denotes the structure of that particular group.
So, in general if G is a group we would like to understand what is the structure and one very
important aspect of that is to see if within the group G , if there are more groups and that
question is important enough to have it is own definition. So, let’s define that. Take any
subset H of G - it is a subset of elements of G and we call this subset is subgroup of G if H
under the same · operation is a group. Examples. it is always good to have a certain
connection of these abstract entities with actual ones. So, for groups it is good to keep two
example groups in mind. One is numbers under addition, the other is numbers under
multiplication and they both are groups, but both have a very different structure as we will
see as we will see later on. Okay. So, examples? Okay, in integers over addition, are there is
subgroup in that?
Student: Even integer.

Even integer yes, excellent yes. 2Z is a subgroup. How about numbers under let’s say
rational numbers under multiplication? You see as a group there?
Student: Positive rationals
Positive rationals is a subgroup under multiplication that’s right; Even Q* is a subgroup where
Q* is defined as a set of nonzero rational numbers. So, we do have subgroups and in fact,
these are not the only subgroups that you can find in these groups. You can have 3Z is also a
subgroup, 4Z is also a subgroup under addition. For rational numbers under multiplication,
you can again create lot of other subgroups. Again let’s see more examples - let’s say this set
{2n | n ∈ Z } , is a is a subgroup of rational numbers under multiplication. Do you see this it
is straight forward - you can multiply any of these two elements it’s closure and all other
property they are also satisfied. So, subgroups of a group tell us a lot about the structure of
that group.
In fact, there is one very important way of writing a group in terms of it is subgroups, because
just imagine the following. You have a group, you identify a collection of subgroups of that
group. Each subgroup is in a sense smaller than the group itself and you keep identifying
further subgroups of that - it is possible that this process can continue forever and it is also
possible that this process stops after sometime. In either case you have a very a nice hierarchy
of subgroups of that a group and that structure of that hierarchy will tell us the structure of the
group itself. So, one very nice way that this hierarchy can be defined is when I can write a
group as a direct product of two subgroups. Let me define that.
So, lets G be a group and H 1 , H 2 be subgroups of G . And now I am going to define what
is called the direct product of H 1 and H 2 , which is written as H 1 × H 2 . It is simply a pair of
elements (a, b) where a ∈ H 1 and b ∈ H 2 . H 1 × H 2 is a group under component wise
multiplication.
The elements of H 1 × H 2 are pairs, so given two such pairs you multiply them component
wise and there is a natural way of doing it because first component belong to H 1 and second
component belongs to H 2 and there is a multiplication operation defined both on H 1 and
H 2 . The nice thing or interesting thing is that this is a group. Why? You can quickly run
through all the properties. Closure trivially there, associativity trivially there because
component wise component wise it is basically we are borrowing properties of H 1 and H 2
and if H 1 and H 2 are commutative then so is this or if they are not then no. Then identity,
what is the identity.
Identity is.
Student: Identity of a pair (Refer Time: 36:55).

Yeah the pair (1,1) which is the identity of H 1 and identity of H 2 , which both happen to be
the same, because they are both subgroups of G. Inverse property is again borrowed from H 1
and H 2 . In-fact everything is just borrowed from H 1 and H 2 . It is clearly, therefore a group.
This in itself does not tell us something very interesting. Just we have seen a way of creating
a new group from given two groups using this direct product. What is interesting is the
following, that is if G can be written as H 1 × H 2 , by that what I mean is every element of G
can be uniquely identified with an element of H 1 × H 2 ; such that when we do the operation
on elements of G that is exactly the same operation that we do on H 1 × H 2 .
So, we already define a group operation on H 1 × H 2 - that operation and the operation of
elements on G they are identical. You see what I am trying to say. Let’s do an example.
Let’s say consider Q* under multiplication and we define H 1 to be {2n |n ∈ Z } and H 2 to
be all rational numbers a/b in Q* where a, b ∈ Z , 2 ∤a, b . Of course, I have to say a, b
not zero also. So, H 2 is all those nonzero rational numbers such that 2 does not occur in the
numerator or denominator as a multiplier, so that is both a and b are odd numbers.. Is this a
subgroup of Q* under multiplication?
Student: Inverse is
b/a .
Student: Yeah
Of the same form odd by odd?
Student: Yes, then identity is 1 by 1.
1 by 1 again same, correct.
Student: And other properties are basically.
The key thing is the closure - a/b · a′/b′ , it’s just that aa′ is odd, so their product is odd. bb′
is product is odd and therefore it is an element in H 2 . So, this is also a subgroup. What is
H 1 × H 2 ? It is a pair of collection (2n , a/b) . Now, comes the key point. Look at any element
of Q* . I can write it as c1 /c2 , where c1 , c2 are integers. Now take out the powers of 2 from
c1 and powers of 2 from c2 , so I can write this as 2m c1 ′/c2 ′ , where c1 ′, c2 ′ are odd and m
is any integer, positive or negative, whatever the case may be.
So, this is in 1 to 1 correspondence with (2m , c1 ′/c2 ′ ) . See this view allows us to transform a
given element of Q* to the corresponding elements of H 1 × H 2 and vice versa. If you are
given an element of H 1 × H 2 , just multiply the components and you get an element of Q* . If
you get an element of Q* , you would do this process of separating out powers of 2 and then
that splits as an element of H 1 × H 2 . So, this correspondence I claim is 1 to 1. Do you see
that? If you have two numbers c, c′ maps to the same element; that means, they are powers of
2 are the same and the odd component are the same then numbers have to be the same.
And similarly, the reverse way. If two elements of H 1 × H 2 map to the same product then
since the product have separate powers of 2 and odd part again they will both be the same
numbers. So, it is a very simple correspondence, which shows that Q* can be viewed as
H 1 × H 2 . One more thing. The group operation on Q* , we should also ensure that it is same
as the group operation on H 1 × H 2 . Group operation on Q* is the multiplication - when you
multiply two numbers of Q* you multiply the powers of 2 of the two numbers then the odd
part also you multiply them, which is exactly what the operation on H 1 × H 2 is.
So, this demonstrates a complete correspondence between Q* and H 1 × H 2 and that is the
situation when we say that a group like Q* can be factor or returned as a product of two
subgroups. So, let us formulize this. This is just an example to motivate this definition which
I am going to give now. The key thing is this correspondence. How do we formally state this
correspondence between the group and the product H 1 × H 2 and that is done through the
notion of an isomorphism.
In fact, I’ll do it in two steps. Let (G,·) and (H,*) be two groups. We say that given two
groups G and H under their own of group operations, a mapping ϕ taking elements of G
to elements of H is a homomorphism if the following property hold. And the property
simply says, that the mapping ϕ respects the group operations. ϕ(a · b) where · is the group
operation of G is same as ϕ(a) * ϕ(b) , where * is the group operation under H.
So, it does not matter whether we have first do the group operation in G and then apply ϕ or
first apply ϕ and then do the group operation of H . This is also in a way saying that ϕ
preserves the group operation while moving from G to H . Then comes the second
definition. In a homomorphism it is not necessary that the mapping ϕ is a 1 to 1 onto
mapping. ϕ is an isomorphism if ϕ happens to be a 1 to 1 and onto mapping from G to H,
then it is called an isomorphism.
So, it is a isomorphism is a very special kind of homomorphism and it is the isomorphism

that we have been talking about earlier when we say that G can be written as H 1 × H 2 , what
I really formally want to say there is a G is isomorphic groups H 1 × H 2 . Because two
isomorphic groups are essentially the same groups. You can put elements into 1-1
correspondence and the group operation is preserved, whatever is the group operation in one
side is the group operation on the other side.
We’ll write ≅ and this designates that the two groups are isomorphic to each other. If we can
write G ≅ H 1 × H 2 , then tells us something very nice about the structure of G , that I can
actually factor G as two subgroups.
Okay, time to stop today, and then we will continue tomorrow.

Modern Algebra
Lecture - 03
Groups: Isomorphism
Yesterday, we saw that we can split some groups into subgroups and write them as a direct
product of two groups.
And, we saw one example, which was Q* , which we split as specifically two sub groups;
where the first one was just powers of two and the second one was all the rational numbers
where numerator and denominator are odd numbers.
Now, one thing that I would like you to observe, which is also an interesting phenomenon is
that if we consider the subgroup H 1 - it is all the powers of two: positive and negative. This
is in itself a group under multiplication. Does this group look familiar to you? Well, when I
say familiar, of course, it is powers of 2. These are very familiar numbers but is it a group
that you have already encountered before in this course? It is sort of addition, you think so?
Why?
That’s right. I mean if you see in the exponent they are integers- positive and negative. And
the group operation, which is multiplication is the addition operation of numbers in the
exponent. So, at least if you think of this group as the operations happening in the exponent,
it is simply the group of integers under addition. But it seems it’s over. It’s not precisely over.
We need to have a more precise correspondence between these two groups if we really want
to claim something about them. And thankfully, the notion of isomorphism is just the one that
we need.
The group which consists of powers of 2 under multiplication is isomorphic to the group of
integers under additions. I should say (Z,+) . Well, for the proof I just need to exhibit a
mapping from H 1 to Z or conversely from Z to H 1 , which is one-to-one onto and
preserves the group operation. And, that’s simple; at least Z to H 1 is easier to describe –
ϕ(n) = 2n . This is a one-to-one onto map from Z to H 1 . Does it preserve the group
operation? Consider ϕ(n1 + n2 ) is by definition 2n1 +n2 and that is 2n1 * 2n2 . This is
ϕ(n1 ) * ϕ(n2 ) . So, it preserves the group operation. It is one-to-one onto map and that’s
precisely the definition of an isomorphism of groups.
So, when two groups are isomorphic, we really have the same group really. The only
difference is the way the symbols are used is different. So, the moment we realize that H 1 is
isomorphic to Z , we can go back to where the H 1 came from and we can write Q* as
isomorphic to Z × H 2 ; where, H 2 consists of rational with odd numerator and denominator.
That’s already a very interesting fact that we found the whole copy of Z in Q* . In fact, we
can find more copies of Z and Q* . If you look at H 2 and take out all powers of 3 from H 2 -
if we define H 3 as all rational numbers, where numerator and denominator are neither
divisible by 2 or by 3; so, this is taking out all powers of 3. So there is another group, which
we have taken out, which is all powers of 3. That is also isomorphic to Z for the same reason
that all powers of 2 are isomorphic to Z . Then we can write Q* as isomorphic to
Z × Z × H3 .
Now, we can continue this exercise. We can replace H 3 by Z × H 3 – you can just continue
with this. So, that’s an interesting fact that once can write Q* as copies of Z multiplied with
each other. The next question is how about Z itself? Can we write Z further? Can we
further divide Z into two groups, a product of two groups rather? And, the answer to that is
no.
And I should say just to be very correct, there are no non trivial groups. I can always define
G1 to be Z and G2 to be just the identity element. And then, obviously, it is a product and
that is really not saying anything interesting.
And this is also quite a remarkable theorem, which says that we cannot write Z as a product
of two sub groups. Proof is very easy actually. And, that proof itself will lead us to something
interesting. Let us prove it by contradiction. Suppose there is an isomorphism. Consider
where does the number 1 go. The number 1 is will be mapped by this mapping ϕ to some
element of G1 × G2 . So, let us say it is (h1 , h2 ) . Then, where does 2 go to?
Student: 2 is 1 plus 1
2 is 1 plus 1. Yes, that’s right. And, because ϕ is an isomorphism, this is same as ϕ(1 + 1) .
And, that means, it is (2h1 , 2h2 ) . Here I am taking some liberty with the notation – 2h1 is
h1 + h1 . And, in general, ϕ(k) would be (kh1 , kh2 ) . And this describes the entire range of
ϕ because if you look at Z , there is just all possible values of k – positive and negative
integer and that’s mapped to these elements.
Now, ϕ is an isomorphism. So the range of ϕ is entirely G1 × G2 . So, that means in this

range, (kh1 , k h2 ) - these elements cover the entire G1 × G2 . Is that true? Is it possible? See
ϕ(1) = (h1 , h2 ) , ϕ(2) = (2h1 , 2h2 ) . Now the fact that h1 , h2 are in G1 , G2 means that 2h1
and 2h2 are also obviously there. Since h2 ∈ G2 , (2h1 , h2 ) ∈ G1 × G2 . So, it must occur in
the range of ϕ . So, (2h1 , h2 ) is (kh1 , k h2 ) and then, this is your two group elements. You
subtract them and get ((k − 2)h1 , (k − 1)h2 ) is 0. Is that possible?
The first thing is k can be either 1 or 2; it cannot be simultaneously both. So, it can only
make one of the two components 0.
So, assume without loss of generality, assume k is not equal to 1. Then, (k − 1)h2 is 0. This
implies that the element h2 is very special. You add h2 to itself a few times you get 0. So,
now let’s see what is the simplest way of showing that, we will get a contradiction. Look at
ϕ(k − 1) ; that is going to be ((k − 1)h1 , 0) . And, how does this helps us. Any suggestions?
Student: Groups G1 and G2 will be finite.
Not necessarily.
Student: If we add the, see G1 consists of only h1 .
Multiples of h1 . G1 considers only multiple of h1 . G2 consists of only multiples of h2 .
Student: And after a point
And after a point h2 multiples becomes 0. So, that’s right. So, G2 is finite group.
Student: Even G1
But, is that a contradiction?
Student: If both of them are finite.
If both of them, then there is contradict; but if G2 is finite and G1 is infinite, then?
Student: One-to-one relationship is violated.
One-to-one relationship is violated?
Student: Because integers are
Why?
Student: (Refer Time: 16:25) Sir, we were saying that G1 is infinite, G2 is finite.
G1 is infinite, G2 is finite, yes.
Student: So we can think h1 into h2 .
h1 into h2 ?
Student: Earlier taken example. (2h1 , h2 )
(2h1 , h2 ) .
Student: So, we can take h1 into h2 .
No, we cannot multiply them. G1 , G2 are two groups when we are just looking at a product
of the groups.
Student: (Refer Time: 17:18) h1 and 2h2 are also a member of G1 and G2
h1 and 2h2 are members of them, right?
Student: So, we will proceed and then we will get 0
Let’s not spend more time on this. Work it out. This is a simple assignment problem. I am
just losing my way somewhere here. It is a very simple proof. So, that tells us basically that
Z cannot be divided further into two smaller subgroups. So, in some sense, Z under addition
is an individual group and Q star is not. So, why is that happening? Why it is that Z is
indivisible? The way this proof goes that should give a hint. We looked at where does 1 go to
because once we know where 1 goes to, all other elements of Z can be decided. So, that
leads me to the definition.
So, we say an element g of a group G is a generator of the group if the entire group can be
written as in terms of g . So, here I am writing group G as multiplicative. So, I am taking
powers of g and writing G as set of all powers. And, that’s the example that we have
already seen. 1 in Z is a generator. Does Q* have a generator? That is, does there exists a
number whose different powers generate the entire Q* . No. On the other hand, 2n is of
course has a generator, which is 2. So, the existence of generator is a difference, which is
making Z indivisible; whereas, Q* is not. And, this observation can now be extended
further.
To a more general definition - group G is called a finitely generated group if there exists
finitely many elements in G – g 1 , g 2 up to g k such that every other element of the group can
be written in terms of these elements – g 1 to g k ; and, this is the general form of the various
elements written in terms of g 1 to g k . Here α1 , α2 to αk are integers. So, what are finitely
generated groups? Well, of course Z plus is finitely generated. It has only one generator. Is
Q* finitely generated?
Actually then number of generators are in Q* is exactly equal to number of primes and that is
infinite. So, the Q* is not finitely generated. The restriction of Q* , where we can say – let’s
say all a/b , where a, b only consists of primes up to r . This is finitely generated, because
now only you have finitely many prime numbers with which all the numbers are written and
that is finitely generated.
Any other example of finitely generated groups you can think of? It is not easy to find out an
example of finitely generated group. There are many, but most likely you have not come
across that. So, let me give you another example without giving more details about it. Maybe
later on if we have time, we will come back to this example. Look at this cubic curve
y 2 = x3 + Ax + B with 4A3 + 27B =/ 0 . These are very technical conditions. And, look at
all rational points which lie on this curve. When I say rational point, this obvious meaning is
that, coordinates of those points must be rational numbers.
Student: Both coordinates?
Both coordinates. And, one can show that there exist infinitely many such points. They form
a group under the certain addition operation. And, that group is finitely generated. But, to
show this is a very nontrivial exercise and it took very long time for mathematicians to prove
it. So, recall that we started with the aim of understanding the structure of groups. And, all
this exercise you have done so far expressing it as a product of smaller groups, etcetera is
geared towards it. Now, I am going to give you a big structure theorem, which is very
general. It applies to all finitely generated groups. And, it completely describes this structure
of these groups.
Let G be any finitely generated group. And, in this course, I have already said yesterday, we
will discuss commutative groups. So, whenever I say group, I mean commutative groups.
This theorem does not hold for non-commutative groups; it is only for commutative groups
we are talking. If I talk about non-commutative groups, I will say I am talking about
non-commutative groups; otherwise, when I say group, it is a commutative group. So, G is a
finitely generated group. Then, there exists two numbers – integers n and m such that G is
isomorphic to Z × Z × ... × Z × G0 ; where the number of copies of Z is m and G0 is a
very small group. It is a finite group with n elements.
This is describing the structure of any finitely generated group and it is complete description,
because Z – we have already seen we cannot further split it and the structure of Z is very
simple. There is just one element 1 that generates this entire group in a very simple way. So,
really we cannot have a simpler group than Z .
And, there are m copies of Z present in it. And then, there is a tiny bit of extra finite group,
which sticks out. We will not prove this theorem; I will just give it to you to show that these
abstraction and then going through all these analysis, we thus have some very interesting
consequences that we can prove a structure theorem like this and which is applicable to wide
variety of groups coming out of various domains. And, once we have proved this theorem, we
do not have to prove it specifically for the groups we have encountered.
Now, let’s continue further our investigation into groups. This is a nice theorem to know, but
this is still not a complete picture of groups, because just look at Z . I have said that we
cannot split it further, but still Z does have sub groups. All even numbers we saw yesterday
is a subgroup of Z . I cannot write unfortunately Z as a product of even numbers times some
other subgroups because it’s not feasible But, we would still like to understand the various
subgroups of Z and how they relate to each other. And, for that, we will use the notion of
homomorphism that I have already defined just before defining isomorphism. A
homomorphism is a mapping between two groups such that the group operation is preserved.
There is no requirement of the map being one-one and onto.
So, consider the following map. It’s just simple map, multiplication by 2, when I write 2Z to
represent the subgroup of Z , which consists of even numbers. So, ϕ is a homomorphism.
It’s trivial to see this. The addition of two numbers is mapped to two times the addition of
those two numbers, which is just two times one number plus two times the other numbers.
The range of ϕ is all even numbers. Right.
And, that’s a subgroup of Z . What does it leave out? ϕ leaves out all odd numbers. In fact,
if you see Z and divid this into even numbers (range of ϕ ) and odd numbers (outside the
range of ϕ ). However, I can write the set of odd numbers as 1 + range(ϕ) . And, this leads
to the following viewpoint that, let us call 2Z to be range of ϕ and define the following
relationship.
We defining a binary relation on Z – the set of all integers. And, the relation says n is
related to m if and only if n − m is an even number. Do you remember equivalence
relations? This relation R is an equivalence relation for obvious reasons. A number n is
related to itself, because n − n , which is 0 is an even number. Then, reflexive property n is
related to m , which also means m is related to n .
And, transitive property – n1 is related to n2 ; n2 is related to n3 ; that means n1 − n2 is

even; n2 − n3 is even - clearly means n1 − n3 is even; so, its relationship is transitive. So, R
is an equivalence relation. This equivalence relation as we know, any equivalence relation
divides a collection of elements into equivalence classes. So, for this particular relation, we
will divide set Z into equivalence classes. What are the equivalence classes? Odd numbers
and even numbers; these are the just two equivalence classes and those are precisely these.
This is one equivalence class; this is other equivalence class. Now, this leads to a more
general observation. Let us consider in general group and a subgroup of that.
And, we do the same thing. Just mimic that exercise – define an equivalence relation. Or, first
define a relation; we will show its an equivalence relation. So, a R b with respect to the
subgroup H . And, just remember what we did; we did the subtraction of the two elements.
That should be in the subgroup. Let us write this group multiplicatively as we would typically
do for a general group. So, there we have to say ab−1 ∈ H ; a − b becomes ab−1 or a/b .
b−1 represents the inverse element corresponding to b .
Now, it’s straightforward to see that this R is also an equivalence relation for exactly the
same reasons – a R a ; okay, maybe it requires a proof. a related to a since aa−1 , which is
the identity. We just write the identity as e and this belongs to the subgroup H . Since H is a
subgroup, it has the identity. a R b ⇒ b R a since a R b means that ab−1 ∈ H ; H is the
subgroup. So, inverse of ab−1 is also in H . What is the inverse of this element? ba−1 . And,
a R b and b R c implies a R c since ab−1 ∈ H , bc−1 ∈ H implies the product is in H
because H is a subgroup, the product is also in the subgroup ac−1 ∈ H . So, it is quite
remarkable that, the notion of equivalence relation seem to fit perfectly with the notion of
groups.
You see the identity of group is corresponding to the reflexive property. The inverse property
corresponds to the symmetry. And, the closure property corresponds to the transitivity. So,
the net result is that the group G is now divided into equivalence classes. And, these are
determined by the subgroup H . You have G ; then you have H as one equivalence class.
The subgroup itself will be a one equivalence class. Then, there will be other equivalence
classes: a1 H, a2 H .
And, I will explain what I am mean by a1 H - it is simply all elements of the form a1 times
an element of subgroup H. Why are any two elements related? If you look at a1 h1 and a1 h2 ,
these are two different elements in this. They must be related to be in the same equivalence
class.
a1 h1 (a1 h2 )−1 should be in H . So, this is a a1 h1 h2 −1 a1 −1 . This is a commutative group. So,

a1 and a1 −1 takes care of itself. This becomes h1 h2 −1 , which is of course in H . So,
commutativity here is essential. If you don’t have commutativity, there is a problem. So,
clearly all elements in here are in the same equivalence class.
How about two elements across? Can they be in the same equivalence class? Let us look at
one element from a1 H and one element from a3 H . Are they in the same equivalence class?
a1 h1 (a3 h2 )−1 , this is a1 a3 −1 h1 h2 −1 ; just rearranging the terms. This is of course in H . So,
this is a1 a3 −1 . That element will determine the equivalence class it is in. So, let us assume
without that, it is in a1 H . I am saying that all of these – each one of this contained an
equivalence class.
The question is can two of them all be contained in one single equivalence class? So, this is
one element from here and one element from here. And, suppose this is contained in the same
equivalence class, which is that, this is in H - that is when it is in the same equivalence class.
h1 h2 −1 is an H . So, this could imply that, a1 a3 −1 ∈ H . This implies that a1 ∈ a3 H . So,
these two classes are equal. If a1 ∈ a3 H , then a1 H ⊆ a3 H . So, it is the other part of
derivation. So, in the end, we have group G is split into disjoint equivalence classes. Each
equivalence class being of the form some a1 H . Now, I will do a little bit of magic.
Let us collect all the equivalent classes and put them in one set. Call it Ĝ . And, let us define
an operation on Ĝ . The · on Ĝ is defined as ai H is one element of Ĝ , aj H is another
element of Ĝ . They are operated on each other. You get ak H which is the third element of
Ĝ . This is ak is that element of G such that ai aj ∈ ak H .
And, let me state the theorem, which I will prove later. Under this operation, this set of
equivalence classes itself is a group. This is, taking some time to absorb, because elements of
Ĝ are not elements of G , they are sets of elements of G . And, we are operating on sets of
elements through this dot operation in creating such different sets; and, these operation
themselves lead – produce group structure on Ĝ .
We will continue tomorrow.

Modern Algebra
Indian Institution of Technology, Kanpur
Lecture - 04
Groups: Quotienting
So, last time we left off at this theorem, where the theorem if you see the screen, we had
defined the set Ĝ as the set of equivalence classes with respect to the relation induced by
these subgroup H on the group G . So, ai H is a representation or a name for one
equivalence class and we defined an operation · on this set Ĝ as the following: ai H · aj H -
this operation between 2 equivalence classes which gives a third equivalence classes ak H
and the definition is that ai aj is an element of the equivalence class ak H . That definition is
precise and now we want to show that Ĝ is a group. So, let us prove this.
What are the conditions we require for a group? First is closure that is by definition because
you see that ai H · aj H is by definition and element of Ĝ . So, this certainly closed under the
operation · . Next associativity, so we have (a1 H · a2 H) · a3 H . What is this element? You
consider this. So, a1 H · a2 H = a′H where by definition a1 a2 ∈ a′H , and then
a′H · a3 H = a′′H , where a′a3 ∈ a′′H . This is also by definition, correct.
Now, put these 2 together. What is the relationship or what is a′′H in terms of a1 , a2 , a3 ? See
a1 a2 ∈ a′H and then a′a3 ∈ a′′H . Okay. So, we can say a1 a2 = a′h1 and a′h1 a3 ∈ a′′H .
Clearly you can remove h1 from here and let us argue this little more carefully. So, we know
that a′a3 ∈ a′′H . This implies a′a3 h1 ∈ a′′H because h1 ∈ H . This implies a′h1 a3 ∈ a′′H
because it is a commutative group and a′h1 is a1 a2 .
So, this is an equivalence class, this product is an equivalence class defined by the element
a1 a2 a3 whichever is the equivalence class which a1 a2 a3 belongs to is this product. Now,
associativity follows immediately because you see that whether we do the operation on
(a1 H · a2 H) · a3 H or do we a1 H · (a2 H · a3 H) that would give us a2 a3 a1 . Now, by again by
commutativity, a2 a3 a1 or a1 a2 a3 they are all in a′′H . So, this completes the proof for
associativity. Are you convinced? Is there a doubt in this? No, okay. It’s pretty
straightforward just that I went through the detail because it is important to formally write
down this details and verify that for a general group this thus hold.
Next, what is the next property? Commutativity, okay. So, we have a1 H · a2 H = a′H where
a1 a2 ∈ a′H . a2 H · a1 H = a′′ H , where a2 a1 ∈ a′′H and these together a1 a2 and a2 a1
and again by commutativity are the same elements. So, they both belong to the same
equivalence class. These two equivalence classes are the same. Next, identity. What is the
identity? H is the identity. Why is that? If you have a H · H , this is a H . Similarly, H · aH
again just use commutativity, it’s also aH .
Inverse. That’s the final property. So, what is the inverse of element aH ? a−1 H . Let’s verify
that? Consider aH · a−1 H . Let’s say it is a′H . Then again by definition aa−1 which is
identity is in a′H and if identity is in a′H , what does this mean? It means a′H is H . That’s
the only equivalence class which contains the identity element and that is the end of the
proof. So, now we can see something quite interesting that has happened. We started with a
group, we identified a subgroup of the group and then we introduce the equivalence classes
with respect to that subgroup.
From the equivalence classes we made a new set and we defined a new operation which is
somewhat related to the group operation, but not quite the same because this operation over
equivalence classes and it turns out of that new set under that new operation is also a group.
This is a very important process, so it is given a name and a notation. So, let’s define that.
So, that new group Ĝ is written as G/H . This is the division sign and in a certain sense this is
what we are doing. We are dividing the group G by the subgroup H because each
equivalence classes are now represented by one element. So, there is certain amount of
division happening and that’s the reason why we write this way and that is why we call it
quotienting.
So, this slash is a quotienting operation and we of course, always need to quotient a group
with subgroup and the result of quotienting is another group. Now, having seen this let us get
back to our motivating example. What we started with was this group of integer under
addition. Then, we identified a subgroup. There are many subgroups - let’s say set of even
numbers that’s a subgroup, now if you quotient Z by the subgroup of even integers by this
theorem that we proved we must get an another group, what is that group?
Firstly, what are the equivalence classes we will get. One equivalence class there will 2Z
itself. Other equivalence classes? Set of all odd numbers. 2Z is the all even number and there
is exactly one more equivalence class which is all old numbers. So, this new group will have
just 2 elements. The 0 or the identity would be 2Z and then there will be the other set of all
odd numbers that’s the other element. So, what would be the operation on this new group?
So, let’s write the odd number as 2Z + 1 as a notation. 2Z is identity. What is the new
operation? How does it look like? So, we only need to work look at the definition of new
operation on 2Z + 1 itself because 2Z is the identity. Anything that you operator on 2Z with
you get back that element. So, what is (2Z + 1) · (2Z + 1) ? It’s 2Z . Everything else is
defined already.
So, that’s it. This completely describes this group Z /2Z . Is this a familiar looking group?
Let’s define another group Z 2 , which is just two number 0 and 1 and the operation is
addition modulo 2. So, for this operation addition modulo 2, 0 is the identity and 1 + 1 = 0
and that’s it. That completely describes this group. So, do you see similarity between Z 2 and
Z /2Z ? What is the similarity? It’s isomorphic to Z 2 . We already have that notion now. This
is isomorphic because except for the notation that is I’m writing 0 by 2Z and 1 by 2Z + 1 ,
nothing else is different. The operation · is really addition modulo 2. That’s excellent.
So, let’s again let me reiterate this. We started with group of integers quotiented with a
subgroup. We got a new group which is familiar group. It is the set of integer modulo 2 under
addition. We can do the same exercise, this is the first example, let’s say.

