Dummit Foote
Dummit Foote
Dummit Foote
If G/Z(G) is cyclic with generator xZ(G). Every element in G/Z(G) can be written as xk z
for some k ∈ Z and z ∈ Z(G). Now, let g, h ∈ G, then g = xa z and h = xb w for z, w ∈ Z(G).
We have gh = xa zxb w = xa+b zw = xb+a wz = xb wxa z = hg.
3.2.4 Show that if |G| = pq for some primes p, q (not necessarily distinct), then
either G is abelian or Z(G) = 1.
3.2.8 Prove that if H and K are finite subgroups of G whose orders are relatively
prime then H ∩ K = 1.
|H||K|
Since H ∩ K ≤ H and H ∩ K ≤ K and |HK| = |H∩K| . This implies that |H ∩ K| | |H|,
|H ∩ K| | |K|. Since |H| and |K| are relatively prime, we have |H ∩ K| = 1, hence H ∩ K = 1.
3.2.10 Suppose H and K are subgroups of finite index in the (possible infinite)
group G with |G : H| = m and |G : K| = n. Prove that l.c.m.(m, n) ≤ |G : H ∩ K| ≤ mn.
Deduce that if m and n are relatively prime then |G : H ∩ K| = |G : H| · |G : K|.
l.c.m(m, n) ≤ |G : H ∩ K| ≤ mn
3.2.18 Let G be a finite group, let H be a subgroup of G and let N E G. Prove that
if |H| and |G : N | are relatively prime then H ≤ N .
Let f : G → G/N with kernel ker f = N . Consider f |H : H → H/N , then |Imf |H | | |G/N | and
|Imf |H | | |H| (image divides both domain and codomain). But |G/N and |H| are relatively
prime. So Im f |H = 1 which implies that H ≤ N = ker f .
3.2.19 Prove that if N is a normal subgroup of the finite group G and (|N |, |G :
N |) = 1 then N is the unique subgroup of G of order |N |.
1
Let H ≤ G with |H| = |N |. Then (|H|, |G : N |) = (|N |, |G : N |) = 1. By 3.2.18 H ≤ N but
|H| = |N |. So H = N .
3.3.3 Prove that if H is a normal subgroup of G of prime index p then for all K ≤ G
either (i) K ≤ H or (ii) G = HK and |K : K ∩ H| = 1.
Consider
MN
M N
M ∩N
In fact, every subgroup of an abelian group is normal. But G is simple, so G has no proper
subgroup. So |G| must be prime, hence G ∼ = Zp .
3.4.5 Prove that subgroups and quotient groups of a solvable groups are solvable.
Similar way shows that G/H is solvable. Since Gi /H ≤ Gi+1 /H and in fact Gi /H E Gi+1 /H.
And (Gi+1 /H)/(Gi /H) is abelian.
3.4.12 Prove (without using the Feit-Thompson Theorem) that the following are
equivalent:
(i) every group of odd order is solvable.
(ii) the only simple groups of odd order are those of prime order.
(i) ⇒ (ii)
Let G be a simple group of odd order. Suppose that |G| = n = pm for some m 6= 1. If G is
abelian, there exists an element x 6= 1 ∈ G of order p, i.e. hxi E G, a contradiction. If G is
non-abelian, since G is solvable and every finite group has a composition series. So we have
1 = H0 E H1 E · · · E Hn = G
By Exercise 1, the length of the composition series must be at least 2. So Hn−1 6= 1. But this
implies that Hi−1 E G, a contradiction since G is simple. Therefore, the only simple groups of
3
odd order are those of prime order.
(ii) ⇒ (i)
3.5.9. Prove that the (unique) subgroup of order 4 in A4 is normal and is isomor-
phic to V4 .
4.1.1 Let G act on the set A. Prove that if a, b ∈ A and b = g · a for some g ∈ G,
then Gb = gGa g −1 (Ga is the stabilizer of a). Deduce that if G acts transitively on
A then the kernel of the action is ∩g∈G gGa g −1 .
Let K be the kernel of the action, i.e. K = {g ∈ G | g · a = a for all a ∈ A}. Show that
K = ∩g∈G gGa g −1 .
On the other hand, let x ∈ ∩g∈G gGa g −1 . Then x ∈ gGa g −1 for all g ∈ G. So for any
b ∈ A we have x · b = ghg −1 · b for some h ∈ Ga (i.e. h · a = a). But G acts transitively on A,
4
i.e. g ·a = b. Thus, x·b = ghg −1 ·b = gh·a = g ·a = b. So x ∈ K. Therefore, K = ∩g∈G gGa g −1 .
4.1.2 Let G be a permutation group on the set A (i.e., G ≤ SA ), let σ ∈ G and let
a ∈ A. Prove that σGa σ −1 = Gσ(a) . Deduce that if G acts transitively on A then
∩σ∈G σGa σ −1 = 1.
Let x ∈ σGa σ −1 . And let x = σgσ −1 for g ∈ Ga . Then x · σ(a) = σgσ −1 · σ(a) = σg · a = σ(a).
Hence x ∈ Gσ(a) . On the other hand, let x ∈ Gσ(a) . So x · σ(a) = σ(a). We have
σ −1 xσ · a = σ −1 (σ(a)) = a. Hence, σ −1 xσ ∈ Ga , i.e., x ∈ σGa σ −1 Therefore σGa σ −1 = Gσ(a) .
Now, if G acts transitively on A, by previous exercise the kernel is ∩g∈G gGg −1 . Let ϕ : G → SA
be the homomorphism associated to the action. Since G ≤ SA , ϕ is injective, the kernel is 1.
Suppose that σ(a) = a for some a ∈ A and some σ ∈ G. This implies that Ga 6= 1 hence
−1 6= 1, a contradiction to the result in preceding exercise.
T
σ∈G σGa σ
For a fixed a ∈ A, and any b ∈ A, there is σ ∈ G such that σ(a) = b. To show this σ is unique,
suppose there is τ ∈ G such that σ(a) = τ (a) = b. Then τ −1 σ(a) = a, so τ −1 σ ∈ Ga = 1.
Thus, σ = τ . Hence we have an injective map ϕa : A → G sending a to σ. Hence |A| ≤ |G|.
Similarly, if we define ψ : G → A by ψ(σ) = σ · a. Then ψ is injective since σ(a) = τ (a) implies
that τ −1 σ ∈ Ga = 1 which again implies thatσ = τ . Therefore, |G| = |A|.
4.2.8 Prove that if H has finite set index n then there is a normal subgroup K of
G with K ≤ H and |G : K| ≤ n!.
Let G act by left multiplication on the set A of left cosets (i.e. G/H). And let πH be the
associated permutation representation afforded by this action
πH : G → SA
Since |G| = pα . Since p is the smallest prime dividing |G|. By Corollary 5, any subgroup of
index p is normal. In particular, let α = 2, then |G| = p2 . By Cauchy’s Theorem, G has an
element x of order p. So H = hxi and [G : H] = p. H has index p, hence H is normal.
4.2.11 Let G be a finite group and let π : G → SG be the left regular representation.
Prove that if x is an element of G of order n and |G| = mn, then π(x) is a product
of m n-cycles. Deduce that π(x) is an odd permutation if and only if |x| is even
5
|G|
and |x| is odd.
Let H = hxi. Since H is cyclic. For each g ∈ G, the stabilizer of g in H is 1 (GH (x) = 1), so
the action is faithful (i.e. the kernel of the action is 1). So each orbit of H in G has order n
and is the cycle containing g in the decomposition of π(x). i.e. for each g,
Hg = (g xg x2 g · · · xn−1 g)
Since |G| = mn, G = Hg1 ∪ · · · ∪ Hgm . Hence the image of x of left regular representation is a
product of m of n-cycles. Now, if |x| is even and |G|
|x| is odd, by Proposition 25 π(x) is an odd
permutation. Conversely, if π(x) is odd and |x| is odd, the number of cycles of even length in
the decomposition is zero, which is even. By Proposition 25 again, π(x) is even, a contradiction.
4.2.12 Let G and π be as in the preceding exercise. Prove that if π(G) contains an
odd permutation then G has a subgroup of index 2. [use Exercise 3 in Section 3.3]
4.2.13 Prove that if |G| = 2k where k is odd then G has a subgroup of index 2. [Use
Cauchy’s Theorem to produce an element of order 2 and then use the preceding
two exercises.]
4.2.14 Let G be a finite group of composite order n with the property that G has
a subgroup of order k for each positive integer k dividing n. Prove that G is not
simple.
4.3.4 Prove that if S ⊆ G and g ∈ G then gNG (S)g −1 = NG (gSg −1 ) and gCG (S)g −1 =
CG (gSg −1 ).
(⊆) Let gag −1 ∈ NG (gSg −1 ), for a ∈ NG (S) = {x ∈ G | xgSg −1 x−1 = gSg −1 }. We have
(⊇) Let x ∈ NG (gSg −1 ). Then xgSg −1 x−1 = gSg −1 . This implies that g −1 xgSg −1 x−1 g = S,
i.e. (g −1 xg)S(g −1 xg)−1 = S. Thus g −1 xg ∈ NG (S), so x ∈ gNG (S)g −1 .
6
(⊆) Let gag −1 ∈ gCG (S)g −1 , for some a ∈ CG (S) = {x ∈ G | xsx−1 = s for all x ∈ S}. Then
for any s ∈ S we have
(gag −1 )gsg −1 (gag −1 )−1 = gag −1 gsg −1 ga−1 g −1 = gasa−1 g −1 = gsg −1
since asa−1 = s. Thus, gag −1 ∈ NG (gSg −1 ).
(⊇) Let x ∈ CG (gSg −1 ). Then xgsg −1 x−1 = gsg −1 for all s ∈ S. This implies that
g −1 xgsg −1 x−1 g = s, i.e. (g −1 xg)s(g −1 xg)−1 = s for all s ∈ S. Thus g −1 xg ∈ CG (S), so
x ∈ gCG (S)g −1 .
4.3.5 If the center of G is of index n, prove that every conjugacy class has at most
n elements.
4.3.6 Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the
fact that hgi ≤ CG (g) for all g ∈ G to show that there is at most one possible class
equation for G. [Use Exercise 36, Section 3.1]
A quicker way: suppose |Z(G)| is prime, by Exercise 36 in Section 3.1, G/Z(G) is cyclic, so G
is abelian, a contradiction. Hence |Z(G)| = 1.
Now, since hgi ≤ CG (g) for all g ∈ G (Exercise 6 in Section 2.2). Let g ∈ G be any non-identity
element in G. Since Z(G) = 1 hgi ≤ CG (g) G. |CG (g)| can only be 3 or 5. Since |G| = 15,
by Lagrange Theorem, we must have hgi = P CG (g). So the number of conjugates in every
conjugacy class is 3 or 5. By Class Equation, ri=1 |G : CG (gi )| = 15 − 1 = 14. Thus, we have
the relation 3a + 5b = 14 where a denotes the number of conjugacy classes of order 3 and b
denotes the number of conjugacy classes of order 5. The only one solution is a = 3 and b = 1.
Therefore, there is at most one possible class equation for G. i.e. |G| = 15 = 1 + 3 + 3 + 3 + 5.
4.3.13 Find all finite groups which have exactly two conjugacy classes.
Every element in an abelian group is a conjugacy of its own. So abelian groups having exactly
two conjugacy classes must have two element hence is isomorphic to Z2 . Now, let G be non-
abelian group. Note that the identity in G forms a conjugacy class of its own. So |Z(G)| = 1.
Since we assume that G has exactly two conjugacy classes, there is an element g 6∈ Z(G) such
that
|G| = 1 + |G : CG (g)|
7
by class equation. And |G : CG (g)| is the number of elements in the other conjugacy class. So
we have
|G| − 1 = |G : CG (g)|
this implies that the number of conjugates in the other conjugacy class divides |G| − 1. But
this happens only when |G| = 2. Hence G is isomorphic to Z2 .
|G|
since > 1.
|M |
4.3.27 Let g1 , g2 , · · · , gr be representatives of conjugacy classes of the finite group
G and assume these elements pairwise commute. Prove that G is abelian.
4.3.29 Let p be a prime and let G be a group of order pα . Prove that G has a
subgroup of order pβ , for every β with 0 ≤ β ≤ α. [Use Theorem 8 and induction
on α.]
Suppose any group of order pa has a subgroup of order pb for every b with 0 ≤ b ≤ a. Let G
be a group of order |G| = pa+1 , want to prove that G has a subgroup of order pb for every
0 ≤ b ≤ a + 1. By Theorem 8, Z(G) 6= 1. Then H = Z(G) is a normal subgroup of G. For
any subgroup A of G containing H, we have the following correspondence:
G/H /G
A/H /A
1 /H
Since |G/H| = pc for some c with 0 ≤ c ≤ b. By induction hypothesis, G/H has subgroup of
order pb for every b with 0 ≤ b ≤ c. Hence by Lattice Isomorphism Theorem, G has a subgroup
of order pb for every 0 ≤ b ≤ n + 1.
4.3.30 If G is a group of odd order, prove for any nonidentity element x ∈ G that
x and x−1 are not conjugate in G.
First, note that x 6= x−1 . since if x = x−1 , then x2 = 1 and |x| = 2, a contradiction since |G|
is even and cannot have an element of even order.
Suppose x and x−1 are conjugate, want to show that CG (x) has even order, a contradiction.
By definition, CG (x) = {g ∈ G | gxg −1 = x}.
|G| = 203 = 7 · 29. Since H is normal, by Corollary 13 |G/CG (H) is isomorphic to a subgroup
of Aut(H). And |Aut(H)| = 6, so |G/CG (H)| must divide 6. Since |G| = 7 · 29, the only
possibility is |G/CG (H)| = 1. Hence G = CG (H). i.e. H ≤ Z(G). So |H| divides |Z(G)|.
Since |P | = 7, we have |Z(G)| = 7 or 7 · 29. This implies that |G/Z(G)| = 1 or 29, hence
G/Z(G) is cyclic. Therefore, G is abelian.
4.5.13 Prove that a group of order 56 has a normal Sylow p-subgroup for some
prime p dividing its order.
9
|G| = 56 = 23 · 7. Since
n2 ≡ 1 (mod 2), n2 | 7 so n2 = 1, 7
n7 ≡ 1 (mod 7), n2 | 8 so n7 = 1, 8
If n7 = 1, the Sylow 7-subgroup is normal in G. Then we are done. If n7 = 8, G has 8 · 6 = 48
non-identity elements of order 8 in these Sylow 7-subgroups. The rest 56 − 48 = 8 are precisely
elements in Sylow 2-subgroup. This implies that G has a unique Sylow 2-subgroup.
4.5.14 Prove that a group of order 312 has a normal Sylow p-subgroup for some
prime p dividing its order.
4.5.15 Prove that a group of order 351 has a normal Sylow p-subgroup for some
prime p dividing its order.
4.5.17 Prove that if |G| = 105 then G has a normal Sylow 5-subgroup and a normal
Sylow 7-subgroup.
(The solution basically follow the proof of the example on page 143 where |G| = 30).
|G| = 105 = 3 · 5 · 7. Since
n5 ≡ 1 (mod 5), n5 | 3 · 7 so n5 = 1, 21
n7 ≡ 1 (mod 7), n7 | 3 · 5 so n7 = 1, 15
Let P = Syl5 and Q = Syl7 . If n5 = 1 and n7 = 1, then we are done. If either n5 = 1 or n7 = 1,
i.e., either P or Q is normal. Let H = P Q. Then H ≤ G by Corollary 15, and both P and Q
are characteristic subgroups of H by statement (2) on page 135 that every subgroup of a cyclic
group is characteristic (H is cyclic by Example on page 143), or it is easy to see that n5 = 1 and
n7 = 1, hence P EH and QEH. So P and Q are characteristic in H. Since H EG by Corollary
5. By exercise 4.4.8 or the statement (3) on page 135, both P and Q are normal subgroup of G.
Now, assume that neither Sylow subgroup is normal, i.e., n5 = 21 and n7 = 15. Then G has
21 · 4 = 84 elements of order 5 and 15 · 6 = 90 elements of order 7. So total elements would be
84 + 90 = 174 > 105, a contradiction. Therefore, one of P or Q, hence both, must be normal
in G.
10
4.5.24 Prove that if G is a group of order 231 then Z(G) contains a Sylow 11-
subgroup of G and a Sylow 7-subgroup is normal in G.
4.5.25 Prove that if G is a group of order 385 then Z(G) contains a Sylow 7-
subgroup of G and a Sylow 11-subgroup is normal in G.
Let P = Syl7 . Since G/CG (P ) ∼= a subgroup of Aut(P ) by Corollary 13. And |Aut(P )| = 6.
So |G/CG (P )| must divides 6. But |G| = 5 · 7 · 11, the only possibility is that |G/CG (P )| = 1.
Hence, G = CG (P ). i.e. P ≤ Z(G).
4.5.27 Let G be a group of order 315 which has normal Sylow 3-subgroup. Prove
that Z(G) contains a Sylow 3-subgroup of G and deduce that G is abelian.
4.5.34 Let P ∈ Sylp (G) and assume N E G. Use the conjugacy part of Sylow’s Theo-
rem to prove that P ∩ N is a Sylow p-subgroup of N . Deduce that P N/N is a Sylow
p-subgroup of G/N
12
Let |P | = pa and |Q| = pb , so a Sylow p-subgroup of G/N has order pa−b . By 2nd Isomorphism
Theorem,
PN P
=
N P ∩N
So |P N/N | = |P |/|P ∩ N | = pa /pb = pa−b . Thus, P N/N is a Sylow p-subgroup of G/N .
G/N G
P N/N PN
1 N
10.1.5 For any left ideal I of R define
X
IM = { ai mi | ai ∈ I, mi ∈ M }
finite
to be the collection of all finite sums of elements of the form am where a ∈ I and
m ∈ M . Prove that IM is a submodule of M .
First, prove that IM is a subgroup. PNote that thePidentity exists since 0 ∈P
M and take m = 0,
then 0m ∈ IM . P For x, y ∈ IM , x = mi , y = bi mi , we have x + y = (ai + bi )mi ∈ IM .
aiP
Finally, let y = −ai mi , then x + y = (ai − ai )mi = 0, so y is the inverse of x. Thus, IM
is a subgroup.
P P
IM is closed under the action of ring elements: let r ∈ R, rx = r ai mi = rai mi ∈ IM
since rai ∈ I. Hence, IM is a submodule.
(2) Prove that Ann(N ) is closed under right and left multiplication by element of r ∈ R.
Let x ∈ Ann(N ), xn = 0 for all n ∈ N . (rx)n = r(xn) = r · 0 = 0, hence rx ∈ Ann(N ).
Similarly, (xr)n = x(rn) = x · 0 = 0.
13
10.1.19 Let F = R, let V = R2 and let T be the linear transformation from V to V
which is projection onto the y-axis. Show that V , 0, the x-axis and the y-axis are
the only F [x]-submodules for this T .
Since the F [x]-submodules of V are precisely the T -invariant subspace of V . We see that
T (0) = 0 ⊆ V , T (V ) ⊆ V , T (x-axis) = 0 ⊆ x-axis and T (y-axis) = y-axis ⊆ y-axis. Hence,
these are F [x]-modules.
In general, for L be any 1-dimensional subspace on R2 , neither x-axis nor y-axis. It is easy to
see that T (L) 6⊆ L. Hence the only F [x]-submodules for T are 0, V , x-axis and y-axis.
10.2.4 Let A be any Z-module, let a be any element of A and let n be a positive
integer. Prove that the map ϕa : Z/nZ → A given by ϕa (k̄) = ka is a well-define
Z-module homomorphism if and only if na = 0. Prove that HomZ (Z/nZ, A) ∼ = An ,
where An = {a ∈ A | na = 0} (so An is the annihilator in A of the ideal (n) of Z).
The homomorphism between them will be determined by where the generator in Z/nZ goes.
Let ϕ : Z/nZ → Z/mZ be a Z-module homomorphism, then ϕ(0) = 0. Hence,
To have a homomorphism, we must have nϕ(1) ∈ mZ and this is true if and only ϕ(1) =
m m
, or its multiples. These elements form a group generated by and is cyclic
gcd(m, n) gcd(m, n)
of order gcd(m, n). Therefore, HomZ (Z/nZ, Z/mZ) ∼ = Z/(m, n)Z.
10.2.7 Let z be a fixed element of the center of R. Prove that the map m 7→ zm is
an R-module homomorphism from M to itself. Show that for a commutative ring
R the map from R to EndR (M ) given by r 7→ rI is a ring homomorphism (where I
is the identity endomorphism).
14
Recall that EndR (M ) =HomR (M, M ). Let ψ : R →EndR (M ) be defined by ψ(r) = rI. For
r, s ∈ R, we see that ψ(r + s) = (r + s)I = rI + sI = ψ(r) + ψ(s). Also, ψ(rs) = rsI = rI · sI =
ψ(r)ψ(s). Therefore, ψ is a R homomorphism.
First, note that ϕ is surjective since each component ϕi : Ai → Ai /Bi is a natural projection
hence surjective (well, I thought it is clear that the product of a surjective map is surjective).
It suffices to prove that ker ϕ = B1 × · · · × Bn .
(1) ker ϕ ⊆ B1 × · · · × Bn
If a = (a1 , · · · , an ) ∈ ker ϕ, then ϕ(a) = (a1 + B1 , · · · , an + Bn ) = (0̄, · · · , 0̄) ∈ (A1 /B1 ) ×
· · · × (An /Bn ). So we have ai + Bi = 0̄ hence ai ∈ Bi for each i. Thus, a = (a1 , · · · , an ) ∈
B1 × · · · × Bn .
(2) ker ϕ ⊇ B1 × · · · × Bn
For b = (b1 , · · · , bn ) ∈ B1 × · · · × Bn , we have ϕ(b) = ϕ(b1 , · · · , bn ) = (b1 + B1 , · · · , bn + Bn ) =
(B1 , · · · , Bn ) = (0̄, · · · , 0̄) in (A1 /B1 ) × · · · × (An /Bn ). Hence, b ∈ ker ϕ.
ϕ
M /N
p q
ϕ̄
M/IM / N/IN
Given n ∈ N , want to show that there is m ∈ M such that ϕ(m) = n by showing that
q(ϕ(m) − n) = 0̄ ∈ N/IN . Indeed, by commutativity of the diagram we have
10.3.9 (a) Show that M is irreducible if and only if M 6= 0 and M is cyclic module
with any nonzero element as generator.
(⇒)
Suppose M is not cyclic, let M = ha1 , · · · , an i for n > 1. i.e. M is finitely generated by
a1 , · · · , an with ai 6= 0 for all i. Then M has a submodule N 6= 0, and N ( M . This is because
we can let A = {a1 , · · · , ak } for k < n, then N = RA = {r1 a1 , · · · , rk ak | ri ∈ R, ai ∈ A} is a
submodule. This proved that M is cyclic.
(⇐)
Suppose M is a cyclic module with any nonzero element as generator. i.e. M = Ra for any
a ∈ M . Want to prove that M is irreducible. Assume that M has a proper submodule N , then
16
0 6= N ⊂ M . Pick a nonzero element n in N , so we have 0 6= Rn ⊂ N ⊂ M . But Rn = M , so
N = M hence M is irreducible.
On the other hand, suppose Tor(M ) = M , take m ∈ M and m 6= 0. By what we just proved in
(⇒), M = Zm. Now, since Z is a PID, so Ann(m) = (k) for some k ∈ Z. Then M ∼ = Z/kZ. We
claim that k is a prime. If not, k = qr for some q, r ∈ Z, then 0 6= Z/qZ ( Z/kZ ⊂ M . Con-
tradiction to the fact that M is irreducible. So k is a prime. Hence we proved that M ∼
= Z/pZ
some some prime p.
(⇒)
Suppose M is irreducible. Define ϕ : R → M by r 7→ rm for m 6= 0, by previous exercise, M
is cylic, i.e. M = Rm for some m 6= 0. So ϕ is actually surjective. Prove that ker ϕ =Ann(m).
For r ∈ ker ϕ, ϕ(r) = rm = 0 implies that r ∈Ann(m). The converse is also clear. Thus we
have R/I ∼ = M where I = ker ϕ =Ann(M ).
Now, prove that I is a maximal ideal. First, I is an ideal since for any s ∈ R, s(rm) = s · 0 = 0.
So SI ⊆ I. To show that I is maximal, suppose I is not a maximal ideal, there is an ideal J
such that I ( J ( R.
R/I ∼
=M R
J/I J
1 I
Then M ∼= R/I has a submodule J/I which contradicts to the irreducibility of M . Hence, I
must be a maximal ideal.
(⇐)
Suppose M ∼ = R/I for a maximal ideal I of R. If M is not irreducible, there is a submodule
N such that N ( M . By Lattice Theorem again, there is an ideal J with I ( J ( R. This
implies that J is not maximal.
17
10.3.11 If M1 and M2 irreducible R-modules, prove that any nonzero R-module
homomorphism from M1 to M2 is an isomorphism.
10.4.4 Show that Q ⊗Z Q and Q ⊗Q Q are isomorphic left Q-modules. [Show they
are both 1-dimensional vector spaces over Q.]
By definition on page 367, since Q is a left Z-module and a (Q, Z) bimodule, hence Q ⊗Z Q is
a left Q-module. Also, since Q is a left Q-module and a (Q, Q) bimodule, thus Q ⊗Q Q.
Now, show that they are isomorphic to 1-dimensional vector spaces over Q.
18
Let P be a Sylow p-subgroup of A with |P | = pk . Define ϕ : Z/pk Z×A → P by (xmodpk Z, a) 7→
xa.
Thus, ϕ is bilinear. Note that A is subgroup of finite abelian group, hence is abelian. So P is
a Z-module. Since Z/pk Z and A are Z-modules and let Z/pk Z ⊗Z A be the tensor product of
Z/pk Z and A over Z. By Corollary 12 there is a Z-module homomorphism Φ : Z/pk Z⊗Z A → P .
Z/pk Z × A
ι / Z/pk Z ⊗Z A
ϕ Φ
'
P
10.4.6 If R is any integral domain with quotient field Q, prove that (Q/R)⊗R (Q/R) =
0.
m1 m2 m1 m2
( mod R) ⊗ ( mod R) = n2 ( mod R) ⊗ ( mod R)
n1 n2 n1 n2 n2
m1 m2
= ( mod R) ⊗ n2 ( mod R)
n1 n2 n2
m1
= ( mod R) ⊗ (m2 mod R)
n1 n2
= 0
since (m2 mod R) = 0.
19
10.4.12 Let V be a vector space over the field F and let v, v 0 be nonzero elements
of V . Prove that v ⊗ v 0 = v 0 ⊗ v in V ⊗F V if and only if v = av 0 for some a ∈ F .
Let {v1 , v2 , · · · , vn } be a basis for V . (Here we assume that V is finite dimensional so that for
any element in V , its expression of a FINITE linear combination of the basis makes sense. If
V is infinite, then let’s choose a finite dimensional subspace W ⊂ V containing v and v 0 to
proceed the proof). So V ⊗F V = hvi ⊗ vj | i, j = 1, 2, · · · , ni. Let v, v 0 ∈ V , then v = Σai vi
and v 0 = Σa0i vi . So we have v ⊗ v 0 = (Σai vi ) ⊗ (Σa0i vi ) = Σi Σj (ai a0i )vi ⊗ vj . Similarly, we have
v 0 ⊗ v = Σi Σj (a0i ai )vi ⊗ vj . Hence we have v ⊗ v 0 = v 0 ⊗ v if and only if ai a0j = a0i aj for all i, j.
Now, for fixed j, we have ai a0j = a0i aj for all i if and only if ai = a0 i aj (a0 j )−1 if and only
if ai = (aj (a0 j )−1 )a0 i since F is commutative and every element in F has a multiplica-
tive inverse. Since this is is true for some fixed j, let’s denote the term a = (aj (a0 j )−1 ),
note that a ∈ F . So for this fixed j, we have ai = a · a0 i , i = 1 · · · , n. Hence we have
v = Σai vi = Σ(aa0 i )vi = aΣa0 i vi = av 0 .