We can do a very similar example again. Let’s say take another subgroup of Z . Sorry.
Student: 3Z
Yes, in general any subgroup of Z will be of the form nZ that is all multiples of the number
n . That’s the general form of subgroup of Z . It requires the little bit of working out, but you
can do that on your own. So, what is Z /nZ ? You can work this out in a very similar fashion.
What will be the equivalence classes for Z /nZ ? nZ itself will be a equivalence class, what
are the other ones? nZ + 1, nZ + 2 to nZ + n − 1 . There will be exactly n equivalence
classes. The group operation would be it’s essentially addition modulo n . So, this is
isomorphic to Z n which is the set of numbers between 0 and n − 1 under addition modulo n
. So, that’s the structure of the subgroup of the group that you have obtained by quotienting a
general subgroup from Z .
One interesting fact about this is none of these group we are getting are subgroups of Z , none
of these Z n is a subgroup of Z , but by quotienting with subgroups of Z we are getting them.
So, which basically is telling us that this quotienting by subgroups is creating new groups and
this is the one very good way of creating new groups.
So, let’s take another example and I would like to seek suggestion from you on which group
should we look at to next. You take a group, take its subgroup, quotient it and see what we
get. So, suggestions for the next group to be considered. Real numbers, okay that’s fine. So,
non 0 real numbers under multiplication, fine. Consider R* which is a set of all non 0 real
numbers under multiplication. So, that’s a group. Find a sub group. Q* . Now, what is R* /Q* ?
Very interesting question, what are the equivalence classes?
Student: It should be R* .
It should be R* only. That is, is it isomorphic to R* . It wouldn’t be R* , but one can ask if it
is isomorphic to R* Firstly, we can worry over it later; let’s identify the equivalence classes.
What are the equivalence classes under this quotienting? So, take any equivalence class here.
One of course one will be Q* itself, but take any other equivalence class. So, let’s consider
some aQ* where a is some non-zero real number in R* . So, what are the elements in a Q* ?
See, by definition elements of aQ* are going to be of the form a times some rational
number. So, elements of aQ* therefore, are all rational multiple of a .
And now, we quotient this out so that means, each equivalence class represents a real number
and all its rational multiples and now we can talk about operation between different
equivalence classes. What would that give us? Each equivalence class is represented by a real
number a , we can say a represents aQ* class and if you multiplied a1, a2 - two real numbers
you will get a1 a2 . Now, is a1 a2 a different equivalence class or can it be the same
equivalence or one of the same equivalence classes? It will be a different equivalence class
because a1 and a2 cannot be rational multiples of each other. Otherwise they would be in the
same equivalence class.
So, in fact, even if a1 , a2 are rational multiple of each other then they are from the same
equivalence class and their product is a1 a2 times all is rational multiples. So, a1 a2 will not
be rational multiple of a1 or a2 .
Student: Unless a1 , a2 are rational.
Unless a1 , a2 are rationals which means that one of the them is identity equivalence class.
So, essentially in this group operation we’ll get will get a subset of real numbers such that no
real number is a rational multiple of any other real number. So, you collect all such real
numbers and then the just define the multiplication operation between them and that is the
subgroup that we will get. So, this is a different subgroup than R* . We can view it as a
subgroup R* but it’s a different one because for example integer do not exist in this. What
would happen to all integers? They are in Q* . So, they all are essentially collapsed into
single identity element that is 1. So, all integers collapse to 1, in fact all rational number
collapse to 1.
So, it’s a strange kind of a subgroup that you get, but it is a something which is not so
obvious to see unless we has the quotienting operation. Here is a question. Is R* isomorphic
to R* /Q* × Q* ? This is not always true.
For example, in integers we know that integers are not isomorphic to Z /2Z × 2Z . Why do
we know this? Because 2Z is subgroup of Z, but Z /2Z is not a subgroup of Z and by
definition, the product structure exist when only if you can split a group into subgroup and
product of two subgroups essentially. So, there is no subgroup of Z that is isomorphic to
Z /2Z . So, it Z is not isomorphic to this. At times it is. Not in this case, but for certain other
groups it is.
Let’s takes the next example. Just consider Q* as the starting group a subgroup of Q* which
is let’s say all powers of 2. We had something to do with this particular subgroup earlier. Let
me write this as H 2 = {2n |n ∈ Z } . This is subgroup of Q* . You have seen this. Quotient
Q* with this. What is this. Again it is something you have seen.
Just think by definition it is saying equivalence class contains all rational numbers that differ
by a power of 2, by multiple of a power of 2 essentially. So, each equivalence class can
therefore, be represented by a rational number which is not divisible by 2. When I say
divisible it essentially its prime factorization has no power of 2 and then again that’s also
subgroup. We saw that last time there is and then I am going to write again using slightly
better notation H 2 where 2 means get rid of 2.
This is set of all rational numbers a/b where a, b ∈ Z Okay so, I shouldn’t write it as
equal to, but it is isomorphic. Because equal I can’t really write equal because Q* /H 2 is a
collection of equivalence classes - it is not really collection of all numbers, but just like we
can establish isomorphism in the integer case, we can establish an isomorphism between
these two and we can write Q* ≌ Q* /H 2 × H 2 . This we saw earlier that Q* can be split as a
product of H 2 and H and H is an isomorphic to Q* /H 2 .
2 2
So, what we have seen or learnt here is that there are distinctions between groups of this kind.
Earlier, we have saw that the distinction between groups - either some can be split as a
product of subgroups and some cannot be split as product of a subgroups. Here, the same
distinction is being rephrased in another terminology, which is using quotienting. That is can
we split a group as isomorphic to this quotient times that subgroup with which we are
quotienting. Sometimes you can do this like here, sometimes you cannot do that like there
and coming back to this question I will leave this as an assignment problem, please work this
out.
You should be able to prove it both ways that is whichever way is correct if it is isomorphic
to this you should able to prove it. If it is not isomorphic that also you should be able to
prove. Okay, so which means that quotienting sometimes gives us new groups. Sometimes it
doesn’t, like in case of Q* /H 2 , we do not get any new group - it’s already present as a
subgroup of Q* , but there are cases like integers where we do get new groups. So,
quotienting is therefore a very interesting and very important operation. We will see its
importance in subsequent lectures as well and we will establish that this is one of the really
important and fundamental operation in algebra.
Now, its importance is further enhanced by its connection with homomorphism. So, actually
if you recall I started with homomorphism, then kind of switch to this and the reason I switch
to this was because I wanted to find out the connection that what do homomorphisms
between two groups have to say about quotienting. Or, what is the quotienting had to do with
a homomorphism? So, and the answer is very interesting.
Let’s consider group G , subgroup of G and quotient. So, here are three groups that we
consider. Define a map from G to G/H as ϕ(a) = a′H where a ∈ G and a ∈ a′H . So,
this map is actually a homomorphism of G on to G/H . It is clear that ϕ is an onto map, that
is there is for every element aH ∈ G/H there is an element in G that ϕ maps it to. So, if
you have an element aH then a ∈ G is certainly mapped to this. Why is this
homomorphism? Well, let’s verify this.
What is ϕ(a1 a2 ) , that is some a′H such that a1 a2 ∈ a′H and what is ϕ(a1 ) · ϕ(a2 ) ? This is
let’s say a′′H such that a1 ∈ b1 H, a2 ∈ b2 H, b1 b2 ∈ a′′H all by definition. So, what about
a1 a2 ? This implies that a1 = b1 h1 , a2 = b2 h2 , b1 b2 = a′′h . This implies, just do this analysis
a1 a2 = a′′hh1 h2 .
So, a1 a2 is contained a′′H . So, ϕ(a1 a2 ) is therefore, equal to ϕ(a1 ) · ϕ(a2 ) . That’s the proof
that ϕ is a homomorphism. It’s not a surprise, I mean the way ϕ was defined - it is was
essentially using the fact already observed that of G/H is a group under that new operation
we have defined and it is really just that restatement of that. So, this says that if we have a
group and a quotient group there is an homomorphism which takes G onto G/H and what
this homomorphism really does is just contraction. Look at the each equivalence class of
G/H is contracted to single element of G/H and that’s the map. What is more interesting is
the converse.
Consider any homomorphism from group G to group Ĝ . Now, define two sets - H to be all
elements a ∈ G such that ϕ(a) is identity and Ĥ to be range of ϕ , and we have this
lemma that H is a subgroup of G and Ĥ which is a range of ϕ is isomorphic to the

quotient group of G/H and this lemma tells us that essentially any homomorphism induces a
quotienting of the group on which it is operating. It quotients that group with a subgroup and
the output on the range produces the quotient group.
Proof, a very simple. It is just a little bit of manipulation. Why is H is the subgroup of G is
very easy to see. There is closure property that is if a1 , a2 ∈ H then a1 a2 ∈ H because if
ϕ(a1 ) is a identity, ϕ(a2 ) is identity and then ϕ(a1 a2 ) by the fact that ϕ is a
homomorphism equals to identity. Then associativity it just follow some associativity of G
itself. Commutativity follows from commutativity of G itself. Identity is the identity that is
identity of G is also in it, because ϕ being homomorphism will map an identity to an
identity. Have you seen this? Why would homomorphism always map an identity to an
identity?
Student: ϕ(ae) = ϕ(a) · ϕ(e) .
Yeah. So, it’s simple, ϕ(ae) = ϕ(a) · ϕ(e) and which means ϕ(a) = ϕ(a) · ϕ(e) . So, ϕ(a) can
be cancelled on both side by multiplying with ϕ(a)−1 and therefore, you get ϕ(e) is identity.
So, identity is the same and inverse - if a ∈ H then a−1 ∈ H because a ∈ H means
ϕ(a) is an identity, that means ϕ(a)−1 is identity and therefore, ϕ(a−1 ) is also identity, that’s
it. So, H is the subgroup G and by definition of H , ϕ maps H to identity element. It’s
precisely the subgroup that is map to the identity element.
Now, let’s look at the range of ϕ . So, let us say, let b ∈ range(ϕ) . Define the set, let’s say
Rb = {a ∈ G | ϕ(a) = b} . So, collect all elements in a ∈ G that are mapped to this
element b ∈ Ĝ . I want to show that this Rb is one of the equivalence classes of G/H and
for that to show I need to just show that if a1 , a2 ∈ Rb then a1 a2 −1 ∈ H . You see this
trivially so. What is ϕ(a1 a2 −1 ) . This is ϕ(a1 ) · ϕ(a2 −1 ) . ϕ(a1 ) = b and what is ϕ(a2 −1 ) ?
ϕ(a2 ) = b ⇒ ϕ(a2 −1 ) = b−1 and so the product is identity.
So, a1 a2 −1 ∈ H and that’s it. That shows that all elements of H that are mapped to the
element b are precisely the equivalence class G induced by H , and therefore, the range of
ϕ which is Ĥ is an image of the G/H . Each equivalence class is represented by one
element of Ĥ . And because ϕ is a homomorphism, the operation on equivalence classes on
the G side is mimicked by the operation on the elements on Ĥ on the Ĝ side. This I am just
stating, I am not writing it down. This will be good exercise for you to go back and write it
out and verify it and convince yourself.
Another thing which I have skipped for homomorphism, it is again a simple fact which I
would ask you to verify yourself is if ϕ is a homomorphism and ϕ(a) = b then
ϕ(a−1 ) = b−1 , verify this. So, that is the relationship between quotienting and
homomorphism. Again what we proved shows that these are again different ways of viewing
the same phenomena that quotienting can be related to or connected with a homomorphism
and vice versa. So, when we say we want to study quotienting of groups, we might as well
want to study homomorphism between groups. So, that makes homomorphism also of equal
importance and again we will see later on many more applications of homomorphism.
We will break for today and we will meet tomorrow.
Modern Algebra
Lecture - 05
Groups: Structure Theorem
So, today we continue our discussion forward. Last time we saw this correspondence between
homomorphism between two groups and the quotienting operation on groups and the new
group that we get.
And as one of the examples we had seen, that when you quotient set of group of integers with
multiples of number n , then you get a new group, which you were calling Z n . This is a
group of numbers between 0 to n − 1 under addition modulo n . Now, this is a group and this
is the type of groups we will now focus on.
This particular group or this is actually a collection of groups, different for varying values of
n . This is what is called a finite group. I think I have defined it earlier also. This is unlike the
typical groups we have encountered - group of integers or rational numbers, real numbers or
matrices and so on. The only example of a finite group we saw earlier was the group of
permutations over 1 to n , that is also a finite group, but this is another example of a finite
group.
And if you recall one more thing in this structured theorem for finitely generated groups,
what I showed was that any such group can be written as a product of infinitely many Z
times one finite group and that finite group I had left undescribed and without giving further
details. So, now we will focus on finite groups and see what is their structure and how do we
further use them in some of the applications. So, first let’s talk about the structure of finite
groups. And again, here a useful example to keep in mind would be Z n . Although we are
talking about general finite group, this Z n is a canonical example of a finite group.
So, we can pose the same question as we had for infinite groups, in fact, that question is for
general group. Can a group is written as a product of two sub-groups and in case of finite
group, the question is can we do that. So, let’s take an example. Consider, let us say, Z 6 . Can
I write it as a product of two subgroups. Firstly, does it have subgroups? If it does not have
subgroups, then you, of course, you cannot write as a product of two subgroups. Sorry.
Student: (Refer Time: 05:14)
What are the subgroups? {0, 3} is a subgroup, wonderful; yes. {0, 3} is a subgroup of Z 6 .
Easy to verify, 3 + 3 = 6 , which is 0 modulo 6 and these are the only two elements. Now, do
you notice something interesting about this subgroup? Can you relate it to one of the groups
we have already encountered? It’s isomorphic to Z 2 , exactly the same structure as Z 2 except
that it has {0, 3} whereas Z 2 having {0, 1} but that just says that 3 in this subgroup plays
exactly the same role as 1 in Z 2 . And, so I can write, it is isomorphic to Z 2 . So, we have
found is Z 2 sitting inside Z 3 .
Student: {0, 2, 4}
{0, 2, 4} wonderful, that is another example; yes. {0, 2, 4} is another subgroup. Now, can
you make a similar statement as we did for {0,3}. It is isomorphic to Z 3 . So, we got two
subgroups in it. So, certainly it is at least possible that Z 6 can be written as a product of two
subgroups. Well, here are two subgroups. Can we write Z 6 as the product of these two
subgroups? What would it entail? Let’s consider the following question.
One simple sanity check we can do when we want to address a question like this for finite
groups is to at least count the number of elements on the two groups. If the counts are
unequal, certainly they are not isomorphic. Z 6 has 6 elements, Z 2 has 2, Z 3 has 3. How
about the product Z 2 × Z 3 ?
Student: 6.
6, 2 times 3, because every element in Z 2 , every element in Z 3 together forms one element
of Z 2 × Z 3 , that’s by definition. So, it has also 6 elements. So, at least the numbers do match,
but that is not sufficient. We have to actually establish an isomorphism between Z 6 and
Z 2 × Z 3 . The answer to this question is, yes, we can establish an isomorphism between Z 6
and Z 2 × Z 3 .
Can you make a guess of what that mapping should be?
Student: Addition
Addition, addition of the pairs. That will not be isomorphic if you look at.
3 plus 4 will become 1, you are right. 3 plus 2 will become 5; 3 plus 0 will become 3; 0 plus 0
is 0; 0 plus 2 is 2; 0 plus 4 is 4. So, one is missing.
Student: 3 plus 4 (Refer Time: 09:30).
Oh, 3 plus 4 mod 6, I am sorry, that is the one is there. It is a good idea to investigate this,
this map, ϕ(a, b) = a + b (mod) 6 where (a, b) ∈ Z 2 × Z 3 . Now, the question is, is this an
isomorphism? So, firstly we want to check if this is a homomorphism and then want to check
if it is one-one on-to.
So, to check if it is a homomorphism, we would want to see if

ϕ((a1 , b1 ) + (a2 , b2 )) = ϕ(a1 + a2 , b1 + b2 ) . This is the definition of addition for the product
group Z 2 × Z 3 . Here, again I am writing the group operation as a plus because that is a more
natural representation. Typically, I am your writing in multiplicatively, but here it is clearly
makes more sense to write it as an addition operation. So, the group operation on the product
is simply the component wise addition here. So, which is what we get here; and by definition
of ϕ this is equal to a1 + a2 + b1 + b2 (mod 6) and this is same as
(a1 + b1 ) + (a2 + b2 ) (mod 6) = ϕ(a1 , b1 ) + ϕ(a2 + b2 ) . That’s pretty straight-forward and.
So, we do get the fact that ϕ is a homomorphism.
Now, to check if it is an isomorphism we just need to show that it is one-one onto. Since it
maps 6 elements to 6 elements, as long as we show that, it is 1 to 1, it is obviously onto. So,
is this 1 to 1? We have just verified that it is. So, there we are. We can indeed write Z 6 as a
product of two smaller subgroups. Fine, that’s a good start.
How about Z 3 or Z 2 ? Can we further split it?
This is the only subgroup of Z 2 or Z 3 is just the 0, which is a trivial one. That is something
we are not interested in at all. So, there is no further splitting of Z 2 or Z 3 . So, in that sense,
Z 6 is a different, has a different structure than Z 2 or Z 3 .
How about Z 5 ? Can we split Z 5 into smaller subgroups? Can we find a subgroup of Z 5 ?
That’s the first question? No. In fact, if you just think about it for a moment you will realize
that you can generalize this observation into the following theorem that for any prime p the
group Z p does not have any nontrivial subgroup. I will prove this theorem, but as a corollary
of an even more general theorem.
So, let’s suspend this discussion or this theorem for a bit and let’s move on to the structure of
subgroups of a finite group. So, if you have a finite group, it may or may not have subgroups.
You have seen examples of both. If it has a subgroup, what should be the size of that
subgroup? Finite group has a size, what would be the size of subgroup?
And the theorem states: If G is a finite group of size s and H is a subgroup of G , let’s say
its size is t , then the number t divides the number s . Okay. And the proof is almost
immediate based on what we learnt last time. H is a subgroup of G , quotient G with H ,
this is a group. What are the elements of this group? Equivalence classes induced by H in
G . So, G/H is also a finite group since G is finite. It is a finite group of size, say l ; that
means there are l equivalence classes in G/H .
What is the size of an equivalence class? Each equivalence class is of the form aH . How
many elements are there in this equivalence class?
Student: Size of H .
Size of H because take any element of H , multiply it with a , that is it. Two such elements
cannot be equal. If a h1 = ah2 . then, you can this cancel a on both sides and get h1 = h2 ,
right. So, these are all distinct elements. Size of aH is exactly t . G consists of l distinct
equivalence classes each of size t ; therefore s is t times l , that’s it. That established the
theorem.
And now, from this theorem the previous theorem follows as a corollary because if we have a
group Z p , its size is p , any subgroup will have a size that will divide p . Now, since p is
prime, the only numbers that can divide p are 1 and p itself. So, a subgroup of Z p is either
Z p itself or the trivial group. So, it has no nontrivial subgroup.
We can now look further into this and exploit the finiteness. Take an element of a finite
group, let’s call it a . We can associate a number with a , which is called the order of a . It is
the smallest number m greater than 0 such that am is identity. Here, I am writing the group
operation as a multiplication. Okay, this number is written as, in fact, to be more specific it is
ordG (a) . So, this represents that it is the group G with respect to which you are looking at a
.
For a finite group G , does the order of an element always exist? That’s the first question one
must ask. The answer is yes and it’s simple to see, just.
Student: Pigeonhole principle.
Sorry, pigeonhole principle, yes. Let size of G is s . So, consider s + 1 powers of a , i.e
a, a2 , ..., as+1 , all are elements of the group G. So, two of them must be equal by pigeonhole
principle and then, you know, that means, you just do the cancellation and you get a to the
some power is identity. So, the answer is yes. So, it makes sense to talk about order of any
element for a finite group.
What, can we establish a relationship between the order of an element and the size of the
group? Again, the answer is yes. So, in fact, it follows as a corollary of the previous theorem
that order of a divides the order of G . Here I am using ord(G) to represent the size of the
group G . The proof is straightforward. Consider the set {am | m ≥ 0} . So, all powers of a ,
this is a subgroup of G and what is the size of this subgroup? Order a , that is, it is order a is
the smallest number for which am is identity. So, all smaller powers of a, a2 , ..., am−1 , they
are all distinct elements of the group and am is the identity. So, there are exactly m elements
and that’s it.
Now, you apply the previous theorem, you get this is corollary. So, we have in the finite
group several interesting facts. Every subgroup will divide the size of the group. Moreover,
every element has an order which is naturally associated with a subgroup of the group and
follows, that the order of every element also divides the size of the subgroup.
Now, we can come to some more definition, at least one more definition and then, I will
define the structure or will give this structure theorem for finite groups. What is the simplest
form of a finite group? Just like the simplest form of an infinite group was Z . Why? Because
it had one generator and that generator produced all elements of that group. So, that in a sense
is the simplest structure in a group.
So, when we talk about the finite group, we can again look at the simplest groups or simplest
finite groups being the ones that have just one generator. And do you know examples of these
groups?
Student: Z n .
Z n is one typical example for any n ; actually Z n is a group with just one generator. And
since these are finite groups and this fact is very useful in case of a finite group, we have a
special name for this.
So, we call a finite group cyclic group if it is isomorphic to Z n . And equivalently, cyclic
groups are precisely the groups with exactly one generator and that’s a lemma. Simple lemma
to prove. One way is straightforward. If G is cyclic, it is isomorphic to Z n and Z n has
exactly one generator, therefore G has one generator. So, that direction is very
straightforward, which is 1. So, that is a forward direction.
What about the reverse direction, which is it? Suppose G has one generator, then you want to
show that it is cyclic. So, let us say, let a be the generator of G , then by definition of
generator what can we write G as? And by definition of generator, since a is generator, its
different powers generate all of G and since G is finite, a has an order, let us say, order of
a is m , then these m powers of a , {a, a2 , ..., am } constitute the entire G .
And now, we want to show, that such a G is isomorphic to Z n for some n . Do you see that
isomorphism? G is then isomorphic to Z m , what is that isomorphism? ϕ(aj ) is mapped to
j (mod m) . It is easy to see that it is a homomorphism and it is a one to one mapped and
therefore, an isomorphism. So, these are in a sense, the simplest form of finite groups, cyclic
groups. Note, that we have already seen, since Z n , for all n is cyclic group, there are still
distinctions within these. Say Z p have no subgroups, whereas Z n for a composite n does
have subgroups.
Now, let me give you the structure theorem for finite groups. Let G be any finite group of
size n and again, I emphasize, that I am talking about commutative group here, not a general
group. And let n , which is a size of this group G be written as in its prime decomposition,
p1 α1 p2 α2 ...pk αk - p1 to pk are distinct prime numbers. Then, G is isomorphic to this product,
Z p1 α1 × Z p2 α2 × ... × Z pk αk . Each one of them is a cyclic group and therefore, all is the simplest
form of a finite group and the group G itself can be written as a finitely many products of
such groups. So, this completely describes the structure of a commutative finite group.
Again, I am not going to prove this theorem, you can look it up. The proof is not very
difficult, just follows the concepts I have introduced, but it’s certainly little involved. Any
questions so far?
So, now, you can go back and describe the structure of finitely generated groups even better,
that is, contains a finitely many copies of Z and then, finitely many copies of cyclic groups.
This subsumes our earlier discussion about Z 6 . Z 6 was written as Z 2 × Z 3 , that follows
from this theorem because 6 is 2 times 3 and therefore, you write it as a product.
Another interesting observation here is that we cannot further subdivide Z p1 α1 and to see this
as an example, consider Z 4 . It has {0, 1, 2, 3} . Does it have a subgroup? What is the

subgroup? {0, 2} is a subgroup, that is, Z 2 . So, the, if Z 4 splits as a product of two smaller
groups, since the sizes have to match, it has to be two times two groups, two groups of size
two each and Z 2 is one of them. So, it has, it appears, that it should be something like being
isomorphic to Z 2 × Z 2 , that’s by the structure theorem it goes further. This is the only way
you can split Z 4 .
But is it true, is Z 4 isomorphic to Z 2 × Z 2 ? If it were, we should be able to establish an

isomorphism between Z 4 and Z 2 × Z 2 . Now, suppose ϕ sending Z 4 to Z 2 × Z 2 is a
homomorphism. Okay. So, let ϕ(1) = (a, b) , then what is ϕ(2) ? This is ϕ(1 + 1) and by
homomorphism property, it is ϕ(1) + ϕ(1) , which is (a, b) + (a, b) , which is (a + a, b + b)
and what is a + a in Z 2 ? (0, 0). So ϕ(2) is 0 for any homomorphism ϕ , from Z 4 to
Z 2 × Z 2 , which means, this mapping is not one to one. It is sending 0 to (0,0) as well as 2 to
(0,0). So, it is a homomorphism, but it is not 1 to 1; this implies that ϕ cannot be an
isomophism.
So, this is the very useful tool. This homomorphism is a very useful tool to establish. Just like
we did, if a mapping is one to one or not and if all we need to do is to see what are the
elements being mapped to the identity.
In fact, let me now describe another definition which is kind of familiar. We are just giving a
name to something we already know. So, we already have come across this set - set of
elements of G that are mapped to identity by the homomorphism ϕ .
We are just giving a name to it, kernel, and denoting it by k er(ϕ) . So, this set is called the
kernel of the map and the observations are that first, of course, this we have already seen,
kernel is a subgroup of G . ϕ is one to one if and only if k er(ϕ) is just one element. And
this is again easy to see. If ϕ is one to one, surely k er(ϕ) will be identity because it says, if
it is one to one it cannot map two elements to identity.
The interesting one direction is the other way, if k er(ϕ) is identity, then ϕ is one to one.
Why? Let me just recall, ϕ is an isomorphism from G/ker(ϕ) to range(ϕ) . We saw this last
time ϕ is essentially like quotienting G with the k er(ϕ) and which gives us equivalence
classes and then, we have an isomorphism between this group and range(ϕ) , which is a
subgroup of H .
Now, if k er(ϕ) is identity what does it imply? G/ker(ϕ) is just G and we establish an
isomorphism between G and range(ϕ) . Clearly, one of the simpler consequences of this is
that ϕ is one to one. Since ϕ is an isomorphism from G to range(ϕ) . So, it is, it is an
isomorphism. So, it has to be one to one.
And this is the tool we just used to show, that there is no isomorphism between Z 4 and
Z 2 × Z 2 and that is an example of Z pα which cannot be split further. In fact, this I will give
you as an exercise. Let me give that as an assignment problem also.
Prove, that Z pα , which is a cyclic group, cannot be written as a product of Z pα1 × Z pα2 for any
α1 , α2 . They have to be nontrivial, that is, at least one. And the proof follows along the
similar lines as I just described for Z 4 , that is why you cannot split in the structure theorem
anything further.
So, we will meet again tomorrow.

Modern Algebra
Lecture – 06
Groups: Applications
Today, I am going to show you one application of groups and it is a very interesting
application, clearly simple one also. There are large variety of problems on which groups can
be applied, but we will restrict our attention to just this one. And we will use groups in
developing this abstraction further from the next lecture.
And this application is related to counting. So, let’s start with an example. Suppose, we have
a square and we want to color the vertices of this square with two colors, let’s say black or
white. How many ways are there to color these vertices with these two colors. But then if you
look at this closely and that is where the complication set in - there are certain symmetries of
coloring.
For example, these two colorings are essentially the same, because you can rotate this square,
and get to the other coloring or vice versa. So, we want to not just color, we also want to
know that how many distinct ways are there to color a square in this fashion. And distinctness
is counted with respect to this rotation, i.e if I have a square and I color the vertices and then
rotate that square and get another coloring that is the same coloring. Now, how many
colorings are possible?
We just count. Number of colorings with respect to rotation symmetry. First is all blacks;
another one is all white; third one is 1 black, 3 white; and it does not matter which of this 4
vertices I color as black, because I can rotate it and get any other vertex colored as black and
all other 3 as white. So, actually there is just one coloring with respect to rotation symmetries
in which we can color vertex with black and all other three as white.
Correspondingly, there is 1 white and 3 black. Then how about 2 whites and 2 blacks, how
many colorings are there? 2, right. We can colour adjacent vertices as black or with diagonal
vertices via black, and then all other possible colorings can be extracted by symmetries. So,
we get 6 different colorings and that is it, there are no any other colorings possible. So, that is
simple enough.
But now let’s me propose a more difficult coloring. Let us say suppose we have a n gon with
say 97 vertices which is basically similar structure with 97 vertices in it. Now how many
colorings are possible? Same 2 colors - black and white. How do you count? Distinct
colorings with respect to this rotational symmetry. So again all black is clearly 1 all white is
another, 1 black is another one, 1 white is another one, and then rotationally they are all
same.
How about 2 blacks? It will be determined by the gap between the 2 blacks, there will be
some number you can count easily. But then 3 blacks, 4 blacks, 27 blacks, now if you have to
count 27 blacks, you have to worry about the gaps between these 27 blacks. Is that going to
be easy? No, that’s not going to be easy at all.
Now, by the end of this lecture, we’ll derive a very convenient way of counting such numbers
So, how do we use the groups? Well, let’s stay with this particular example, although we will
generalise it very soon. We have to factors out the rotational symmetries.
Let’s try to represent the rotational symmetries in a abstract way. When I say rotational
symmetry, what does it really mean? A rotation is a mapping of or a rotation of by certain
number, there are 97 vertices so how many different rotations will there be? 97 with different
notations. You can take these vertices and keep it to itself that is a trivial rotation - there is no
rotation. Or send it to the first one or second one or third one. Once you fix mapping of a
vertex to the other vertex all other are automatically fixed, so there therefore 97 notations
possible.
Each rotation can be viewed as a permutation of 1 to 97 correct. So, if I start labelling this
vertices by numbers, so 1 to 97, each rotation I can represent as sending this number 1 to 97
to a permutation of these numbers and that permutation will be determined by where 1 is
mapped to say 1 is mapped 4, then 2 would be mapped to 5, 3 would be mapped to 6, 4 would
be mapped to 7 and so on, 97 would be mapped to 2 and so on right, it is very simple.
But this abstraction now allows me to view these rotations as maps, and these are not just
maps, collectively these rotations form a group. Such that π j (i) = i + j − 1 . So, π 1 keeps
does not rotate at all, π 2 shifts by 1. Also, when I map 97 and add something to this, I will
reduce it by taking out that folding back.
For mod 97, I will need to do this numbering slightly differently, I should have number this
with 0. I should have started with let us say the number not 1, but 0, 1, 2, 3 and up to 96. And
then I would say that instead of 1, I will call it π 0 , π 1 , up to π 96 . In that case, I can now say
that π j (i) = i + j (mod 97) .
Now let G be this collection of {π 0 , π 1 , ..., π 96 } . This is a finite group under the usual
operation, which is works for permutations that’s the composition. And this is simple proof.
π i · π j = π i+j (mod 97) , because π i shift by i and π j shifts by j which is same as π i+j (mod 97) . So,
there is a closure property this shows. It has associativity trivially, it is not commutative. Or
is it? It is commutative also in this case. Then there is identity which is π 0 , and there is
inverse. What is inverse of π i ? 97 − i or 96 − i .
97 minus i plus not plus i plus 1.
Student: π i and π 97 − i .
Yes.
Student: 97 minus i plus i.
So, you get 0 that is right, so the inverse is π 97 − i . So, for these reasons, now we are seeing
some interesting structure emerging. Now we can apply our knowledge about groups to
understand the number of symmetries or each group element here captures a one type of
symmetry, and then we can use it to count. It requires a little bit of cleverness. So, let me
describe that. So, we have these group elements, each one of them we apply it to a picture
like this, and rotate it to get another such figure.
So, let me explain a little bit further. So, we have this group of symmetries. Each one of this,
I am going to view it as follows. I am going to view each of these elements as operating on a
set of elements we call it X which is notationally I write it as π ∈ G takes an element of X
and produces another element of X . What are these elements? These are the elements which
we really want to count. What are the elements you want to count? We have this n gon, 97
gon, we color each vertex, give some coloring to vertices.
Set X is the collection of all possible colorings of 97 vertices which is 297 . Each vertex can
be either black or white, there are 97 vertices. So, that is 297 , so that is set X . When I apply
π ∈ G on one element of this, it just rotate it and produces another element of set X . This is
an important point or rather important view, it is not very difficult to understand, but it’s
important view to keep in mind that these rotations denoted as π takes one coloring of this 97
vertices and sends it to another coloring of this 97 vertices. Okay. Any question on this, no?

Now, what is it that I want to count. I can put it in a very nice way. Let’s say here are all the
circles. These are all elements of the set X each of the point here represents one coloring of
the 97 vertices. Okay. Now each coloring, if you look at it, and apply a π on it, it would be
map to another coloring, let say this is map by π 1 to this, may be by π 2 to this and so on.
Keep in mind that is there are some π ’s, which will map this point to itself. For example, π 0
will map an element to itself, it does not change the element. But that is not the only π that
may map an element to itself.
For example, if you look at the element which is all black. And you apply π on it, where
does it go to. It stays here, no matter which π you apply. So, all π will just take it to itself.
So, for different elements here, we will get different structure that some π will take it to
some other nodes and some π will map it to itself. So we create this picture, we keep this
picture in mind - we have these element, we create this arrows corresponding the different π
taking that element or node to another node.
And now let us define a relation. a R b if there exist a π ∈ G , such that π (a) = b , this is
for a, b ∈ X . So, when two nodes a, b ∈ X are related, what does it say about a and b . It
says that there is a rotation of a that gives me b and therefore a and b are from this
rotational symmetry. So, we do not want to count in our original counting a and b
separately. We want to count them together as just one element. So, this relation is what we
want to enforce on this set, and then count how many distinct unrelated elements are there, so
exactly that is what I am driving at but for that we have to first show that R is an equivalence
relation and that is easy.
Well, firstly a is related to a , π 0 (a) = a . Second a R b implies b R a , since π (a) = b

implies π −1 (b) = a . We know that π 0 belongs to the group G , it is an identity map. We
know that if π ∈ G then π −1 ∈ G that is due to group property.
And then a R b, b R c implies a R c , since π (a) = b and π ′(b) = c implies π ′ · π (a) = c .

And π ′ · π is again by the closure of the property belongs to the group. So, here is another
demonstration that there is a close relationship between equivalence relation and groups. So
this group G induces an equivalence relation on these elements.
So, now the picture looks much nicer, because now there are equivalence classes here with
respect to this relation. By the way, this is an equivalence class of its own, this all blacks,
because every element in G takes it to itself, so this is a solitary equivalence class.
And, now the problem is the number of equivalence classes. Each equivalence class gives me
one distinct element, because an element of an equivalence class can be transformed to each
other via this rotations. So, we just need to count, how many distinct equivalence classes are
there and that will give me exactly the number of different colorings.
Are you with me so far? There is a slight problem because different equivalence classes will
have different number of elements. You saw example - here is an equivalence class that are
just one element in it, whereas some other equivalence class will have many more. So, we
have to do a slightly careful counting, we cannot count it very easily. So, how do we count?
Well, here is a trick we use.
So, in general, we are given a finite group G and a set X such that π is a mapping X to X
for π ∈ G . Then G induces an equivalence relation on X and we want to count the number
of equivalence classes. So, now, let I am removing this connection with this rotation group.
Consider a finite group, in fact it need not even be commutative could be any group. We are
given a finite group and a collection of elements X such that every element of finite group
maps elements of X to X . This induces an equivalence relation on X and then you want to
got the number of equivalence classes.
How do we do that? For that define a set Y as follows. Y is a set of pairs, one component of
a pair is π ∈ G , and second component to the pair is a ∈ X . And (π, a) pair belongs to
the set Y if π (a) = a ; that means, this element π of the group G does not disturb a . It
sends a to itself; π need not be identity, π could be any mapping. But it just happens that it
keeps the element a to itself.
What is a size of Y , well, I will write it in two ways. Size of Y , first way I will write it as it
is ∑ |X π | , where X π is set of elements a ∈ X , such that π (a) = a . This is trivial to

π∈G
observe. So I am not doing something special here, I am just saying that Y contains a pair
(π, a) for all π and all a satisfying this equation, you just split that and write it as sum over
all π ∈ G and X π where X π is all the elements a ∈ X , such that π (a) = a .
So, this set Y , I am dividing it into subsets one corresponding to each π , and then summing
size of X . The other way I will write size of Y as ∑ |Ga | . So that is a other split. You fix
a ∈X
one for each a ∈ X , you just count the number of π that map a to itself. So that is a
slightly different split of Y . Now I am splitting Y in subsets, one for each a ∈ X and then
counting how many π map that a to itself, and clearly both although is summations give me
the same number which is size of Y .
Therefore, we can write equality between these two sums, pretty straight-forward so far, but
the cleverness here is already used which is to define this set Y and count it into two
different ways. Once we set this equation up, the rest is pretty straight-forward. Keep in mind
the final target. What is the final target? I want to count the number of equivalence classes.
Let us focus on this sum.
This sum I am going to write slightly differently and for that let us go back to the picture.
Keep this sum in mind, what is the sum? This is sum of for every element a ∈ X the number
of mappings or number of π ∈ G that keep a to itself. If you go back to this picture, what
we are talking about is here is an element a , how many of π map this element to itself? So
that is how many arrows are in this picture going from this element to itself that is Ga . If this
element is a , |Ga | in this picture is the number of arrows that send a to itself. Correct? Feel
free to stop me and ask a question for clarification.
Now, I will just make one final observation, and then we are done. If a R b , then size of Ga
is same as size of Gb . That means, if I draw a sub picture of that picture say this is a , and
this is b , a R b , so that mean there is a π that maps a to b b. And now I want to see how
many self loops are there in this picture, which are arrows which send a to itself. Similarly, I
want to count how many self loops are there on b and I want to related these two numbers
and the relationship is that the numbers are the same. Why? Well, let us prove this. Suppose,
π (a) = b , then for σ ∈ Ga which send a to itself, and I am going to transform this map into
a map that is map b to itself. How do I transform it?
Student: Composition with π .
Composition with π and that wouldn’t be sufficient. I want π σπ −1 ∈ Gb . Why? This is

another map in G . What happens when you apply this map on b . Apply on b, π −1 will take
b to a , then σ will take a to itself and that π will take a to b . Therefore, π σπ −1 (b) = b .
So, we have established a map from Ga to Gb and that map, let’s give it a name ψ .
Yes.
The relative position the colors are the same, you are right.
You are very right that is the intuition and I am just putting that intuition in formal symbols
that is all otherwise there is nothing very special happening here. It is easy to see that ψ is a
1 to 1 and onto map. Therefore, size of Ga is same as size of Gb , ψ maps Ga to Gb is a 1
to 1 and on to map, both are finite sets so their size are the same. And once we have the same
sizes I am done. We will just go back to this.
Let us rewrite this as splitting this further into equivalence classes. So, O is an equivalence
class summation a ∈ O size of Ga . So, I am writing this sum and splitting it. First sum
going over all equivalence classes and then next I am going over all elements within that
equivalence class, so in the inner sum size of Ga is the same for every element. So this I can
write as size of O times size of Ga .
Now, what is size of O times size of Ga . Go back to this picture, here is an equivalence
class. Size of Ga is the number of π that map a to itself. How many elements are there in
the equivalence class. That number we do not know, but what we know is that if we take any
π ∈ G , it will keep either a to itself or send a to some other elements in the equivalence
class.
And so there are size of G different elements in G . Each one of those elements corresponds
to one arrow in this equivalence class; and size of Ga times size of O is exactly equal to
number of arrows in this equivalence class which is size of G . Why is that? To prove this, I
will just appeal to you to think about the last lecture or the lecture before and observe the
following.
G is a group, first observation is Ga is a subgroup of G . And we then when we quotient G

with Ga , you get equivalence classes. Each of this equivalence class corresponds to one
elements in the equivalence class. I leave you to proof this, it is not difficult just need to
observe properly. So, number of equivalence classes is precisely 1 over size of G times
∑ |X π | .
π∈G
What we just proved is called Burnside’s theorem that number of equivalence classes equals
1/|G| × ∑ |X π | . This theorem is applicable in general for all groups and all X where G
π∈G
operates on those X . These are finite groups of-course, and it is a very useful tool in
counting, why? Let us get back to our original example of 97 vertices. How many π are
there? 97, in the group G there are 97 π . Let us count X π for each one of them. What is the
X π0 size of this? π 0 maps an element to it each node to itself, there are 97 nodes in that
figure. π 0 maps each node to itself. So, how many colorings are there which are preserved by
π0 .
Student: 297 .
All 97; How about X π1 ? π 1 sends one vertex to the next one and so on. How many colorings
would be preserved by π 1 ? If the first vertex, vertex number zero has say color black, π 1 will
map 0 to 1 in order for that colors to be preserved one also must have colored black; one is
goes to 2, the 2 also must have the color black, 2 goes to 3, so all of them must have the color
black or all white, so that just 2. How about X π2 ? Here 0 goes to 2; 1 goes to 3. So, 0 and 2
must have same color, 1 and 3 must have same color. So, 2 goes to 4. 0, 2, 4, 6, 8 - this even
number must have the same color; 1, 3, 5, 7, 9 - odd number must have the same color create
counted mod 97.
So the next observation is there it is actually one, when you start moving from 0, 0 to 4, 8
when you come to 98, that is just as one.
And then.
No, see it would I am claiming that it is all white or all black only, because suppose 0 is
black, then 2 must be black, 4 must be black, 6 must be black and so on 94 must be black 96
must be black, 98 must be black and 98 is 1. And if 1 is black then 3 must be black, then 5
must be black, 7 must be black, so all must be black.

So this is also two. And in fact, all of them are 2, and this falls out of the fact there since 97 is
a prime number. In case 97 was not prime, suppose we had in 9, then there should be
different because 0, 3, 6, 9, 0 that would be one.
Student: That is what I was saying.
That is what you are saying, OK perfect, so that I did not understand that and that is it. So,
the number equivalence classes is 1 by size of G which is 97, and this sum inside is
297 + (2 * 96) . So, therefore, we have an exact count of the number of equivalence classes.
If the G has a large number of elements, then you have more difficulty in counting.
Then you would have to do a little more involve counting. It will depend on number of
factorization of whatever that number is. The key gain here is that this count is now one for
each elements of G . G which is set of symmetries, we will have a small number of
elements. So, you can do this counting much more easily compared to earlier where we were
looking at all possible ways that is really too many.
So, we are done with the groups and I will mail you the assignment also, the second one on
groups today, and send you a submission deadline.
Modern Algebra
Prof. Manindra Agarwal
Lecture - 07
Rings: Introduction
We finished with the groups last time, and I hope I managed to convey the nice properties of
the abstraction we get, starting from the simple arithmetic and coming to groups. Now, today,
I will start on the next abstraction which is called rings. Rings are perhaps the single most
important abstract object in whole of mathematics. It is extremely fundamental, extremely
useful, and is applied in a very wide variety of domains.
One of the reasons for why it is so important is simply that, it very nicely abstracts almost the
whole arithmetic. If you go back to our motivating example of numbers, we know numbers
admit two operations, addition and multiplication, and then, when we abstracted out groups
from them by focussing on one of those two operations. Now, in case of rings, we abstract, or
try to abstract the full arithmetic, which means, simultaneously abstract out both the
operations, and the way they interact with each other.
Now, if you see the set of numbers, the addition and multiplication operation satisfies certain
properties and we can express those properties in terms of the abstraction we already have.
For example, in a set of numbers, let us say X , X is a group under addition, whether we start
with X being a set of integers, or a set of rational numbers, or set of real numbers, etcetera, it
is a group under addition. And again, when I say group, I mean a commutative group.
In fact, that is the reason why we abstracted out the group structure to begin with. Now, what
can we say about the set under multiplication? For some of these sets, like real numbers or
rational, the non-zero numbers form a group under multiplication, we have also seen that.
But, it is not true in case of integers. In case of integers, we do not get a group under
multiplication. And at this point, I would like to leave out integers from the picture because I
want to abstract out this arithmetic in a most general possible way. We can do arithmetic on
integers, but it is somewhat more restricted, in terms of division not being there, or only part
of the division being there. We can divide 4 by 2, but we cannot divide 4 by 3.
So, in that sense, one would like to abstract out the most general arithmetic structure and
because of that I would not say that this set of numbers is a group under multiplication but it
does satisfy all other group properties, namely X * means X ∖{0} is closed under
multiplication. It is associative, has an identity, and it is also commutative. The only property
missing is the inverse. So, that is the structure of the numbers under multiplication.
We already know the structure of the numbers under addition but there is one thing that is
still missing, that is the way addition and multiplication interact with each other and that is
captured by the next property. This is called Distributive property. This property captures the
way, when we do addition and multiplication simultaneously. So, a * (b + c) , this number is
same as a * b + a * c . These three are the properties we can abstract out, when working with
any set of numbers and this is what is a ring.
This exactly the same set of property that I had listed earlier with one minor difference,
which is that I have removed the commutative property for the multiplication operation. And,
depending on whether multiplication is commutative or not, the ring will be a commutative
ring or not. Sorry, plus star, of course. The structure R under addition and multiplication is a
commutative ring if the multiplication operation is commutative.
Note that, addition operation must always be commutative. We are not leaving any option for
it to be not commutative. And again, for the purpose of this course, we will be interested
essentially in commutative rings, not in other types of rings. So, whenever I say a ring, I will
normally mean a commutative ring. So, that is the definition. Pretty straight forward. And
now, we can immediately ask examples of rings.
Of course, integers, rationals, reals, complex numbers, under addition and multiplication is a
commutative ring. That is straight forward. That is where the definition comes from. How
about other examples; you have any other suggestions? Matrices, absolutely yes; if you look
at, and there is this notation that is used here. Let us use M n ×n to represent the set of all
n × n matrices, under addition and matrix multiplication is a ring. It is not necessarily a
commutative ring. Because, under multiplication, the matrices do not commute. So, it is not a
commutative ring. So, that is a general ring.
Any other examples? So, what you are saying is, a + i√2 . What is i ? √− 1 . So what is the
set of numbers you are talking about? a is from integers, and i√2 ? a + i√2 will not make
sense. Probably, you mean a + b√2 , where a and b are both integers.
So, that is indeed a ring. It is given a specific name Z [√2] , and it contains numbers of the
kind a + b√2 , where a and b are integers. Why is this a ring? Firstly, it is group under
addition. That is, that is clear. The additive identity is zero, and the inverse of a + b√2 is
− a − b√2 . So, it is clearly a group under addition. How about multiplicative property?
Identity is 1, for multiplication; it is closed under multiplication. When you multiply two such
numbers, you will get another number of the same kind. It is clearly associative. And
similarly, the distributivity properties also hold. So, it is a ring.
In fact, instead of √2 , I can have something else also; I can have √3 .
Any irrational number, exactly. In general, I can have Z [α] is the set of numbers a + bα ,
a, b ∈ Z , α is an irrational number. That is great. Okay. That is great progress. What about
other examples? How about a set of polynomials? Let us say, polynomials in one variable.
So, shall we say Z [x] . This is just a collection of all polynomials, with integer coefficients, in
variable x . Is this a ring? Pretty obviously yes, because you can add polynomials, you can
multiply polynomials.
Under addition, it clearly forms a group. Under multiplication, again, the expected properties
of multiplication hold, and multiplication distributes over addition. That is, if you have
polynomials p, q and r , then, p × (q + r) , is the same as p × q + p × r . So, again, a very
simple, but nice example of a ring. And, the good thing is we can already see this abstraction
generalizing and giving or spanning a number of different types of structures - polynomials,
matrices, numbers, and numbers of different kind, you have this integer, rational, irrational
number as well.
Student: In polynomial, how multiplication will be done?
Multiply two polynomials? Sorry?

Yes, yes, d is not fixed; d can be any positive integer. So, multiplying two polynomials of
degree d1 and d2 , will give us a polynomial of degree d1 + d2 , that is right. And, you are
very right that if you restrict or put an upper bound on the degree then it ceases to be a ring
because of the same reason that if you multiply two polynomials, their degree might exceed.
So, these are some very typical examples of rings.
And for rings, the central question again is, what is the structure of a ring? Just like we had
asked this central question about groups, what is the structure of a group. And then, we put in
a large number of cases, explained the structure, that a group is several copies of Z . Not all
groups, but, finitely generated groups consists of several copies of Z plus a small piece
which is the finite group part and that also has copies of the kind Z pα , for different primes p .
So, in the same fashion, we can ask what is the structure of a ring. And again, we will not be
able to explain the structure of all possible rings, that can be very bizarre. But, we will be
able to explain the structure of a fairly large class of rings, and the clue to that again starts
from this - the real, original example of a ring, which is the ring of integers. Just like the
group under addition for integers, in a sense gave us the canonical example of a group and we
could express other groups into form of several copies of Z , or Z quotiented with a
subgroup. In the similar fashion, we should ask in what way is the structure of set of integers
reflected or found in other rings.
Are you following me? No? So, in case of groups, we found after some investigation that the
integers under addition is perhaps the most basic type of group. It is the simplest type of
group with one generator. And then, we can express any finitely generated group as several
copies of Z , right. So, in that sense, the group of integers was the simplest group possible,
and other more complex groups can be expressed in terms of this. So, the structure of the
group of integers is found in or is repeated in any finitely generated group.
And, we asked the same question for the ring of integers, that firstly, what is the structure of
ring or integers? Structure of group of integers is pretty simple. Under addition, it is a single
generator 1 that generates the entire group. So, now that we have two operations, we will
again ask, what is the structure of the ring of integers, and then, ask the question, is this
structure found in other rings as well? So, first question that we must ask is, what is the
structure of the ring of integers?
Now, we have integers with both addition and multiplication operation. And, there is a very
nice structure that has been found in this ring, which is called the fundamental theorem of
arithmetic. This is something you have come across already that every integer can be
uniquely expressed as a product of prime numbers. So, what is the proof of this? Do any of
you remember? Firstly, one has to define what prime numbers are? They are the numbers
which have no divisors, except 1 and the number itself.
So, we are talking about the notion of division, which only exists partially, in case of integers.
And, prime numbers are those which cannot be divided by any other number, except 1 and
itself. Firstly, we can write every number as the product of prime numbers; that is the first
step one needs to carry out to prove this. Can you do that? That should be easy; I will leave
that to you.
Similarly, one can show that, if a number has more than one expressions as a product of
primes, the primes in these expressions are the same. There cannot be two distinct ways of
writing a number as a product of prime numbers. There is one thing which I forgot. Every
m ∈ Z , which is positive (for a negative number, one has to multiply with -1, which is not
exactly a prime number; and that, we have to also use in the product). So, every m ∈ Z ,
m > 0 , can be uniquely expressed as a product of prime numbers.
So, that is the basic structure of the ring of integers. And, we would like to know, if in a
general ring, such a structure, or a similar structure exists. Let us take an example. Firstly,
this structure will only exist in rings, where division is not fully available. If you look at a
ring of rational numbers, there are no prime numbers. Why? Or, let us look at a ring of real
numbers. I claim there are no prime numbers in it. See, the proof is very simple.
Suppose, the number a , which is a real number, is a prime, then a/2 is also a real number;
and a = 2 × a/2 ; so, it is certainly not a prime. So, there exists no real primes or in a ring of
reals, there are no primes. In a ring of complex numbers, there are no primes. In a ring of
rational numbers, there are no primes. And, the reason is the same in all three of them that,
the division is available completely.
So, you can divide any number by any other number, and get a third number. So, that is
something we will not really be looking at because that’s does not quite mimic the structure
of the ring of integers. So, let us restrict the attention only to the rings where division is not
fully available. And further, I will restrict attention to only commutative rings.
Okay, now with this setup, let us go back to the examples we had listed. What were the
examples? 1 is already ruled out; 2 is not, because 2 is not commutative. What about 3? Yes,
3. Well, 3 is a subset of 4. So, 4 and 5, these are the two rings; in both these rings division is
not fully available. So, let us consider Z [√2] to begin with.
What is the structure of numbers in here? Does fundamental theorem of arithmetic hold?
Well, in order to see if the fundamental theorem of arithmetic holds, we have to define the
prime numbers. What are prime numbers in this ring? Is 3 a prime number in this ring? Or, let
me ask you - is 2 a prime number in this ring? 2 should be a prime number in this ring, why?
So, in order to show that 2 is prime, by definition, prime number is a number which cannot be
divided by any number except 1 and itself. Now, for 2 to be a prime number, we have to
prove that in this ring, 2 cannot be divided by any other number, except 1 and 2. Is that true?
It is not true. 2 is √2 × √2 ; that is it is the same number squared, and √2 is a number in this
ring. So, 2 is not a prime in this ring. √2 is a prime; that is a good point. Is √2 a prime?
Suppose, √2 can be written as (a + b√2)(c + √2) ; that is a general form of 2 numbers. And,
in order to show that the number is prime or not, you have to just write it in this fashion, and
see, if you can find values of a, b, c, d . This implies that, √2 = ac + 2bd + (ad + bc)√2 .
This implies that, ac + 2bd = 0 , and ad + bc = 1 .
So, if this is equal to √2 , then the integral part must be zero, and this multiplier of √2 must
be 1. So, what do we get out of this? This tells us that, d = − ac/2b , and here we get
− (a2 c/2b) + bc = 1 ; and, this gives us (2b2 − a2 )c = 2b ; and a, b, c are integers. So,
firstly, the right hand side is an even number. So, left hand side must also be an even number,
which means a2 c is an even number, which means either a or c is an even number.
So, now, we have to do some bit of slightly messy argument. If a is an even number, then
a2 is a multiple of 4; correct? And that is ok, that is not a problem. You are saying a and d
are 0, b and c are 1; so that is a solution. Sure, that is a solution. That is good to know.
Is that the only solution? That is the question. And, what I am claiming, and I will leave that
to you to workout is that the only solution in integers is a = d = 0 , and b = c = 1 . Or, the
other way; or a = d = 1 , and b = c = 0 . And, this will show that, √2 is indeed a prime
number. So, prime numbers do exist, but they look different; because, 2 is no longer a prime.
Similarly, 3 is not a prime number, because, 3, I can write as (√2 + 1)(√2 − 1) . 4 is certainly
not a prime.
How about 5? Seems a prime? Seems a prime, yes; but, this is getting to be painful, in the
sense that we have to check every element-wise if it is a prime or not. It is going to be very
difficult to really understand the structure of this. Can we come up with a more general result,
saying that this is a list of all primes. So, that is one question. I do not expect an answer from
you immediately. But, fortunately it is possible to list down the list of all primes in this.
Again, we have to do it, listing first for Z [√2] , then for Z [√3] , then, for Z [√5] . Again, that
will be a painful process. Ideally, for any Z [√n] , we would like to have a list of prime
numbers; and, that is indeed possible. We can list down, come up with the expression for all
the primes in such rings. That is, would resolve at least one question. But, the other question
still remains unsolved, which is. So, suppose we have the list of all primes. Is the
fundamental theorem of arithmetic true? That is, every number can be uniquely written as a
product of prime numbers. The answer is not necessary.
For example, let us consider Z [√− 5] . In this, one can show that, 2 and 3 are prime numbers.
It is same as Z [i √5] . So, it is a subset of complex numbers. In this ring, 2 and 3 happen to be
prime numbers; and, the proof is again of the same kind; you just express it in terms of a
general product, and see that the solutions are only trivial ones. But now, I can write 6 as 2
into 3. So, 6 is a product of two primes 2 and 3; I can also write 6 as (1 + √− 5)(1 − √− 5) .
So, there is the problem.
Look maybe, these are not prime numbers, one could say; (1 + √− 5) may not be prime. If it
is not prime, it will factor further into prime numbers. And, this may also, will also factor
further into prime numbers. Eventually, all those terms, it will have at least 2 factors of
primes. This will also have at least 2 factors of prime, so there will be at least four prime
factors, whereas, in here, there are only 2 prime factors; but there is no way these 2
factorizations are the same. They are equal, but they are not the same.
So, the fundamental theorem of arithmetic breaks down here, and then, suddenly, there is a
problem. Because, then it is not clear, what exactly is a structure here. We have prime
numbers here, but we can write each number in multiple ways as products of prime numbers
and that creates a problem. These are the questions that immediately arise. These are not very
strange rings. These are simply, slight generalizations of the ring of integers. You put in one
additional number, irrational number, in the ring of integers that gives you another ring,
which is a slightly bigger set of numbers.
And there itself, the structure that existed in integers no longer exists. And, there is a major
problem that arises, because that structure does not exist anymore, studying rings of this kind
becomes a challenge because having a unique prime factorization is a very nice property. Any
number you can then write in terms of prime numbers and study those prime numbers, prime
factorizations separately. But here, there are multiple prime factorizations, then the one prime
factorization may give you one thing, and another prime factorization will give you another
thing.
So, it is not clear what one should want. So, these are the challenges that immediately comes
about when we study the ring of integers. And, there is a very nice history, for which I have
run out of time. So, I will tell you about it next time, about why and how these rings of
integers, this additional higher rings of integers were discovered. Initially it was believed,
without thinking much about it implicitly that, fundamental theorem of arithmetic will hold
everywhere.
But, that lead to certain consequences which people believed to be true. And then, at some
point, it was discovered that fundamental theorem of arithmetic actually breaks down. And, if
it breaks down then all the earlier consequences were false.
So, that will form the beginning of the next lecture.

Modern Algebra
Lecture - 08
Rings: Failure of Unique Factorization
Let us start today with bit of a history, which is also the story of exactly how and why these
kind of rings that I described yesterday. So, this is related to Fermat’s Last Theorem.
Now, how many of you have heard about this Fermat’s Last Theorem? Anyone?
What?
You have heard of it. It states that, there are no solution of the equation in integers whenever
n is greater than equal to 3. So, this was a statement that was or a hypothesis, which was
made by Pierre de Fermat, who was a mathematician in seventeenth century. This is a very
interesting statement, because when n = 2 , then x2 + y 2 = z 2 has lots of solutions.
32 + 42 = 52 being one of them; where as soon as n becomes 3 or higher number, there are
no solutions; that is a claim by this.
And, to such a nice and simple statement and so intriguing that it attracted a lot of attention.
In fact, while Fermat was studying a book by Euclid, in the margin of that book he wrote this
statement and also wrote that I have found a very ingenious proof of this statement, but this
margin is too small to write that proof down and soon after he died. So, he could never write
down his proof. And, this became a big mystery to all other mathematicians, exactly what
was that proof of Fermat that he could not write down.
And, over the centuries, number of mathematicians tried to find the proof but none of them
succeeded. This theorem was finally proved after more than 300 years in 1995 about 20 years
ago using some very advanced mathematics. But, that is not really what I am talking about
today. I am talking about a proof that was also found long time ago around nineteenth
century. And, that proof went like this. Let us consider ring – So, you take this complex
number and then just like we did last time that, introduce or add this into the set of integers
and look at all the numbers we can form with this. These are going to be a subset of complex
numbers.
Now, if you look at this xn + y n – the left hand side of this equation, I can write it as - here I
can just take y n out in common – (x/y)n + 1 . And, a number raised to the n plus 1; this will
factor in this ring in a very nice way. What is the property of ω n ? ω n is ei π/n . Then, ω n n is
ei π . And, what is ei π ? − 1 , so ω n n is − 1 . And, ω n 2n is 1 . So, this will factor as one of
the values of x/y would be ω n . But, there will be many values. In fact ω n 3 also will satisfy
this property, that is, n-th power is − 1 .
So, in fact, you can see ω n , ω n 3 , ω n 5 , essentially all odd powers of ω n , will satisfy this, that
is, n-th power is − 1 . All even powers will satisfy that their n-th power is 1 . So, I can factor
this as – and therefore, equal to (x − ω n y)(x − ω n 3 y)...(x − ω n 2n−1 y) . And, each one of these is
a number in this ring. And now, since I have this being equal to z n , this would mean that this
product equals – z is an integer, which is multiplied to itself n times. These are also
numbers. So, z is an integer in this ring and these are also in numbers from this ring. So, the
same number is being written as a product in two different forms now.
If unique factorization holds in this ring; then, the only way it is possible is that, the numbers
further get factored out. Each of this number gets further factored into other numbers.
Similarly, z here also gets factored into other numbers. Essentially when you write z n in
terms of product of primes and the other side also in terms of product of primes and it comes
out to be same.
And then, it was shown, then that is not the case. It was shown that the factorization of one
side was different than this integer z factoring into primes in this ring. And, this would hold
whenever n was 3 or more. Now, what does that mean? That means this equation cannot be
satisfied. There is no x, y , z , which satisfies this equation, because if it did exist, then we will
have this two factorization, two distinct factorizations of the same number in this ring. And,
that is not possible.
So, mathematician of 19th century implicitly assumed that in this ring there will be unique
factorization. And, based on this assumption, this was a short proof that Fermat’s Last
Theorem is true. And, many actually thought that, this is the proof that Fermat also formed.
The only problem was, in fact, when this was published, everybody were very happy and then
suddenly somebody found, rather observed that, this ring has a problem, that is, a number can
be factorized in this ring in two different ways just like we had that, Z [√− 5] two different
ways of factorizing. And suddenly, the whole proof broke down because if you can factorize
number in two completely different ways, then this argument does not work.
So, in summary, there is no unique factorization for n greater than or equal to 3. So, this
brought to the notice of the community that, there are rings of numbers like rings of integer of
a more general kind, where the unique factorization into primes does not work. This was a
very, very unusual situation. It made a lot people very uncomfortable that how can this be and
what are these? These are certainly are numbers, but numbers must have this nice property of
unique factorization and these rings do not have. So, a lot of effort was made trying to restore
this unique factorization property. And, the mathematician, who was finally able to do, was
Martin Kummer I believe, who introduced what are called ideal numbers in these rings of
numbers, using which he restored the unique factorization property.
Now, ideal numbers, which I will describe shortly, got more nicely captured as or shortened
as ideals. And, ideals have since then come to acquire a very, very important place in the
theory of rings. So, that is a story. And, that is how this general ring of numbers was initially
considered and this structure of this ring needs to be understood because we would like to see
exactly how these rings behave. As you can see that, this ring is intimately connected to a
very nice question about integers, in order to resolve this question about integers; we have
naturally introduced this ring.
So, that is kind of a another example of what I claimed earlier right in the beginning of this
course that, these abstractions – once you do it and once you study them, will help us in
proving things about – things that are of real interest to us. Problems about integers are
course interesting. So, one would like to prove them and this is one way of doing it.
Let us take another example: of another type of ring. Let us say, we want to answer the
question - find solutions of the equation x2 = y 2 + 2 . So, essentially, you want to know the
pair and this is solutions over integers. This answers the question of or addresses the question
of whether two squares are separated by number 2. And, you want to find the solutions of
this. How would you go about finding a solution? You have a strategy in mind? Yes?
Student: We can factorize x2 − y 2 .
Factorize x2 − y 2 , so (x − y )(x + y ) , OK?
Student: then, 2 is (x − y )(x + y ) .
So, simple solution. You have x2 − y 2 , which is (x − y )(x + y ) , which is 2. So, we have two
integers which was product is 2. So, one of them has to be 1 and other has to be 2, and then
you just write down all these solutions. Excellent. Next example: Just made x2 to x3 . This is
asking the question, if a cube and square deferred by 2. Now, that factorization trick would
not work; x3 − y 2 you cannot factorize. So, what can one do? Well here, why not factorize
the right-hand side – y 2 + 2 ?
Student: (y + √− 2)(y − √− 2) .
That is right, (y + √− 2)(y − √− 2) . Sorry.
Imaginary number, is not allowed to has a value of x and y . That is correct; we have to find
integers values. Now, we want to find a way to derive these solutions. So, in order to find that
way, there is no restrictions, we can use any way. So, this idea of if you can factorize both
sides and then you can equate the side as the factors; that is the idea, which was used here in
(x − y )(x + y ) = 2 . That is very simple factorization got it as a result very quickly; here I am
trying to use the same idea.
The only difference here is that, now, I have moved from the domain of integers to a larger
ring of integers. This is the ring Z [√− 2] . That is a ring that had occurred last time also. Of
course, we again come to this issue of unique factorization being there and can this be
factorized, can there be unique factorization in this ring. It turns out that, in this ring, the
unique factorization holds. It requires certain effort to prove, but I will not make that effort.
You have to just believe me. It is not very difficult to prove, but it just requires some
calculations to show. Then, unique factorization does work in this ring.
The whole point being that, the knowledge or extensions to these bigger rings will help or
does help us in answering questions about integers. The questions of the kind we are
interested in. So, gives us a reason why we would like to study this larger rings of numbers in
general rings. So, let us continue a bit more towards this. So, since unique factorization holds,
we will have x3 equals these two products. Now, of course, we have to see and it is possible
that, these numbers further factor into something. That is always possible. Then, that will
depend on the value y , which is what we need to find out.
However, there is one thing we can do, which is to look at these two factors and ask the
question – do they have a nontrivial GCD? So, suppose a number t divides y + √− 2 and t
divides y − √− 2 ; they have a common factor. Remember t is in this ring, because this is the
ring over which we are now operating. Then what? Then, it follows that, this implies that, t
divides 2y and t divides their difference, which is 2√− 2 . So, this limits the possibilities of t
. Now, of course, I have to see what are the primes in this ring. So, let me give you a trick to
find out a number whether a number is prime in a ring like this. And, that trick involves using
norms.
So, let us just consider this following: a + b√− 2 . Define norm of this to be a2 + 2b2 . And
now, the claim is if norm of a + b√− 2 is prime in Z ; then, that number is prime in Z [√− 2] .
This is a very simple proof. Let us see if we can discover it. Suppose norm is prime and
further suppose this is not prime. I should give you in, I cannot expect you to prove it
immediately because there is one more fact that we must know the following. Let us denote
N (a + b√− 2) to be the norm of that number then, norm happens to be multiplicative.
This can be verified very simply. You just multiply this out; compute the norm on both side.
It will turn out that, it is actually multiplicative, which is a very nice property particularly
with respect to finding if a number is prime in this ring, because if now suppose that
a + b√− 2 is not a prime; it factors as (c + d√− 2)(e + f √− 2) . Then, norm of this equals
product of norms of this; norm of this is a prime. So, since this is a prime over integers, these
two are also integers. So, product of these two integers is a prime; that means one of these
two integers must be equal to that prime and the other must be equal to 1; whichever it is.
Now, let us see suppose N (c + d√− 2) = 1 . By definition, norm is c2 + 2d2 = 1 . Now, c, d