10.4.15 Show that tensor products do not commute with direct products in general.
∞
N M
Consider Q Z ( Z/2i Z), this implies
i=1
Q ⊗Z ( Z/2i Z) ∼= (Q ⊗Z Z/2Z) ⊕ (Q ⊗Z Z/22 Z) ⊕ (Q ⊗Z Z/23 Z) ⊕ · · ·
L
Since for each i, Z/2i Z is a torsion abelian group (i.e. every element has a finite order), by
exercise 10.4.8(d) (prove later) each term on the right
L handi side is 0, hence the direct product
is 0. However, the left hand side is not zero since Z/2 Z is not a torsion abelian. This is
because there is no integer m such that m( Z/2i Z) = 0 when i goes arbitrarily large. Hence,
L
we conclude that tensor products do not commute with direct products in general.
Now we prove 10.4.8(d): If A is an abelian group group, show that Q ⊗Z A = 0 if and only if
A is a torsion abelian group (i.e. every element of A has finite order).
(⇒)
Suppose Q ⊗Z A = 0. For a ∈ A and q ∈ Q, q = m n for m, n in Z and n 6= 0, we have
0 = q⊗a = m n ⊗ a = m · 1
n ⊗ a = 1
n ⊗ m · a. Since 1
n 6
= 0, so m · a must be zero. So a has finite
order m. Since a ∈ A is arbitrary, so every element in A has finite order hence A is a torsion
group.
(⇐)
Suppose a ∈ A has finite order, say m. i.e. ma = 0 with some m ∈ Z+ . Let q ∈ Q. Then we
q q q
have q ⊗ a = m · m ⊗a= m ⊗m·a= m ⊗ 0 = 0. Hence Q ⊗Z A = 0.
20
(⇐)
Suppose P1 and P2 are projective modules, by Proposition 30(4), P1 and P2 are direct sum-
mand of free R-modules. i.e. P1 ⊕ Q1 and P2 ⊕ Q2 are free for some modules Q1 and Q2 . But
free module is isomorphic to a direct sum of regular modules, so P1 ⊕ Q1 ⊕ P2 ⊕ Q2 is also
a direct sum of modules, hence is free. Since P1 ⊕ Q1 ⊕ P2 ⊕ Q2 ∼ = (P1 ⊕ P2 ) ⊕ (Q1 ⊕ Q2 ),
therefore, P1 ⊕ P2 is a direct summand of a free module hence is projective.
(⇒)
ϕ
Suppose P1 ⊕ P2 is projective. For R-modules M, N , let M −→ N −→ 0 be an exact sequence,
and f : P1 −→ N an R-module homomorphism. Want to prove that f lifts to an R-module
homomorphism into M . i.e. f lifts to F : P1 −→ M .
Now, want to prove that f lifts to an R-module homomorphism F : P1 −→ M and the small
diagram commutes. i.e. f = ϕ ◦ F .
(a) By Example (3) on page 391, free Z-modules have no nonzero elements of finite order so
no nonzero finite abelian group can be isomorphic to a submodule of free module. Since A is
abelian, A is not isomorphic to a submodule of a free module. So A is not a direct summand
of a free module. Hence A is not projective.
21
Let {x1 , · · · , xk } be a basis for F . Suppose x ∈ ∩∞ n=1 nF . Note that {nx1 , · · · , nxk } be a basis
for nF . So if x ∈ ∩∞ n=1 , x ∈ F , x ∈ 2F , x ∈ 3F, · · · , hence
x = a1 x1 + a2 x2 + · · · + ak xk
= a21 (2x1 ) + · · · + a22 (2x2 ) + · · · + a2k (2xk )
..
.
= an1 (nx1 ) + · · · + an2 (nx2 ) + · · · + ank (nxk )
(Note: the notation aji , here j is an index, not an exponent!)
Thus, 0 = (a1 − nan1 )x1 + · · · + (ak − nank )xk for all k. Hence a1 = nan1 , i.e. a1 = 2a21 =
3a31 = 4a41 = · · · . So n | a1 for all n. Similarly, ak = nank which implies that n | ak for all n.
Therefore, a1 , · · · , ak must be zero, hence x = 0.
(a) For ϕ ∈ HomZ (R, M ), r, r0 ∈ R, define rϕ : R −→ M by rϕ(r0 ) = ϕ(r0 r). First prove that
rϕ ∈ HomZ (R, M ).
22
For r1 , r2 ∈ R, and a ∈ R, we have
- (rϕ)(r1 + r2 ) = ϕ((r1 + r2 )r) = ϕ(r1 r + r2 r) = ϕ(r1 r) + ϕ(r2 r) = rϕ(r1 ) + rϕ(r2 )
- (rϕ)(ar1 ) = ϕ((ar1 )r) = ϕ(a(r1 r)) = aϕ(r1 r) = arϕ(r1 )
by the definition of rϕ and that ϕ is a homomorphism.
Now, prove that this action of R on HomZ (R, M ) makes it into a left R-module.
Let ϕ, ψ ∈ HomZ (R, M ). First, by Proposition 2 in 10.2, define ϕ + ψ by (ϕ + ψ)(r) =
ϕ(r) + ψ(r), then ϕ + ψ ∈ HomZ (R, M ) and with this operation HomZ (R, M ) is an abelian
group.
(3) r(ϕ + ψ) = rϕ + rψ
r(ϕ + ψ)(r0 ) = (ϕ + ψ)(r0 r) = ϕ(r0 r) + ψ(r0 r) = rϕ(r0 ) + rψ(r0 ) = (rϕ + rψ)(r0 )
since ϕ + ψ ∈ HomZ (R, M ).
(4) 1 · ϕ = ϕ
1 · ϕ(r0 ) = ϕ(r0 · 1) = ϕ(r0 ).
For b1 , b2 ∈ B, r, s, ∈ R, we have
- F 0 (b1 + b2 )(r) = F (r(b1 + b2 )) = F (rb1 + rb2 ) = F (rb1 ) + F (rb2 ) = F 0 (b1 )(r) + F 0 (b2 )(r)
- F 0 (sb)(r) = F (r(sb)) = F ((rs)b) = F 0 (b)(rs) = sF 0 (b)(r)
by the definition of F 0 and the fact that B is an R-module, F is an R-module. The last equality
holds because F 0 (b) ∈ HomZ (R, M ) hence by (a) we have F 0 (b)(rs) = sF 0 (b)(r).
10.5.16 Prove that every left R-module M is contained in an injective left R-module.
(a) Show that M is contained in an injective Z-module Q.
(b) Show that HomR (R, M ) ⊆ HomZ (R, M ) ⊆ HomZ (R, Q).
(c) Use the R-module isomorphism M ∼ = HomR (R, M ) and the previous exercise
to conclude that M is contained in an injective R-module.
10.5.25 Prove that A is a flat R-module if and only if for every finitely generated
ideal I of R, the map A ⊗R I → A ⊗R R ∼ = A induced by the inclusion I ⊆ R is again
∼
injective. (or equivalently, A ⊗R I = AI ⊆ A).
10.5.26 Suppose R is a PID. This exercise proves that A is a flat R-module if and
only if A is torsion free R-module (i.e. if a ∈ A is nonzero and r ∈ R, then ra = 0
implies r = 0).
(a) Suppose that A is flat and for fixed r ∈ R consider the map ψr : R → R defined
by multiplication by r : ψr (x) = rx. If r is nonzero show that ψr is an injection.
Conclude from the flatness of A that the map from A to A defined by mapping a
to ra is injective and that A is torsion free.
(b) Suppose that A is torsion free. If I is a nonzero ideal of R, then I = rR for
some nonzero r ∈ R. Show that the map ψr in (a) induces an isomorphism R ∼ =I
ψ η
of R-modules and that composite R −
→I −
→ R of ψr with the inclusion : I ⊆ R is
1⊗ψr 1⊗η
multiplication by r. Prove that the composite A ⊗R R −−−→ A ⊗R I −−→ A ⊗R R
24
corresponds to the map a 7→ ra under the identification A ⊗R R = A and that this
composite is injective since A is torsion free. Show that 1 ⊗ ψr is an isomorphism
and deduce that 1 ⊗ η is injective. Use the preceding exercise to conclude that A
is flat.
Proof
12.1.5 Let R = Z[x] and let M = (2, x) be the ideal generated by 2 and x, considered
as a submodule of R. Show that {2, x} is not a basis of M . [Find a nontrivial
R-linear dependence between these two elements.] Show that the rank of M is 1
but not free rank of 1.
Since x · 2 + (−2) · x = 0 for x, −2 6= 0. Thus {2, x} is not linearly independent, hence is not a
basis of M . Now, prove that the rank of M is 1. Since
where ϕ : (2, 0) 7→ (1̄, 0) and (0, x) 7→ (0, 1). Note that Z[x]/(x) is a torsion submodule since
for any 0 6= m ∈ Z[x]/(x), m ∈ Z, we have x · m = 0 because (x) is an ideal. And Z[x] is torsion
free. By exercise 12.1.1(b), rank(M ) = rank(Z[x]) = 1. Hence the rank of M is 1. Finally,
suppose M is free of rank 1, R can be seen as a R-module. Then R/M = Z[x]/(2, x) ∼ = Z2 is
a torsion R-module. By exercise 12.1.1(b), R has rank 1, a contradiction!
For r ∈ Ann(B), then rb = 0 for all b ∈ B. Let (b) be a cyclic submodule generated by b ∈ B,
Ann((b)) ⊆ R is an ideal. Since R is a PID, Ann((b)) = (s) for some s ∈ R. By assumption,
p ∈ Ann((b)) = (s), so p = sr for some r ∈ R. But p is a prime. In PID, prime p is irreducible.
Since s cannot be a unit, thus r is a unit. Hence, s = pr−1 ∈ (p). This implies that (s) ⊆ (p).
Thus, Ann(B) ⊆ Ann((b)) = (s) = (p).
12.2.3 Prove that two 2 × 2 matrices over F which are not scalar matrices are
similar if and only if they have the same characteristic polynomial.
Let A and B be two 2 × 2 matrices. Suppose A and B are similar, then A = P BP −1 for some
invertible 2 × 2 matrix P . Let λ be an eigenvalue of A. We have
A − λI = P BP −1 − λI = P (B − λI)P −1
Hence
det(A − λI) = det(P (B − λI)P −1 )
= det(P )det(B − λI)det(P −1 )
= det(P )det(P −1 )det(B − λI)
= det(P P −1 )det(B − λI)
= det(B − λI)
Hence A and B have the same characteristic polynomials.
Conversely, if A and B have the same characteristic polynomial p(x). If p(x) is irreducible, then
p(x) is the invariant factor for both A and B. So A, B are similar. Suppoe p(x) = f (x)g(x)
with f, g linear. If f, g have different roots, then p(x) is the minimal polynomial for both
A and B. Then A and B have the same invariant factor, hence they are similar. Now, if
p(x) = (x − a)2 , then the minimal polynomial is f (x) = (x − a). Then A, B are scalar matrices,
a contradiction. Hence, A, B are similar.
12.2.4 Prove that two 3 × 3 matrices are similar if and only if they have the same
characteristic and same minimal polynomials. Give an explicit counterexample to
this assertion for 4 × 4 matrices.
(⇒)
Let A and B be two 3 × 3 matrices. Suppose A and B are similar, then A = P BP −1 for some
invertible 3 × 3 matrix P . Let λ be an eigenvalue of A. We have
A − λI = P BP −1 − λI = P (B − λI)P −1
Hence
det(A − λI) = det(P (B − λI)P −1 )
= det(P )det(B − λI)det(P −1 )
= det(P )det(P −1 )det(B − λI)
= det(P P −1 )det(B − λI)
= det(B − λI)
Hence A and B have the same characteristic polynomials.
(⇐)
Suppose A and B have the same characteristic polynomials p(x). Consider the following cases:
A and B have the same invariant factors and rational canonical form, hence they are similar.
If m(x) = (x − a), A and B are scalar matrices since the invariant factors
(x − a), (x − a), (x − a)
and the rational canonical form:
a 0 0
0 a 0
0 0 a
27
Hence A and B are similar.
Example of an 4 × 4 matrix that have the same characteristic polynomial but are not similar.
Let
1 0 0 0 0 −1 0 0
0 1 0 0 1 2 0 0
A= , B=
0 0 0 −1 0 0 0 −1
0 0 1 2 0 0 1 2
pA (x) = pB (x) = (x − 1)4 and mA (x) = mB (x) = (x − 1)2 . But A and B are not similar since
they do not have the same rational canonical form.
is
xn + an−1 xn−1 + · · · + a1 x + a0 .
x 0 ··· 0 a0
−1 x · · · 0 a1
. ..
0 −1 . . . a2
. ..
.. . x an−3
0 0 ··· −1 x + an−2
x 0 ··· 0 a0 x 0 ··· 0
−1 x · · · 0 a1 −1 x · · · 0
. .. . ..
= x 0 −1 . . . a2 + a0 0 −1 . . .
. .. . ..
.. . x an−3 .. . x
0 0 ··· −1 x + an−2 (n−1)×(n−1) 0 0 ··· −1 (n−1)×(n−1)
So
p(x) = det(xI − A)
= x(xn−1 + ak−1 xk−2 + ak−2 xk−3 + · · · + a2 x + a1 ) + a0 (−1)k−1 · (−1)k+1
= xk + ak−1 xk−1 + · · · + a2 x2 + a1 x + a0
as desired.
12.2.10 Find all similarity classes of 6 × 6 matrices over Q with minimal polynomial
(x + 2)2 (x − 1).
This is because the matrix is 6 × 6, the product of all invariant factors (which is the char-
acteristic polynomial) must have degree 6. Also each invariant factor divides the next. i.e.
a1 (x) | a2 (x) | · · · | an (x). Hence, the rational canonical form are below:
−2 1
−2
1
−2 1
1. 2.
0 0 4
0 0 4
1 0 0 1 0 0
0 1 −3 0 1 −3
1 −2
0 2 0 2
1 −1 1 −1
3. 4.
0 0 4
0 0 4
1 0 0 1 0 0
0 1 −3 0 1 −3
29
−2 0 0 4
0 −4
1 0 0
1 −4 0 1 −3
5. 6.