are integers. And, everything is positive here. So, what would this mean? This means d is 0;
d cannot be anything other than 0. And, c is ± 1 ; which means that one of the two factors is
simply ± 1 ; which shows that, this number a + b√− 2 is a prime in that larger ring.
So, this is the trick, which is very commonly used to establish if a number is prime in the
larger rings or not. Now, let us get back to our proof here t divides 2y and t divides 2√− 2 .
So, what can t be? See t divides 2√− 2 ; and, 2√− 2 = − (√− 2 )3 .
Since t divides 2√− 2 and this happens to be equal to − (√− 2 )3 which means that t is
2 3
basically some power of √− 2 , i.e √− 2, √− 2 , √− 2 . These are three possibilities for t.
And, t divides 2y as well. So, that would that implies that, if t were equal to 2; then of
course, this is perfectly fine. t is a √− 2 ; that is also fine. It divides this trivially.
3
If t was √− 2 ; then, it would imply that, y itself is a multiple of √− 2 . So, y is an integer;
so, y must be even. So, that is what we have to write basically. Either if t is 2 or √− 2 ; then,
it is fine and if it is not, and then it is going to be that y is going to be even. And then, you go
back to this. And, I think I am just going to leave this proof at some point, but let us continue
until we have time.
Let me ask the question - can y be even? Suppose y is even; then, right-hand side is even;
which means left-hand side must also be even; which means x is even. If x is even; then, x3
cube is a multiple of 4 actually multiple of 8, but it is certainly a multiple of 4. So, left-hand
side is a multiple of 4; which means right-hand side must also be a multiple of 4. Since y is
even, y 2 is a multiple of 4; then, 2 should also be a multiple of 4; which it is not. Therefore,
y must be odd. And, since y is odd; now, let us come to this. We know now that y is odd
and t divides y + √− 2 .
Look at the norms. So, y + √− 2 can be written as t times something else. Take the norms.
What is the norm of y + √− 2 ? It is y 2 + 2 . So, that means y 2 + 2 can be written as norm of
t times something else. So, what is norm of t ? Only possibilities for t are some power of
√− 2 .
In whatever power it is, the norm of t must be even; whereas, y 2 + 2 is an odd number. So,
that is also not possible. So, the only possibility for t therefore is that, t is 1. That is the only
thing possible. And, the fact that, t is 1 implies going back that, these two: y + √− 2 and
y − √− 2 are – do not have a common divisor. They do not have a common divisor, yet their
product is a cube. What does it mean? It means that, individually they must be cubes;
otherwise, it is not possible.
Both factors must be cubes themselves. And then, we can carry this argument further and
derive from this the possible values of y ; I am not going to give you the complete details, but
this is the type of argument one employs when finding out integral solutions of equations.
And, as you see, it very naturally lends to this larger rings. And, those larger rings actually
help finding the solution like this property for example that we have just derived, we could
not have found without going into this larger rings. I will just leave it at this point.
Just work it out. It requires a little bit more work. In fact, we can quickly see that, c must be
equal to a and d must be equal to − b , because there is product of these two is x3 and x is
an integer. So, that means, product of these two must be an integer. So, that is the only way it
is an integer is that when you multiply this and this, you must get a2 + 2b2 .
It will also imply that x is equal to a2 + 2b2 and y + √− 2 is equal to this. So, we have now
got expressions of both x and y in terms of x and b . And then, we can further argue; find
out what values of a and b are possible. So, this example also illustrates the same point that,
these rings are extremely useful in solving the problems that we encounter over integers. One
of the key things, which I use without proving in this ring, admits unique factorization;
otherwise, this whole argument breaks down. So, ring admitting unique factorization is very
important.
So, the next question is which of these rings admit unique factorization? And, a lot of work
has actually gone into that. And, we now know a lot about it, but not the complete picture.
So, let me give you what we know about this. Then, I think we have 7 and 4 more. There are
4 other rings of numbers, which are larger and I do not recall them now. But, in the square
root of a negative number there are 8 rings, which admit unique factorization. All other
square root of negative numbers do not admit unique factorization. In fact, last time we saw
an example of Z [√− 5] ; that did not admit unique factorization.
What about the positive side? The positive side we know much less. We know some rings
that admit unique factorization. Some that do not, but exact situation is not known. So, that is
the state of current knowledge. And, that already shows that, there are number of interesting
problems, which are still a challenge to mathematicians to address.
For this course, we will not get into those challenges. We will stay with what is very
well-known. And so, coming back to these rings, we have seen examples of how and why
these rings are useful. We also have seen why these rings having a unique factorization, is
important. And, we have also seen why some rings do have a unique factorization, some rings
do not.
Now, the next step is to like I said, come up with a notion, which allows unique factorization
in all these rings. And, that is where either it mentioned already is - that is given by the theory
of ideals. And, that is something we will start in the next lecture.
Modern Algebra
Lecture - 09
Rings: Birth of Ideals
So, today we will learn from the examples of last time and try to develop a theory which
will allow us to understand how things actually factorize in rings and to set it up formally,
I will go to the more general settings and then define various quantities of interest that we
have.
So, let us start with a general ring and as I have said that we will be focusing only on
commutative rings. The first thing we are interested in division that is what are the
elements that are divisible by what other elements. That essentially prime factorization of
any number and that is what we want to see happening in this ring.
First thing we need to understand is what are the elements that are divisible by something
else. Now, one type of elements divide just about everything for example, if your ring R
is a set of real numbers, in that case every number divides every other number. So,
divisibility there is very general and therefore, there is no concept of prime numbers there
and so on. We have seen earlier as well. So, it is important to understand first that which
elements have an inverse present in the ring because if an element has an inverse present
in the ring, in that case that element would divide every other element. So, let us formalize
that.
So, we say that an element a in the ring is a unit if there is another element b in the ring,
such that ab = 1 which is another way saying that b is the inverse of a . Now, if an
element is a unit, then that element would divide every other element, that is take any
element c , c divided by a is same as c times b . So, there is a division available, so it is
only with respect to non-unit elements that we can talk about factorizing them. So, let’s
c ∈ R be a non-unit.
Now, I want to see if I can factorize c into, let’s say c1 c2 . That is whether c = c1 c2 .
There would be c which I can factorize, there would be c which I cannot factorize and
for the case of integers that is the difference between a prime and composite numbers. So,
let us make a distinction of these two types of non-units. So, this non unit c we call an
irreducible element if whenever we can write c as c1 c2 , one of c1 , c2 is a unit.
Now, in integers, when we define a prime number, we say that whenever c is a prime and
write c as c1 c2 , then c1 or c2 is 1. Here, we are not insisting that c1 or c2 be 1. We are
saying that it should be a unit. Do you see a reason for that? There is a very simple reason.
So, basically if a ring has only 1 as unit or rather in integer case, there are two units; 1 and
− 1 . If the ring has no other unit, then this definition is the same. Then calling it unit or
call it 1 or − 1 is the same, but if the ring has units other than 1 and − 1 , then we must
use this definition otherwise I can always write, take any c , I can always write it as
c (ab) where a is a unit because ab = 1 .
So, c is c (ab) and is equal to (ca) b . So, I have managed to write c as product of two
elements ca and b neither of them is 1 or -1. So, that definition will completely fail to
make any useful contribution. So, we must have a more general definition and this is the
right one.
The second difference is I am not calling c a prime element. I am calling it an irreducible

element and there is a reason for it which is that prime numbers, I would keep associated
with unique factorization situation. As prime numbers we understand that in every number
can be uniquely factored a product of prime numbers. So, I want to keep the name prime
to be a reserved for such structure.
In a ring, in general that structure may not exist, that unique factorization as we have seen.
So, instead of calling such element c in this definition a prime element, we will call it a
irreducible element and once we have this unique factorization, then whatever numbers
we have which admits such unique factorization or whatever irreducible elements which
give rise to unique factorization we will call them primes.
Now, we know that certain rings have unique factorization property and canonical
examples are, leaving aside integers, we have Z [√− 2] . This has unique factorization and
Z [√− 5] does not have. And the target remains the same. How do we restore or bring in
that unique factorization property in all of these rings and more, not just these rings.
So, I describe last time, this problem arose because initially people thought that all such
rings have unique factorization which came up with the problem in terms of proof of
Fermat’s last theorem. So, then all such rings were relooked at, some rings had unique
factorization, some did not and there is a mathematician I mentioned last time, Kummer
who was the first one to address this issue and he wanted to see how can we restore
unique factorization.
So, what he said was that for example in Z [√− 5] , we have 2 × 3 = (1 + √− 5)(1 − √− 5)
and these happened to be irreducible elements. Now, I will use the terminology I just
defined - 2, 3, (1 + √− 5), (1 − √− 5) are irreducible elements in this ring. So, it cannot be
factored further. What Kummer said was that is because there are some other numbers
into which these number factor, but those numbers have not present in this ring Z [√− 5] .
So, he postulated these additional numbers and said that these numbers are irreducible in
this ring. They do factor in terms of these other numbers and then, the unique factorization
property occurs. So, he developed that theory and gave this mysterious other numbers.
The name given is ideal numbers. But what are these ideal numbers have in it? Where are
they? They are not in this ring Z [√− 5] . So, one has to bring them in from outside. What
are their properties? All these sorts of question arose. What Kummer said was that look,
these numbers are numbers that divide 2, 3, (1 + √− 5), (1 − √− 5) for example.
These numbers are of the kind where they satisfy the typical divisibility properties. What
are typical divisibility properties? The two primarily properties, first one that if c divides
a and c divides b, then c divides a + b and second if c divides a , then c divides ab ,
for any b .
Then his work developed this theory to show that the unique factorization property would
hold when you include such ideal numbers, but the whole thing was kind of unsatisfactory
because you can always ask where exactly are these ideal numbers coming from, what
really is their connection with the actual numbers we start with. It seems that these
concepts has been pushed to just fix this unique factorization property. So that therefore, it
was not very satisfactory.
Now, after Kummer another mathematician also very famous Dedekind, sure you may
have heard of his name. Dedekind made a simple change of perspective and with that
simple change of perspective, he could very nicely set the whole thing up. He could
reconcile what really means by ideal numbers and with what are the actual numbers, how
do they interact with each other, how are they related to each other, exactly where are they
coming from, everything was very nicely explained and this all followed by a very simple
change of perspective which was following. What Dedekind said which I will call
Dedekind’s view. Since the point of interest in divisibility primarily, his view was treat
each number not as a number per say. Of course, it is number, but he adopted a different
view which is to view each number as a collection of numbers which are multiples of this.
For example, take the number 2 over integer. I can view it just as a number two.
Alternately, that is the Dedekind’s view that represent number 2 as a set of all even
numbers because that set is precisely set of all numbers which are divisible by 2. Similarly
number 3 is viewed as set of all number which are multiples of 3, 4 with the set of all
numbers which are multiples of 4.
So, it seems like a very unnecessarily complicated way viewing a number which it is in
case of integers, but this view helps resolve this mystery of ideal numbers very nicely. So,
at the cost of introducing this additional complexity, we are able to gain something. So, let
us adopt that view. Now this definition leads itself to a very nice generalization which is -
we call a set I which is a subset of elements of the ring R , an ideal.
If it satisfies these two properties that if two elements a, b ∈ I , then their sum
a + b ∈ I . If an element a ∈ I , then ab ∈ I , for any b ∈ R . So, this set I actually
captures the divisibility properties. So, the set I , therefore is representing a set of
numbers which have let us say in some nice way. The advantage here is that in definition
of I, we do not have to say that all the elements are multiples of some c .
If all the elements are multiples of say c , then certainly it satisfies these two properties,
but this is more general and in fact that difference is very important in our calculations
later on. So, what are ideals of a ring. Let’s see some examples.
In Z the ring of integers, all multiples of a number are ideals. So, I will denote this
notation (a) to represent the set of all multiples of a and this notation I will use in
general for general rings, not only for Z . Are there other ideals in Z . The answer is no.
These are the only ideals in Z and the proof is also straight-forward. Take an ideal I in
Z , there is a smallest number in this ideal I . Consider the smallest number, let a ∈ I be
the smallest. Let say positive number. Now, I claim that this ideal is simply all multiples
of a .
Suppose there is a number b which is not multiple of a in I and in that case take the gcd
of a and b , that is a number smaller than a . The gcd by extended Euclidean algorithm if
you remember can be written as aα + bβ . Now, since a and b are both in the ideal aα is
in the ideal and bβ is in the ideal and their sum is also in the ideal. So, that means the gcd
of a and b is also in the ideal I which is number smaller than a . That is not possible.
So, it follows that the ideal is simply all multiples of a .
In Dedekind’s view we are identifying number with all the multiples of that number which
is ideal and in in integers the ideals are in one to one correspondence with numbers. Let us
go to a more interesting case. How about Z [√− 5] . Say let us look at these numbers
1 + √− 5 and 1 − √− 5 . Let me consider the following case. Let us define an ideal I to be
the following. So that is the slightly complex definition.
Student: One of.
1 is not in this, 2 is in this. So, it contains all even numbers, surely. It contains all multiples
or even multiple of √− 5 . That is also true. So, that is nice observation. We can write the
ideal also as {2α + 2β √− 5 | α, β ∈ Z [√− 5]} . Why? We want to show that any element
in this set is in this set and element in this set is in this set. That you can derive very
easily, I don’t think I need to explain how to be that. This further I can simplify it even
more. It is {2α + 2β √− 5 | α, β ∈ Z } .
So, this is a more presumably restricted set than this. Here, α and β can be any numbers
in Z [√− 5] , here α, β are only integers, but it follows pretty simply by the fact no matter
what, you substitute that in here, multiply it out, collect the coefficient of √− 5 on one
side, coefficient of the integral part on the other side.
Both these numbers would be even why because everything here is even. There is a
multiplier of 2 here. There is a multiplier of 2 here. So, it is 2α + 2β √− 5 where α, β are
also in Z [√− 5] . Any arbitrary element you can get by this factor of Z [√− 5] , but since
there is a multiplier 2 sitting outside, it makes every component an even integer. If this is
not clear you work this out. I will leave this as a very simple exercise for you. What is
important right now to observe is that this ideal. Why is this an ideal?
Two numbers of this form when you add, they still remain in this form. That’s simple. So,
this is an ideal and we can simplify this ideal to this expression. Now, this is an ideal of
Z [√− 5] . Next question I am going to ask is - Is this ideal expressible as all multiples of
some number of a ?
Student: 3 + √− 5 - that number is not in this set.
3 + √− 5 , not in this.
Student: Second information is 2α + 2β √− 5 . α should be 3/2 .
Yeah, that is not allowed.
So are you saying that these two equalities do not told?
Student: Two elements (Refer Time: 31:33) minus 3.
(1 + √− 5) is not here, so what I have done is wrong, sorry. So, let me erase all of these. I
had something else in mind that I wanted to show here. Sorry good that you pointed out.
So, let us get back to this idea. Let just try to understand this. First question does this ideal
contain all elements of this ring.
This question is same as showing whether 1 is in the ideal I because if ideal contains the
entire ring, then of course 1 is in the ideal. If 1 is in the ideal, then all multiples of 1 are in
the ideal and all multiples of one are precisely the whole ring. So, is 1 in this ideal I ? If 1
is in the ideal I , then we will have 1 equal to α(1 + √− 5) + β(1 − √− 5) which is
(α + β ) + (α − β )√− 5 , right. So, this may α = β .
Student: α, β ∈ Z [√− 5] .
I am treating them as integer which is wrong. So, we will have to express α itself as an
element of Z [√− 5] which is α0 + α1 √− 5 and let’s just collect the coefficients together -
α0 + β 0 − 5α1 + 5β 1 + (α0 + α1 + β 0 − β 1 )√− 5 . So, does this have a solution over
integers? No, first look at this; this must be equal to 1. How can this be equal to 1? (
α1 + β 1 )5 is a multiple of 5 and this is α0 + β 0 .
Student: Sir. So, assume that β 1 = α0 + α1 + β 0 .
Okay
Student: Yeah.
OK.
Student: Then the real part becomes even.
Let us do that. It is a good point, so not this.
So, the conclusion is that β 1 = α0 + α1 + β 0 and

1 = α0 + β 0 − 5α1 + 5β 1 =− 4α0 − 4β 0 − 10α1 .
Student: Sir, in the above equation there should be
Sorry.
The something.
Student: There should be − β 0 + β 1 .
This one.
Say that again this should be plus.

This one, you cannot see it there? There now you cannot see there. So, which are you
talking about? So, just read out the line.
Student: Minus 5.
The imaginary part, yeah, α0 + α1 .
+ β0 − β1 .
− β 0 , why would that be minus?
Student: Beta.
β 1 should be plus, − β 0 + β 1 , you are right. So, make the required changes in the
equations. And that is not possible. There is no solution possible because this right hand
side is even number left hand side is 1. So, surely this ideal contains many elements but
does not contain 1.
Now, the next question which is what I am leading to is can this be express as multiple of
some a . Let us try to work this out. Suppose, a = c + d√− 5 , then 1 + √− 5 is
(c + d√− 5)(u + v √− 5) . Both 1 + √− 5 and 1 − √− 5 are multiples of c + d√− 5 , correct.
Now, let us see what the best way of arguing this. If you take the norms, the norm on this
side would be 6. So, the norm of this must divide 6 and since this is 6, norms of these two
are the same. Norm of this must divide 6. So, let just see what are the divisors. Norm of
this is c2 + 5d2 this divide 6. So, what are the possibilities. Very few. It could be d = 0
and then, c = 1 , then it is fine or d = 1 in that case c = 1 .
So, if d = 0 , then a = 1 which is now, which is certainly not the case. That is this ideal I
is not all multiples of 1 which is the entire ring, that’s not true. We have already seen it. 1
1 means it is this number itself 1 + √− 5 . So, 1 + √− 5 , sure I can write it as 1 + √− 5
times this and we are saying that ideal I is simply all the elements generated by 1 + √− 5 .
Then, we come to this point, how about this? Can 1 − √− 5 be written as 1 + √− 5 times
s + t√− 5 . Now, is this possible? See again if you go back take the norm, this is 6. So, this
norm should be 1 which is s2 + 5t2 must be 1 which means t must be 0 which means s
must be ± 1 .. This is certainly not true. That is false and therefore, we can conclude.
So, here is an ideal which is slightly funny one. It does not quite correspond if you look at
the Dedekind’s view. This ideal does not correspond to any number in the ring because
numbers correspond to the ideal which has collection of all multiple of that number. Here
is an ideal. It is an ideal in the ring, but does not correspond to any number. So, what is
this and it turns out this is what in the Dedekind’s view this is what Kummer ideal
numbers were. That is those ideal number that he brought in from outside with a certain
property, to restore the unique factorization property. Once you to adopt the Dedekind’s
view, those ideal numbers are ideal of this kind which are not multiples of a single
number and now keeping with the Dedekind’s view, we can write any number. Now, a
number in the Dedekind’s view is an ideal of certain kind.
So, any number slash ideal has a product in the unique way of certain type of number slash
ideals. So, I need to define the product of ideals and this is just an essentially translate the
arithmetic over numbers to arithmetic over these ideal which is Dedekind’s view that I
will do next time. We will focus particularly on this example of Z [√− 5] because we
know that unique factorization does not hold here. You know that 2 times 3 is equal to 6
factorizes in two different ways, so will see how using these ideals we can restore the
unique factorization property once we define the arithmetic over ideals and once we
understand that, then I will just state more general structured theorems for enlarged class
of rings which are called in the, to honor the work of Dedekind they are called Dedekind
rings. Those are rings which precisely admit this unique factorization in terms of ideals.
So, that is the plan for the next class.

Modern Algebra
Lecture – 10
Rings: Ideal Arithmetic
Let us continue from the last lecture where I left off by highlighting some questions that we
need to resolve about ideals before we can start doing arithmetic in terms of ideals. In order
to do arithmetic over ideals we have to define arithmetic over ideals, let us do that first.
As usual we start with a commutative ring and let us collect all the ideals of this ring in the
set RI . So, now we are treating each ideal as an element, and now we would like to do
arithmetic exactly as we do over ring. So, we again have to go through the properties of the
ring and see those properties hold here. It must be commutative group under addition, that is
the first thing we need to ensure. So, the first step is to define addition of ideals.
Let us say you have two ideals in RI . And the definition is as you would expect in the most
natural way, that addition of the two ideals is simply addition of elements of ideals. I 1 + I 2
contains all the elements of the form a1 + a2 , where a1 ∈ I 1 and a2 ∈ I 2 . Now the result
must be an ideal of the ring which is very important. So the question is, is this set an ideal of
the ring. Think about it, answer is straightforward. It is indeed an ideal of the ring, why? For
ideal we have to satisfy two conditions, that if you have two elements from I 1 + I 2 their sum
must also be in the same ideal. If a1 + a2 and a1 ′ + a2 ′ are elements of this ideal then their
sum is (a1 + a1 ′) + (a2 + a2 ′) , where (a1 + a1 ′) ∈ I 1 , (a2 + a2 ′) ∈ I 2 and so their sum is in
I1 + I2 .
The second property is that, if you multiply any element b from the ring to an element
a1 + a2 of this set the result must be in I 1 + I 2 . And this also seen very easily. b (a1 + a2 ) by
distributive property of multiplication is ba1 + ba2 . Now since a1 ∈ I 1 , so ba1 ∈ I 1 ,
similarly ba2 ∈ I 2 , since I 1 , I 2 are ideals. So this is in I 1 + I 2 . So, we have defined
addition of ideals very simply.
So, this shows the closure property, that is, not only defines the addition but also shows the
closure property of addition. Now we have to verify other properties - commutative is straight
forward, is just because ring addition itself is commutative so I 1 + I 2 is same as I 2 + I 1 .
Then identity, what is the identity here? Identity must be an ideal, the additive identity was 0.
So, what should that ideal be?
Student: Ideal of 0.
Ideal of 0; what is ideal of 0? Yes. What does it contains.
Student: Singleton set.
Single Singleton just a 0, absolutely.

And so I + (0) , (0) is the ideal of 0 - this is the notation I defined earlier, right. This is I
because you are just adding 0 to every element of I which is I , that is done. So, identity is
defined, and now how about the additive inverse; inverse of?
Student: I (Refer Time: 06:25).
I is inverse of itself.
Student: (Refer Time: 06:29) element (Refer Time: 06:38).
Yes.
Why?
Student: Since that I 1 + I 2 contains I 1 .
I 1 + I 2 contains I 1 .
Student: Because 0.
Because 0 belongs to every ideal, is that correct.

Student: but then it inverse if we want if you want to find an inverse.
Now there is a problem with inverse, 0 does belongs to every ideal. You see 0 times an
element is 0 and that must belong to the ideal. Similarly your − a must also belong to the
ideal, because when you multiply − 1 with every element and that should belong to ideal so if
a belong to ideal, − a also belong to the ideal. So, then is I + I = 0 ? It is not.
I + I is not 0, it will contain 0, it is not 0. Right, So we have a problem. We have a problem

that this is not really a group under addition, it has closure, it has commutativity, associativity
also pretty much follows, it has an identity, but it does not have any inverse. So, that is
something we lose already.
Fortunately it is not a big loss, because you see why we are interested in these, looking at the
ideals, to get the unique factorization. So, the situation here in terms for ideals is same as the
situation of positive integers. Set of positive integers or non negative integers I should say
they also do not have inverse, but the prime factorization does hold over the set of positive
non negative integers. Isn’t it? So, we do not really need the additive inverse for the carrying
out the prime factorization kind of operation. So, we do not worry about it. We have the
required property that we need. So, that is what the addition word says.
Now multiplication of ideals, how about that? How do we define multiplication of ideals? If
you try to define it, let us say I 1 · I 2 in the exactly the same way as a1 a2 , where a1 ∈ I 1
and a2 ∈ I 2 and let us ask the question is this an ideal?
Student: (Refer Time: 10:31) I 1 is (Refer Time: 10:34).
I 1 is closed under multiplication. So this is contained in I 1 surely, it is also contained in I 2

that is very correct. This is contained in I 1 intersection I 2 . That is the good observation, but
is it an ideal.
Student: from I 1 , the minimum element is within the ideal (Refer Time: 11:01) into the
minimum element is within the ideal (Refer Time: 11:06) multiplication of that (Refer Time:
11:08).
If there is a minimal element representing ideal, but we have seen last time that is not true for
all ideals.
Yes, exactly, that is the problem. You multiply any element of the ring to this you stay within
this that is not a problem, but if we add two elements of this a1 a2 + a1 ′a2 ′ is not clear whether
it is in this set. So, this definition does not quite work. So, how do we define the product of
two ideals? Suppose we define it in this way which is the smallest ideal containing
{a1 a2 | a1 ∈ I 1 , a2 ∈ I 2 } this is cheating actually. This is forcing the product to be an ideal
just by saying I would like I 1 · I 2 contain all products of the form a1 a2 . But that set may not
be ideal, so let us pick up the smallest ideal that contains this set and define that to be I 1 · I 2 .
Question is, is this definition sensible, is there really any smallest ideal or are there multiple
ideals containing this in the set which are not really comparable to each other. Fortunately,
we have a lemma which says that if I 1 , I 2 ∈ RI and their intersection is also an ideal. Why?
Again proof is very simple.
Student: If 2 element (Refer Time: 14:12).
If two elements are in the intersection I 1 ⋂ I 2 , their sum is also in the intersection, because
these two elements are in I 1 as well as in I 2 so their sum should be in I 1 as well as in I 2 .
Similarly, take any element b in the ring and multiply it with an element of I 1 ⋂ I 2 , since it
is in I 1 so the product will be in I 1 , and since it is in I 2 the product will be in I 2 .
Therefore, it is in both I 1 and I 2 . The proof is immediate. And this lemma ensures that there
is indeed a smallest ideal that contains this set, because if there are more than one ideal
containing this set their intersection will also contain this set and that is also an ideal which is
a smaller ideal than these two individual. So, you can keep taking intersections, essentially
take all ideals that contain the set take their intersection that is an ideal which is the smallest
ideal containing this.
There is another very nice characterization of this ideal I 1 · I 2 . It is the elements of the
following form which are finite sum of product of elements of I 1 I 2 . And now you can
directly verify that it is an ideal. Take two elements of this form, add them that will also be
elements of this form. Multiply any arbitrary element of the ring to an element of this form
that will be an element of this form. It certainly is an ideal. It certainly contains the set and is
the smallest set containing this because if this is contained in any ideal, all finite sums of
element of this form must also be contain in that ideal. Because that ideal must be closed
under addition, so certainly this is also contained in that ideal. Now this proof is also
immediate. So, that is multiplication of ideals.
That is a good question. Not always, this is always contained in I 1 ⋂ I 2 . Why? Because this is
contained in I 1 ⋂ I 2 which is an ideal and this is the smallest ideal containing I 1 ⋂ I 2 , and
therefore this ideal is contained in I 1 ⋂ I 2 . But, it may be equal to I 1 ⋂ I 2 it may not be equal
to. So we have now the definition of multiplication of ideals. Now again we need to verify
the properties of multiplication. For multiplication the closure is already defined,
commutativity is clear, associativity is clear, how about the identity, what is the identity here?
The whole ring, the whole ring is. Yes, actually I multiplied with 1 is I . This ideal is equal
to the whole ring. Whenever, 1 is in an ideal then that ideal equals to whole ring because then
you multiply every element of the ring you get those elements. So, this is the unit of
multiplication. And for multiplication we any way do not need an inverse so we are quite fine
with that. The final property is a distributivity of multiplication over addition, right. So, we
have to see that I 1 · (I 2 + I 3 ) , what this? Does this property hold?
Yes. So, let us just write down what is the general element of this. General element of this
would be ∑ a1,j (a2,j + a3,j ) .

j
Just multiply out and you get ∑ (a1,j a2,j + a1,j a3,j ) . This is an element of I 1 I 2 and this is an
j
element of I 1 I 3 , so all the elements of this are contain in this. How about the other way
round, same way? So let us take a general element here, let us use a different colored pen.
This is ∑ a1,j a2,j + ∑ a1,i a3,i . Can I write it in this form?

j i
Student: Answer is one element (Refer Time: 21:48) is not necessary (Refer Time: 21:50)
That element is common here and here, but that is the simple trick to around this. Just write
this ∑ a1,j (a2,j + 0) + ∑ a1,i (0 + a3,i ) and then combine these two sums. This is an element of
j i
I 2 + I 3 , this is also an element of I 2 + I 3 , and then you are multiplying elements of I 1 and
summing up. It is just a very trivial way of ensuring and this is a point. So that establishes
this distributivity property and that is very crucial.
Good, so we have now established arithmetic over ideals, okay. And since we have
established arithmetic over ideals we can talk about given an ideal writing it as a product of
other ideals. We now in order to complete the picture, we also would need to define the
notion of prime ideals. That is what we would want that prime ideals should be one which are
cannot be factored into smaller ideals. So, what should that definition be? Make a guess.
Yeah, whenever.
Student: (Refer Time: 24:25) I 1 or I 2 is a ideal of 1.
That does make sense. It mimics the property of prime numbers. There is another way of
defining prime ideals, which is I is a prime ideal if for every ideal J which contains I ,
either J is equal to I or J equals the whole ring. This is somewhat not so intuitive
definition of a prime ideal. What is the relationship if any between these two, can you think
about it.
Student: when we saying I equal to I 1 + I 2 then I 1 and I 2 are supersets of I .
I 1 and I 2 are.
Student: I mean I 1 and I 2 are superset.
I 1 times I 2 .
Student: And then we are saying that if I is equal to J K then (Refer Time: 26:46) J equal
to I (Refer Time: 26:49).
If I is equal to J K then.
Student: J equal to I or R .
Yes.
Yes.
I will give this as a small exercise for you to work on. So, in the next class we will come
back to this, because I want you to think a bit about these types of reasonings and see what
you come up with.
Work it out. Think about it at leisure, we are meeting tomorrow again so we will continue
with this tomorrow morning. So, we have in either case we have defined the notion of prime
ideals, and now we post that same question. In terms of ideals, what does the question
translate to with respect to the factoring. Let us look at a definition.
I have been holding on to that definition for a while. So, I found a name to this special kind of
ideal which are generated by single element of the ring. These are called Principal Ideals.
And, now the question of uniquely factoring an element of the ring in terms of prime
elements translates to uniquely factoring a principal ideal which corresponds to the element
as a product of prime ideals. And that, as I have been saying can be shown for certain kind of
rings which called Dedekind domains. Every ideals, not only principal ideals, every ideal can
be uniquely expressed as a product of prime ideals. That is a theorem which I will not prove,
so just stated it I will write it down also but let us more interestingly look an example.
The one we got stuck with and that is where the whole discussion started from, the ring
Z [√− 5] . In this ring we had 2×3 equals (1 + √− 5)(1 − √− 5) right and
2, 3, (1 + √− 5), (1 − √− 5) are irreducible, we have seen this. I think I have not given a proof
that these are irreducible element but I did ask you to work this out. And using the notion of
norms you can very easily show that all of these elements are irreducible, okay. So, these are
irreducible elements. In this ring, the unique factorization property break downs. And now let
us look at the corresponding ideals and see what we can say.
So, first let us look at the ideal of 2. Ask the question, is this a prime ideal? Answer is no.
The principal ideal 2 factors has the ideal (2, 1 + √− 5)(2, 1 + √− 5) , so it is a square of an
ideal let us see this. What are the elements of this product ideal? All finite additions of
products of this kind what is the general element of this ideal; 2α + (1 + √− 5)β , you
multiply this with another element from this that is the same ideal here so let us say
2γ + (1 + √− 5)δ , what is this equal to? Just multiplying out everything and you see this.
Now notice that every term here is an even multiple. Now inside this square brackets we have
an element of the ring, not sorry not the ring but it is actually an element of this ideal, correct.
This part is even, so that is a 2 times of something and this is just a multiple of 1 + √− 5 . So,
this is an element of this ideal. It actually it does not have to be, it is actually, it is enough that
it is a element of the ring because then if there is a multiplied two outside which puts it inside
this ideal.
Now any finite sums of this form will also be even multiples and therefore that will also be an
element of the ideal, principal ideal generated by 2, okay. So this shows that all elements of
this product ideal are contained in this ideal. How about the converse? That is converse is
trivial. Why?
For the converse I have to show that every even number that is this principal ideal is
contained in this ideal. In particular let us try to show that 2 is contained in this ideal. If I
show that 2 are contained in this ideal then I am done. So, why is 2 contained in this ideal?
Student: take β , γ , δ to be 0.
Take β , γ , δ to be 0 and this is 0.
Student: Actually that 2 is contained in this we will take already we have 2 common so we
require 2αγ − 2βδ to 1.
That should equal to 1.
Student: And it will take 2 common then that should be equal to 1, no that should be equal to
1 by 2.
Yes right. That is not possible, actually that is true. So, my claim is wrong that means. I have
missed out something let us see in this calculations. What shows is that the principle ideal of
4 is contained in this but that is not equal to that any way. So, what is going wrong here?
So let us stop here and I will leave this also as a bit of an exercise for you and tomorrow
when we meet we will fix this and we will also discuss those two alternative definitions of
prime ideals.
Modern Algebra
Lecture - 11
Rings: Special Ideals
So, did anyone work this out, this place we got stuck yesterday?
Student: We have worked out it (Refer Time: 00:22).
What? Its not possible to write principle ideal 2 as a factor of 2 ideals. You can prove that?
Okay, then that is contrary to what I can show, so let us see who is right.
Student: Equation is somewhat wrong, last equation.
That is wrong, that is clearly wrong that you showed last time and that cannot be, how about
this?
Is this true?
Let us see that. I think this should be true. So, we have to show inclusions two ways. Let’s
just do it fresh today. So, an element here which is (2α + (1 + √− 5)β)(2γ + (1 − √− 5)δ) that
is a general element of this product.
Of course, there are sums also but each term of the sum has this form. And if you multiply
this out, this equals 4αγ + 2αδ(1 − √− 5) + 2βγ(1 + √− 5) + 6βδ . And this is in principle ideal
2 because each term is a multiple of 2.
Student: What is the whole ring here?