0 0 4
0 0 4
1 0 0 1 0 0
0 1 −3 0 1 −3
12.2.11 Find all similarity classes of 6 × 6 matrices over C with characteristic poly-
nomial (x4 − 1)(x2 − 1).
The characteristic polynomials is p(x) = (x4 −1)(x2 −1) = (x2 +1)(x2 −1)2 = (x2 +1)(x−1)2 (x+
1)2 . The possible m(x) are (x2 + 1)(x − 1)(x + 1), (x2 + 1)(x − 1)2 (x + 1), (x2 + 1)(x − 1)(x + 1)2 ,
(x2 + 1)(x − 1)2 (x + 1)2 . Since p(x) and m(x) have the same roots and m(x) | p(x). Thus, the
corresponding invariant factors are
Since the matrix has order 4, i.e. A4 = 1, the minimal polynomial divides x4 − 1 = (x2 +
1)(x + 1)(x − 1). The possibilities for m(x) are
x + 1, x − 1, (x + 1)(x − 1), (x2 + 1), (x2 + 1)(x + 1), (x2 + 1)(x − 1), (x2 + 1)(x + 1)(x − 1).
But the order is precisely 4, so we exclude x + 1, x − 1 and (x + 1)(x − 1). Furthermore, A is
a 2 matrix, degree m(x) ≤ 2. So the only possibilities are (x2 + 1), hence
0 −1
A=
1 0
Over C, x2 + 1 = (x − i)(x + i). So the possibilities for m(x) are
x + i, x − i, (x + i)(x − i), (x ± i)(x ± 1).
30
For example, by computation we see that
4 i 0
A = =I
0 i
So the 2 × 2 matrix with m(x) = x + i has order 4. Same for other three cases. Therefore, the
corresponding rational canonical forms are
i 0 −i 0 0 −1 ±1 0
, , , .
0 i 0 −i 1 0 0 ±i
Note that the second and the third are similar.
By computation, we see that they both have characteristic polynomial p(x) = (x − 1)2 (x + 1).
Since A − I 6= 0 and A + I 6= 0, the minimal polynomial mA (x) must have degree at least two.
But by computation, (A − I)2 6= 0 and (A − I)(A + I) 6= 0. Thus, mA (x) = (x − 1)2 (x + 1)
has no distinct roots, hence A is not diagonalized.
Since A3 = A, the minimal polynomial m(x) divides x3 − x = x(x + 1)(x − 1). Since m(x) has
no repeated roots, A can be diagonalized. However, if charF = 2, take
1 1
A=
0 1
It is easy to see that A3 = A but A is not diagonalized.
31
√ √
(x + 1−2 3i )(x + 1+ 3i
2 ) because A is a 2 × 2 matrix and A 6= I. Hence the possible invariant
factors will be
√ √
1− 3i 1− 3i
(1) (x + 2√ ), (x + 2√ )
1+ 3i 1+ 3i
(2) (x + 2 ), (x + 2 )
√
(3) (x − 1)(x + 1−2 3i )
√
(4) (x − 1)(x + 1+2 3i )
√ √
1− 3i 1+ 3i
(5) (x + 2 )(x + 2 )
(1)
" √ #
− 1−2 3i 0√
0 − 1−2 3i
(2)
" √ #
− 1+2 3i 0√
0 − 1+2 3i
√ √ √
1− 3i
(3) Since (x − 1)(x + 2 ) = x2 − ( 1+2 3i )x − 1−2 3i , we have
" √ #
0 1−2√3i
1 1+2 3i
√ √ √
1+ 3i
(4) Since (x − 1)(x + 2 ) = x2 − ( 1−2 3i )x − 1+2 3i , we have
" √ #
0 1+2√3i
1 1−2 3i
√ √
1+ 3i 1− 3i
(5) Since (x − 2 )(x − 2 ) = x2 + x + 1, we have
0 −1
1 −1
Proof
33
p
Let α = in lowest term, q 6= 0 and (p, q) = 1. So we have f (α) = 0 which is
q
αn + an−1 αn−1 + · · · + a1 α + a0 = 0
Subtracting a0 from both sides we obtain
αn + an−1 αn−1 + · · · + a1 α = −a0
p
Since α = , so we have
q
pn pn−1 p
+ a n−1 + · · · + a1 = −a0
qn q n−1 q
Factoring out p we get
pn−1 pn−2
1
p + an−1 n−1 + · · · + a1 = −a0
qn q q
n−1 n−2
Since there’s no factor 1
p in the term p qn + an−1 pqn−1 + · · · + a1 1q and a0 is an integer. Hence
p must divide a0 .
p
On the other hand, rearrange f ( ) = 0 in different way we have
q
n pn−1
p p
= − an−1 n−1 + · · · + a1 + a0
qn q q
Multiply q n−1 on both sides, we obtain
pn
= − an−1 + an−2 pn−1 q + · · · + a1 pq n−2 + a0 q n−1
q
since the there’s no factor q on the right hand side. Also the right side is actually an integers
since all p, q, ai are. Hence q must be 1.
p p
This proves that α = = = p which is an integer.
q 1
13.1.8 Prove that x5 − ax − 1 ∈ Z[x] is irreducible unless a = 0, 2 or −1. The
first two correspond to linear factors, the third corresponds to the factorization
(x2 − x + 1)(x3 + x2 − 1).
Case 2 : If f (x) = (x2 + bx + 1)(x3 + dx2 + ex − 1) = 0, i.e. f (x) can be factored into the
product of two irreducible polynomials in Z[x] of degree 2 and 3. (Remark, the product of the
34
constant terms in these two factors is the constant term in f (x) which is −1, so they must be
either 1 or −1).
b+d=0
1 + e + bd = 0
be + d − 1 = 0
From the first one we have b = −d, then plug into the second and the third we get 1+e−d2 = 0
and −de + d − 1 = 0. The former one gives e = d2 − 1, plugging into the latter one we obtain
d3 − 2d + 1 = 0 which gives (d − 1)(d2 + d − 1) = 0. So d = 1 since d is an integer. Then we
also get e = 0. Now, since a = e − b, so we have a = 0 − 1 = −1. Contradiction!
Case 3 : If f (x) = (x2 + bx − 1)(x3 + dx2 + ex + 1) = 0, similar to Case 2 except swap the
negative sign on the constant terms of the two irreducible factors.
b+d=0
e + bd − 1 = 0
be − d + 1 = 0
Similarly, we get b = −d from the first equation, the plug into the second and the third. But
by solving them simultaneously we get d3 + 2d − 1 = 0 which has no integer root since neither
1 nor −1 is a root of this equation (by plugging ±1 into it).
Therefore we proved that this f (x) = x5 − ax − 1 is irreducible over Z[x] except at x = 0, 2, −1.
√ √ √ √ √ √
13.2.7 Prove that Q( 2 + 3) = Q( 2, √ 3). Conclude
√ that [Q( 2 + 3) : Q] = 4. Find
an irreducible polynomial satisfied by 2 + 3.
√ √ √ √
First, prove that Q( 2 + 3) = Q( 2, 3). (⊆) is clear, so we prove (⊇).
√ √ √ √ √ √ √ √ √ √
Denote 2+ 3 by α. So α( 2− 3) = ( 2+ 3)( 2− 3) = −1. This gives 2− 3 = − α1 .
√ √ √ √
Combining this equation with α = 2 + 3 and solving for 2 we get 2 2 = α − α1 , hence
√ √ √ √
2 = 21 (α − α1 ) so 2 ∈ Q(α). Similarly, solving for 3 we get 3 = 21 (α + α1 ), hence
√ √ √ √ √
3√is also
√ in Q(α). √ √So we proved that Q( 2, 3) ⊆ Q(α) = Q( 2 + 3). Therefore,
Q( 2 + 3) = Q( 2, 3).
√ √ √
Now √ that [Q( 2 + 3) : Q] = 4. Consider the extensions of fields Q ⊆ Q( 2) ⊆
√ prove
Q( 2, 3).
35
√ √
Q( 2)/Q is degree 2 since the minimal √ polynomial for 2√ over Q is x2 − 2 (monic, irreducible
by √ Eisenstein
√ at
√ p = 2 and having 2 as a root) so [Q( 2) :2 Q] = 2. On the other hand, √
Q( 2, 3)/Q( 2) has degree √ at most
√ 2 and is 2 if and √ only if x −√3 is irreducible over Q( 2).
It remains to prove√that 3 6∈ Q( 2). If it √ were, then 3 = a + b 2 for a, b ∈ Q. So we have
2 2
3 = a + 2b + 2ab 2. If a, b 6= 0, solve for 2 we get
√ 3 − a2 − 2b2
2=
2ab
√
a√ contradiction
√ since this implies that √ 2 ∈ Q which is not√ true. Now, if a = 0 we have
3 = b 2, multiplying both side
√ by 2 we found that 6 is a rational, a contradiction.
If
√ b = 0,√we would have that 3 = a a rational, again a contradiction. This proved that
3 6∈ Q( 2).
√ √ √
Hence Q( 2, 3)/Q( 2) √ has
√ degree 2, then by Theorem 14, the multiplication
√ √ rule √ for fields
√
extensions,√we have
√ [Q( 2, 3) : Q] = 4. But we just proved that Q( 2 + 3) = Q( 2, 3).
Hence [Q( 2 + 3) : Q] = 4, as desired.
√ √
Finally, find an irreducible polynomial satisfied 2+ 3.
√ √ √ √
Let α = 2 + 3. Square both sides we get α2 = 5 + 2 6, so α2 − 5 = 2 6. Square both
sides again we have α4 − 10α√
2 + 25 = 24 which is α4 − 10α2 + 1 = 0. Hence the polynomial
√
x4 − 10x2 + 1 satisfies α
√ √ = 2 + 3. This polynomial is monic and
√ has
√ α as a root. Since we
just proved that [Q( 2, 3) : Q] = 4 and by Proposition 11, [Q( 2, 3) : Q] = deg mα (x).
This implies that the minimal polynomial for α has degree 4 so it is actually the polynomial
x4 − 10x2 + 1. This implies that x4 − 10x2 + 1 is irreducible.
13.2.13
√ Suppose F = Q(α1 , α2 , · · · , αn ) where αi2 ∈ Q for i = 1, 2, ·, n. Prove that
3
2 6∈ F .
Consider the fields extension F ⊆ F (α2 ) ⊆ F (α). Since α satisfies the polynomial x2 − α2 over
F (α2 ) which has degree 2. This implies that [F (α) : F (α2 )] ≤ 2. Assume that [F (α) : F ] is
odd, we must have [F (α) : F (α2 )] = 1. i.e. F (α) = F (α2 ).
13.2.17 Let f (x) be an irreducible polynomial of degree n over a field F . Let g(x)
be any polynomial in F [x]. Prove that every irreducible factor of the composite
polynomial f (g(x)) has degree divisible by n.
Without loss of generality we assume that f (x) is monic (if not, we could divide f (x) by its
leading coefficient to obtain a monic polynomial). Let p(x) be an irreducible factor of f (g(x)),
so p(x) divides f (g(x)). So p(x) is monic and has degree k ≤ n. Let α be a root of p(x), we
have [F (α) : F ] = deg minα (x) = k.
Since α is a root of p(x), so α is also a root of f (g(x)). So f (g(α)) = 0 and this implies that g(α)
is a root of f (x). So f (x) is the minimal polynomial for g(α). Hence we have [F (g(α)) : F ] =
deg ming(α) (x) = n. Now consider the field extensions
F ⊆ F (g(α)) ⊆ F (α)
By the multiplication rule of fields extension, we see that [F (g(α)) : F ] divides [F (α) : F ]. i.e.
n | k.
13.2.18 Let k be a field and let k(x) be the field of rational functions in x with
P (x)
coefficients from k. Let t ∈ k(x) be the rational function with relatively prime
Q(x)
polynomials P (x), Q(x) ∈ k[x], with Q(x) 6= 0.
(a) Show that the polynomial P (X) − tQ(X) in the variable X and coefficients in
k(t) is irreducible over k(t) and has x as a root.
(b) Show that the degree of P (X) − tQ(X) as a polynomial in X with coefficients
in k(t) is the maximum of the degrees of P (x) and Q(x).
P (x)
(c) Show that [k[x] : k(t)] = [k(x) : k( )] = max (deg P (x), deg Q(x)).
Q(x)
37
(a) First, P (X) − tQ(X) is irreducible over (k[X])[t] since it is nothing but a polynomial of
degree one in t. But because (k[t])[X] = k[t, X] = k[X, t] = (k[X])[t], and P (X), Q(X) are
relatively prime. So P (X) − tQ(X) is irreducible over (k[t])[X]. (If not, P (X) − tQ(X) =
F (x, t)G(X), note that F and G can’t be both in t since their product has degree one in t.
But this implies that G(X) divides both P (X) and Q(X), contradiction!). Now, since k[t] is a
UFD, by Gauss Lemma P (X)−tQ(X) is irreducible over its field of fractions which is (k(t))[x].
P (X)
Furthermore, if we plug x into P (X) − tQ(X) we get P (x) − tQ(x). But t = Q(X) , so we have
P (X)
P (x) − tQ(x) = P (x) − Q(X) Q(x) = 0 hence P (X) − tQ(X) has x as a root.
Hence degree of P (X) − tQ(X) is n which is the maximum of the degree of P (x) and Q(x).
P (x)
(c) The first equality is clear since t = Q(x) by assumption. On the other hand, in part (a) we
assumed that the polynomial P (X) − tQ(X) is monic, and proved that it is irreducible and
has x as a root hence is the minimal polynomial of x over k(t). Also part (b) proved that it
has degree n. Hence we have [k(x) : k(t)] = deg minx (X) = max (deg P (x), deg Q(x)) = n.
13.2.22 Let K1 and K2 be two finite extensions of a field F contained in the field K.
Prove that the F -algebra K1 ⊗F K2 is a field if and only if [K1 K2 : F ] = [K1 : F ][K2 : F ].
Let {αi } be a basis for K1 over F and {βj } be a basis for K2 over F . Then {αi ⊗ βj } is a
basis for K1 ⊗F K2 over F . Define a map ϕ : K1 ⊗F K2 → K1 K2 by ϕ(αi ⊗ βj ) = αi βj , and
extend it by linearity. It is easy to check that ϕ is an F -algebra homomorphism. The map ϕ
is surjective because the elements αi βj span K1 K2 as an F -vector space.
(⇐) By assumption the F -vectors spaces K1 ⊗F K2 and K1 K2 have the same dimension over
F , namely [K1 : F ][K2 : F ]. Thus by linear algebra ϕ is injective and hence an isomorphism.
Therefore, K1 ⊗F K2 is isomorphic to the field K1 K2 .
13.4.1 Determine the splitting filed and its degree over Q for x4 − 2.
√ √ √ √
Since
√ f (x)
√ = x4 − 2 can be factored as (x − i 4 2)(x + i 4 2)(x − 4 2)(x + 4 2), the roots are
± 4 2, ±i 4 2.
√
First, prove that the splitting field for f (x) is Q( 4 2, i).
Let K be the splitting field for f (x). By definition of the splitting field, K contains all roots of
38
√ √ √ √
f (x) hence contains 4 2 and
√ the ratio
√ of two roots i 4 2, 4 2 which
√ is i. So K ⊇ Q( 4 2, i).
√ On
all roots ± 4 2, ±i 4 2 clearly lie in the field Q( 4 2, i), so we have K ⊆ Q( 4 2, i).
the other hand, √
Hence, K = Q( 4 2, i).
√
Now, prove that [Q( 4 2, i) : Q] = 8
√ √
Consider the extensions of fields: Q ⊆ Q( 4 2) ⊆ Q( 4 2, i).
√
(1) The extension Q( 4 2)/Q has degree 4 since it satisfies the polynomial x4 − 2 which
√ is ir-
4 4
√ at p = 2. So x − 2 is the minimal polynomial for 2 over Q.
reducible over Q by Eisenstein
And it is monic. Hence [Q( 4 2) : Q] = deg min √
4 (x) = 4.
2
√ √
(2) The extension Q( 4 2, i)/Q( 4 2) has degree at most 2 since it satisfies the polynomial x2 + 1
and is precisely 2 if and only
√ if x2 + 1 is irreducible.
√ Indeed,
√ x2 + 1 is irreducible over Q (none
4 4 4
of its roots i, −i are in Q( 2). So we have [Q( 2, i) : Q( 2)] = 2.
13.4.2 Determine the splitting filed and its degree over Q for x4 + 2.
First, find the roots of this polynomial. If α is a root of this equation, then α4 = −2, then
(ξα)4 = −2 where ξ is any 4th root of −1. Hence the solutions of this equation are
√
ξ a 4th root of − 1
4
ξ 2,
π
To compute√ξ explicitly,
√
since x4 + 1 = (x2 + i)(x2 − i), so x2 = i = e 2 i hence x =
π π
±e 4 i = ±( 22 + 22 i). On the other hand, we have x2 = −i = e− 2 i which implies that
π
√ √ √ √
x = ±e− 4 i = ±( 2
2√ − 2
2 i). Hence ξ = ± 2
2
± 2
2 i. The roots of the polynomial x4 + 2 are
√
2 2
√
4
therefore (± 2 ± 2 i) 2.
√
Prove that the splitting field for the polynomial x4 + 2 is Q( 4 2, i).
Let K be the
√ splitting field for √x4 +√2 over Q. K must√ contains all roots of this√ polynomial.
√ √ √ √ √ √ √
Let α1 = ( 2 + 2 i) 2, α2 = ( 2 − 2 i) 2, α3 = (− 2 + 2 i) 2 and α4 = (− 22 − 22 i) 4 2.
2 2 4 2 2 4 2 2 4
√ α1 α4 3 √
Hence 4 2 = (α1 + α4 )/(α1 α4 ) and i = (α1 + α3 )( ) . So K contains 4 2 and i. i.e.
√ α1 + α4
K ⊇ Q( 4 2, i).
√ √ √ √
On the other hand, all roots lie in the field Q( 4 2, i). For example, 4 2( 22 + 22 i) is a product
√ √ √ √ √ √
of 2 4 and 22 + 22 i, but the latter is just a linear combination of 2 and i, where 2 = ( 4 2)2 .
√ √ √ √
Hence the root 4 2( 22 + 22 i) lies in the field Q( 4 2, i). Similarly all other roots also lie in the
√ √ √
4
field Q( √ 2, i), so K ⊆ Q( 4 2, i). Hence, the splitting field for x4 + 2 over Q is Q( 4 2, i). i.e.
K = Q( 4 2, i).
Now, compute the degree extension. The degree is actually the same as the exercise 13.4.1
since the polynomials x4 − 2 and x4 + 2 have the same splitting field over Q. So I will just do
39
the copy and paste:
√ √
Consider the extensions of fields: Q ⊆ Q( 4 2) ⊆ Q( 4 2, i).
√
(1) The extension Q( 4 2)/Q has degree 4 since it satisfies the polynomial x4 − 2 which
√ is
irreducible over Q by Eisenstein at p = 2 (Remark: although we just proved that Q( 4 2, i) is
the splitting field for the polynomial x4 + 2. To compute the degree
√ of extension, we should
know that the minimal polynomial would be x4 − 2 over√Q since 4 2 is a root of x4 − 2 but
not √of x4 + 2). So x4 − 2 is the minimal polynomial for 4 2 over Q. And it is monic. Hence
4
[Q( 2) : Q] = deg min √ 4 (x) = 4.
2
√ √
(2) The extension Q( 4 2, i)/Q( 4 2) has degree at most 2 since it satisfies the polynomial x2 + 1
and is precisely 2 if and only if x2 + 1 is irreducible.
√ Indeed,
√ x2 + 1 is irreducible over Q (none
4 4
of its roots i, −i are in Q). So we have [Q( 2, i) : Q( 2)] = 2.
13.5.4 Let a > 1 be an integer. Prove for any positive integers n, d that d divides
n if and only if ad − 1 divides an − 1. Conclude in particular that Fpd ⊆ Fpn if and
only if d divides n.
40
13.5.7 Suppose K is field of characteristic p which is not a perfect field: K 6= K p .
Prove there exist irreducible inseparable polynomials over K. Conclude that there
exist inseparable finite extensions of K.
13.5.8 Prove that f (x)p = f (xp ) for any polynomial f (x) ∈ Fp [x].
By Proposition 35, for any a1 , a2 ∈ Fp , (a1 + a2 )p = ap1 + ap2 . Suppose (a1 + · · · + an−1 )p =
ap1 +· · ·+apn−1 , by induction we have (a1 +· · ·+an−1 +an )p = (a1 +· · ·+an−1 )p +apn = ap1 +· · ·+apn .
13.6.15 Let p be an odd prime not dividing m and let Φm (x) be the mth cyclotomic
polynomial. Suppose a ∈ Z satisfies Φm (a) ≡ 0 (mod p). Prove that a is relatively
prime to p and that the order of a in (Z/pZ)× is precisely m.
Y Y
Note that xm − 1 = Φd (x) and can also be written as Φm (x) · Φd (x). Let p be an
d|m d|m,d<m
odd prime and p - n. Suppose a ∈ Z satisfies Φm (a) ≡ 0 (mod p), then p divides Φm (a). Hence
p divides am − 1. i.e am − 1 ≡ 0 (mod p) or am ≡ 1 (mod p). Thus a is relatively prime to
p (if not, then a = pb for some integer b, then we would have bm pm ≡ 0 (mod p), contradiction.).
To compute the order of a ∈ Z/pZ, note that the order of a (mod p) must be less than or
equal to m from what we just proved. Suppose a has order less than m, then al ≡ 1 (mod p)
for some Qinteger l with l < m and l | m. This implies that al − 1 ≡ 0 (mod p). So we have
l
a − 1 = d|l Φd (a) ≡ 0 (mod p). This implies that Φk (a) ≡ 0 (mod p) for some integer k ≤ l
and k | l. Now, Φk (x) has a as a root (mod p) with k ≤ l < m and k | m. This implies that
xm − 1 would have a as a multiple root mod p. Since we know that xm − 1 is separable which
means it has distinct roots (example (2) on page 547). Contradiction! Hence the order of a
(mod p) is preciesly m.
13.6.16 Let a ∈ Z. Show that if p is an odd prime dividing Φm (a) then either p
divides m or p ≡ 1 (mod m).
41
Hence we proved that either p divides m or p ≡ 1 (mod m).
13.6.17 Prove there are infinitely many primes p with p ≡ 1 (mod p).
We will prove by assuming 13.6.14. Consider Φm (1), Φm (2), · · · , Φm (n), · · · ∈ Z[x]. These are
monic polynomial in Z[x] of degree at least one, so by 13.6.14 there are infinitely many distinct
primes dividing Φm (i) for i = 1, 2, · · · . But in 13.6.16 we just proved that either p divides m or
p ≡ 1 (mod m). Since only finitely many of them are divisors of m because m is finite, hence
there are infinitely many primes p with p ≡ 1 (mod m).
√ √
14.1.4 Prove that Q( 2) and Q( 3) are not isomorphic.
√ √
Suppose Q( 2) and Q( 3) are isomorphic. Want to draw a contradiction.
√ √ √ √
√ 2) →
Let√σ : Q( √Q(√ 3) be an isomorphism, since σ√is a√homomorphism, we have σ( 2· 2) =
σ( 2)σ( 2) = 3 3 = 3. On the other hand, σ( 2 · 2) = σ(2) = σ(1) + σ(1) = 1 + 1 = 2
by the fact that an homorphism maps the identity to identity. Hence we obtain 2 = 3, a
contradiction!
Here is alternative
√ less-technical
√ proof:
Let σ : Q( 2) −→ Q( 3) be an isomorphism. Then σ takes 1 to 1 i.e. σ(1) = 1, it follows
that σ(a) = a for a in the prime subfield Q. Hence σ fixes the prime subfield Q.
√ √
Now, since Q( 2) is the field generated by√ 2 over Q. √And the isomorphism
√ σ fixes
√ Q, so σ
is completely determined by its action on
√ 2, i.e. by σ( 2). So let σ( 2) = a + b 3 for some
a, b ∈ Q. The minimal polynomial for√ 2 over Q is x2 − 2 (since
√ it is monic, irreducible by
Eisenstein Criterion at p = 2 and has 2 as a root). Then σ( 2) must also be a root of this
polynomial
√ (any automorphism
√ permutes the roots of an irreducible polynomial). However,
σ( 2) of the form a + b 3 cannot be its root.
√ √
If it were, then (a + b 3)2 − 2 = a2 + √
2ab 3 + 3b2 − 2 = 0. If a = b = 0, we have −2 = 0,
contradiction (or, in this case, σ maps 2 toqzero which means that σ is a zero map, contra-
diction). If a = 0, b 6= 0, 3b2 = 2, then b = 23 ∈
6 Q, contradiction. If a 6= 0, b = 0, then we
√ √ 2 −3b2
have a = 2, a contradiction. Finally if a 6= 0, b =6 0, then we have 3 = 2−a2ab , again a
√
contradiction since 3 is not rational.
√ √ √
Hence
√ σ( 2) cannot be of the form of a + b 3 which implies that Q( 2) is not isomorphic to
Q( 3).
√ √
14.1.5 Determine the automorphisms of the extension Q( 4 2)/Q( 2) explicitly.
√ √ √ √ √
Note that 4 2 is satisfies the polynomial x2 √
− 2 over Q( 2) so the [Q(√
4
2) : Q( √2) has degree
√
at most 2 but is precisely 2 since the root 4 2 is not contained
√ in√Q( 2). So Q( 4 2)/Q( 2) is
a quadratic extension.
√ Any automorphism
√ √ σ ∈ Aut(Q(√4 2)/Q( 2) is completely determined
by its action on 4 2. i.e. σ sends 4 2 to − 4 2 fixing Q( 2).
42
14.1.7 Determine Aut(R/Q).
(a) Prove that any σ ∈ Aut (R/Q) takes squares to squares and takes positive reals
to positive reals. Conclude that a < b implies σa < σb for every a, b ∈ R.
1 1 1 1
(b) Prove that − m < a−b < m implies − m < σa − σb < m for every positive integer
m. Conclude that σ is a continuous map on R.
(c) Prove that any continuous map on R which is the identity on Q is the identity
map, hence Aut(R/Q) = 1.
(b) Since any automorphism σ ∈ Aut(R/Q) fixes Q. For every positive integer m, we have
1 1 1 1
− = σ(− ) < σ(a − b) = σ(a) − σ(b) < σ( ) =
m m m m
Now, we conclude that σ is continuous on R. Given any > 0, for any positive integer m and
1
let δ = m < . If | a − b |< δ, we have
1
| σ(a) − σ(b) |< <
m
since is arbitrary, hence we conclude that σ is a continuous map on R for every a, b ∈ R.
(c) In real analysis we know that for any real number a ∈ R, there exists a sequence of rational
numbers {qn } such that {qn } → a as n → ∞ (Q is dense in R). Now, let σ be any continuous
map on R which is the identity on Q. We have
a = lim qn = lim σ(qn ) = σ( lim qn ) = σ(a)
n→∞ n→∞ n→∞
since σ fixes Q and note that limit can exchanges with the continuous function σ.
Hence σ(a) = a. i.e. Any automorphism from R to R fixing Q must be the identity.
14.1.8 Prove that the automorphisms of the rational function field k(t) which fix
at + b
k are precisely the fractional linear transformations determined by t 7→ for
ct + d
a, b, c, d ∈ k, ad − bc 6= 0.
Let σ : k(t) −→ k(t) be an automorphism fixing the field k. Since σ is completely deter-
p(t)
mined by its action on t, let σ : t 7→ for polynomials p(t), q(t) in k[t] with q(t) 6= 0. i.e.
q(t)
p(t)
σ(t) = . Want to prove that p(t) and q(t) are both linear.
q(t)
Now, want to use exercise the result of 13.2.18 in previous homework, by adapting suitable
variables in this exercise. We have
p(t)
[k(t) : k(σ(t))] = [k(t) : k( )] = max (deg p(t), deg q(t))
q(t)
43
p(t) p(t)
As proved in exercise 13.2.18, k(t) is an extension of k( ). So k( ) ⊆ k(t). But σ is
q(t) q(t)
p(t) p(t)
an automorphism mapping t to , hence [k(t) : k( )] must be 1. This implies that max
q(t) q(t)
(deg p(t), deg q(t)) is 1 which means that both p(t) and q(t) has degree no greater than one.