Whole ring is Z [√− 5] . Okay, so that is one way it is clear. How about the other way. Like I
said last time it is enough to show that 2 belongs to this product. Does 2 belong to this
product?
Just multiply (1 + √− 5)(1 − √− 5) − 2.2 is 2 . So, this sum of products is an element of this
product ideal, because this is an element of the first ideal, this is an element of the second
ideal, this is an element of the first ideal, this is an element of second ideal. And in the
product ideal the elements are finite sums of pair wise product and this is 6 and this is 4 then
difference is 2. So, therefore, the answer to this question is yes. So, the proof that you worked
out that it is not factorizable, there is a problem in that proof I guess.
Similarly, the principle ideal 3 factors as in a very similar way in this fashion, i.e
(3) = (3, 1 + √− 5)(3, 1 − √− 5) , and you can see almost identically the connection or why
this is happening. If you take a generic element of this product, which will be of the same
kind except that instead of 2 will have 3. And you multiply this out instead of four 4, we have
9, 3, 3 and this 6 will remain.
So, in all of this multiple of 3, so any element of this belongs to the principle ideal generated
by 3; and conversely, I can generate 3 in this, basically the same thing 3 into 3 minus
(1 + √− 5)(1 − √− 5) is 3, so that is it. So, that also shows that the principle ideal of 3 also
factors. So that is the factorization of 2 and 3. And if you recall, we had this equation, so let
us see the right hand side of this equation, how about 1 + √− 5 and 1 − √− 5 .
1 + √− 5 I will claim factors as (2, 1 + √− 5)(3, 1 + √− 5) . Again look at the general element
of this. What is that? Go back to the same picture, in the general element there will be 2 here
and 3 here, 1 + √− 5 , 1 + √− 5 . When you multiply them out 2 times 3 would be 6, all other
terms will be multiples of 1 + √− 5 , and 6 is 1 + √− 5 times 1 − √− 5 , this 6 is also multiple
of 1 + √− 5 . So, this therefore is contained in this principle ideal. Conversely, is 1 + √− 5
contained in this?
Student: Yes.
Yes, why?
Student: 3 into 5 (Refer Time: 07:40).
3 into.
Student: 1 + √− 5 - 2 into (Refer Time: 07:51).
Wait a minute. Now we have to show an element of a here times an element of a here, and
may be some finite sums of such elements is equal to 1 + √− 5 . So, what is that equation you
are saying?
Student: 2 × (1 + √− 5)
1 + √− 5 = 3(1 + √− 5 ) − 2(1 + √− 5)
This is an element here and this is an element here, so this times this minus this times this,
and 1 − √− 5 in a very similar way will factor as (2, 1 − √− 5)(3, 1 − √− 5) and now you we
can explain this seemingly different factorizations. Because when you look at these in terms
of principle ideals they will factor as non-principle ideals and we are still yet to decide on the
definition of what a prime ideal is, but whatever that definition is these are prime ideals as it
turns out I will not show that. I will leave it for you to work out. And then 2 into 3 is these
two ideals time these two ideals and (1 + √− 5)(1 − √− 5) is these ideals times these ideal
both the product are same.
(2, 1 + √− 5), (2, 1 − √− 5), (3, 1 + √− 5), (3, 1 − √− 5) all these four are prime ideals. And
now you see the reason why we get this equality that 2 times 3 is (1 − √− 5)(1 + √− 5 ) is that
just a rearrangement of this prime ideal factorization. So that is an example of how this
unique factorization property gets restored once we look at the ideals.
So, if you see this equation in the ring does not admit unique factorization
2, 3, (1 − √− 5), (1 + √− 5) , they are all irreducible elements in this ring so you cannot be
factored further, but there are 2 different ways of factorizing the number 6. So, unique
factorization property in this sense does not hold in the ring. Now when instead of looking at
these as numbers if you look at these as principle ideals then we have unique factorization
property again. As principle ideals this is the principle ideal 6 is same as principle ideal 2
times principle ideal 3. Similarly, it is same as principle ideal (1 + √− 5) times (1 − √− 5) .
Now moving to the ideal world allows us to further factor these principle ideals into prime
ideals which is what happening here.
And now in terms of these principle ideals being factored as prime ideals, the unique
factorization property is restored, so that was an original motivation why the ideals were
defined. Like I said Kumar who was the mathematician who identified the problem and then
tried to resolve it by defining ideal numbers, which were some hypothetical numbers, which
helped resolve this or reinstate the unique factorization property.
And then later on Dedekind said these ideal numbers instead of forcing them in if we just
look at ideals, just by changing the view like I said earlier that do not look at a number as a
number itself or view the number as all multiples of that, i.e look at the principle ideals
corresponding to that number. And then this unique factorization property is brought in.
These ideal numbers of Kummer are at least in this case are these prime ideals. So, their
presence brings about the unique factorization. So, this resolves this example, but we are still
have not looked at the general case.
But before we look at the general case, we still have to resolve that certain amount of
confusion about the definition of what a prime ideal is. I gave two definitions last time. Did
you think about this? The first one is quite natural. The second one seems strange, but let me
suggest to you why this is also a possible definition of a prime ideal. We defined ideals as, at
least principle ideals as multiples of numbers.
So, in case of ring of integers, if you look at a principle ideal, all ideals there are principles,
look at a prime number, look at the principle ideal of that prime number which are all
multiples of that prime number. That ideal is a prime ideal in this second definition sense.
Because there is if there is an ideal which contains this, so let us look at number 7, look at all
multiples of number 7, that is a principle ideal. If there is any ideal that contains all multiples
of 7 what could that be?
It has to be ideal principle ideal 7 or principle ideal 1, cannot be anything else because 7 is a
prime. So that is this definition this is second one, that is there is no ideal between a prime
ideal and the whole ring. So, it does make sense the second one also. So, any comments about
or thoughts about these two possible definitions are they equivalent?
Sorry.
Student: Could not prove that equivalent.
Could not prove that equivalent actually, they are not equivalent in general; in fact, this
factorization of ideals itself gets messy if the rings are not of certain kinds. And I should say
that if the rings are not nicely behaved or not of a particular kind, then these two definition
become different, even factorization of ideal itself is kind of not very nice.
So, for prime ideal they are such basic concepts that we would like a definition that is
universal, it holds for all rings, and then it can be done in a nice enough way. Since
factorization of ideals is not very nice for certain rings, we will not use this factorization
definition of prime ideals, so that brings us to the second definition. The second definition
also becomes restricted for certain types of rings; we will see examples of this later not now.
For now, you just have to believe me. So, we cannot use the first definition we would not
want to use second definition, so which definition do we use, we use a third definition.
This is a new way of stating the definition ideal I is a prime ideal if for every a, b in the
ring, if the product ab is in the ideal then either a is in the ideal or b is in the ideal. Now
again if you go back to the setting of ring of integers, this is also very sensible definition.
Look at a principle ideal generated by a prime number, and we are saying that if two numbers
a, b where the product of two number is in that principle ideal then either a or b is in that
principle ideal which is equivalent to the fact that if a product is divisible by prime number
then one of the two elements in the product must be divisible by a prime number. So that is
one of the fundamental properties of prime number. And this definition captures that
property, so hence this can be also used as a reasonably good definition of the prime ideal,
and this is the one that we use across all the rings.
Now, it turns out that for certain types of rings, all the three definitions are equivalent. I will
not show it, but I will just kind of highlight a bit, but the second definition that I had earlier
given that is still interesting enough or it can be generally defined for all rings. So, I will give
it a different name, not call those prime ideals, but call them with a different name. Same
definition execpt the name I give to such ideals is not the different, these are called maximal
ideals.
Let me just shows one connection that can be proven very easily for all rings, which is that
every maximal ideal is prime ideal. And this is good to know because this tells us that the our
definition of prime ideals is a more general one than the initial one that I gave. So why is very
maximal ideal prime?
Well, suppose I is maximal, and we want to show that I is prime. So, consider ab ∈ I .
Now let us create two ideals I 1 being (a) + I , I 2 being (b) + I , both are ideals. Both
contain the ideal I , and since I is maximal, what does is it mean? It means that they are
either equal to I or they are equal to R . Suppose, if any of them is equal to I , what does
this mean? That a ∈ I , and if that is a case then we are already done.
On the other hand, if (a) + I = R , then we have αa + β = 1 , R contains 1 and this sum
of ideal is R , so there would be α in the ring R , β in the ideal I such that αa + β is one.
Multiply this equation with b ; now this implies that b is an ideal, because β is in I , so b β
is in I ; ab is in I by assumption, so αab is in I therefore, b is in I , has a pretty
straight-forward proof showing that every maximal ideal must be prime.
Now the target is to be able express uniquely every ideal as a product of prime ideals. And I
have been saying that this only holds for certain types of rings, and the types of rings where
this holds is are the ones where all, but at least these definitions coincide for the prime ideals.
So, let us define those settings. And in order to define those settings, if you recall going back
to the our discussion on groups, we were able to express certain types of groups as a sum of
copies of Z and plus some finite group.
What was the condition impose on those groups? Finitely generated. If you go back to your
that result about the structure of groups, there are theorems stating that every finitely
generated group can be expressed as sum of copies of Z plus a finite group which further can
be expressed as sum of Z pα , right. So that structure theorem only was true for finitely
generated groups not for other kinds of groups. And I have been saying that it is this structure
theorem for rings that I have been aiming for only hold for certain types of rings.
So, how do we connect this notion of finite generation with rings? With groups we had seen.
What is the simplest finitely generated group. The one with just one generator, and then you
all powers of that generator generate that group, so in case of group being finite it is a cyclic
otherwise it is just a copy of Z . So, what could be analogues notion of rings, a finitely
generated ring let say a ring with one generator?
Student: Principle
A principle ideal, a principle ideal like yes that is seems like a natural thing, principle ideal
there is just one generator, all multiples of one element make the principle ideal. But then
every ring has one generator only, because whatever ring that we are talking about is equal to
principle ideal of one. So, we cannot limit our attention to saying that a ring is finitely
generated or not.
Well, in all rings in this sense are finitely generated. In fact they are the simplest possible
ones.. So, what we will have to do and this is a slight leap of imagination that instead of
looking at whether rings are finitely generated, we look at whether ideals are all finitely
generated. A principle ideal is surely finitely generated it is just generated by a single
element. Not all ideals are principle we have seen that.
In fact, if you look at this ideal (2, 1 + √− 5) , this is not principle. But it is sum of two
principle ideals (2) + (1 + √− 5) . In fact, this was just a short hand to write this. So, this ideal
is also finitely generated. So, in general, we can say an ideal is finitely generated, if it can be
written as a sum of finitely many principle ideals and that is a very natural definition. And a
ring is called a Noetherian ring, if all its ideals are finitely generated.
Now the name Noetherian is as often the case is named after the mathematician, who first
studied these kinds of rings. The unique part of this name is that it is a woman, Emmy
Noether, she was one of the earliest woman mathematicians known at least probably amongst
the prominent mathematician, and she was probably the earliest woman mathematician. So
examples of Noetherian rings, we have seen examples of many ring.
Student: Integers.
Integers surely noetherian right, all ideals are principle, so they are the extreme example of
Noetherian rings. How about these extensions?
Z [√d] . 1 comma √d will generate everything, but what we need to show is that every ideal
is finitely generated. Every element of this is of the form an integer plus an integer times √d .
So, if an ideal has two generators of this kind then by taking appropriate differences, we will
get in that ideal one element being pure integer and other element being an integer times √d.
And now individually on these you can apply this property of Z . So, if there are two integers
- pure integers in that ideal, and then their gcd will also be in the ideal. So, there will be a
unique minimum integer in that ideal; and every integer in that ideal will be written as a
multiple of this. Similarly, for integer times √d , and then put all these arguments together,
you will find that any ideal in this ring can have at most two generators.
How about this Z [x] ? This will behave exactly the same way as this. So 1 and x are the two
natural components right. Let us say if you have an ideal so elements of these are
polynomials, if you have an ideal with two polynomials inside that ideal the gcd of those
polynomial will also be in that ideal.
Of course, taking gcd, we have to be a little careful, because coefficient are integers, so we
have to show that coefficients are integers, but you can work this out. I will leave this small
exercise to you, and you can work this out and find that this will also have maximum two
generators I believe. I am not sure, it may be that every ideal here may also be principle. You
need to just check this out so either one or two, no more than two generators are needed for
this.
How about any other example of a ring that you know of. We are ruling out non-commutative
rings, we are just focusing on commutative rings; in fact, lot of examples exists which have
this property that the ring is Noetherian. So, instead of giving more example of this, let me
give you an example of a non-noetherian ring. And this example is also of quite a bit of
interest, because we will develop on this subsequently.
So, let us define an interval let say [− 1, 1] . We are considering continuous mappings from
the interval [− 1, 1] to real numbers. Collect all continuous mappings in the set F . I will
define arithmetic on this set and claim that under that arithmetic, it forms a ring. What is the
arithmetic, very simple, addition take two mappings and define the sum of those two
mappings as a new mapping.
Will that be continuous? Yes, that will be continuous. The, what is the identity? This is also
commutative and associative. How about the identity of this operation? The zero map which
maps everything to zero that’s the identity of this operation. Then inverse, if you have
mapped g then − g is the inverse map because g + (− g ) this map will map the entire
interval to zero. So, this forms a commutative group addition
Now, you defined multiplication of mapping also. So, g 1 times g 2 is also a continuous map
from the same interval. The identity of multiplication exists which is the map, which sends
everything to one, this is commutative, and it is associative. The distributive property of
multiplication holds over addition. So that means F is a ring under addition and
multiplication. Now I claim that this ring is not Noetherian and to prove that what would I
need to do? I would need to show an ideal in this ring which is not finitely generated. Let us
do that.
Consider ideal I of F defined as I contains all mappings g ∈ F , and g (x) = 0 on

certain interval around the number 0. So, essentially you the origin, this is minus 1, this is
plus 1. And any map which is of this kind that it is 0 in this part and then it is continuous
everywhere else that is fine.
There must be a non-trivial interval around origin on which this map must take the value 0.
There will be lot of such continuous maps. Is this an ideal? If g 1 ∈ I , and g 2 ∈ I , this
implies that g 1 + g 2 is in I because if there is a interval around origin on which g 1 vanishes
and interval around origin of g 2 vanishes then intersection of those intervals on that g 1 + g 2
will vanish so that is fine. Now g ∈ I , and g ′ ∈ F implies g g ′ ∈ I , because if there is a
interval on which g is 0, then g g ′ is also 0 on that interval. It maybe 0 on a larger interval,
but certainly 0 on that interval. So, I is an ideal certainly.
Student: It could make the full interval trivial.
It could make the full interval trivial that is also possible, yes.
Student: We are not allowing α to be 1.
Let us make it one, good point, now I is an ideal of the ring F and this ideal is not finitely
generated, why, well suppose I can be written as (g 1 ) + (g 2 ) + ... + (g k ) or some
g 1 , ..., g k ∈ F . In fact g 1 , ..., g k not only would be in F it would be in I also, has to be by
the very definition, then there is an interval on which g 1 vanishes, there is an interval on
which g 2 vanishes there is an interval on each one of them vanish, take the intersection of all
those intervals and half it.
And cut it down by half or reduce it further basically. Now define a continuous map which is
0 only on that reduced interval and nonzero everywhere else. There will exist such a map in
F and that map will be on I also, but any mapping in this sum is going to be 0 on a larger
interval around 0. Whereas, that in new map that I have defined will be 0 only on a smaller
interval, so it is not possible.
So, these are the rings which will cause problems to us if you are looking for that structured
theorem. So, we will not be able to prove as nice as structured theorem about such rings, so
for now we will let us just focus on Noetherian rings, but it is not unfortunately noetherian
rings are not sufficient for us to prove that structured theorem. We have to bring in some
additional conditions.
So we say that ring R is a Dedekind domain if R is Noetherian, R is an integral domain I

have not defined what integral domain is which I will do, but unfortunately even these two
are not sufficient, we have to push in one more condition which I do not even want to write
down. So, you just keep in mind if some small technical condition which is almost always
true, so it is not something very limiting or problematic.
It is just that if I want to write down that condition fully I will have to define two more
things, which I do not want to do, but I do want to define what integral domains are. So, we
call a ring R integral domain, if whenever a product ab is 0 in the ring then either a is 0 or
b is 0, this is a very obvious looking condition. Of course, this is certainly true for integers or
numbers in general, but this is not true for all rings. In fact, go back to the ring I just defined
above, this one this is not true for this. Why?
Student: In a noetherian (Refer Time: 50:54).
That the product it vanishes everywhere, and that is a 0. So, again non integral domains are
very funny kind of a strange kind of a rings which are very interesting also, but not useful for.
Student: Can be in matrix also.
Can be matrix also, very right.
Yes.
Yeah, that is right absolutely. So, while non-integral domains are interesting rings and we
will come across some later, but as far as numbers and their factorization are concerned, we
do not want them around, because they can’t really handle factorization there. So, that is why
we bring in this condition also and then there is additional technical condition all put together
the rings that satisfy them are called Dedekind domains and then we have the theorem for
Dedekind domains. Now we can prove the theorem about Dedekind domains, which I will do
next time.
Modern Algebra
Lecture – 12
Rings: Dedekind Domains
Today, we partially prove this theorem.
Let R be a Dedekind domain, and I be any ideal in the ring; then I can be uniquely written
as a product of prime ideals. I will not give you the complete proof of this because it requires
some efforts. So, and I want you to make that effort on your own. So instead, what I will do
is, I will give a partial proof and leave the rest to you. So, the key property that we are going
to make use of is the following. This is the property I am going to make use of. I will not
prove this property, and it is a very interesting statement. It says that, if you have any ideal I
in a Dedekind domain, then there exists another ideal J in the ring such that I J is a principal
ideal.
So, let us assume this to be true, and proceed with proof. One very interesting consequence of
this is the following. In fact, I will show you two consequences. First one is cancellation -
there is 2 products, I I 1 and I I 2 which are equal, and these are ideals; then, I 1 = I 2 .
Essentially, it says that I can cancel I from both sides of the equation. It works out exactly
the same way as we can cancel numbers. This is not at all obvious, but by using the property
above we can make sure of that. How? Well, if you have I I 1 = I I 2 , and corresponding to
ideal I , we have an ideal J , such that, I J is a principal ideal. So, I I 1 = I I 2 implies
J II 1 = J II 2 . This implies, since J I is principal ideal (a) , (a)I 1 = (a)I 2 . And, this implies
I 1 = I 2 because you can easily cancel principal ideal multiplication. Why? Take any element
b ∈ I 1 , then abI 1 is in (a)I 1 , ok.
Let me write it in a different ink. Correct? Now, a Dedekind domain has the property that, if
it is an integral domain, which is that, ac is zero, then either a is zero or c is zero. In this
case, a is surely not zero. This is, of course, follows from this fact and so b − b′ is zero. So,
this shows that whatever is in I 1 is in I 2 ; b was in I 1 , and therefore is in I 2 also. And
vice-versa. So, that shows that I 1 = I 2 . So, that is the first interesting consequence.
The second one is that if I is contained in I ′ then I can be written as a product of I ′ with
another ideal J ′ . The converse of this is always hold. We have seen that. If I is the product
of two ideals then I is contained in both the ideals. In fact, I is contained in the intersection
of these two ideals. This we saw all time ago. This shows the converse that.
In case, R is a Dedekind domain then containment is sufficient to imply that I can be

factorized as I ′ times something. How does this work? So, let us start with this that, we have
I J is a principal ideal. No. In fact, we should use the J corresponding to I ′ . So, I ′J is a
1
principal ideal generated by a . And now I am defining J ′ to be a
IJ and I need to explain
that. See, I is contained in I ′ , which is I ′J equals principal ideal, generated by a . This is
another way of saying that, all elements of I ′J are multiples of a . So, since I is contained
1
in I ′ , all elements of I J are also multiples of a . And, a
simply means that they just divide
all those elements by a . So, the elements in here are multiples of a . So, just cancel out, or
take out a from this, and you get elements of I J divided by a . I am not saying that a is
invertible; I am simply saying that if ab is in I J , then b is in J ′ .
So, this is defined as set of all b such that ab ∈ I J . And, since we know that every element
of I J is of the form ab , so, this definition makes sense. J ′ is an ideal because if b1 , b2 ∈ J ′
, then ab1 , ab2 ∈ I J . This implies that ab1 + ab2 ∈ I J since I J is an ideal. And, this
implies that b1 + b2 ∈ J ′ . And, similarly, if you have b ∈ J ′ then ab ∈ I J , implies
cab ∈ I J , since I J being an ideal; this implies that cb ∈ J ′ . So, that shows that J ′ is an
ideal of R . And now, let us look at what is I ′J ′ . What is this equal to? This is I ′. 1a IJ . Just
rearranging, because this is commutative multiplication and I ′J by definition is, (a). What is
1
a
times principal ideal (a) ? Its R itself; or, this is just principal ideal 1. And, any ideal times
principal ideal 1 is just I itself. This shows that, I factors as I ′J ′ . Two very interesting
properties follow by that result and we will make use of both of these.
Okay. So, now, let us consider I and R ; which is an arbitrary ideal in the Dedekind domain.
First, I will show that, I is a product of prime ideals. Let us assume, for the sake of
contradiction that it is not. Let us collect all ideals of R which are not products of prime
ideals. Let this collection be M . I am going to show that this collection is empty; and, that
will prove the lemma. Well, since our assumption is that I is not a product of prime ideals
then, M is not empty; clearly, because I is in M .
Now, consider a maximal element in M ; that is, start with I . See if this collection M has an
ideal, which is a super set of ideals; and keep taking larger and larger sets, containing the
previous ideals; larger and larger ideals. And, by the fact that the ring is Noetherian, one of
the consequence of that is that any such chain of increasing ideals is going to be finite
because all ideals are finitely generated. When you look at an increasing chain of ideals, you
are basically saying that you are increasing the number of generators, essentially, one by one.
And, so since there are only finitely many generators in the ideal, any ideal is only finitely
generated. So, there will not be many or only a finitely many ideals in this chain. That is a bit
of a intuitive proof, not a formal proof, but we can formalize it along these lines.
So, consider an increasing chain of ideals which starts with I contained in I 1 contained in
I 2 ,.., I l and I l being the maximal in this chain. So, any ideal that strictly contains I l is
outside. Now, I l it is not a maximal ideal of R . I l is maximal in the chain. That is, there is
no ideal bigger than I l in this chain, but if you look at the ideal I l , it is not a maximal ideal
in the ring R ; because of the property we proved that, every maximal ideal is prime. So, if an
ideal is a prime ideal then of course, it can be written as a product of prime ideals. So, I l
cannot be maximal. So, let us say, let J be a maximal ideal containing I l . So, J contains I l .
Now, invoke the property number two, which we just showed. That means I l can be written
as J J ′ , where J contains I l , J ′ will also contain I l . Can J ′ be equal to I l ? Then,
(1)I l = J I l . J is a maximal ideal, by definition is non-trivial; in the sense, it is not equal to
the full ideal; then, by cancellation, (1) = J not possible. Therefore, J ′ not equal to I l .
So, I l is strictly contained in J ′ , and I l = J J ′ . Now, J is a prime ideal in itself. J ′ is a

strictly larger ideal than I l . So, it is not in M , which means, J ′ can be written as a product
of prime ideals; which means, I l can also be written as a product of prime ideals; all the
prime ideals of J ′ times J , which is a prime ideal. It is a clever proof.
And, this is a contradiction. So, that is the proof of the lemma which shows that every ideal
can be written as a product of prime ideals. And now, we need to show that this expression is
unique. And, this will follow pretty simply; which is, let us say, I 1 I 2 ...I k = J 1 J 2 ...J l . So, let
us consider two products of prime ideals which are equal. This implies that I 1 contains this
product. So, I 1 is a prime ideal and it contains product of prime ideals. What would this
imply? I claim, this implies that I 1 equals one of these J i . Suppose, J t is not a subset of I 1 ,
for 1 ≤ i ≤ l . This implies that there exists at ∈ J t ∖I 1 .
So, if J t is not in I 1 , then, there is an element of J t which is not in I 1 . But, this product
a1 a2 ...al is in I 1 . Now, I 1 is a prime ideal. What is the property of prime ideal? By
definition, an ideal is a prime, if whenever ab is in the ideal, one of a or b is in the ideal.
So, if this product a1 ...al is in the ideal, one of at is in I 1 . And, that shows that this
assumption that none of the J t are contained in I 1 was wrong. So, what we have managed to
show after this is that if a prime ideal I 1 contains a product of prime ideals, then, this prime
ideal is actually contains one of those prime ideals itself; not just a product, but one of the
prime ideals.
Now, J t is also a prime ideal; I 1 is a prime ideal, and J t is contained in I 1 . Now, we use
another property of Dedekind domains, which is that those conditions that I listed for
Dedekind domains imply that every prime ideal is maximal. In fact, the three definition of
prime ideals that I gave, I think I mentioned this last time; the three definitions of the prime
ideal I gave earlier, all of the three coincide for Dedekind domains. So, every prime ideal in
particular is maximal. J t is a prime ideal, and which is contained in ideal I 1 which is also a
prime ideal. So, J t has to be maximal, I 1 also have to be maximal which means J t is equal
to I 1 . And now, go back to the product. Now, from this reasoning, what we learn is that, I 1
occur on the right hand side also. And now, I can renumber, or rearrange the right hand side,
so that, J 1 is J t . Now, use the cancellation property. This implies that I 2 ...I k is equal to
J 2 ...J l and repeat. So, you just keep canceling the identical ideals from both sides, and
eventually, we get that, firstly, k = l , and J t = I t . Good. So, this gives you a flavor of the
arguments that one can use while working with this abstract ideals. I showed you the entire
proof except this key lemma which is for every ideal I , there is an ideal J , such that I J is a
principal ideal. This really provides the heart of the proof and proof of this is a little tricky,
but, completely elementary; that is, you can follow it from the concepts you know. So, try to
think about it.
Any questions on this? This is probably the longest proof I have done in this course, and quite
abstract also. This seems like manipulation of symbols without much intuition; if you think
about it, you will find that there is a clear intuition there. Basically, one is trying to use
cancellation which is a very powerful tool of ideals, which you can use to prove the
uniqueness. And, the containment being equivalent to factorization, we use to show that,
there exists a prime factorization of every ideal. Any questions? So, this is a story of the
development by Kummer, and then by Dedekind. So, after significant amount of effort,
which I just outlined earlier, they were able to restore this unique factorization property for of
course, a limited class of rings. And, if you recall, this entire thing originated from that
Fermat’s last theorem that and the proof of Fermat’s last theorem used unique factorization
implicitly, which broke down in a particular ring. But now that we have restored unique
factorization, we can try going back to the proof, and see how it works out. Unfortunately,
that proof now breaks down. Because, that proof does not work when you are using ideals; it
only works when you are using numbers only.
So, of course, attempts continued to prove Fermat’s last theorem for another 100 plus years.
But, what we got in place of proof was this notion of ideals. Now, they have certain utility in
terms of what I just showed, but it has turned out that, they have far more reaching utility,
than just this. So, let us discuss that aspect of ideals. And for that, I will again go back to the
group theory we developed. We had this notion of subgroups in groups and then we defined
the notion of quotienting a group with a subgroup, which corresponded with a
homomorphism between groups. And there is a nice correspondence between quotienting a
group with a subgroup and the homomorphism from that group to another group. And, we
will establish something very similar for rings, and we will see that the ideals play the role of
subgroups.
So, let us first define the notion of a homomorphism for rings. So the notion of
homomorphism generalizes very naturally into rings. In groups, we have one operation and
homomorphism preserved that operation. And in rings, we have two operations. So, we want
homomorphism to preserve both the operations. So, if we say ϕ(a1 + a2 ) = ϕ(a1 ) + ϕ(a2 ) and
ϕ(a1 a2 ) = ϕ(a1 )ϕ(a2 ) . All such mappings are homomorphisms. And further, if ϕ is a 1 1
onto map, then ϕ is an isomorphism; that is, again, exactly the same as for groups. And, the
notion of isomorphism allows us to deduce which rings are identical. If two rings are
isomorphic then essentially, they are the same ring, okay.
Now, some properties of homomorphisms for rings. Firstly, you should expect that, since it
has preserved two operations, then it should satisfy more properties than the properties of
homomorphisms for groups. And, that is certainly true. For example, what is homomorphism
of zero? If ϕ is a homomorphism, what is ϕ(0) ? This is always zero. Why? Yes, it follows,
because, ϕ(0) is ϕ(0 + 0) , which is ϕ(0) + ϕ(0) . And then, you can cancel one of them, and
then deduce ϕ(0) is 0. ϕ(1) = 1 - second property,.
Let us see that. What you are saying is that ϕ(1.a) = ϕ(1)ϕ(a) . And, this implies
ϕ(a).(ϕ(1) − 1) = 0 . Is this is what you are saying?
Student: So, ϕ(1.a) = ϕ(a) .
ϕ(1.a) = ϕ(a) , yes.
Student: So, we have phi of a on both sides.
Yes. So, that is what, I am taking the right hand side to the left, and writing that in this
fashion. Now, does it imply that ϕ(1) = 1 ?
Student: can be 0 or 1.
For integral domains, yes: very right. If ring is a integral domain, then one of them must be
zero. So, either ϕ(1) is zero, for all a . So, that is the homomorphism; that is a trivial
homomorphism; every element is mapped to zero; or, ϕ(1) is 1. Even if it is not an integral
domain, if for any element a , it is mapped by ϕ , to a unit of the ring R2 , then also we can
say that ϕ(1) = 1 . Recall the definition of a unit. Unit was an element which has an inverse
within the ring. So, for most of the rings, ϕ(1) = 1 . So, we will stick to this assumption that
for homomorphism ϕ(1) = 1 , because, that just simplifies certain things.
How about ϕ of a unit? What happens to this, when a is a unit of R1 , is ϕ(a) is also a unit
of R2 ?
We assume ϕ(1) = 1 .
Student: ab = 1 .
ab = 1 , yes.
Student:. So, ϕ(ab) is
So, ϕ(1) = ϕ(ab) so ϕ(a)ϕ(b) = 1 , and that shows that ϕ(a) is also a unit. So, this is a
nice property, which follows just by the fact that ϕ(1) = 1 . Now, consider kernel. Again,
exactly as defined as for groups. Consider all elements of the ring R1 , which are sent to 0 by
ϕ . Actually, here, we can define it in probably two different ways. Since there are two
identity elements either consider ϕ sending elements to zero, or to 1. But, defining it for zero
makes a lot more sense, as you will see. What can we say about kernel of ϕ . Kernel of ϕ for
groups was a subgroup. Here, it is an ideal. Why? See, if ϕ(a1 + a2 ) is ϕ(a1 ) + ϕ(a2 ) , if
ϕ(a1 ) and ϕ(a2 ) both are zero, then ϕ(a1 + a2 ) is also zero; if ϕ(a) = 0 , then
ϕ(ba) = ϕ(b)ϕ(a) = ϕ(b)0 = 0 . So, that implies that the kernel is an ideal of the ring R1 .
And, what does the kernel do, or rather, what does the map ϕ do? It takes the ideal kernel of
ϕ within R1 , and sends precisely that ideal to zero; other elements, it does not set to zero.
So, this is again, defining that notation of quotienting, in a very natural way. And again, we
can look at the equivalence classes that are created by ϕ within R1 . And, what are those
equivalence classes like? Use the relation R which is a and b are related, if ϕ(a) = ϕ(b) .
And, if you look at the set of the equivalence classes; one would be the kernel.
Let us give a name to the kernel. Let I 0 be kernel of ϕ ; then, there will be an I 0 as one
equivalence class; other equivalence classes would be of the kind c + I 0 , which means, this
has elements of the form c + a , whenever a ∈ I 0 . And, you can see that easily;
ϕ(c + a) = ϕ(c) + ϕ(a) = ϕ(c) . And, whenever ϕ(c) = ϕ(d) , then, ϕ(c − d) is zero. And,
therefore, this are precisely the equivalence classes that one gets.
Now, once you get these equivalence classes, let us again continue with our analogy with
groups, and try to define a quotient ring. If you recall, the quotient groups, we defined by
taking these equivalence classes, and defining a group operation on these equivalence classes.
And, we showed that is a quotient group. Can we do the same here? Let me stop here, and
leave this for the next class. You think it over. It is very natural, we can define the quotient
class very easily; but I want you to give it some thought, and we will continue tomorrow.
Modern Algebra
Lecture – 13
Rings: Quotient Rings
Let us continue from where we left off yesterday.
We started with a ring and looked at the one ideal of the ring, which is in this case was a
kernel, defined by the homomorphism. And then, we realized that this ideal induces
equivalence classes in the ring. And, now the question was that can we define another
ring made of this equivalence classes. So, did any of you think about it? No? What
would be your guess? Can we get another ring of these equivalence classes, like we got
for the groups? Yes, we can actually.
Let’s say S to be this collection of equivalence classes and define operations addition
and multiplication as follows. I am defining the sum of equivalence classes.
Now, although I am using plus sign here, but it is being used in two ways; one is the plus
within the ring R1 which is what is meant by this plus and this plus. And, this is the
addition of the equivalence classes. So, you must keep this in mind while you are looking
at such an expression. This is defined as simply c1 + c2 + I 0 . So, both the pluses are
within the ring R1 . So, you simply add these two elements c1 and c2 and plus I 0 is
whatever is the equivalence class that we get. And, similarly (c1 + I 0 ) * (c2 + I 0 ) is
defined as c1 c2 + I 0 . So, this is the definition of these operations on the set S .
Now, we need to verify that these operations on the set S makes it a ring. For that, we
have to see that it is a commutative group. So, let us look under addition. So, firstly the
closure under addition is by definition. Associativity is also clear, again by definition.
Commutativity is clear by definition. How about the identity? Well, I 0 is identity
because if when you add I 0 to any c + I 0 , by definition we will get this c + I 0 . And
the inverse, c + I 0 will have as inverse − c + I 0 , whatever that equivalence class is.
So, every element of the ring R1 is in one of the equivalence classes. So, look at
whatever equivalence class, the element − c belongs to that is same as − c + I 0 , and
then you add these two. The other properties follow. Now, for multiplication again the
closure is clear, associativity is clear, it is also commutative. The identity is only thing
one has to worry about. What is the identity? 1 + I 0 is the identity. The other properties
follow: the distributivity of multiplication over addition that also follows.
(c1 + I 0 ) * ( c2 + I 0 + c3 + I 0 ) is equal to c1 * (c2 + c3 ) + I 0 . And then, this