That is, p(t) = at + b, q(t) = ct + d for a, b, c, d ∈ k and p(t), q(t) are relatively prime, i.e.
bc at + b
ad 6= bc (If ad − bc = 0, then ad = bc, rearrange we have a = , plugging into we obtain
d ct + d
bc b
dt+b (ct + d) b
= d = which means at + b and ct + d are not relatively prime). Therefore,
ct + d ct + d d
at + b
the automorphism σ is determined by t 7→ for a, b, c, d ∈ k and ad − bc 6= 0.
ct + d
at + b
On the other hand, if ad − bc 6= 0, then σ(t) = defines an automorphism. Clearly σ
ct + d
p(t)
is well-defined since it is independent of choice of t. And it maps a rational function to
q(t)
p( at+b
ct+d )
a rational function at+b ∈ k(t) by just simplifying by doing multiplication on the fraction
q( ct+d )
dt − b
and use common denominator ct + d. The inverse is given by σ −1 (t) = , by taking the
−ct + a
inverse on the matrix
a b
c d
An easy verification can be done as below:
at + b d( at+b
ct+d ) − b
adt+bd−bct−bd
ct+d t(ad − bc)
σ −1 σ(t) = σ −1 ( )= = −act−bc+act+ad
= =t
ct + d −c( at+b
ct+d ) + a ct+d
ad − bc
Similarly, we have
dt − b a( dt−b ) + b adt−ab−bct+ab
t(ad − bc)
σσ −1 (t) = σ( ) = −ct+a
dt−b
= −ct+a
cdt−bc−cdt+ad
= =t
−ct + a c( −ct+a )+d −ct−a
ad − bc
Hence σ is a bijection. Also, it is a homomorphism since for f (t), g(t) ∈ k(t), we have
at + b at + b at + b
σ(f + g)(t) = (f + g)( ) = f( ) + g( ) = σ(f (t)) + σ(g(t))
ct + d ct + d ct + d
at + b at + b at + b
σ(f g)(t) = (f g)( ) = f( )g( ) = σ(f (t))σ(g(t))
ct + d ct + d ct + d
at + b
Therefore σ(t) = for ad − bc 6= 0 defines an automorphism on k(t).
ct + d
14.2.3 Determine the Galois group of (x2 − 2)(x2 − 3)(x2 − 5). Determine all the
subfields of the splitting field of this polynomial.
√ √ √
The extension K = Q( 2, 3, 5) is Galois over Q since it is the splitting field of the polyno-
mial (x2 − 2)(x2 − 3)(x5 − 2) since all roots are in K. On the other hand, any field containing
these roots√contains
√ K.√Any automorphism σ is completely√ √determined
√ by its action on the
generators 2, 3 and 5 which must be mapped to ± 2, ± 3 and ± 5 respectively. Hence
44
the only possibilities for automorphisms are the maps
√ √ √ √ √ √
√2 → √ − 2 √2 → √ 2 √2 → √2
σ: 3 → √3 τ: 3→√ − 3 δ: 3→ √ 3
√ √ √
5→ 5 5→ 5 5→− 5
It’s clearly to see that σ 2 = τ 2 = δ 2 = 1. Moreover,
√ √ √ √ √ √
√2 → −√2 √2 → √− 2 √2 → √ 2
στ = τ σ: 3→√ − 3 σδ = δσ: 3→ √ 3 τ δ = δτ : 3 → −√3
√ √ √
5→ 5 5→− 5 5→− 5
That is, three maps commute with each other. Furthermore,
√ √
√2 → −√2
στ δ: 3 → −√3
√
5→− 5
So the Galois group G is isomorphic to Z2 × Z2 × Z2 .
subgroups fixed
√ √fields√
{1} Q( 2,√ 3,√ 5)
{1, σ} Q(√3, √5)
{1, τ } Q(√2, √5)
{1, δ} Q(√2, √3)
{1, στ } Q(√ 5,√ 6)
{1, τ δ} Q(√2, √15)
{1, σδ} √ √ Q(√ 3, 10) √ √
{1, στ δ} Q( 6, 10, 15) ∼
√ = Q( 6, 10)
{1, σ, τ, στ } Q(√5)
{1, σ, δ, σδ} Q(√3)
{1, τ, δ, τ δ} Q(√ 2)
{1, σ, τ δ, στ δ} Q(√15)
{1, τ, σδ, τ σδ} Q( √10)
{1, δ, στ, δστ } Q(√ 6)
{1, στ, τ δ, σδ} Q( 30)
hσ, τ, δi Q
which is the correspondence between
√ the
√ subgroups
√ of the Galois group Z2 × Z2 × Z2 and the
fixed fields. And |Gal(K/Q)| = [Q( 2, 3, 5) : Q] = 8.
45
√
By example in 13.4 on page 541, the roots of the √ polynomial xp −√2 are ξ p 2 where ξ is the pth
root of unity. The splitting field for xp − 2 is Q( p 2, ξp ) and [Q( p 2, ξp ) : Q] = p(p − 1).
√ √
p − 2 (all roots ξ p 2 are
Since Q( p 2, ξ) is the splitting field for the
√ separable polynomial
√ x
distinct), so it is Galois over Q. Hence [Q( p 2, ξp ) : Q] =| Gal(Q( p 2, ξp )/Q) |= p(p − 1).
τ p−1 = 1.
√ √ √ √ √ √
a
p
2 → p 2 → ξa p 2 p
2 → ξ p 2 → ξa p 2
Also σ τ : and τ σ:
ξ → ξa → ξa ξ → ξ → ξa
So σ a τ = τ σ.
14.2.5 Prove that the Galois group of xp − 2 for p a prime is isomorphic to the
group of matrices
a b
0 1
where a, b ∈ Fp , a 6= 0.
√ √ √ √
The roots of xp − 2 are p 2, ξ p 2, ξ 2 p 2, · · · ,√ξ p−1 √
p
2 where ξ is a primitive pth root of unity.
p p
Hence the splitting field can be written as Q( 2, ξ 2) (example in 13.4, p541). Any automor-
phism σ ∈ G maps each of these two elements to one the roots of xp − 2 hence can be defined
by √ √
p
2 → ξb p 2
σa,b :
ξ → ξa
√
Since every element in G must map these generator √ ξ and p 2 to their√conjugate so there are
at most p(p − 1) choices. And p(p − 1) = |Gal(Q(ξp , p 2)/Q)| = [Q(ξp , p 2) : Q]. So σa,b where
a, b ∈ Zp , a 6= 0 are precisely the elements of G.
- f is injective: if σ ∈ ker(f ), f (σ) is the identity matrix (note that here the operation of the
matrix group is a multiplication). This implies that a = 1, and b = 0. Hence σ1,0 is an identity
46
in G.
- f is clearly surjective by the way it is defined.
- f is a group homomorphism: for σa,b , σc,d ∈ G, since
√ √ √
p
2 → ξ b p 2 → ξ bc+d p 2
σc,d · σa,b =
ξ → ξ a → ξ ac
So we have σc,d · σa,b = σbc+d,ac .
ac bc + d c d a b
Hence f (σc,d · σa,b ) = f (σac,bc+d ) = = = f (σc,d )f (σa,b )
0 1 0 1 0 1
Therefore we proved that f is an isomorphism.
First, find the roots of this polynomial. If α is a root of this equation, then α8 = 3, then
(ξα)8 = 3 where ξ is any primitive 8th root of 1. Hence the solutions of this equation are
√
ξ a 8th root of 1
8
ξ 3,
Compute ξ explicitly, first note that x8 − 1 = (x4 − 1)(x4 + 1)
√
= (x√− 1)(x + 1)(x2 + 1)(x4 + 1).
And exercise 13.4.2 proved that the roots for x + 1 are ± 2 ± 22 i. In addition, ±i are the
4 2
√ √ √ √
2 2 2 2
roots of x2 + 1. Hence ξ is precisely (± 2 ± 2 i)i which is again ± 2 ± 2 i.
the other hand, all roots are clearly contained in Q( 3, ξ). Therefore, K = Q( 8 3, ξ).
8
√ √ √ √
2
√
2
Now prove that Q( 8 3, ξ) = Q( 8 3, 2, i) where ξ is a primitive 8th root of 1, i.e. 2 + 2 i.
√
⊆ is clear. The other containment ⊇ is also true√because
√ 2 = ξ + ξ 7 and i = ξ 2 . Hence we
8
proved that the splitting field field is actually Q( 3, 2, i).
√ √
(2) Prove that [Q( 8 3, 2, i) : Q] = 32.
√ √ √ √ √
Consider the extension of fields Q ⊆ Q( 8 3) ⊆ Q( 8 3, 2) ⊆ Q( 8 3, 2, i)
√ 8
For the first containment, Q( 8 3) has degree
√ √ √x − 3 √
8 over Q since it satisfies which is monic,
irreducible by Eisenstein at p = 3 and has 3 as a root. Also, [Q( 3, √2) : Q( 8 3)] has degree
8 8
at most 2 and
√ is precisely
√ 2 if and only if x2 − 2 is irreducible over Q( 8 3). And this is true if
8
and only if 2 6∈ Q( 3).
√ √ √ √
So we√want to√prove that 2 6∈ Q( 8 3). First, prove that 2 6∈ Q( 4 3). We√know already √
4
that
√ √ 2 ∈
6 Q( 3) (use
√ the same
√ argument in√example(2) √on page √526). So if 2 ∈ Q( 3) =
4 4 4
Q( 3)( 3), then 2 = a + b 3, for a, b ∈ Q( 3). But Q( 3)/Q( 3) is a quadratic extension
hence is a degree
√ 2 Galois
√ extension. So there are two automorphisms: √ an identity and σ,
4
say. Hence σ : 2√7→ − 2. Since √ an automorphism will maps√ a + b 3 to
√ its conjugate, so
4 4 4
we have σ : a + b 3 7→ a − b 3. But we just assumed that 2 = a + b 3, so this implies
47
√ √
σ: 2 7→ a − b 4 3. In addition, automorphism σ fixes Q. Combine all these we have
√ √ √ √ √ √ √
σ(2) = σ( 2 2) = σ( 2)σ( 2) = (a − b 3)(a − b 3) = (a − b 3)2
4 4 4
√ √
So these√two should be equal. Hence we have (a + b 2)2 = (a − b 2)2 . After simplifying we
get 2ab 4 3 = 0. So either a or b must be 0.
√ √ √
4
If b = 0, a 6= 0, we have2=√ a, a contradiction. If a = 0, b 6
= 0, we get 2 = b 3. Raise
2 √
to both sides, we have b = √
4
, a contradiction since b ∈ Q( 3). Hence we proved that
√ √ 3
2 6∈ Q( 4 3).
√ √
8
√
Now,
√ proceed√ to the
√ next stage.
√ Want to
√ prove that 2
√ ∈
6 Q( 3). Assuming that 2 ∈
8 4 8 8 4
Q( 3) = Q( 3)( 3), then 2 = a + b 3 for a, √b ∈ Q( √ 3). Use the same method above,
8
we √can √get a contradiction,
√ hence prove that 2 ∈
6 Q( 3). Therefore, we proved that
8 8
[Q( 3, 2) : Q( 3)] = 2.
√ √ √ √ √ √
Finally, Q( 8 3, 2, i) has degree 2 extension over Q( 8 3, 2) since i 6∈ Q( 8 3, 2) and satisfies
the degree 2 irreducible polynomial x2 + 1.
Thus, by the multiplication rule of field extensions, we have
√
8
√ √
8
√ √
8
√ √
8
√ √
8
√
8
[Q( 3, 2, i) : Q] = [Q( 3, 2, i) : Q( 3, 2)][Q( 3, 2) : Q( 3)][Q( 3) : Q] = 2·2·8 = 32
Any automorphism
√ √ fixes the prime field Q and is completely determined by its action on the
8
generators 3, 2 and i.
√ 8
√
8
√8
√
8
√8
√
8
3 → ξ 3
3 → 3
3 → 3
i√→ i √ i√→ −i√ i√→ i √
Define σ: τ: δ:
2→ 2
2→ 2
2→− 2
ξ→ξ
ξ→ξ 7
ξ → ξ5
√ 8
√
8
√ √
7 83→ 83
3 → ξ 3 → ξ
i√→ i → √ −i →√−i √
So we have στ σ:
2 → 2→ 2→ 2
ξ → ξ → ξ7 → ξ7
48
√ 8
√ √ √ √ √
3→ 83→ξ83
8
3→ξ83→ξ83
i√→ i → √i i√→ i →√ −i
σδ : √ , δσ : √
2→− 2→− 2 2→ 2→− 2
ξ → ξ5 → ξ5 ξ → ξ → ξ5
and √ √ √ √ √ √
8
3→ 83→ 83
8
3→ 83→ 83
i√→ i → √−i i√→ −i → √ −i
τδ : √ , δτ : √
2→− 2→− 2
2→− 2→− 2
ξ → ξ 5 → ξ 35 = ξ 3 ξ → ξ 7 → ξ 35 = ξ 3
|H||K|
| HK |=
|H ∩ K|
if H and K are finite subgroups of G (note that here H ∩ K = 1). But G has order 32, hence
we must have HK = G.
Now since both H and K are normal subgroup and H ∩ K = 1, by Theorem 9 in section 5,4
we have G = HK ∼= H × K. But H ∼= D16 and K ∼= Z2 , hence we have G ∼
= D16 × Z2 .
14.2.13 Prove that if the Galois group of the splitting field of a cubic over Q is the
cyclic group of order 3 then all the roots of the cubic are reals.
Let K be the splitting field of a cubic polynomial f over Q. Let G denote the Galois group
Gal(K/Q). Suppose G is a cyclic group of order 3. We claim that in this case all roots are
real. Suppose not, since complex roots are conjugate so must be in pair. Let β1 and β2 be the
conjugate complex roots and α the real root. The conjugation induces a transposition that
permutes these two roots. And it is an automorphism of K fixing Q. So the Galois group G
contains an element of order two. But this is a contradiction since by assumption G is cyclic
of order 3 that can’t contain an element of order 2.
Since f (x) is an irreducible polynomial over Q√ hence separable. Find all roots of f (x). Let
f (x) =px4 − 4x2 + 2 = 0, we have x2 = 2 ± 2. Taking the square root both sides we get
√
x = ± 2 ± 2 and these are the roots of f (x).
49
p √
Now, show that Q( 2 + 2) is the splitting field of f (x).
p Let√K be the splitting field of f (x).
K must contains all roots hence apparently contain Q( 2 + 2). On the other hand, need to
p √ p √ p √ √2+√2
prove that all roots are in K. Let β = 2 − 2, since β = 2 − 2 = 2 − 2 √ √ =
√ √ 2+ 2
2 √ √ 2 α 2−2 2
p √ . But α2 = 2 + 2 gives 2 = α2 − 2. Hence, β = p √ = = α− ,
2+ 2 p 2+ 2 α α
√ p √
so β is in Q( 2 + 2). Hence all roots ±α and ±β are in K. Thus, K = Q( 2 + 2) is the
splitting field of f (x).
Since K is the splitting field of the separable polynomial f (x), so K/Q is Galois extension. Let
G = Gal(K/Q). Any automorphism in G is completely determined by its action on α. Define
2 σ 2 σ 2 2 α2 − 2 2
σ : α 7→ β(= α − ). So we have α −→ α − −→ (α − ) − 2 = − α2 −2 =
α α α α− α α
√ 2
√ √ α
2 2α 2 − 2α 2 − 2(2 + 2) 2+2 σ 2 2
−√ = √ = √ p √ = −p √ = −α −→ −α − = −α + =
α 2 2α 2( 2 + 2) 2 + 2 −α α
√ √
−α2 + 2 2 2 p √ σ p √
=− =p √ = − 2 − 2 −→ 2 + 2 = α
α α 2+ 2
In summary, we have
σ 2 σ σ σ
α −→ α − −→ −α −→ β −→ α
α
i.e.pσ 4 = 1. Therefore, the Galois group G is isomorphic to cyclic group of order 4. i.e.
√
Q( 2 + 2) is a cyclic quartic field.
√ √
14.3.6 Suppose K =√Q(θ) = √ Q( D1 , √D2 ) with D1 , D2 ∈ Z, is a biquadratic extension
and that θ = a + b D1 + c D2 + d D1 D2 where a, b, c, d ∈ Z are integers. Prove
that the minimal polynomial mθ (x) for θ over Q is irreducible of degree 4 over Q
but is reducible modulo every prime p. In particular show that the polynomial
x4 − 10x2 + 1 is irreducible in Z[x] but is reducible modulo every prime.
(1) Prove that there are no biquadratic extensions over finite fields.
Prove that the minimal polynomial for θ over Q is irreducible of degree 4 over Q.
Since the biquadratic field extension has degree 4 (exercise 13.2.8), so [Q(θ) : Q] = deg
mθ (x) = 4. And by definition, the minimal polynomial is irreducible. Hence mθ (x) is ir-
reducible of degree 4.
50
However, mθ (x) is reducible modulo every p. Suppose not, suppose it is irreducible mod p.
But since mθ (x) is the minimal polynomial, so we have [Fp (θ) : Fp ] = deg mθ (x) = 4 which is
not possible since we just proved that there are no biquadratic extensions over finite fields.
(2) Show that x4 − 10x2 + 1 is irreducible in Z but not irreducible modulo every prime.
4 2
√ we have proved that the polynomial x − 10x + 1 is
In exercise 13.2.7 in the first homework,
√
irreducible in Q[x] and has ± 2 ± 3 as roots. By Corollary 6 (the corollary of the Gauss
Lemma) in Section 9.3, then x4 − 10x2 + 1 is irreducible over Z[x].
14.3.7 Prove that one of 2,3 or 6 is a square in Fp for every prime p. Conclude
that the polynomial
x6 − 11x4 + 36x2 − 36 = (x2 − 2)(x2 − 3)(x2 − 6)
has a root modulo p for every prime p but has no root in Z.
If 2 or 3 is a square in Fp , then we are done since the polynomial has a root mod p. If neither
2 nor 3 is a square in Fp , want to prove that 2 · 3 = 6 is a square in Fp .
But we assume that 2, 3 are not squares, so m, n are both odds. Thus, we have 2 · 3 = σ m σ n =
σ m+n which is a square in Fp since k + m must be even. This implies that 6 is a square in Fp .
i.e. the polynomial has a root mod p for every prime p.
√ √ √
However, this polynomial has no root in Z[x] since all roots ± 2, ± 3 and ± 6 are not in Z.
p √
14.4.1 Determine the Galois closure of the field Q( 1 + 2).
p √ √ √
Let α = 1 + 2, then α2 = 1 + 2. So α2 − 1 = 2. Squaring both sides we have
(α2 − 1)2 = α4 − 2α2 + 1 = 2 hence α4 − 2α2 − 1 = 0. So α satisfies theppolynomial
√
f (x) = x4 − 2x2 − 1 over Q. To find all roots of f (x), solving for x we get x = ± 1 ± 2.
p √ p √
Now, α is in R, however the other root 1 − 2 is in C. So the field Q( 1 + 2) is clearly
not a splitting field of f (x) hence not a Galois extension of Q.
14.4.3 Let F be a field contained in the ring of n × n matrices over Q. Prove that
[F : Q] ≤ n.
Proof
By exercise 13.2.19, the ring of n × n matrices over Q contains fields of degree n over Q. Let
F denote one of such fields. Since F sits inside an n × n matrix ring over Q, F/Q is a finite
extension. By Theorem 25, F/Q is a simple extension, so F = Q(θ) for some θ ∈ F .
Now, since [Q(θ) : Q] = deg mθ (x). And the minimal polynomial must divide the characteris-
tic polynomial which has degree n for an n × n matrix over Q. Hence, the degree of minimal
polynomial can not be greater than n. i.e. [Q(θ) : Q] =degmθ (x) ≤ n. Therefore, we proved
that [F : Q] ≤ n.
14.5.4 Let σa ∈ Gal (Q(ξn )/Q) denote the automorphism of the cyclotomic field of
nth roots of unity which maps ξn to ξna where a is relatively prime to n and ξn is a
primitive nth root of unity. Show that σa (ξ) = ξ a for every nth root of unity.
σ : ξn 7→ ξna is a homomorphism with (a, n) = 1. Let ξ be any nth root of unity, then ξ = ξnm
for some positive integer m. Hence we have
σa (ξ) = σa (ξnm ) = σa (ξn )m = (ξna )m = ξnam = (ξnm )a = ξ a
Therefore, σa (ξ) = ξ a for every nth root of unity.
√
14.5.10 Prove that Q( 3 2) is not a subfield of any cyclotomic field over Q.
√
Suppose that Q( 3 2) is a subfield of a cyclotomic field over Q. i.e. there is a field extensions
√
3
Q ⊆ Q( 2) ⊆ Q(ξn )
for some positive integer n. Since the Galois group of the cyclotomic field extension Q(ξn )/Q
is isomorphic to the multiplicative group (Z/nZ)× which is a cyclic group of order ϕ(n) hence
is abelian. So every subgroup of the Galois group is normal subgroup.
√
3
Let H be the corresponding
√ subgroup of the field Q( 2), so H is normal
√ in G. By Galois
3
Theory, the Q( √ 2) is Galois over Q. But this is a contradiction since Q( 3 2) is not Galois over
√
Q because the 3 2 satisfies the irreducible polynomial f (x) = x3 − 2 over √ Q, however, Q( 3
2)
3
is not a splitting field of f (x) since the root ξ3 is not contained in Q( 2).
√
Therefore, Q( 3 2) is not a subfield of any cyclotomic field over Q.
52
14.5.11 Prove that the primitive nth roots of unity form a basis over Q for the
cyclotomic field of nth roots of unity if and only if n is squarefree.
By the discussion on the bottom of page 598, suppose n = pa11 pa22 · · · pakk is the decomposition
a2 a
p ···p k
of n into distinct prime powers. Since ξn2 k is a primitive pa11 th root of unity. The field
K1 = Q(ξpa1 ) is a subfield of Q(ξn ). Similarly, each of the fields Ki = Q(ξpai ), i = 1, 2, · · · , k
1 i
is a subfield of Q(ξn ). Hence K1 K2 · · · Kk ⊆ Q(ξn ). On the other hand, the composite of the
the fields contains the product ξpa1 ξpa2 · · · ξpak , which is the primitive nth root of unity, hence
1 2 k
contains Q(ξn ). i.e. K1 K2 · · · Kk ⊇ Q(ξn ). Thus, the composite field is Q(ξn ). i.e.
Q(ξn ) = K1 K2 · · · Kk = Q(ξpa1 ) · · · Q(ξpak )
1 k
Since the extension degrees [Ki : Q] = ϕ(pai i ), i = 1, 2, · · · , k and ϕ(n) = ϕ(pa11 )ϕ(pa22 ) · · · ϕ(pakk ),
the degree of the composite of the fields Ki is precisely the product of the degrees of the Ki .
i.e.
[Q(ξ n ) : Q] = [Q(ξpa1 ) · · · Q(ξpak ) : Q] = [Q(ξpa1 ) : Q] · · · [Q(ξpak ) : Q]
1 k 1 k
Now, prove that the primitive nth roots of unity is linearly independent over Q if and only if
n is squarefree.
(⇐)
If n is square free, then n = p. In this case, {1, ξp , ξp2 , · · · , ξpp−2 } is linearly independent since
its minimal polynomial over Q, ξp has degree p−1. Multiplying by ξp we get {ξp , ξp2 , · · · , ξpp−1 },
this is still linearly independent hence they also forms a basis of Q(ξp )/Q.
(⇒)
Suppose n is not square free, i.e. n = pa for a ≥ 2. Then the primitive pa -th roots of
a−1
unity is ξpka where (k, p) = 1. Let ω = ξppa , then ω p = 1 which implies that ω p − 1 =
(ω − 1)(1 + · · · + ω p−1 ) = 0. But ω 6= 1, hence we have
a−1 a−1 a−1 a−1 a−1 (p−1)pa−1
0 = 1 + ξppa + (ξppa )2 + · · · + (ξppa )p−1 = 1 + ξppa + ξp2pa + · · · + ξp a
Multiplying ξpa gives
a−1 +1 a−1 +1 (p−1)pa−1 +1
ξpa + ξppa + ξp2pa + · · · + ξpa
i.e.
a−1 a−1 (p−1)pa−1 +1
ξpa = −(ξppa +1 + ξp2pa +1 + · · · + ξpa )
a
This proves that the primitive p -th roots of unity for a ≥ 2 is linearly dependent. There-
fore, the primitive nth roots of unity {1, ξn , ξn2 , · · · , ξnn−1 } is a basis if and only if n is squarefree.
is separable (guarantee all roots are distinct?). And X n − 1 is separable if and only if it is
relatively prime to nX n−1 , which is true if and only if nX n−1 is nonzero in Fp , i.e. p - n, or
(p, n) = 1.
Let K√be the √ splitting field of f (x) over Q. √K contains all roots hence contain the ratio of two
4 4
roots 7,√ i 7 which is i. Hence
√ K ⊇ Q(i, 4 7). On√ the other hand, all roots above are in this
field Q(i, 4 7). So K ⊆ Q(i, 4 7), hence K = Q(i, 4 7).
σ 2 = (1234)(1234) = (13)(24)
σ 3 = (1234)(13)(24) = (1432)
τ σ = (24)(1234) = (14)(23)
τ σ 2 = (24)(13)(24) = (13)
τ σ 3 = (24)(1432) = (12)(34)
In summary, we have G = {1, (13), (24), (13)(24), (12)(34), (14)(23), (1234), (1432)} which is
isomorphic to D8 .
15.1.2 Show that each of the following rings are not Noetherian by exhibiting an
explicit infinite increasing chain of ideals:
(a) the ring of continuous real valued functions on [0, 1].
(b) the ring of all functions from any infinite set X to Z/2Z.
(a) Let R be the ring of continuous real valued functions on [0, 1]. And let Jm be the collection
1 1
of closed sub-intervals on [0, 1] such that Ji = [ 21 − 2i 1
, 2 + 2i ], then J1 ⊃ J2 ⊃ · · · Jn ⊃ · · · .
Now, let Ii be subsets of R as follows:
I1 = {f ∈ R | f = 0 on J1 }
I2 = {f ∈ R | f = 0 on J2 }
..
.
In = {f ∈ R | f = 0 on Jn }
..
.
Prove that all Ii , i = 1, 2, · · · are ideals of R.
Now, the collection of {Ji } is an infinite descending chain of sub-intervals in [0, 1]. This gives
us a collection of ideals vanishing on {Ji } which is an infinite ascending chain. i.e.
I1 ⊂ I2 ⊂ I3 ⊂ · · ·
is an infinite ascending chain of ideals.
(b) Let R be the ring of all functions f : X → Z/2Z. And let Ji be the collection of points in
X such that Ji = X − {xα1 , xα2 , · · · , xαi } where xαj ∈ X and αj ∈ I an index set. Then we
55
obtain an descending chain of subsets J1 ⊃ J2 ⊃ · · · Jn ⊃ · · · in X.
I1 = {f ∈ R | f = 0 on J1 }
I2 = {f ∈ R | f = 0 on J2 }
..
.
In = {f ∈ R | f = 0 on Jn }
..
.
Same argument as in (a) will prove that all Ii , i = 1, 2, · · · are ideals of R.
Now, the collection of {Ji } is an infinite descending chain in X. This gives us a collection of
ideals vanishing on {Ji } which is an infinite ascending chain. i.e.
I1 ⊂ I2 ⊂ I3 ⊂ · · ·
is an infinite ascending chain of ideals.
⊇ is clear.
⊆ For an element a ∈ ker ϕn ∩imageϕn , there exists an element b ∈ M such that ϕn (b) = a.
Also, ϕn (a) = 0. This implies that ϕn (ϕn (b)) = ϕ2n (b) = 0. So we get b ∈ ker ϕ2n = ker ϕn
since M is a Noetherian R-module so that the ascending chain of ideals ker ϕ stabilize for
sufficiently large n.