is same as c1 c2 + I 0 + c1 c3 + I 0 . The heart of all of this, I mean this might appear a
little strange or funny to you. The reason why all of this go through is at its core is the
following. Let’s say look at these two equivalence classes, take any element in this
equivalence class and any element in this equivalence class. Add these two elements in
the ring. That would belong to this equivalence class. c1 plus let say a1 ∈ I 0 plus c2
plus a2 ∈ I 0 is same as c1 + c2 + a1 + a2 and a1 + a2 is in I 0 . So, clearly it
belongs to this equivalence class.
So, this addition respects the equivalence classes. Two equivalence classes on the whole
get added and make it third equivalence class. Similarly, in the case of multiplication
when you look at c1 + a1 and c2 + a2 here, what do you get? You get c1 c2 + c1
a2 + c2 a1 + a1 a2 . Now last three, they all belong to I 0 . So, even multiplication of
the equivalence classes respects this equivalence class. And that is why all these
properties follow very straight forward. So, we do get another ring by quotienting. And
then, we will use a very similar terminology now, for this quotienting with a ideal.
R / I will represent that ring. And then, it follows essentially as a corollary that if ϕ is
homomorphism from R1 to R2 , then if you look at range of ϕ which is all the elements
in R2 that they are mapped to is also a ring.
We can just simply borrow this; that is, subring of R2 . And, this R1 quotiented with
kernel of ϕ is isomorphic to that subring of R2 . This is again exactly the same way as
we did for groups. The same theorem extends here. So, quotienting with ideals allows us
to create new rings. Again, just as it allowed us to create new groups.
So, let us see some examples. So, start with Z and pick an ideal of Z . Any ideal of Z is
a principle ideal, which is let’s say multiples of n ; and, you quotient with this principle
ideal (n) . We get another ring. What is that ring? Yes. So, this is again this will be all
equivalence classes formed by this principle ideal (n) . And, there will precisely be n
equivalence classes. And that is simply Z n numbers from zero to n − 1 , and addition is
modulo n , multiplication is modulo n ; because quotienting by the principle ideal (n) is
equivalent to taking numbers and operating on them and dividing by n and taking the
remainder.
So, this is again occurring. I mean, again in for groups also we had Z n , but there only
had addition modulo n , there, when this was the sub group nZ . Now, we have an ideal
which is something that has more structure. So, we get actually both the operations; all
this right.
Let us go ahead. Some other examples, what are the rings have we come across? Ring of
polynomials. And, let us look at Z [x] . What could be an ideal in this? Take any
polynomial and principle ideal polynomial generated by some polynomial. And that is
equal to what? There is a small caveat here it is simpler to describe if p is a monic
polynomial degree of d ; that is the coefficient of xd is 1. If it is not 1, then it becomes
slightly messy. It can still be described in exactly the same way, but a bit little bit of
twist.
d−1
So, in case it is monic, then it is simply { ∑ αj x j | j α ∈ Z } . Well coefficients will have
j=0
to be integer because they are all be multiples of that whatever the leading coefficient of
p , I would not say there will be multiples or with kind of will be become a little trickier
to describe. It will still be integer, but one has to be little careful when defining that.
In fact, let us take an example when p is not monic. Just think out about that. And that is
the third example we are saying. How about Z [x] and let me say divide by (2x) . What is
this? See, Z [x] contains arbitrary polynomials with integer coefficients, whenever their
integer coefficient is even, then that will be in this ideal (2x) , except for degree zero
coefficients. And then, we can take that out. But, the remaining ones can survive.
So, that would mean that this is {α0 + ∑ αj x j | αj ∈ {0, 1} f or j ≥ 1 and α0 ∈ Z } .

j≥1
That is, arbitrary degree polynomials will survive here, as long as coefficient of any
power of x is 0 or 1, then it does not fall in this ideal generated by (2x) . And, so those
polynomials or those higher powers of the x will survive. And that is what I am writing
here. So, it becomes a kind of messy to describe this ring as opposed to that ring.
Student: (Refer Time: 16: 26) 2 x or + 1 minus.
Yes, it will be more complicated too. So, but the moment it is a monic polynomial, it
becomes very easy to describe. So, that is one observation that we can make.
Student: What do we make; the any one will be one. Then, we can say that it is correct
expression. We put the condition that for j greater than 1, it should be as a; coefficients
will be 0 or 1. We can put the condition that any of the coefficients is 1 then, it will not
come in the form of 2x, all other will.
Any one of the coefficient, why not all coefficients?
Student: If like x2 + .
Take a concrete example.
Student: x2 + 2x .
Yes.
Student: So, according to what we have written, it will not come.
It will not come. Yes.
Student: But, it should come in the quotient.
No, why? x2 + 2x is same as x2 ? Is it? Yes
Student: If we do Z [x] quotient x .
Z [x] quotient x ? Yes.
Student: Then, it will become Z .
That is only Z . Yes. That is right. There is nothing else there. You can write that
Z [x] / (x) that simply isomorphic to Z . More examples; you have seen different other
types of rings as well. Yes, we have spent quite a bit of time in this larger class of
integers.
So, Z [√− 5] , for example. What could be an ideal there? We have seen examples of
ideals. That is right. Of course we can look at a principle ideal. And, let say just a
multiple of 2, 3 and that would probably not be as interesting. So let’s see, some different
ideals. How about? Sorry.
Student: (2, 1 + √− 5) .
That is one possibility. Yes, we can look at (2, 1 + √− 5) . This if you remember is a
prime ideal. What is this quotient equal to. Let us see what the equivalence classes we
get out of this when we quotent it. a + b√− 5 is an arbitrary element of Z [√− 5] . When
you quotient with this, what happens? Firstly, there is a 2 here.
So, we can reduce a and b modulo 2. So, all such multiples of two will go inside this
ideal. So, what will be remaining is a can be 0 or 1 and b can be 0 or 1. Only four
possibilities are left. So, there would be at most this four equivalence classes. No more
than that four. Further, then we have 1 + √− 5 . So, we can get rid of 1 + √− 5 . We can
replace square root of √− 5 with − 1 because we you can say a + b(1 + √− 5 − 1) .
And, that is a − b + b(1 + √− 5) . So, this goes away into the ideal.
Now, we are just left with the integer a − b . Now, all multiples of two of this will also
go in the ideal. So, eventually what will be left with would be just two equivalence
classes corresponding to number 0 which is the ideal itself and number 1.
So, take a general element a + b√− 5 . And, I will write it as this. So, I get
(a − b) + b(1 + √− 5) . Second term is in the ideal. And, I can further write this as some
residue r + 2α + b(1 + √− 5) . Because a − b is an integer, so you just write it as
depending on if it is even or odd as this form. So, r is either 0 or 1. 2α + b(1 + √− 5) is
in the ideal.
Student: It will go to the 0.
This will go the 0. So, there will be only two equivalence classes left or two elements left
in this quotient ring, which correspond to r being 0 and r being 1. So, these are the two
equivalence classes. So, we get a ring with just these two elements 0 and 1. And, what
are the operations on this ring? Well, very naturally 0 + 0 is 0, 0 + 1 is 1, 1 times 1 is 1
and so on. So, what we get really here is this is isomorphic to Z 2 .
Student: Suppose, we instead of 1 + √− 5 , we did it with 1 − √− 5 , then also we will

get Z 2 .
Z2 .
Student: But, these two rings different because.
The rings are different, but they are isomorphic.
Student: For (3, 1 + √− 5) we will get Z 3
Let see, let us.
Student: Plus something and that.
(3, 1 + √− 5) . This will be isomorphic to Z 3 . Yes. Let us look at this. Although, I had
initially said ignore this, but now we can come back to this and see what this means. All
multiples of 2 taken out, so this will leave just 4 elements. How about the operations?
This is a ring with four elements. What are the operations or the addition and
multiplication? How do they go in this ring?
Student: a1 + a2 modulo 2.
So, the addition is simply component wise. So, a1 + a2 and b1 + b2 mod 2; how about
multiplication? See, if I write, let say, this element as the following simpler form (0, 1)
that means, a is 0, b is 1 plus (0, 1) . That is (0, 0) . How? What about (0, 1)
multiplied with (0, 1) ? What is this equal to? There is √− 5 multiplied with √− 5 . There
is − 5 . − 5 modular 2 is 1. This is (1, 0) . What is this? (0, 1) . (1, 0) is simply number
1. So, 1 multiplied by anything. So, this is the multiplicative identity (1, 0) in the ring.
(0, 0) is the additive identity. And, the how about (1, 1) multiplied with (1, 1) ? This is
1 + √− 5 times 1 + √− 5 ; that is equal to?
Student: (0, 0) .
(0, 0) . So that is a funny ring. It is very simply defined. But, suddenly you find that it is
a very strange kind of a ring, which even has elements whose product or whose square is
0, non-zero element. Non?
Student: It is not an integral domain.
It is not an integral domain, certainly, not. Clearly it is; all right. So, it is a ring of four
elements but with a strange structure; Any other example that you can think of? Let me
go back to this. Some lectures ago, I had created a new example; continuous map. F is a
set of all continuous maps from [− 1, 1] interval to R .
Let us look at that. This, we have earlier seen already that it is a ring. What are ideals in
this ring? Certainly, if you collect all maps which is 0 in a particular interval, this
collection totally forms an ideal. So, if you let I be the collection of maps in F that are 0
in a given interval [α, β] . In that case, I is an ideal of F because addition of two such
maps lies in I , multiplication of such a map with any other map in F also lies in I .
What is F quotiented with I ? So, you have to see what are the equivalence classes
induced by this quotienting.
Student: All non-zero maps in this ring.
All non-zero maps in this ring?
What are the? No. We get equivalence classes. Each equivalence class will contain a set
of maps. What is the property of those set of maps? Equivalence classes will be of the
form g + I , where g is a any map. And all maps in that equivalence class will be of the
kind g + a where a is a map in I . So, what is common property of these maps?
Student: They have same value in that interval.

Exactly, they have exactly the same value in that interval.
Not necessarily constant. They all take on every point between α and β . They take one
value. For different point within α and β , the values may be different.
So, the function need not be constant in the interval [α, β] . They can be varying, but
every function in equivalence class has exactly the same value sequence or value series
in the interval [α, β] . There will be infinitely many equivalence classes.
Absolutely.
Student: Each equivalence class will have infinitely many functions
Each equivalence class will have infinitely many functions. So, it is a very much bigger
ring than we have seen so far. What this quotienting here is doing is let us focus only on
the interval [α, β] . Not care about outside this interval. Now, give me all continuous
functions, which are distinct in the [α, β] . So, the quotienting is just collecting all the
functions and making equal all those functions that agree on the interval [α, β] because
for the purpose of my analysis within the interval [α, β] those functions are distinct.
They are been made the same. And, then we have just looked at the different behaviour
within the interval [α, β] .
This is a very important observation because this allows us to, and we will see that later,
to look at the functions. Firstly we restrict our attention to a certain domain only and look
at the functions that behave differently within that domain. And these functions are
simply obtained by quotienting operation.
Let us stay with the same example. And, let me define a specific map I α, β . So, I collect
all the maps that take zero value on these two points α, β . They may take values zero on
other points. This is also an ideal. Do you agree with me? This is also an ideal. So, I can
consider F quotienting this ideal.
How does this ideal look? This has all the maps that take the same value on these two
points alpha and beta. Or, rather the equivalence class in this are those maps which have
the same value on the points α and β .
Now notice that I α, β I can write as I α * I β . See, this is an ideal in the ring F . And, all
these three are ideals in the ring F . Any element of this product has the form, and it is a
finite sum of a map in I α and a map in I β . Finite sum of products of this form - a map
in I α , a map in I β . So, any such product of a map in I α and a map in I β would be 0 on
both α and β . So every map in this would be 0 on α and β . But, there only shows one
ring. How about taking a map which is 0 on both α and β ? Can I write it as in this
form?
If a map is 0 on both α and β , can I write it as a product of two continuous maps? One
of them is 0 on α and other is 0 on β . If you look at a map f in I α, β which is 0 on both
α and β , you write it as, let say, map is g . And, look at the map; square root of g . That
is also 0 on both α and β . It is also continuous. g is continuous, and then square root of
g is also continuous. So, g is square of g times square root of g where
student: If f is negative.
Yes, if f is negative then there is a problem. So, square root of g will not always work.
Student: Over C .
But, over R ? You can do something simple enough actually. You have α here and β
here. And, for f what we know is that wherever f is coming from, f is 0 at α , not f ,
some g here. Then, it just goes all over. But, it stays 0 on β . And then, it again goes
over.
So, if you define a map which is continuous, so let us say take this simple map. This one,
this may not work right away. I think it is a very good exercise. So, instead of me doing
it let me give it you to work out an example. It is very simple. You do not have to try
anything complicated, just construct. See, you have a lot of flexibility. You have to
define any two maps whose product is equal to this map.
One of them should vanish on α , other should vanish on β . Make one of them vanish on
α , make one of them vanish on β . And, define this value on the remaining, on the
remaining point, so that the product is the values that we want. You just have to ensure
that there is continuity.
Fine, so I will leave it at that and have you work it out. So, that is it for today.
Modern Algebra
Lecture – 14
Fields
So we were at this point last time where the ideal I α,β was defined as the product of I α
and I β and I had given this is an exercise.
Now, let us look at as our next example - I α which is all functions that vanish on α .
This is an ideal. And I claim that this is a maximal ideal of this ring F . Do you see the
proof of this? So, what do you need to show to conclude that I α is the maximal ideal?
Any ideal that contains properly I α is the entire ring which means it contains one as
well.
Students: That contains I α .
No, it is contained in I α ; I α, β is actually contain I α intersection I β . So, it is a very
simple proof but it is a constructive proof. So, let us say let some g ∈ F , and g ∉ I α
which means g (α) =/ 0 . Now, consider this ideal (g, I α ) . So, this is an ideal which
certainly contains I α , and it contains this one extra function g , and whatever is the
ideal generated by that.
Now let us pick up f ∈ I α such that f (α) = 0 and f (β) = g(β) − g(α) . In this ideal,
let us say ĝ = g − f . That is simple calculation that ĝ (∖beta) is a fixed number, which
is nonzero, everywhere in the interval [− 1, 1] . So, ĝ is a function that is not zero
anywhere in the interval and actually is the fixed value. ĝ is in this ideal by its definition
1
is g − f . And 1 = g(α)
gˆ . So, this shows that the I α is the maximal ideal, so there is a
nice connection that the ideal of all function that vanish on one point is a maximal ideal
and we will extend this a little later.
But for now let us stop here with examples and let us look at two special kinds of
quotients particularly they start with maximal ideal.
And let us start with any ring R . And then you look at the quotient R by I . This
quotient ring has very special structure, because I is a maximal ideal in R . And this
special structure is that every nonzero element of this ring is a unit. Typically in a ring
units are few. In most of the examples we have seen most of the elements are non units
but this is a ring where every nonzero element is a unit. How do you prove this? Let us
pick up an element, let a + I ∈ R / I . This is a typical element. So, this, a + i is a
non-zero element in this quotient ring. The element I is that corresponds to the 0
element.
Now, I is a maximal ideal element in the ring R , and a is not in I , which means there
if you add a to the ideal I , and look at the bigger ideal - this ideal this is the whole ring
which means in particular 1 = ba + t where b ∈ R, t ∈ I . Since these two ideals
are same that means the element 1 which is in this ideal is also in this ideal. The element
1 being in this ideal simply means that it is of this form ba + t .
So, now, consider this product (b + I)(a + I) . Now (b + I ), (a + I ) are elements of the
quotient ring what is their product? It is ba + I . What is ba + I . Look at this equation,
this is an element of ba + I , so that means 1 is in this. It is 1 + I . It is a simple proof
but it is making use of the properties of this quotient ring that if two elements differ by
the any element in the ideal I then they belong to the same equivalence class.
So, here 1 and ba differ by t which is in ideal I, so the equivalence class ba + I and
1 + I are the same. This is the identity in the quotient ring. So, this is all we need to
prove that a + I is a unit because there is an element b + I whose product with this is
1 + I . This is an existential argument. Well it will be can be formulated in terms of
constructive argument, but then we will have to go into this the generators of the ideal I
itself and how do you construct the general element of that ideal because then you use
the generators of the ideal I and then write everything in terms of that.
It depends on the ring R , how complex is this ring. If the ring R is simple, then finding
inverse is easy. If it is not simple then it is going to be more difficult. So, this proof
generalises to all rings. It does not matter what the ring is and that is why by the
existential argument is better. This also means that in this quotient ring, there are no
non-trivial ideals. Every ring has this two ideals - the ideal of zero which just contains
zero, and the ideal one which contains the whole ring. This quotient ring has no other
ideal because any non zero element if it is there in the ideal then since it is a unit then
entire ring will be in that ideal, so that the ideal is one.
Now these are very special kind of rings and these are called fields. Examples, set of
rational numbers is a field, all non-zero elements being units is equivalent to saying that
division is possible in the ring. Because if you have an element a then 1/a is also in the
ring because since a is unit, so there is a b such that ba = 1 , so b = 1/a . So, if you
have to divide by a , you simply multiply by 1/a . Hence, so complete division is
possible and we have already know that in rational complete division is possible except
division by 0 is not possible, division by any non-zero element is possible; other
examples.
Student: Real numbers.
Real numbers, complex numbers absolutely are all fields. Any other example you can. Z
is not a field. Z p is a field yes. Z p is a field where p is prime. Why Z p a field?
Student: Every element maybe congruent to 1 mod p .
You can invoke that but we invoke the previous theorem also.
Student: (p) is prime ideal, it is maximal ideal.
(p) is a maximal ideal. See Z p is isomophoric to Z quotient with the ideal generated by
prime number p . Now this is a maximal ideal if p is prime and so the quotient of the
ring Z with this is field; any other example.
Student: Non-singular matrices.
They do not form the field; not just commutative.
If you add two non singular matrices the result may not be non singular but then it is a
not field, it has to be a ring first. Here is an interesting polynomial. Polynomials won’t
be a field, but a quotient of two polynomials, if you collect all, which are called rational
functions, they form a field. Yes, so for example, let me write that. For example, you can
take any field F , and you take F (x1 , ..., xn ) this stands for all functions of this form
p(x ,...,xn )
{ q(x1 ,...,xn ) | p, q ∈ F [x1 , ..., xn ], q =/ 0} .
1
So, take polynomials p and q in this ring. This ring you know F [x1 , ..., xn ] this is the
ring of polynomials where coefficients are coming from the field F and the variables
being x1 to xn . Take two polynomials p and q and take their quotient p/q , q should
not be 0, this is a field. Why, this is a ring because you can add two such functions then
you get a function, it is commutative group under addition, you can multiply also. And
there is a multiplicative inverse here. p/q has multiplicative inverse as q /p , as long as
this is not a zero polynomial. Otherwise it is a 0 element. If p is 0 then it is a zero
element. All constants, that is in the field F anyway.
Then it would not be, if this was a ring R and there is a problem yes. And we are taking
care of it very trivially. This is exactly the same way we go from integers to rational
numbers. It is a same process, so these are called as somebody pointed out that these are
called rational functions, ratio of two polynomial.
Very good question. He asked is there a ring whose quotient with maximal ideal is this
field. That is the very good question. And the answer is in one sense it is trivial but if you
do not get into triviality then it is not always true. For example, in rationals we cannot
always say that the rationals can be viewed as a ring quotiented by its maximal ideal.
Well, that is trivially because we are starting with Q that is what I wanted to eliminate,
otherwise Q[x] quotient with maximal ideal generated by x of course that is Q but that
is cheating because we have already defined Q . In fact, there are three ways of creating
fields from rings, one is take a maximal ideal quotient that gives you a field. The other
two ways they are specialized, they do not hold for all rings. This Q and F (x1 , ..., xn )
the fourth example they are constructed using that way and that is through the following.
Suppose we start with an integral domain as a ring, then we let

F = { ab | a, b ∈ R, b =/ 0} . So, just take the quotient of pairs of elements. This is a
a1 a2 a1 b2 +a2 b1
field. Why? Just verify the additivity multiplicativity. b1
+ b2
is b1 b2
, numerator is
an element of the ring, b1 b2 is an element of the ring. And since R is an integral domain
b1 b2 is not 0, since we started with b1 and b2 being nonzero.
a1 a2 a1 a2
Similarly b1
× b2
is b1 b2
, again b1 b2 is not zero, so it is an element of F and all other
properties follow pretty straight forwarded way. This is the only two places where I need
the fact that R is an integral domain. And this is exactly how we formed rationals and
this field of rational functions. Z is an integral domain, F [x1 , ..., xn ] is also an integral
domain and that is why we could do this. And the third way is to extend a field. Look at
C . We derive the motivation from there. How do you create C ?
Student: pi n (Refer Time: 21:43) polynomial.
It is falls back to that quotienting, so it is basically R[x] quotiented with x2 + 1 this is

isomorphic to C . So, this way actually is the same, one could say that as quotienting.
How do you create R ? That is trickier. How do you create R ?

Yeah, so basically we start with, let us say Q and we want to create R from Q . If you
can do that then we will have a complete picture that fundamentally we having integers.
And then from starting with those integers, we are now creating by certain operations all
of this Q is that fractional field and then from Q we can go to R , R go to C via this
quotienting. Yeah that is the question. How do we do that? In fact, what are real number?
Do any of you know the definition of real number?
But there is a slight problem with that definition. So, for example, the number 0.999999
forever and 1.00000 forever are these distinct numbers? Why?
n
9
Let us look at that. Let us say an = ∑ 10
n and bn = 1 . This represents that 1.0 all
j=1
zeros, an is 0.999 n nines are there, that is what this definition is. So, we have two
infinite sequences, a1 , a2 , a3 up to all the way this is an infinite sequences, and b1 , b2 to
be infinity, these are infinite sequence. What is bn − an . 1 minus 9 by 10 is 1 by 10, 1
minus 0.99 is 0.01, 1 minus 0.999 is 0.001, so this is the general form of the difference.
Then what is the limit of the difference as the n tends to infinity. In fact, I do not even
need to take the absolute value because this is always positive, so I can simply write this
as limit of the difference.
So these two are the same number because this number an is an approximation of the
number we had really are looking at, 0.9999 forever. So, to construct that number, I have
to look at n going to infinity and when I am goes to infinity this difference is 0. So, there
is this confusion that in the sense that we have this decimal notation of real numbers does
lists all the numbers, but it seems to be listing a number more than once.
So, then we have to worry about which numbers are equal in this sequence, which are not
equal and only the non-equal numbers are the ones that we want. It is not a very neat
representation of real numbers. Although we use it very all the time but if we think about
it we only uses rational numbers, we never use real numbers. So, but if you really want
real numbers, then this representation is not a very good one, so the defining real
numbers in the sense, in this way is also not very appropriate.
Say in fact, this problem troubled the mathematicians of 17 and 18 century because that
is when they realized that real numbers exist, that is irrational numbers. So firstly, they
believed that there only a rational numbers, then it was pointed out that √2 is not a
rational number. You can prove that very easily and that was considered as a huge wrong
because until then nobody had even thought about this. And then people started saying
square root 2 certainly is a number, so what kind of number it is. That is when irrational
number and real numbers were introduced but then one needed a very clean definition of
real number and that was given by Cauchy.
So, since we have got into this although I did not want to, let me complete this story.
What Cauchy did was he defined Cauchy sequences. It is a sequence of infinitely many
rational numbers. Slightly complicated looking definition. But if you think about it the
meaning will become very clear.
This infinite sequence of rational numbers an is called a Cauchy sequence, if for every
ε > 0 , no matter how small ε is, there exist a large enough N . So I am dropping
essentially the first N − 1 numbers from this sequence, and considering the numbers
after that. The difference between aN and am+N for any m is at most ε . Which
essentially saying that if you first drop the first N − 1 numbers, look at the difference of
any pair of numbers in the remaining sequence that difference is very, very small it is
smaller than ε . And this ε we can make as small as follows. Any small ε , there will be
a large enough N , beyond which the difference between any pair of numbers is less than
ε.
So, one of the conclusion of this is if you look at the limit of this sequence, informally at
least converging to some number, because all the numbers become closer and closer to
each other, and so it is starts converging to a particular number. And the number it
converges to is a real number. So, an example of this is just this for example, this { 101n } ,
1 1
this is a Cauchy sequence. Why, pick any ε , and then let say | − |, this
10m+N 10N
1
difference in absolute value is at most, how much? Atmost . So, for any ε , I can
10N
always choose a large enough N , so that the difference is less than ε . What does this
sequence converges to? 0, it is going towards 0.
On the other hand, if you have this Cauchy sequence, let say
1 14 141 1415
{3, 3 + 10
, 3 + 100
, 3 + 1000
,3 + 10000
...} . So if you have this sequence, does this
look familiar? Firstly, this would be a Cauchy sequence. The numerator here is the
decimal expansion of π , so this is 3, 3.1, 3.14, 3.141, 3.1415 whatever is the expansion
of π is. The difference between any two after beyond a point the numbers is very tiny, it
generally goes to become smaller and smaller, and this converges to π . That is why the
expansion will be infinitely long. For every successive digit in the decimal expansion of
π will have one number here. Each one of this number is the rational number, and it is
getting closer and closer to the number π .
Which definition?

Yes, yes that problem exist here also. Yes, but then we have to classify that which of
these numbers are not unique, and they are equal, so we have to create equivalence
classes of these real numbers, which corresponds to a unique number. After all when we
says real numbers is a field, we want one element to represent a unique element.
Whereas, in this representation there is no uniqueness property.
So of course, we understand what real numbers are. There is a real line and every number
in there is a real number but that is an informal understanding. And you want to formally
define it, we would like to do it in a very precise way which gives a unique
representation for every number. There is lacking in that representation, so that is what
Cauchy did.
I will continue with this tomorrow.

Modern Algebra
Lecture – 15
Cauchy sequences and real numbers
Yesterday, we defined Cauchy sequences.
And, I am going to use this to define real numbers.

Consider any Cauchy sequence. Take the limit lim an where an is in the sequence. This
n→∞
limit exists by virtue of the fact that successive numbers in the sequence are closer and
closer to each other.
So, it is going to converge to a number, and I am defining that number to be a . And, one
can show again by using the property that its successive numbers are closer and closer to
each other and so this limit is unique, i.e if there are two numbers with the same limit,
then their difference will go to 0. And, this number a is we call a real number.
OK.
Some of it is doesn’t converge. But if you just look at the harmonic, an is 1/n then take
the limit, it goes to 0.
Student: Term goes to 0.
Term goes to 0. So, that limit of harmonic series kind of Cauchy sequence is 0. So, what
is the advantage of defining a real number in this way? One thing, we can say that are all
real number can be expressed in this fashion. And everything that is expressed in this
fashion is a real number. That intuitive notion of real number does match with this
definition.
But, what is clear is that there are multiple Cauchy sequences converging to the same
real number like 1/n , that limit converges to 0, 1/n2 , limit converges to 0, 1/n3
converge to 0. So, that is not really solving the problem that I stated earlier. But, it solves
one or two other problems about real numbers, which I am fairly sure you did not pay lot
of attention to whenever you came across real numbers.
So, let me ask you. If I give you two real numbers, can you add them? If yes, then how?
How to start with? Where is the right most? There is no right most. It is forever for
infinitely long sequence.
It is not computationally; I am not asking to design an algorithm. Of course, you cannot

design an algorithm. But, if it is conceptually, can you figure out a way of adding them?
OK.
That you can write. Sure. Given real number, let say this kind of infinite size inputs can
be viewed as following that you are given access to a black box, where if you put number
m , you get the m th digit of that real number; m th digit after the decimal point, let say.
So, if you are given such an access to two real numbers and you want to, now you can
actually talk about an algorithm. This algorithm takes a input, of course these two black
boxes and also a digit number k . And, produces the k th digit of the sum after the
decimal point. How do you do that?
Student: k th digit
That is what the decimal representation of real number means that it should be able to
produce, given a k , it should be able to produce the k th digit of the number after the
decimal point.
Any number, whatever, not just k th digit, but we can say give me k plus first digit, k
plus second digit, all the digit we can access it. But, the question is what are the carries?
How do you know?
We won’t know the carry because it goes for forever. And, there is no pattern in that in
general. So, you can’t really add two real numbers by this decimal representation. Of
course, we cannot multiply them either.
Student: I can give an error bound.
You can give an error bound. Yes. But, nothing beyond. So even addition or
multiplication on real numbers in that representation is not well defined. And, that is a
bigger problem with that representations and somehow it is intuitive to an extent but it
does not quite work for real numbers. It only works for rational numbers. So, that is why
you need an alternative representation to define such operations and Cauchy sequences is
perhaps the best such representation. The simplest such representation because a Cauchy
sequence represents a real number. So, now addition of two real numbers is as follows.
So, we are given Cauchy sequences {an } and {bn } . Then can defined as this sequence
{an + bn } . Just component-wise addition and this is a Cauchy sequence {an + bn } .
Why? We just need to verify that property that given an ε , there should be a large
enough N , such that the difference of aN + bN with aN +m + bN +m is smaller than ε .
And that is done simply. Let me show it for this. And then, I will not do for
multiplication. You can do it yourself.

So, pick that N a for an sequence such that the difference is atmost ε/2 after N a and
similarly N b for bn such that the difference is also less than ε/2 . And, let N be the max
of N a , N b . Then, |an+m + bn+m − an − bn | ≤ |an+m − an | + |bn+m − bn | . Look at this
difference; that is bounded by ε . So, we have shown that same property for this sum
sequence. And, similarly we define multiplication.
Just the component wise product and the same property can be shown that this is also a
Cauchy sequence. And, it converges to the product numbers. And that is easy to say.
So, now it is very easily defined. And, that is one great advantage. Now at least we know
what addition and multiplication means over the Cauchy sequences. Now, that we have
it. If we define, let say R to be set of all Cauchy sequences. And, we have this lemma
that R is a ring under addition and multiplication; because if you look at the addition, all
0 sequence which is the identity of addition, the − an and this is inverse of an . And, for
multiplication all one sequence is the identity of multiplication. And, the multiplication
distributes over addition. So, all those properties are satisfied and therefore we get a
commutative ring.
Student: It is an infinite sequence.