15.1.7 Prove that submodules, quotient modules, and finite direct sums of Noe-
therian R-modules are again Noetherian R-modules.
0 → ker ϕ → Rn → M → 0.
The other direction is clear since from the definition of a Noetherian module, an R-module
that is Noetherian has to be finitely generated.
15.1.11 Suppose R is a commutative ring in which all the prime ideals are finitely
generated. Proves that R is Noetherian.
(a) Prove that if the collection of ideals of R that are not finitely generated is
nonempty, then it contains a maximal element I, and that R/I is a Noetherian
ring.
(b) Prove that there are finitely generated ideals J1 and J2 containing I with
J1 J2 ⊆ I and that J1 J2 is finitely generated.
(c) Prove that I/J1 J2 is a finitely generated R/I−submodule of J1 /J1 J2 .
(d) Show that (c) implies the contradiction that I would be finitely generated over
R and deduce that R is Noetherian.
(a) The proof is basically similar to the proof of Proposition 11 in 7.4 which shows that in a
ring with identity every ideal is contained in a maximal idea. Let S 6= φ be the collection of
S by inclusion. If C is
ideals of R that are not finitely generated. Then S is partially ordered
a chain in S, defined I to be the union of all ideals in C. i.e. I = A∈C A. The same proof
as Proposition 7.4 shows that I is an ideal and actually I is a proper ideal. Hence each chain
has an upper bound in S. By Zorn’s Lemma, S has a maximal element I. So I is not finitely
generated.
This implies that all ideals in the quotient R/I are finitely generated. Hence by Theorem 2,
R/I is Noetherian.
57
(b) By assumption all prime ideals of R are finitely generated, so the maximal element I in S
found in (a) is not finitely generated hence is not prime. So there are two elements a, b of R\I
such that ab is contained in I. Let J1 = I + (a) and J2 = I + (b), then J1 , J2 contains I. Since
both J1 and J2 are not contained in I, so they are not in S hence are finitely generated. So
their product J1 J2 is also finitely generated.
(c) Since (b) proved that J1 J2 ⊆ I ⊆ J1 ⊆ R, then by Lattice Isomorphism Theorem we have
Now, since in (a) we proved that R/I is Noetherian, then R/I is a Noetherian module over
itself. Hence its quotient (R/I)/(J1 J2 /I) ∼
= R/J1 J2 is Noetherian by exercise 15.1.7. Then
from above, I/J1 J2 is a submodule of R/J1 J2 , hence again Noetherian. Thus, I/J1 J2 is a
Noetherian R/I-submodule of J1 /J1 J2 . By exercise 15.1.8, I/J1 J2 is finitely generated.
(d) In (c) we prove that I/J1 J2 is a finitely generated R/I-submodule of J1 /J1 J2 . Since J1 J2
is finitely generated by (b). But this implies that I is a finitely generated R-submodule by
exercise 10.3.7. i.e. I is a finitely generated ideal of R. This is a contradiction since I is an
ideal in the set S consisting of all ideals that are not finitely generated. Hence we conclude
that R is Noetherian.
(a) ϕ is a morphism since there are polynomials f1 = t3 , f2 = t4 and f3 = t5 ∈ k[A1 ] such that
ϕ(t) = (f1 (t), f2 (t), f3 (t)) = (t3 , t4 , t5 ) for all t ∈ A1 .
To prove that ϕ is surjective, assume that (x, y, z) 6= (0, 0, 0) (if one of x, y, z is zero, the other
two must be zero. hence V = A3 ). So let t = xy . Then we have
y y y3 y4 y5
3 y 4 y 5
ϕ( ) = ( ) , ( ) , ( ) = , ,
x x x x x3 x4 x5
y3 (xz)y zy x3
= = = =x
x3 x3 x2 x2
y4 x2 z 2 xz y2
= = = =y
x4 xyz y y
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y5 x2 z 2 y
5
= =z
x yzx2
Hence we have
y
ϕ(t) = ϕ( ) = (x, y, z)
x
so that ϕ is surjective.
(c) Note that the image of ϕ̃ is the subalgebra k[x3 , x4 , x5 ] = k + x3 k[x] of k[x]. In particular,
ϕ̃ is not surjective (for example, x2 is not in k + x3 k[x]), hence ϕ̃ is not an isomorphism. Then
by Theorem 6(4), ϕ is not an isomorphism.
15.2.3 Prove that the intersection of two radical ideals is again a radical ideal.
Lemma Let I and J be ideals in the ring R. Then rad (I ∩ J) =rad I∩ rad J.
So now, let I, I be two radical ideals. So rad I = I and rad J = J. Prove that I ∩ J is a radical
ideal. i.e. prove that rad(I ∩ J) = I ∩ J. Indeed, rad (I ∩ J)=rad I∩ rad J (by Lemma) =I ∩ J.
First prove that I is a radical ideal, i.e. prove that rad I = I. It’s obviously that I ⊆ rad I.
It remains to show that
Now, rad (I) = rad (m1 m2 ) = rad (m1 ) ∩ rad (m2 ) = m1 ∩ m2 since m1 , m2 are radical.
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If i = j = 0, i.e. the polynomial is a constant 1, then 1 = (x − 1) + x + 2 ∈ (x, y) + (x − 1, y − 1).
On the other hand, x ∈ (x, y) + (x − 1, y − 1) and y ∈ (x, y) + (x − 1, y − 1). Hence m1 = (x, y)
and m2 = (x−1, y −1) are comaximal i.e. m1 +m2 = R so m1 m2 = m1 ∩m2 by exercise 7.3.34.
Finally, the ideal (x3 − y 2 ) is prime in k[x, y] for any field k hence prime in F2 [x, y]. Therefore,
(x3 − y 2 ) is a radical ideal in F2 [x, y].
15.2.9 Prove that for any field k the map Z in the Nullstellansatz is always sur-
jective and the map I in the Nullstellansatz is always injective. Give examples
(over a field k that is not algebraically closed) where Z is not injective and I is
not sujective.
(1) Prove that the map Z is always surjective and the map I is always injective.
Let I1 = (1) and I2 = (x2 + 1) be ideals of R[x]. Then I1 6= I2 . However, Z(I1 ) = Z(I2 ) = φ.
Hence Z is not injective.
15.2.28 Prove that each of the following rings have infinitely many minimal prime
ideals, and that (0) is not the intersection of any finite number of these.
(a) the infinite direct product ring Z/2Z × Z/2Z × · · · .
(b) k[x1 , x2 , · · · ]/(x1 x2 , x3 x4 , · · · , x2i−1 x2i , · · · ), where x1 , x2 , · · · are independent vari-
ables over the field k.
(a) Consider the ideals I1 = h(0, 1, 1, · · · )i, I2 = h(1, 0, 1, · · · )i, Ii = h(1, · · · , 1, 0, 1, · · · )i. i.e. Ii
is an ideal generated by an element with all 1s on all coordinates except 0 in the ith coordinate,
in the ring R = Z/2Z × Z/2Z × · · · . These ideals are prime since R/Ii = Z/2Z is an integral
domain. Also, these ideals are minimal. If there is a prime ideal J with J ⊂ Ii and J 6= Ii ,
then J must have two or more of zeros, for example, if J has two zeros in the coordinates. i.e.
J = (1, · · · , 1, 0, 1, · · · , 1, 0, 1, · · · ) then R/J ∼
= Z/2Z × Z/2Z which is not an integral domain
since there is a zero divisor when in the product (i.e. a = (1, 0) and b = (0, 1) are nonzero in
Z/2Z × Z/2Z, but ab = 0). Similar for the case that more than two zeros in the coordinate
of ideal J. So J can not be prime, a contradiction. Therefore, Ii for i = 1, 2, · · · are minimal
prime ideals in R.
Now prove that (0) is not the intersection of any finite number of these ideals. i.e. (0) 6= N
T
i=1 Ii
for some positive integer N . Indeed, the element (0, · · · , 0, 1, 1, 1, · · · , ) with first N position
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all 0s is contained in the intersection. Hence (0) is not the intersection of any finite number of
these ideals.
In general, let Ik = (xk12 , xk34 , · · · ) where xk12 is a choice from {x1 , x2 } and xk34 is chosen
from {x3 , x4 } and so on. So there are infinitely many combinations of choices hence there are
infinitely many such ideals in R = k[x1 , x2 , · · · ].
Ik are all prime since if we let J = (x1 x2 , x3 x4 , · · · , x2i−1 x2i , · · · ), by the Third Isomorphism
Theorem we have
R/J ∼ R
=
Ik /J Ik
which is an integral domain (for example, k[x1 , x2 , · · · ]/(x1 , x3 , x5 , · · · ) ∼
= k[x2 , x4 , x6 , · · · ] is an
integral domain). Hence Ik /J is a prime ideal in R/J.
Now, claim that Ik /J are minimal. Since Ik is an ideal such that each coordinate was chosen
form the pair x2i−1 , x2i , so suppose there is an ideal I in R such that I/J strictly contained
in Ik /J, i.e. I/J ⊂ Ik /J, then I/J must have zero in at least one of the coordinates, say x2i−1
x2i = 0. But note that the quotient I/J just means that J is contained in I. And we know
that J = (x1 x2 , x3 x4 , · · · , x2i−1 x2i , · · · ). Hence this is a contradiction since J can not sit inside
I. So we proved that Ik /J is the minimal ideal.
Finally, clearly (0) is not the intersection of any finitely many of these ideals Ik /J. Suppose
(0) = N
T
I /J, but we know the product of finitely many ideals is contained in their product.
QNk=1 k TN QN
i.e. k=1 k /J ⊆
I k=1 Ik /J. Hence we have k=1 Ik /J = 0 which is not possible since no
coordinate in Ik /J is zero for all k.
15.2.33 Let I = (x2 , xy, xz, yz) in k[x, y, z]. Prove that a primary decomposition of
I is I = (x, y) ∩ (x, z) ∩ (x, y, z)2 , determine the isolated and embedded primes of I,
and find rad I.
(1) Prove that (x, y), (y, z) and (x, y, z)2 are primary.
(3) Prove that I = (x2 , xy, xz, yz) = (x, y) ∩ (y, z) ∩ (x, y, z)2 .
X
The ideal I1 = (x, y) = f x+gy for f, g ∈ k[x, y, z] and it consists of elements {x, y, xy, yz, xz
x , y , xyz, · · · }. Similarly, the ideal I2 = (x, z) consists of elements {x, z, xz, xy, yz, x2 , z 2 , xyz, · · · }
2 2
and I3 = (x, y, z)2 consists of elements {x2 , y 2 , z 2 , xy, yz, xz, xyz, · · · }.
To see the intersection I1 ∩I2 ∩I3 , note that x is contained in I1 , I2 but not I3 , hence is not con-
tained in the intersection. Similarly, elements y, z are not contained in the intersection. Now
look at the possible degree-two elements, we see that xy, yz, xz are clearly in the intersection.
In addition, x2 is also in the intersection, but not y 2 nor z 2 . All degree 3 or higher monomial
can be generated by these elements. Hence we conclude that the intersection of these three
ideals is {x2 , xy, yz, xz}. i.e. I = (x2 , xy, xz, yz) = (x, y) ∩ (y, z) ∩ (x, y, z)2
The radical ideal of I is the intersection of the isolated primes of I by Corollary 22. Hence
radI = (x, y) ∩ (x, z) which is (x, yz). The reason (x, y) ∩ (x, z) = (x, yz) is similar to (3).
We had (x, y) = {x, y, xy, yz, xz x2 , y 2 , xyz, · · · } and (x, z) = {x, z, xz, xy, yz, x2 , z 2 , xyz, · · · },
hence (x, y) ∩ (x, z) = (x, yz) since all degree 3 or higher elements can be generated by these
two generators.
15.2.39 Fix an element a in the ring R. For any ideal I in the ring R let Ia = {r ∈
R | ar ∈ I}.
(a) Prove that Ia is an ideal and Ia = R if and only if a ∈ I.
(b) Prove that (I ∩ J)a = Ia ∩ Ja for ideals I and J.
(c) Suppose that Q is a P -primary ideal and that a 6∈ Q. Prove that Qa is a P -
primary ideal and that Qa = Q if a 6∈ P .
Now for any element c ∈ R, and r ∈ Ia , we have (ac)r = a(cr) ∈ I since cr ∈ I. So cIa ⊆ Ia
for any c ∈ R. Similarly we can prove that Ia c ⊆ Ia . Hence Ia is an ideal.
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= {r ∈ R | ar ∈ I} ∩ {r ∈ R | ar ∈ J} = Ia ∩ Ja
by exercise 15.2.2(c).
(d) In a minimal primary decomposition, no primary ideal contains the intersection of the
remaining primary ideals, so forTeach i we can pick an element a in the intersection of Qj for
all j 6= i but a 6∈ Qi . i.e. a ∈ j6=i Qj but a 6∈ Qi . Then by (b), for this element a we have
rad(Ia ) = Pi .
(e) From all above, we proved that the associated primes Pi for a minimal primary decom-
position are precisely the collection of prime ideals among the ideals rad(Ia ) for a ∈ R, i.e.
{rad(Ia ) | a ∈ R} = {Pi }. Hence these prime ideals are uniquely determined by I independent
of the minimal primary decomposition.
16.1.8 Let M be a maximal ideal of the ring R and suppose that M n = 0 for some
n ≥ 1. Prove that R is Noetherian if and only if R is Artinian.
we have
0 −→ M i+1 −→ M i −→ M i /M i+1 −→ 0
0 −→ M −→ R −→ R/M −→ 0
Now, if R is Noetherian, its submodule M and quotient module are also Noetherian by exer-
cise 15.1.7. Hence by this exercise again the submodule M 2 of M and the quotient M 2 /M are
Noetherian. Inductively, we have all M i and M i /M i+1 Noethterian for all i = 1, · · · , n − 1.
But we just proved that M i /M i+1 is Noetherian if and only if they are Artinian. Then by
exercise 16.1.6 the short exact sequence in Artinian version we can prove that R is Artinian.
The other direction is exactly the same argument.
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Therefore we conclude that R is Noetherian if and only if R is Artinian.
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