It is an infinite sequence. Yes.
Student: That is also (Refer Time: 13:18).
Formal power is a infinite sequence. Yes, but there the coefficients need not have a
particular pattern of the kind we want. So, here we want very strong pattern, that is,
successive numbers must get closer and closer to each other. So, we have a ring but we
still have not resolve the problem of multiple representations. In fact, let us look at all
representations, all Cauchy sequence that represent the number 0.
So, let I be the set of all Cauchy sequences which converges to 0. The name gives it
away. This is going to be an ideal. Not only any ideal, it is a maximal ideal. Firstly, why
is it an ideal? It is a commutative group under addition. That is closure under addition. If
you have a two Cauchy sequence both converging to 0, then their sum also converges to
0.
If you have the Cauchy sequence in I and multiple any other sequence to this that would
also converge to 0 because if an converges to 0, then an * bn , as long as bn are bounded
above by some number, it also converges to 0. So, clearly I is an ideal. Why is it a
maximal ideal? That is the question. Well, let us pick up any Cauchy sequence that is not
in I . What does this mean that bn is not in I? It means that bn converges to; this
converges to let say b and b is not equal to 0.
Now, let us define another sequence or numbers; these are rational bˆn which take the
value bn , whenever n ≤ N and 1/bn whenever n > N and N is number chosen, so that
after N , bn is not 0. Such a N exist follows from the fact that the bn sequence
converges to a non-0 number.
So, there will always be a large enough N , beyond which bn is not never going to be 0;
because it comes so close to b that it can never become 0. Now, let me give you as an
exercise that {bˆn } is also a Cauchy sequence converging to 1/b . Once we have shown
this, then the rest is straight forward. Even this exercise is not difficult at all. But once we
have this, then just consider the Cauchy sequence which is obtained by multiplying bˆn
and with bn . This is a Cauchy sequence. How does this look like? Equals some numbers
up to first N numbers, after that it is all ones, directly by definition.
Student: Multiplication of bˆn and bn .
Yes. See, if you look at the definition of bn , above N it is 1/bn . So, 1 over bn times bn
is just 1. So, all the numbers beyond N will be 1. The numbers below that may or may
not be 1. Infant let me write this. Given name to this let us call them c1 , c2 to cN . These
are some rational numbers.
Also, now note that this number or this sequence which is first N numbers is c1 to cN .
And, all the subsequent numbers are 0s. This is a Cauchy sequence clearly because it is
converging to 0 and beyond a point, everything is 0. All differences are 0. And, this
belongs to the ideal I .
Now, what is the difference between these two? So, let us call this sequence C . Then,
{bˆn } * {bn } − {c1 , ..., cN , 0, 0, ....} . This is - the first N numbers are c1 − 1 , c2 − 1 , up
to cn − 1 . All other numbers are 0. It is still a Cauchy sequence. Actually, Cauchy
sequences has the property that first few numbers can be arbitrary, it does not matter. As
long as after certain point, everything is starts converging. So, this is a Cauchy sequence.
It converges to 0. So, it is in the ideal I and the difference between this is all ones.
This is the one. So, now, we have the bn was Cauchy sequence which is outside the
ideal. This is in the ideal. So, in the ideal generated by bn and I , we get the one, which
is therefore it is an entire ring. And therefore, this shows that I is a maximal ideal. Add
any other additional, any new element to I , you get the whole ring. Now, we can define
our class of real numbers.

R = R / I . This will have a unique number or unique element corresponding to every
real number. And, since I is the maximal ideal, it is a field. So, this is the formal
definition of real numbers. Somewhat non-intuitive or little more complicated. But, this
is the way you can properly define real number. The school definition is very informal.
Sorry.
When you are actually working with numbers, then there are always rational numbers.
So, that is why that is fine.
Student: Now, additional representation will be in this quotient ring.
In the quotient ring, yes.
And, now multiplicative inverse also exists because this being a field.
So this process, the whole thing that I went through, you know, starting with rational,
defining the Cauchy sequences. Then, quotienting the ring of Cauchy sequences with a
maximal ideal to get another field. So, of course this also involves that quotienting a ring
with the maximal field together to get a field, but there is something else also done. You
start with the field, you introduce Cauchy sequence of that field and then do this way.
This process is called a completion of a field. And this, we can define for many other
field, not just with rationals.
As long as we have a notion defined of nearness; right, two numbers are close to each
other. And, they get closer and closer to each other. So, that notion must be available on
that field and we can define completion.
So, in particular, for example, we can ask what is the completion of R itself. That is, you
start with R , define Cauchy sequences with elements of R , get that ring, maximal ideal,
quotiented it and get a field what do you get. And that is a theorem. You get back R . R
is therefore, in the sense the ultimate or the final point where we can extend this
completion starting from rational sequence.
That is right. That is what the completion is. That is how the name derives that you
complete this gap between ratios. That is right. And, this whole process if you see
Cauchy sequences, etcetera, that is trying to fill that gap. So, that is all I will talk about in
this completion.
Let me get back to the fields. And, we were discussing the ways of getting fields from
rings. One way was quotienting with maximal ideal and another way was starting with
the integral domain and taking the fractions; that is also a field.
Now, in order to do that we have to start with integral domains, which not all rings are.
But, starting with the general ring, we can actually get very nicely integral domains. That
is a next theorem.
Take a ring R . Take its ideal I and quotient R with I . The ring you get is an integral
domain if and only if I is a prime ideal. Proof is straight forward. Let us say a, b be
elements of the ring, not in I . Now, consider the quotient ring. The quotient ring will
have elements of this form; this is equal to ab + I . Now, if in fact why we start, so let
start with a general two ring elements. And, in general this is true. Now, if ab + I is I ,
this implies that ab ∈ I . Infact, this is if and only, goes both ways.
And, now let us say suppose R / I is not an integral domain. What is that mean? That
means that there exist a and b , so that (a + I ), (b + I ) are non-0 elements of this
quotient ring, whose product is 0. So that means, (a + I ), (b + I ) , which is ab + I is 0,
which means ab ∈ I . right. And, a + I and b + I are non-0, which is equivalent to
saying that a ∉I and b ∉I , such that ab ∈ I . And, these imply that I is not prime. That
is by definition. Conversely, if I is not a prime ideal; that means, there exist
a ∉I and b ∉I such that ab ∈ I . And, this implies ab + I is equal to I . And, this
implies?
Student: It is not an integral domain.
It is not an integral domain, a pretty straight forward. So, now we have this way of
creating fields as well. Start with any arbitrary ring; take any prime ideal, quotient with
the prime ideal, you get a integral domain and then take the field of fractions.
Now, the fields are very special kind of rings, of course. They are very interesting
because the full arithmetic can be done here. They are the only structures where the
complete arithmetic is possible. All four operations; addition, division, multiplication,
subtraction can be carried out completely. And therefore, they are extremely useful.
So, we would like to understand their structure as well. We have tried to understand the
structure of groups and rings. So, let us try to understand structure of fields. Firstly, you
first realise that fields have more structure than rings because division is also available.
They have to satisfy. Now, everything is the unit, so those strange elements which are
non-units do not exist in a field. So, it is expected that it should have more structures.
And, it is indeed true that fields have more structure. More things can be proven about
fields than for rings. And, one of the key thing or key parameters associated with the
field is its characteristic.
The characteristics of a field is the smallest number n such that when you add 1 to itself
n times, you get 0. Now, this 1 and 0 are the identities of the field. There would be fields
where no such sum would ever be 0. For example, for rational numbers, you keep adding
one; the number keeps growing up. It will never become 0. So, let me add that. If no such
n exists, then F is said to be of 0 characteristic, otherwise of characteristic n .
0 characteristics because it is does not quite fit into that but really this is if you add one
to itself 0 times, of course you get 0. That is true with everything. But that is the only
way possible in some fields. So, that is why we call 0 characteristic. So, examples;
Characteristic 0, these are rationals, reals, complex numbers, field of rational functions
over F . These are all characteristics 0. Wait a minute; where this field where F is of 0
characteristic. Otherwise, finite characteristic Z p ; this is a field which is obtained by Z
quotienting with prime or maximal ideal (p) . And, this is of finite characteristics
because in this field if we add one to itself p times; that is 0.
This is more often written as F p to highlight the fact that they are fields of
characteristics p . Then F p (x1 , ..., xn ) ; this also has finite characteristic, in fact, another
simple to prove theorem, but important one.
That characteristic of the field can be either 0 or a prime number. I leave the proof to
you. It is very easy to show. So, what is the structure of fields, in general? The answer is
not straight forward at all. Even though fields have more structure, there is all kind of
fields, particularly when we talk about infinite fields. They are of various kinds, I want
to highlight one special kind, which is algebraically closed.

But, before that let me highlight one very important property of fields, which is called
fundamental theorem of algebra. Have you come across this, the fundamental theorem of
algebra? No. I am sure, you have come across it. When I state it, you will realise.
So, take a field F and take a polynomial P in one variable x with coefficients coming
from the field F , such a polynomial is called a polynomial over the field F because its
coefficients are from the field. Let degree of P is d . Then P (x) has atmost d roots in
F . You have seen this? That is the fundamental theorem of algebra.
You must have seen this for specific fields; reals or complex numbers. But, this holds in
general over all fields. And, the proof is same as the proof for this special case, in case
you recall it. So let, we can prove this by induction. So, for d equals one,
P (x) = cx + c′ . This implies that there is exactly one root. Correct. And, what is that?
Notice that c is not 0 because it is of degree one.
Now, induction step. Assume for d − 1 . Now, you want to prove for d . Let α ∈ F be a
d d
root, then P (x) = P (x) − P (α) , which is same as ∑ ci xi − ∑ ci αi = (x − α) Pˆ (x) .
i=0 i=0
Since P (α) = 0 , by assumption. So, I can just write it in this way. Now, if you look at
this two difference, you can take term by term, a common factor of x − α and collect
with the rest in this Pˆ (x) . Now, Pˆ (x) is a polynomial of degree d − 1 .
Now, the rest is straight forward. Now, you just invoke the induction. This will have at
most d − 1 roots, Pˆ (x) by induction. And, what are the roots of P ? Any root of Pˆ (x)
with α that is possibly the dth root and no other. Any other value of x , where x − α is
non-0 and Pˆ (x) is non-0 will result a non-0 value because it is an integral domain. So,
product of two non-0 value will remain non-0. In fact, this properly holds not only for
field, but for any integral domain. So, that is all for today. We will continue tomorrow.
Modern Algebra
Lecture – 16
Properties of Fields
Now, let us continue our discussion about fields. Last time we proved this so called
fundamental theorem of algebra. In fact, there is another fundamental theorem of
algebra which basically depends on who calls which one the fundamental theorem,
which I will discuss little later in the lecture. But like we saw last time that every field
has this nice property that a d degree polynomial will have at most d roots in it.
The question next would be exactly how many of those d roots are actually in the field.
It is possible that there are 0 roots in the field. For example, look at polynomial x2 + 1
over the reals. It can have at most two roots, but actually it has no real roots. So, the
numbers of roots will be lying between 0 and d and that is a very important part of
investigations about polynomials over fields and that also gives us a way to create more
fields or extend a field. So, let us look at that.
So, let us say, you start with a field F and we consider F [x] . This is the ring of
univariate polynomials over field F . This ring has some nice properties. For example,
F [x] is an integral domain. Every ideal of F [x] is principle. Moreover, the unique
factorization property holds in F [x] . Note that here I am not saying that F [x] is just a
Dedekind domain where the unique factorization over ideal holds, but element wise
also, just like over Z , element-wise unique factorization property holds, similarly here
not only over ideals. So, in that sense, F [x] is very similar to the ring of integers. See,
that every ideal is principle, unique factorization property holds, it is an integral
domain, which are all very nice properties of integers.
The first one is quite clear, that is, a(x), b(x) ∈ F [x] and they are non-zero. Then,
a(x) · b(x) , this polynomial will also not be 0. The reason is very simple. You take the
highest degree term of a , highest degree term of b , their product will give the highest
degree term of ab and that coefficient will be non-zero. So, this is simply borrowing
the fact that F is an integral domain and therefore, F [x] will also be an integral
domain, right. So, that takes care of the 1st property.
The 2nd one, every ideal in F [x] is principle. So, take an ideal of F [x] and take two
elements of this ideal. Now, we have this operation available in F [x] , that is, which
was taking GCD of the polynomials. Just like we can take GCD of the numbers, we can
take GCD of the polynomials and let that be c(x) and then c divides a(x) and b(x)
both, right. Also, we can write c(x) = α(x)a(x) + β(x)b(x) , right. So, this implies that
c(x) is also in the ideal I .
This equation is, essentially, just GCD algorithm that you give you this. So, c(x) is in
the ideal and both a(x) and b(x) are multiples of c(x), right. So, we can say that
a(x), b(x) are in the ideal generated by c(x) . So, with these two properties we are done
because now you start with in the ideal I , the smallest degree polynomial that is
present in the ideal I .
Now, take any other polynomial of the ideal I . So, let us say, a(x) ∈ I is the smallest
degree. Then, what can we say about c(x) ? b(x) is any other polynomial in the ideal
I , c(x) must have the same degree as a(x) and c(x) divides a(x) . That means, c(x) is
just a constant multiple of a(x) , right, which means b(x) is in the ideal generated by
c(x) and c(x) is just a constant multiple of a(x) . So, principle ideal of c(x) is same as
principle ideal of a(x) . So, b(x) is in the principle ideal of a(x) and b(x) was any
arbitrary polynomial in the ideal I . So, this shows, I is simply the principle ideal
generated by a(x) .
And the third one, the unique factorization, well, let us assume, that
a1 (x)b1 (x) = a2 (x)b2 (x) . So, there are here two factorization of one polynomial .
Before that I need to define the notion of irreducible polynomial, that is okay, but that
is already defined. Remember, I had defined the notion of irreducible element of a ring,
which generalizes the notion of a prime number in integers. Irreducible element was an
element such that whenever you write it as a product of two elements, one of those two
elements is a unit.
So, suppose we can write a polynomial as a product of two irreducible elements, a1, b1
is one set of irreducible elements, a2 , b2 is another set of irreducible elements. So, now
I want to show that this is the same, that is, a2 is equal to either a1 or b1 and
similarly, b2 either equal to a1 or b1 .
Consider GCD of a1 , a2 . What is this going to be? Observe both a1 , a2 are irreducible;
Whatever the GCD polynomial is, that divides both a1 , a2 . Because they are
irreducible, that GCD is either equal to a2 or 1. So, if GCD is equal to a2 , that means,
a1 is equal to a2 except a constant multiplier. So, that is fine. Then, of course, we can
argue the same way about b1 , b2 and we have done. So, if GCD equals a2 , we are
done. If GCD equals 1, then, then consider GCD of b1 , a2 . This is the same thing.
Again, this also will either be a2 or 1.
Student: a2 will divide b1 .
Yeah, if it is a2 , then a2 divides b1 , since both are irreducible are just constant
multiples of each other. Otherwise if this GCD is also 1, then that is a bad case for us.
Then, what happens? We will show that it is not possible. How do we show that it is
not possible? GCD of a1 , a2 is 1 and GCD of b1 , a2 is 1. Let us look at what is GCD
of a1 b1 with a2 and I claim, that GCD of a1 b1 with a2 is 1. Why? Well, it is just a
bit of calculation, GCD of a1 , a2 equal to 1 means what?
αa1 + β a2 is 1, GCD of b1 , a2 is 1 means what? γb1 + δ a2 is one. So, just combine

these two, just multiply this with b1 . So, we get αa1 b1 + βa2 b1 is b1 and multiply
this with γ , we get αγa1 b1 + βγb1 a2 + δ a2 is 1. This is b1 and b1 you substitute in
this, you get this equation and this gives me αγa1 b1 + (βγb1 + δ)a2 is 1. So, this show,
that GCD of a1 , b1 and a2 is 1. And that is not possible because clearly, a2 divides
a1 b1 by the assumption and that is it, and proves the theorem.
So, this ring F [x] behaves very much like the ring of integers.
Now, let us consider a general polynomial P (x) be in F [x] . This is polynomial degree
d and let us goes back to that the fundamental theorem of algebra where we are
looking at the roots of this polynomial P (x) in the field F . Since unique factorization
holds, I can write P (x) as, let us say Q1 (x)Q2 (x)...Qk (x) , where Qj (x) is irreducible
and this can be written uniquely. Now, when can we say that P (x) has the root in F in
terms of Q1 (x) to Qk (x) .
Q has.
Qi is in F [x] , each one of them is in F [x] .
See, the fact that I can write here the product of this irreducible polynomial, each one of
Qj is in F [x] .
Student: (Refer Time: 21:23) P(x) will be (Refer Time: 21:25).

Root of P will be root of one of them, yes, clear, right. If P is 0, then one of them will
be 0 that is one conclusion we can derive. But suppose, α is a root of P and α is in F
, can we say something about the nature of this, at least one of these Qj ’s? See, just go
back to this proof. If α was a root of P , then P I can write as this and which is
(x − α)Pˆ (x) .
Student: Linear factor.
Linear factor, so if α is a root of P and α is in F , then one of these Qj ’s will be of
degree 1 and it will be actually of the form x − α where α is a root.
Student: Linear is irreducible.
Any linear polynomial is irreducible, of course.
Student: Up to multiplication by unit.
Up to multiplication by unit; that multiplication by units is fine. So, that tells us

whether by looking at this factorization of P we can see how many roots of P are in
F that equals the number of linear factors. So, there is no guarantee that all of them are
linear factors. So, some of them may not be linear factors. So, let us say, suppose Q1 (x)
has degree greater than 1. So, it is not a linear factor.
Now, consider this ring, F [x] quotiented with the principle ideal Q 1. What can we say
about this quotient ring?
Sorry.
Elements of this, it depends on the degree of Q1 .
Let degree of Q1 be d1 . Elements of F̂ have this form where each element is a

polynomial of degree at most d1 − 1 with coefficients being in F because you are
quotienting with Q1 , which is a polynomial degree d . Quotienting with Q1 , take any
element in F [x] , you can divide this again, just the GCD trick will give you a
polynomial.
After dividing you will get a residue, which will be of degree less than d1 and that is
the element in this. Strictly speaking, the element is this plus the ideal. The elements of
this quotient ring have the form a + I . So, this is the a and plus the ideal. In that plus
ideal I am not writing to simplify this, but you must keep in mind that the elements of
this quotient ring are not polynomials, they are equivalence classes. What else can we
say about F̂ ? We can say that F̂ is a field.
Student: Q1 (x) is irreducible.
Q1 (x) is irreducible, that is, that is by factorization, yeah, it follows because this ideal
generated by Q1 (x) is a maximum ideal. We have seen, it is a principle ideal of course,
but it is also a maximal ideal because Q1 (x) is irreducible. So, if you take any
polynomial that is not a multiple of Q1 (x), its GCD with Q1 (x) will be 1 because
Q1 (x) is irreducible, which means, that this principle ideal generated by Q1 (x) is
maximal. And so, F̂ is a field.
We do not need degree greater than 1, but what if Q1 (x) had degree 1, what is F̂ ?
Then, F̂ is just isomorphic to F because then, elements are of the form α0 plus the
ideal Q1 (x) that is just F. Whereas, if Q1 (x) has degree greater than 1, then we get a
field F̂ , which is not equal to F , it certainly contains F , but it contains elements

more than F . Why are we sure that it contains element more than F ? Simply because
all such elements in F̂ which are polynomials of degree less d1 and two such
polynomials are not same in F̂ . In particular, a root of Q1 (x) is in F̂ . Do you see

that?

What plus 0, did not get that.
Classes corresponding to 0 that is not root of Q1 (x) .
Student: Q1 (x) , since you are quotienting with Q1 (x) , that is kernel so it will map to 0.
The Q1 (x) will map to 0, so that is fine, but in F̂ I am claiming. That is an important
point.
Let me, instead of doing it with x , let us redefine F̂ as F [y] / (Q1 (y)) . It is the same
field, only thing that has happened is, instead of variable x , I am using variable y . So,
in the same field, in the sense, this field is isomorphic to the field that earlier I defined,
fine.
Now, I ask, what is a root of Q1 (x) in F̂ ? Consider Q1 (y + (Q1 (y))) . This is an

element of F̂ . Now, I am writing it properly with the clear understanding, that this is
an equivalence class. This equivalence class is an element of F̂ . So, if you replace x
by this element of F̂ in Q1 , what do you get?

This would be Q1 (y) plus a multiple of Q1 (y) . You do not see that, let us do it more
slowly; let me erase this. Let Q1 (x) be this polynomial and do the substitution. Let us
write I for this ideal. The moment I do this substitution, Q1 (y + I) is now becomes an
element in the field F̂ because y + I is an element in F̂ and I am just doing some

addition, multiplication using that element.
So, this is element in field F̂ , which element is this? Let us carry out this arithmetic.
This is a field, you can do the arithmetic and we know exactly how do you do
arithmetic on the quotient rings. This is same as; multiplication would mean, (y + I)j
is same as y j + I , correct. And this sum would be just the sum of this part plus I . This
is Q1 (y) + I and this is I because Q1 (y) is contained in I . So, therefore what we got
was that Q1 was this element y + I is I in F̂ . So, y + I is a root of polynomial Q1 (x)
in the field F̂ .
So, therefore we conclude, that Q1 has a root in F̂ , which means Q1 is no longer
irreducible in F̂ . So, Q1 further factors in F̂ . So, I can write Q1 as its irreducible

factorization in F̂ . At least one of those factors is linear but there may be still some
higher degree factors. So, we can repeat this whole exercise with the higher degree
factors. So, this iterative process will eventually give me a field. Field, let us call it F ′
such that all roots of P (x) lie in F ′ . There is a name for this particular field, is called as
splitting field of the polynomial P .
All these fields, that we get starting from F̂ and continuing all way up to F ′ , these
fields are called algebraic extensions of the field F . We started with F , we took a
polynomial, which does not factor completely in F and used that polynomial to create
a larger field and successively larger and larger fields till the point that that polynomial
factors completely. So, all the field that we get in this sequence are algebraic
extensions. Yes.
Student: (Refer Time: 42:28) polynomial x (Refer Time: 42:30) minus x square plus 1
(Refer Time: 42:33).
Yes.
Yes, that is right, that is right.
Student: So, those which were already linear in F , so what will happen to them?
They will stay, those root stay. See, any extension contains the smaller field, contains
again in the isomorphic sense, right, because now an element α of F in F̂ becomes α
+ I , that is the corresponding element in F̂ , right. F̂ is quotient field, so the every
element is in equivalence class. So, it is corresponding to α in F , the element in F̂ is

α + I . So, this is what an algebraic extension is.
One of you asked in the last class how do elements of algebraic extension look like?
This is exactly what they look like. They are polynomials of a certain degree in the base
field, the starting field. The key thing is that we must start with a field, look at the ring
polynomials, quotient it with the irreducible polynomial of degree more than 1, we get
extension field which contains the original field. Its elements are polynomials in the
base field. Again, when I say polynomial, this is strictly speaking not true because they
are equivalence classes of polynomials. But I am just going to continue saying they are
polynomials because it is kind of easier to describe that way.
Student: Isomorphic does not mean they are equal.
Two fields are isomorphic we take them to be equal. Not equal, you are right, they are
not identical. But if two fields are isomorphic then you can treat them as identical.
Student: Can a ring be isomorphic to a field.
If two rings are isomorphic and one of them is a field, other has to be a field, it has to
be that. See, the isomorphism means, both additions, multiplication are preserved. So, if
a × b is 1 in 1, then ϕ(a) × ϕ(b) would be 1 in the other, so that means, the inverse
will also exist in the other. So, in that sense isomorphism essentially captures the same
field. In fact, we have to use a notion of isomorphism.
Why do we consider isomorphism? Well, let me define ring of integers as this. Let me
define the following. This is a collection of different set of elements, the symbols I am
using are different, but the operations in both these rings go in the essentially the same
way, ai + aj = ai+j , ai aj = aij .
Would you say that this is also ring of integers? Instead of choosing symbol 0, 1 to
represent numbers, I am choosing different symbols to represent the ring, but they
represent essentially the same entity, and the only way I can establish the equivalence
between them is through isomorphism. They are not equal, but they are isomorphic and
that is why, when algebraic objects are isomorphic, we treat them as identical, alright.
Yes.
Student: How we see algebraic extension as a vector space over the field.
How do we see it? Now, you tell me.
Go back to the example, consider F̂ , what were the elements of F̂ ? F̂ had as

elements polynomials in y of degree less than d1 plus I , of course. This I we will
ignore, that is basically the zero elements. Now if you just look at polynomials in
variable y of degree less than d1 , this forms a vector space over F . I can view F̂ as a
d1 dimensional vector space over the field F with basis being 1, y , y 2 , ..., y d1 −1 . Any
element of F̂ can be written as a linear combination these and these are linearly
independent also. So, this falls out of the way we have created extensions.
Student: Scalar multiplication (Refer Time: 51:30).
Scalar multiplication is through F . Any element of F is treated as scaler here. So,

now, the next question is, we now understand the algebraic extension of a field, are
there other types of extensions of a field? So, extension is basically you start with a
field and create a bigger field. So, algebraic extension we have seen how to do that.
There is one more very important type of extension of a field, which is called
transcendental extension.
So, again start with a field F and let F̂ be the field of all rational functions in the
variable x over F. This field F̂ is a transcendental extension of F; this is clearly not an
algebraic extension. Firstly, it is an extension because F̂ contains F, but there is really
no polynomial involved in this extension. Another way of seeing is that F̂ is not a

finite dimensional vector space over F because it contains polynomials of arbitrary high
degree as elements. So, non-algebraic extensions which include an infinite dimensional
extension are called transcendental extension, but this is not of much interest for us. We
will not really worry too much about this.
Let us get back to the algebraic extension. So, ask the following, is it that we can do an
algebraic extension from any field, that is, take any field, can we find an algebraic
extension of that field which is larger than this? The answer is no for the simple reason,
take a field, consider F , let us say starting, take a polynomial and go to the splitting
field of F with respect to that polynomial. Then, take another polynomial over F ′ and
go to the splitting field of that polynomial with respect to F ′ and keep doing this
process. So, essentially just keep of course, this is an infinitely long process, but you
will eventually have a field which contains roots of all polynomials in it, okay.
So, add roots of all polynomials. This is a little informal that notion, that I am giving
you because I am not precisely defining it. So, let me, instead of that let us just go to
the definition. So, we have a field F where every polynomial over F has a root in F ,
then such a field is called algebraically closed field because you cannot extend this
anymore algebraically. You take any polynomial that has a root in F , take that root
out, you get another polynomial of lower degree that will also have root in it. So, every
polynomial of degree d will have exactly d roots in such a field F , and so, it is not
possible to algebraically extend such a field.
And earlier, what I informally talked about, keep extending it, keep extending it forever
gives a way to visualize that there exist an algebraically closed field, but that is kind of
not a very concrete example. So, let me just end up with a theorem.
The field of complex numbers is algebraically closed. This is also called fundamental
theorem of algebra.
Sorry.
Student: We can find d roots in C.
Yes, every polynomial of degree d has exactly d roots over complex numbers.
Modern Algebra
Lecture – 17
Finite Fields
So, in last lecture we defined algebraic closure of fields and this theorem I said, I am not
going to prove. It is just a statement that said complex numbers which is a field, is
algebraically closed which means that all polynomials defined over complex numbers
have the roots within that. But this does not mean that C is the biggest field that there is.
There are fields which are bigger than complex number and the examples are easy to
construct.
The fact that C is algebraically closed only rules out one way of extending a field. We
can simply look at this field. This is the field of all rational functions in variable x with
coefficients coming from complex numbers. This is the field and is certainly is bigger
than complex numbers because there is no corresponding complex number to the
variable x .
And you can formally prove it by showing that there is no isomorphism between C [x]
and C and then you can continue C [x1 , x2 ] , this is even bigger and so on. So, there is
really no such end or there is no such largest field because any field that you have, you
can always attach a fresh variable to it, look at the rational field with respect to that and
then that is a bigger field. So, this is a chain which keeps on going, but these are two
(Refer Time: 02:28) fields so to speak for us to consider.
We, in this course, we will now step back a bit and let me define a very interesting class
of fields, which is called finite fields.
So, as the name suggests, this is fields with finite number of elements. This is not the
usually field that you encounter, but we know that there exist fields which have finite
number of elements, F p where p is a prime number is one such field, right. So,
examples would be F p which is simply Z quotiented with the prime ideal generated by
prime number; that is certainly one. Can you think of any other finite field?
Student: F p / (x2 + 1) .
F p upon.
Student: Z p upon irreducible polynomial.
Yes, so, that is, I think your idea is correct. Look at the ring of polynomials in variable x
over F p . Take a maximal ideal here. So, this is ring of polynomials in variable x . This
ring I think we have already seen, every ideal is a principle ideal, right. So, if every ideal
is principle ideal, then any prime ideal is maximal, that is very easy to see. So, take any
prime ideal here. Since it is a principle ideal, essentially it means, take an irreducible
polynomial over F p and take the principle ideal generated by that polynomial. So, P (x)
is a polynomial with coefficients coming from F p and it is irreducible, it does not factor
in this ring F p [x] , then the principle ideal generated by polynomial P (x) is a prime
ideal and maximal ideal. So, if we quotient this ring with the maximal ideal, you get a
field.
What are the elements of the field? We have known that as well. The elements are, if
P (x) is a polynomial, let us say, of degree d , then elements of this field will be
polynomials of degree less than d , all such elements. So, how many such elements are
there? pd ; every coefficient can take p values and there are d possible places. So, this is
pd . So, this has therefore, pd elements and we write this field as F pd , just to highlight
the fact that it has pd elements.
And then, there is a big theorem, which says, that for every prime p and number d
greater than 0, there exists a unique field, which call F pd of size pd . So, these fields that
I just listed out, these are unique, that is, there is exactly one field of size pd . This is
unlike again the infinite case. When you look there are many fields with infinite size,
rational, reals and they are not isomorphic, whereas this saying all fields of size pd are
isomorphic to each other. So, there is essentially only one. Further, every finite field is of
this type; that is, there is no other finite field. So, this completely describes all finite
fields.
It is not difficult to prove except the uniqueness part. See, you can easily prove that every
finite field must be of this type. Firstly, look at the characteristics of a field. We have
already seen characteristic must be a prime number in the sense, it is finite, the
characteristic cannot be 0 and if we look at 1 plus 1 plus 1 plus 1, eventually after finite
number of additions you will repeat and therefore, that sum will become 0. So, its
characteristic has to be a prime number.
Secondly, since the field is finite, you said the characteristic is p , then you start with F p
, which lies inside the finite field and then every other element of the field is algebraic
over F p , again because it is finite. If you look at every other element, say, α and look at
α, α2 , α3 and successive powers, after finitely many powers, you will have again a
repetition, that is, a power will be a linear combination of previous powers and so every
other element is algebraic. Therefore, this field is an algebraic extension of F p and then
one can show, that there is an irreducible polynomial in F p [x] so that if you quotient
that irreducible polynomial with this, you will get the field.
So, that is the broad outline of the proof, but I do not want to give the full proof. It is
really not necessary, important thing is this theorem, allows us to completely characterise
finite fields.
Now, the finite fields behave in a way, which is at times very different from normal
fields. One prime example of this is this group F p * . This, if you recall, is the set of all
non-zero elements of F p and when we look at this set under multiplication, it forms a
group under multiplication, it is a finite group, what is the structure of this? Now, just
bring in your knowledge about finite groups into play here. We had this, right, when we
were discussing group theory I gave this theorem which describes a structure of finite
groups. Do you remember that theorem? A theorem said that the any finite group is
isomorphic to.
Student: Groups of prime.
Yes. So, basically direct sums of Z p ei for various primes and numbers, right. So, let us
i
start from that fact. Since F p * is a finite group, we have F p * is isomorphic to

Z p1 e1 ⊕ ... ⊕ Z pk ek . I am writing this additively, whereas this is the multiplicative group
but I hope you understand the meaning. The group operation is treated as the
multiplication of (Refer Time: 13:21).
Now, the next question, that I am going to ask is, are these primes distinct, p1 to pk ? In
the finite group structure they need not be distinct, but in this case are they distinct? And
the answer to that is yes, they are distinct. Why? Well, let us, here is a very simple trick.
Suppose p1 = p2 , then I am going to pick two elements; α and β are the following
kind. If you look at this group, Z p1 e1 , this is a cyclic group with one generator. The
generator being number one and the size of this is p1 e1 . So, then there exist an element
in this whose order is p1 , namely that element would be in this group, would be p1 e1 −1 ,
that is one such element. So, instead of calling it p1 e1 −1 , I am just calling it α and if
there is an element α whose order is p1 in this. Similarly, in Z p2 e2 , there is an element
because p1 = p2 there is an element β with that order as p1 .
Now, consider, consider these elements (aα, bβ, 0, 0, ..., 0) , that belong to this direct
sum where a, b ∈ [0, ..., p1 ] . How many such elements are there; a and b both takes
exactly p1 values from 0 to p1 − 1 . So, there are exactly p1 2 such elements and each
element has order p1 and this is isomorphic to F p * .
So, now, let us switch attention to F p * and let me ask the following question. Consider
this equation Q(x) = xp1 − 1 , how many roots does this polynomial have in F p * ? Since
this is isomorphic to this, we can ask the analogous question here that how many
elements are here whose order is p1 . So, this xp1 − 1 = 0 means, xp1 = 1 , which means,
whatever x that satisfy as an order p1 in the multiplicative group.
So, analogously we ask, since this is isomorphic to this, how many elements of this have
order p1 or at least p1 2 ? And say there is an isomorphism, so each one of those p1 2
elements will give rise to a unique elements here in F p * . This means Q has at least p1 2
roots in F p . Is this possible? The fundamental theorem of arithmetic, this is a field, this
is a polynomial degree p1 , can have at most p1 roots, it cannot have p1 2 roots.
Fundamental theorem of algebra, hence p1 is not equal to p2 . So, F p * is now, we can
say, isomorphic to Z p1 e1 ⊕ ... ⊕ Z pk ek , pi =/ pj .
Now, consider the following element, γ = (1, 1, 1, ..., 1) , what is the order of γ in this
group?
Student: 1.
Order means, the identity of this group is all 0s. So, order means, that how many times
does this element need to be added to itself to get a 0?
Student: multiplication of all primes.
If, let us say, if aγ = 0 , then we have (a, a, ..., a) = 0 . This implies, that
a = 0 (mod pi ei ) ; that is the only way because that is a very simple way. And since each
pi is a distinct prime, this implies that a = 0 (mod ∏ pi ei ) . And what is this product
i
size? This is exactly the size of the whole group. So, this implies, that γ is a generator of
the whole group and since γ is a generator of this group, there is a corresponding
isomorphically generator of F p * , which means, that F p * is cyclic.
So, there is only one generator that will generate the entire F p * , which viewed in terms
of F p * means, that there is an element, let us say, g ∈ F p * such that when you
consider successive powers of g , list outs all non-zero elements of the field. This is in
contrast with, let us say, Q* . The Q* does not have finite number of generators itself.
Since there are infinitely only primes, each prime and its power, essentially, is one
generator; each prime is one generator. So, its successive powers are handled, but there
are infinitely many generators, whereas for a finite field, exactly one generator in the
multiplicative group. So, finite fields therefore, have a much simpler structure compared
to the infinite fields.
And one particular type of finite fields has been used very widely in computer science,
which is this. Take p equals 2, F 2d . So, fields with exactly 2d elements. And the
reason they are of special interest is, because in computer science, whenever you write
any data, you write it as a bit string. So, since this field has 2d elements exactly and if
you think each elements is essentially a polynomial of degree less than d with
coefficients being 0 or 1, that is exactly analogous to a d -bit strings. So, this is
analogous to set of d -bit strings and often it is very useful to associate this meaning with
d -bit strings that they are elements of F 2d because then, you can do arithmetic over it.
So, let me give you one example of this, which is an application without which modern
life would not be possible. What is one of the cornerstones of the modern life?
Communication, right, all kinds of communication, which happens over radio waves and
microwaves and all kind of waves; not only that, another very important facet is storage
of data. You know, we store enormous amount of data.
Now, one problem common to both of these applications is, like if you are storing a data,
what if some part of the data gets corrupted? Similarly, if you are sending data from one
place to another over, let us say, radio channel, what if along the transmission, some part
of that gets corrupted? Then, at the receiving end you have that corrupted data which is
not that useful at times. Similarly, if you are reading the data which is stored and it is
corrupted, then may not be that useful. In fact, in both the application it happens
repeatedly. In communication channel, radio waves, there are always going to be
disturbances, you know, electric disturbances and all kinds of disturbance, which will
bring in some amount of distortion in the data.
In the storage, I mean, if you look at the CDs, DVDs, where the data is stored, I mean I
am sure each one of you has put scratches on those, the back surface where the data is
stored; putting a scratch means the data stored there is erased. And so, it is important,
that in such a situation we can still recover exactly what was sent or what was stored. So,
how is done? Do you have any idea that comes under the broad field of error correction;
the question is how do we do error correction?
So, we can model it as the entire data is a long string, let us call it X of size n and that is
to be either stored or sent. Now, at the receiving side we may have errors or after reading
we may have errors. So, how do we get rid of this? And let us start with the simplest
case, that there is one bit of error. Now, my target is to recover from that error. So, how
should we store this X or how should we send that X so that finally, that error is
recovered, one bit? Say some solution.
I will give you a very simple solution, send X again. That is two copies of X . So,
instead of sending X or storing X , we store X X . So, then, one bit of error means, one
copy of X gets corrupted. So, suppose, suppose 5th bit of X gets corrupted, but the
other X will have the correct 5th bit. So, can we recover from this, the correct 5th bit
value? We would not be able to because then, what when we read, we will see the 5th bit
in two copies, in one of them it is 0 and one of them it is 1. The question is, which is the
right value? We do not know which actually bit got corrupted. So, send, storing two
copies of X is not good enough.
But let us say, send 3 copies of X and then one bit gets corrupted. Now, we are fine
because then, we have 3 copies of X , look at the three copies of 5th bit, two of them will
be uncorrupted, one of them will be corrupted. So, two values will be original values, the
3rd value will be the wrong one. So, we will either have 0, 1, 1 or 1, 0, 0, so pick the
majority occurring bit value that will correct the 1-bit error; that is very simple scheme.
So, that is one way of recovering from one bit of error.
See, the important thing is here is that after receiving we do not really know where the
error is occurred, the only thing we know is that some error may have occurred. And we
are, let us say a priory told that look at most 1-bit of error can occur, then we can
recover. But this is really a toy example, I mean, we should expect much more than one
bits of error to occur.
Student: It can be like error can be at same position. First message error can be at i th
position, second message error can be at i its position.
Yes. So, that means more than one bit of error. So, if let us say, two bits of error has
occurred, then X X X would not work, because.
Student: We know one bit of error.
OK.
Student: Getting in two messages.
Yes.
Student: Both the messages come at the same place.
OK.
Student: So, (Refer Time: 31:52) taking the majority.
Yes.
Why? I mean X X X is my message and we are assuming that one bit of error will occur
somewhere here.
Student: Same bit position error occurs in both of them.
How can it occur? Only one bit of error can get corrupted. In X X X , almost one bit will
go wrong. I am not saying that in each copy of X one bit can go wrong, but this is, like I
said, a very toy example. In real life, the errors will be many more. So, if you say, two
bits of error, which is still a toy example, even with two bits this X X X strategy would
not work. Just like you pointed out, if two bits in the same location of two copies of X
that gets corrupted, then we would not. So, to recover from two bits of error, we have to
actually use 5 copies of X .
So, in general, using this strategy, recovering from t bit errors requires 2t + 1 copies of
X because in the worst case, those t errors can be in the exactly the same position of t
copies of X and then we need t + 1 more copies, which is the majority copies where that
error has not occurred and that can recover. So, this is a possible scheme. The only thing
is, that this is not really a practical scheme.
Let us take for our calculation the example of a DVD. What is the capacity of a DVD?
About 5 GB, right, typically and suppose we want to recover from and one scratch.
Although it may be thin, because the data is so densely packed, one scratch can really
erase large number of bits, tens of thousands of bits actually and there can be multiple
scratches also.
Let us say in the example case, DVD of capacity 5 GB, number of errors is say 1
megabyte. So, multiple scratches are allowed in here, 1 megabyte is a reasonable
number, still compared to 5 GB is a very tiny number. So, N for us is 5 into, gigabytes
is 109 into 8, it is a byte, so number of bits would be 4 * 1010 bits, and t is 8 * 106
bits. So, these many bits can go wrong and this is the total storage capacity available.
Then, if we implement this scheme which I just described, this is the final capacity. So,
what would be n ?
If you have total capacity is given and number of errors bits is given and we are
implementing this earlier strategy, n is equal to capital N /(2t + 1) , correct. n < 104 ,
that is, 10000 bits. So, if you implement this scheme, you can, in the whole DVD of 5
gigabyte size, you can store data worth 10000 bits, which is a little more than 1 kilobyte.
So, this is a completely worthless scheme if you want to practically implement this. And
the same situation will occur in transmission. I mean, you transmit enormous amount of
data in order to send some small amount of real information; that is unacceptable. So, we
need a better scheme.
What is the ideal situation? So, we have n bits of data, t errors we want to tolerate, what
would be the ideal situation? If t bits get erased and you still want original n bits to be
present. So, you need at least n + t bits. So, ideal situation is, that capital N is n + t ,
cannot hope to achieve better than this at all, but can even come close to this. That is the
question and the answer is yes and it is only thanks to finite fields, that we can come,
very close to this.
In fact, we can make N to n + 2t , just a little bit more. So, how do we achieve that? So,
for that let us use some more knowledge that we have developed. We have been talking
about polynomials quite a bit and the fact about the roots of them in the fields, and so on.
So, let us do the basic set up in this lecture and then we will continue tomorrow.
First thing first, that let us first chop X into small pieces, a0 a1 ...am−1 with each length of
each ai being equal to l , which is, since there are going to be m pieces, it is n/m . Now,
define a polynomial Q(x) = a0 + a1 x + ... + am−1 xm−1 . The coefficient of the
polynomials are coming from the message, ok. Now, we let α0 , α1 , ..., αk−1 be distinct
elements of a field F . We chose F so that each of ai ∈ F .
Now, we let bi = Q(αi ) and then, output b0 , b1 , ..., bk−1 . So, the original string or
message was a0 a1 ...am−1 . I have done this transformation of that message and eventually
got this b0 , b1 , ..., bk−1 and this I am claiming is the coded message. So, this is what is
transmitted or stored.
Now, t of them will get corrupted. Suppose t of bi are corrupted and still I want to
recover the original message, which is a0 a1 ...am−1 . That was associated with this
polynomial, so that is equivalent to the polynomial Q because coefficient of polynomial
Q gives exactly the message. So, now let the question, that I need to ask is, I am given
this sequence with up to t of them corrupted, is there a polynomial Q such that this
evaluations of Q on a0 a1 ...am−1 produces values such that at most t of them differ from
the given values, that is, if I can complete the polynomial, then I have recovered the
messages unless there are more than one polynomials. If there are two polynomials or
three polynomials of this property, then I have a problem because then I would not know
what that original message is. So, let me ask the following.
See Q1 , Q2 differs from the given sequence in at most t locations. Then, then if you
look at the Q1 sequence and Q2 sequence, there they will differ on at most 2t locations.
So, therefore, Q1 − Q2 will be 0 on at least k − 2t locations. When I say location, I
actually mean these values, b0 , b1 , ..., bk−1 , on at least so many locations. So, this is the
polynomial, P is the polynomial of degree m − 1 and it is 0 on at least k − 2t values,
which means that k − 2t is less than m or in other words, k is less than m + 2t .
So, which means, if I choose k = m + 2t , then there cannot be two polynomials. See k
has to be less than m + 2t in order for two polynomials, Q1 and Q2 to differ with the
given sequence at most t locations. And if I choose k = m + 2t , there cannot be two
polynomials, there will be a unique polynomial and that unique polynomial will be Q
which is the original one. So, I can recover that. So, therefore, we need to choose
k = m + 2t , fine.
So, let me stop here.

Modern Algebra
Lecture – 18
Application of Fields
So, we left at this point yesterday.
This polynomial P was defined which is we know since Q1 , Q2 differ on at most t

locations with a fixed string, so this polynomial is 0 on at least k − 2t locations. And
that gives us, if we choose k = m + 2t , it will guarantee that will only be a unique
polynomial within t errors of the received string. So from this now let us try to see, what
is the size blow up that happens? We started with n bits, if you remember.
And in this algorithm what we did was that we choose m such that m = n/l , l is the
length of each aj we chopped the message into l bit blocks. And k was the number of
points on which we evaluate the polynomial Q which is of degree m − 1 .
So, let us just put it all together continuing from there, we have k = n/l + 2t ; t is the
number of errors we wish to tolerate, and this tells us the number of points k on which
we need to evaluate. Now this is dependent on two parameters; one is l , and one is k . If
the number of point on which we evaluate the polynomial is k and we concatenate all
the evaluations what is going to be the resulting size of the string we get.
Look at this. This is a polynomial Q , we evaluate this polynomial on αj and we get bi

which is Q(αi ) and then concatenate all the bi . What is the size of this output? It is
about k times the size of each bi . Of course, different bi may have different size, but if
you assuming that they are approximately the same then approximate estimate would be
k times of the size of bi . So, what is the size of bi ? Well, depends really on the field on
which we are working. Let us say begin with integers, that is the most natural starting
point that each ai I treat as the integer between 0 and 2l − 1 and then evaluate this
polynomial and then see what is the result we get.
So, what is going to be the size of the number bi when you evaluate polynomial Q on
αi . It is a degree m − 1 polynomial.
Student: Maximum possible αi
Let us say maximum αi . In fact, since we have to choose k αi it could simply be 1 to

k . We could have done with any k αi , so maximum is k . Then look at this polynomial,
so for x will replace by k . We will get k m−1 here multiplied with am−1 ; this will be the
biggest number, all the other numbers will be smaller. So if we just take this, the
magnitude of this number am−1 has magnitude about 2l and this will have magnitude
about k m−1 .
So, overall magnitude would be, you can say bi is roughly 2l + mlogk . That means, how
many bits will each bi require. l + mlogk bits. So, the total size even if we ignore log k
factor, let us say l + m bits. So total size here would be k (l + m) , that many bits would
be needed. In fact, there will be more bits needed, but this is a very conservative estimate
of the bits required. And the value of k we derived from here was this.
So, if we use integers, size of output is (n/l + 2t)(l + nl) , and this is
n + 2tl + n2 /l2 + 2tn/l . This is the output size. And if you notice typically we want l
to be a small, we do not want to be dealing with too big numbers. So, the dominant term
in all of this is going to be this; n2 /l2 , because t is also comparatively much smaller than
n , l is comparatively much smaller than n . So, n2 /l2 is the dominant term. So which
tells us that for starting with n bit of data we have to roughly go to about n2 bits of data
in order to recover from t bits of error.
Depending on what the value of t is, this may or may not be improvement on the trivial
scheme. There it was about 2tn here is about the n2 , it really depends on the value of t
that this may be better or even be worse than the this previous scheme. So, it does not
seem to be gaining us much until we realize or notice that all of this calculations we did
could have been done over any field. There was no necessary, it was not necessary to do
it over integers or rationals.
(Refer Time: 07:40)
And the problem with all this that arises over integers is that the size of this bi becomes
humongous, very large. We started aj are l bit long and suddenly we have see bi
become l + m bit long so there is a whole value m gets added which is a pretty large
number. And these arise as a result of arithmetic operations over aj . You know a way of
keeping the size of bi limited, that is can be choose a field of operations so that the bi
will also has small size
Yes.
Different values. Field should be large enough, but that is certainly true. My question is
can we reduce the size of bi so that the output size is small. When we choose everything
over integers or rational, when you talk about to field this size is blowing up.
Student: F p .
Exactly, do it over finite field because no matter how much arithmetic you do over F p
you stay within F p and the size of an element of F p is only log p bits long. In fact,
going back to the particular finite I earlier talked about it very naturally fits in here. See
each ai is l bit long. So we treat ai in F 2l and everything. Now these will also be F 2l ,
but here the important point is that we need k distinct αj so 2l must be at least k that is
important. So, l cannot be chosen very small. We have to choose l slightly large enough
so that it is atleast k.
And now, what is the size of b when we do this evaluation? Well, bi is also an F 2l , so
in it is just an l bit long number. So, instead of this we now get this to be exactly equal
to k l bits, each bi l bit long and there are k of them. So, k l bits. So now, let us go to
this.
Again, let me just start from here again and do a calculation. Size of output is
k l = l (n/l + 2t) = n + 2tl . Now we get something very significant. This is telling us n
is the size of input, size of output is n + 2tl . Just an additive part. And notice that n is
going to be very large on the order of giga bytes when you talk about storing in a DVD.
t is may be 1 MB? If you remember, 1 MB right. I picked number of errors 1 MB. So
about 1 mega bytes here and choice of l is ours. So we choose it to be small enough to
take k values, so it would not be very large.
So, overall if you see the addition in here it is going to be much smaller compared to n ,
and that is what gives us a real error correction with very little overhead; Ideally, the
overhead was again I mentioned here n + t , start with n and t is the increase in size.
Here I am getting n + 2tl so is not quite the best, but it comes very close to being best
specially for CD’s DVD because the kind of error that occur there, a scratch it would not
erase a bit. A scratch is many bits long in width. So, we will basically remove large
number of contiguous bits. Now if you store each element here in l contiguous bits. So
when we remove a bunch of contiguous bits and say t bits contiguous in the extreme
case, it basically means we are removing t/l elements. And so this factor of l will go
away.
So, the lose is simply n + 2t in that case. It is a rough argument that I am giving, but it
does come down to very small. And that is why the code that I described is variation of it
is the one that is used to store the data on CD’s and DVD’s because it gives excellent
performance with very little additional space required for correcting this. How did this
become possible? Finite fields. If finite fields we were not there it would just not be
possible. In fact, all communication, this is the code that I describe is used for storage in
DVD’s CD’s, but for communication over microwaves, radio waves, a variation of this
algorithm is used but that also critically uses finite fields. In fact, just over any error
correction algorithm uses finite fields and for the same reason. Otherwise, the size after
doing arithmetic keeps blowing up.
Although, seemingly innocuous and also seemingly very abstract entities, finite fields are
at the heart of model life I would say, model life will not be possible because if you do
not have error correction if you call somebody you would not be able to hear what the
other side is saying. No data storage may be possible proper, there will be errors in the
data. Communication, internet would not exist, nothing will exist. And we arrived at
finite field just by this pure abstract algebra.
We said, let us abstract out, right in the beginning, the key concepts of arithmetic and see
whether we can generalize it, we generalize it and generalize it and we say here are rings,
quotient rings, fields and then we found here, there are whole bunch of kind of fields
some of them are extremely useful like finite fields. So, I hope this convinces you that
the algebra that is we is studied is really fundamental and it required this process of
abstraction can lead to unexpected benefits.
So, this ends one example, I will end with very quickly of describing another example
which again something I promise probably in the first class, geometry. So far most of
this course has been spent on looking at arithmetic numbers, in what way we can study
number by generalizing it and abstracting it out. The other fundamental objects of study
in mathematics are; geometric objects, curves, surfaces, and you would like to study
them also. Studying them is more difficult than studying numbers. The numbers they
have nice structure we have already identified. A lot of excellent structure and a lot of
tools have been developed to analyze them.
Now, on the other hand geometric objects have a problem that you first have to visualize
them. When you want to talk about a circle you have this picture in mind that is a circle
and then you have a tangent and trying do something. For with simple geometry objects
or curves it is fine, but any time you go a little more complex objects, let us say high
order curves. Or even a circle over complex numbers, can you visualize it? x2 + y 2 = 1 ,
where x and y are complex numbers. You cannot, because it is no longer a two
dimensional object. In fact, if you look at this curve, this also came up somewhere in one
of the analysis. And in order to solve for this we had to go up to not just integer, but we
have to extend the ring of integers to this algebraic integers and argue over that.
I mean, there is a complex number also that came in there right, some part of i , was it
√− 2 . Did we have √− 2 ? So, basically there is a some complex numbers also come in,
so we are actually looking at this curve over complex numbers. But now, we didn’t use
geometry at all, we just simply use the basic numbers or arithmetic to derive the solution
that you want; but we found only one solution.
Student: We (Refer Time: 19:50).
Right we gave one solution and we can say that other solutions cannot exist. But curves
like this for this specific curve we can do it, but in general curves of this kind if you want
to analyze we are not always able to do or solve it as simply as we can do it for this
curve. So, there it becomes important to analyze the geometry of this curve. But how do
you analyze geometry of this curve when it is over complex numbers. Over reals, how
does this curve look like. When x3 is between 2 or negative, in fact x less than cube root
of 2 it does not have any solution and x above cube root of 2 it will have solutions, so it
will look something like this I think. Something like symmetric around x axis, will look
something like this, that is over reals.
How does it look like over complex numbers? Hard to even imagine. Now with some
more advance analysis, we can figure out what it looks like over complex numbers. It is
nothing that you would imagine. This looks like a donut or tyre. So, it is not a two
dimensional object any anymore, it is a three dimensional object, because now you have
complex numbers so there are four independent variables in the complex number,
because there y and x both are complex so shape of this curve is like donut shape.
It is a three dimensional because one dimension is canceled by the equation. You see this
is two dimension equations but there are equation so it eventually becomes a single
dimensional object. So, when you have four terms and there is equation setup and there
are two equation setup; real part and complex part. So that reduces a the dimension. It
becomes a two dimensional object but it is rendered in a three dimension space just like
this is a single dimensional object but rendered in a two dimensional space, that is totally
unexpected. There is really no way can make a guess that it would look like this. So the
question therefore is how do we study the geometry of such curves, imagination just
gives up not possible. And there again the algebra that we have developed comes to our
help.
So, let us look at a simple curve which I will just use an example, may be the simplest
possible; a circle. It is nice over reals we can visualize it, but over complex number like
I said, hard to visualize. What I am going to do is abstract out all the geometric
properties in an algebraic object. So let us start building up this, we will just call a
dictionary between geometry and algebra. So, what are geometric things - curves, points,
tangents these are geometric entities and will build a corresponding concepts on the
algebra side.
So first, it is a simple two dimensional space that we are working on. So, let us first build
that the base and it will be useful to start with an algebraically closed fields so we will
just stay with complex numbers. The two dimensional space over complex numbers, that
is where this curves exist. This is not too difficult, this is something we already have
been doing. We have been studying geometric objects like circles with algebraic
equations like this, and this curve is simply a polynomial which lies in this ring. This is
all polynomials in x and y which correspond to all curves over C 2 two dimensional
space. So, that is a simple correspondence.
And in the school this is where we stopped and a little bit more, but let us goes beyond.
Points. A point in C 2 what does it look like. Evaluation, no we will just stay in this ring.
Just like this is the space encompassing everything all objects we are interested in.
Similarly this is now the ring where everything that we want should be there. So, this
contains point.
Student: Constant polynomial.
Constant polynomial; what constant polynomial? So it should correspond to let us say p

is (α, β) , what object do we associate here? We think in terms of curves you will not be
able to set up a correspondence, think in terms of ideals.
Student: Quotient ring.
No, just look at this principle ideal; (x − α, y − β ) . And I say that these correspond to
each other and this is not a very easy fact. One side is clear that if you look at the point p
and look at this ideal. Every every polynomial in this ideal is 0 at p , because every
polynomial in this ideal can be written as something times x − α plus something else
times y − β and at (α, β) it is 0. So, one way is clear. How about the other way that is,
in order to establish this exact correspondence between these points and ideals, we say
here is this point here is an ideal which corresponds to this point. I should also say in
order to make this correspondence unique and there is no other ideal that corresponds to
this point.
In other words every polynomial that is 0 on this point lies in this ideal. So, if you think
about it is not at all simple to conclude.
Well, it is not quite like that. What we are shown is one way that every polynomial here
vanishes at point p , but can there be any other polynomial not in this ideal which
vanishes at point p . This requires theorem to prove and it is a very famous theorem. Was
proved by Hilbert very famous mathematician, it is called Hilbert’s Nullstellensatz. This
theorem was the beginning of algebraic geometry, because this is what allowed
mathematicians to establish this dictionary completely. The theorem has a general
statement and a more specific statement can be given in terms of this. Let I be an ideal
in this ring and corresponding to ideal you define a set of points p so that all
polynomials a(x, y ) in ideal I vanish at p . So collect all such points. And let b(x, y ) be
in C [x, y ] such that b(p) = 0 . Take any polynomial b which is also 0 on all points in
V.
Then there is a power of b that belongs to I . This is one connecting, so this rules out
what I just asked earlier. That if you start with an ideal which vanishes on a set of points,
take any polynomial that also vanishes on the set of points that polynomial may not
necessarily lie in that ideal but some power of that polynomial lies in that ideal, and this
is the best you can say anyway So, this establishes correspondence between ideals and
set of points.
And now going back using Nullstellensatz, we can say that any polynomial that is 0 on
point p , some power of that polynomial must belong to this ideal. And since, this is a
maximal ideal in this ring. I think we have shown this, this is a maximal ideal in the ring,
If I have not shown it is easy to show. This is a maximal ideal in the ring. Since this is a
maximal ideal, if power of a polynomial lies in the ideal then that polynomial also lies in
the maximal ideal, that is again using maximality we can prove it.
So, now putting everything together we get that this is precisely the set of ideal of all
polynomials that are 0 on this point. So, this ideal uniquely identifies this point and vice
versa. So points correspond to maximal ideals. In fact, we can go further and show every
maximal ideal in C [x, y ] has the form (x − α, y − β ) . This also requires some bit of
work. So now, we have a very nice correspondence that points correspond to maximal
ideals. How about curves? With curves we have to be little careful. A curve can simply
be union of two lines for example, that is a reducible curve that is you can factor that
curve into two lines.
So, when I talk about curves I will talk about irreducible curves, which cannot be
factored like; circle, ellipse, parabola, they cannot be factored further into 2 curve. And
again using Nullstellensatz, we can prove that this correspond to prime ideals. So prime
ideal corresponding to a curve, if the curve is let us say c(x, y ) , it would be (c(x, y )) , it
can be shown that the curve is irreducible if and only if its ideal is prime. But this is very
preliminary because really what we want to study is curves and points on that curve, the
property of points and various other things on a particular curve. Right now we have all
points to correspond to maximal ideal, curves corresponding to prime ideal.
So let us go further; points on curve c(x, y ) = 0 , they correspond to maximal ideals of

ring C [x, y ]/(c(x, y )) . Take the quotient ring, quotient with ideal generated by c(x, y ) .
Now maximal ideal of this quotient ring correspond to points on this curve c(x, y ) = 0 .
Student: when we say curve then how do you identify the curve.
We can say since we already have we can always write a curve in algebraically that is
y 3 = x2 − 2 or whatever that equation is. So that much of a correspondence we already
had, but going beyond that points on that particular curve correspond to maximal ideal of
this string. If you want to study properties of points of a curve will just need to study the
maximal ideal of this ring.
Now, something quite magical happens once we are up to this. This is actually called the
coordinate ring of curve. Since this c(x, y ) is prime, this quotient ring is in an integral
domain, we have seen this. So, since it is an integral domain, we can form the field of
fractions, of any integral domain you can have field of fractions.
So the field of fractions of this curve is represented as C (c) . Now, with a little bit more
work, which is not very difficult but it requires some bit of algebra. Of course, these
fields of fractions also have the all the maximal ideals there. It is now become field of
fractions, so we have to be little careful in defining what that maximal ideal looks like
here. But all those sets are because the coordinate ring is subset of field of fractions, so
the maximal ideal also are here and this also happens to be a Dedekind domain. In fact,
this is also a Dedekind domain. So, which means that every ideal here can be uniquely
factored as product of prime ideals, this is the property of the Dedekind domain.
Now there is one more property that it satisfies that it is also a principle ideal domain,
which means that every ideal in this ring is principle, which in turn means that every
prime ideal is maximal. So, every ideal therefore in this ring can be written as a product
of maximal ideals, maximal ideals correspond to points. If you think of this as a ring of
numbers, it is just a ring with lot of nice properties, so you can view them as a ring of
numbers. So, the prime numbers in this ring are points on the curve. And since this is a
ring we can do full algebra on the ideals here, that in turn corresponds to doing certain
very interesting operations on points of the curve. Now, then you have this field of
fractions associated with it this is the full arithmetic you can do in this field of fractions.
So, in this sense we have associated with a curve, a field which is just a collection of
numbers of certain kind where full arithmetic is possible, and the arithmetic on this field
has some very tight connection with the operation on the curve. And since arithmetic we
understand and we can do like I just mentioned earlier that we can do arithmetic in a
much more nicer way, we have many more tools to do arithmetic, so we can understand
that curve and its properties by arithmetic in this field.
And this I indeed resulted in a lots of properties of curves which otherwise we would not
have been able to prove. Some simple example being let us say, two curves of degree,
one curve of degree 2 another curve of degree 3, you understand what degree of curve is,
the highest degree term in the curve. And how many points can they intersect at max? 6,
why? See a line and a degree two curve can intersect in at most two points, in fact a line
and degree d curve can intersect in at most d points that is simple.
Student: Higher (Refer Time: 42:20).
But higher degree let us say degree d1 curve and degree d2 curve what is the maximum
number of points they can intersect.
Student: Maximum of d1 and d2 .
Maximum of d1 and d2 ?No. It is d1 × d2 . Take a circle and an ellipse, you can the max
they can intersect at 4 points. That is just an example, but in general any degree d1 curve
and any degree d2 curve, as long as they do not have a factor in term I mean both the
curves have a line in common then it is a trivial thing. But otherwise the maximum
number of points they can intersect is d1 × d2 .
You cannot eliminate a variable; see if you can eliminate a variable yes. But, the curve
could be any arbitrary, may not allow you to eliminate variables. And that is proven
using algebraic geometry that is take out the curve look at this coordinate rings and then
argue over this. That is a very simple fact. There are many more interesting facts you can
prove. They are also very useful in finding out integer solutions of equations like, one
example was y 2 = x3 − 2 you are looking an integer solutions of that and we saw that
you have to go to higher algebraic number and find a solution, but that was not very ad
hoc way. We can study solutions on families of curves of certain kind by again looking
at the geometry through the coordinate ring.
So, I will stop here.

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MODERN ALGEBRA

Prof. Manindra Agrawal

S.No. Topic Page No.

10 Rings: Ideal Arithmetic 118

12 Rings: Dedekind Domains 146

17 Finite Fields 209

(Refer Slide Time: 00:20)

(Refer Slide Time: 07:05)

Student: The curve

Student: There is some abstraction.

(Refer Slide Time: 23:47)

Student: (Refer Time: 26:49).

Student: (Refer Time: 27:06).

Student: Let’s assume some properties of addition.

Student: Any object that is countable.

Student: (Refer Time: 29:20).

(Refer Time: 29:24).

Student: (Refer Time: 29:26).

Student: (Refer Time: 29:32).

Either power set of count countable (Refer Time: 29:36).

Student: (Refer Time: 29:37).

Student: It is an ordered set with addition and multiplication.

(Refer Slide Time: 34:26)

(Refer Slide Time: 41:20)

Student: Multiplication is repeated addition.

(Refer Slide Time: 00:13)

Then the associativity holds for multiplication, multiplication satisfies commutativity

Student: Distributive(Refer Time: 08:19)

(Refer Slide Time: 09:50)

Student: If you are (Refer Time: 11:00) then it is valid.

0 × a = 0 , that is another interesting observation. Is it true it is true for numbers, but is it

By definition of 0, 0 = a − a . Now use distributivity property of multiplication over

(Refer Slide Time: 24:21)

Student: Even integer.

Student: Positive rationals

Student: (Refer Time: 36:48).

Student: Identity of a pair (Refer Time: 36:55).

Of the same form odd by odd?

Student: Yes, then identity is 1 by 1.

1 by 1 again same, correct.

Student: And other properties are basically.

So, it is a isomorphism is a very special kind of homomorphism and it is the isomorphism

Okay, time to stop today, and then we will continue tomorrow.

(Refer Slide Time: 00:29)

Student: (Refer Time: 01:44).

(Refer Slide Time: 08:27)

Now, ϕ is an isomorphism. So the range of ϕ is entirely G1 × G2 . So, that means in this

Student: (Refer Time: 13:40).

(Refer Slide Time: 14:09)

Student: Groups G1 and G2 will be finite.

Student: If we add the, see G1 consists of only h1 .

Multiples of h1 . G1 considers only multiple of h1 . G2 consists of only multiples of h2 .

Student: And after a point

But, is that a contradiction?

Student: If both of them are finite.

Student: One-to-one relationship is violated.

One-to-one relationship is violated?

Student: Because integers are

G1 is infinite, G2 is finite, yes.