Solutions to Problems

Chapter 1
Section 1.1
1. Multiply the equation by an−1 to get

a−1 = −(cn−1 + · · · + c1 an−2 + c0 an−1 ) ∈ A.

2. Since A[b] is a subring of B, it is an integral domain. Thus if bz = 0 and b = 0, then

z = 0.
3. Any linear transformation on a finite-dimensional vector space is injective iff it is
surjective. Thus if b ∈ B and b = 0, there is an element c ∈ A[b] ⊆ B such that bc = 1.
Therefore B is a field.
4. Since P is the preimage of Q under the inclusion map of A into B, P is a prime ideal.
The map a + P → a + Q is a well-defined injection of A/P into B/Q, since P = Q ∩ A.
Thus A/P can be viewed as a subring of B/Q.
5. If b + Q ∈ B/Q, then b satisfies an equation of the form

xn + an−1 xn−1 + · · · + a1 x + a0 = 0, ai ∈ A.

By Problem 4, b + Q satisfies the same equation with ai replaced by ai + P for all i. Thus
B/Q is integral over A/P .
6. By Problems 1-3, A/P is a field if and only if B is a field, and the result follows. (Note
that B/Q is integral domain (because Q is a prime ideal), as required in the hypothesis
of the result just quoted.)

Section 1.2
1. If x ∈
/ M, then by maximality of M, the ideal generated by M and x is R. Thus there
exists y ∈ M and z ∈ R such that y + zx = 1. By hypothesis, zx, hence x, is a unit. Take
the contrapositive to conclude that M contains all units, so R is a local ring by (1.2.8).
2. Any additive subgroup of the cyclic additive group of Z/pn Z must consist of multiples
of some power of p, and it follows that every ideal is contained in (p), which must therefore
be the unique maximal ideal.
3. The set of nonunits is M = {f /g : g(a) = 0, f (a) = 0}, which is an ideal. By (1.2.8),
R is a local ring with maximal ideal M.

1
2

4. S −1 (g ◦ f ) takes m/s to g(f (m))/s, as does (S −1 g) ◦ (S −1 f ). If f is the identity on

M , then S −1 f is the identity on S −1 M .
5. By hypothesis, g ◦f = 0, so (S −1 g)◦(S −1 f ) = S −1 (g ◦f ) = S −1 0 = 0. Thus im S −1 f ⊆
ker S −1 g. Conversely, let y ∈ N, s ∈ S, with y/s ∈ ker S −1 g. Then g(y)/s = 0/1, so for
some t ∈ S we have tg(y) = g(ty) = 0. Therefore ty ∈ ker g = im f , so ty = f (x) for some
x ∈ M . We now have y/s = ty/st = f (x)/st = (S −1 f )(x/st) ∈ im S −1 f .
6. The sequence 0 → N → M → M/N → 0 is exact, so by Problem 5, the sequence
0 → NS → MS → (M/N )S → 0 is exact. (If f is one of the maps of the first sequence,
the corresponding map in the second sequence is S −1 f .) It follows from the definition of
localization of a module that NS ≤ MS , and by exactness of the second sequence we have
(M/N )S ∼ = MS /NS .

Section 2.1
1. A basis for E/Q is 1, θ, θ2 , and

θ2 1 = θ2 , θ2 θ = θ3 = 3θ − 1, θ2 θ2 = θ4 = θθ3 = 3θ2 − θ.

Thus
 
0 −1 0
m(θ2 ) = 0 3 −1
1 0 3

and we have T (θ2 ) = 6, N (θ2 ) = 1. Note that if we had already computed the norm of θ
(the matrix of θ is
 
0 0 −1
m(θ) = 1 0 3
0 1 0

and T (θ) = 0, N (θ) = −1), it would be easier to calculate N (θ2 ) as [N (θ)]2 = (−1)2 = 1.
2. The cyclotomic polynomial Ψ6 has only two roots, ω and its complex conjugate ω. By
(2.1.5),

T (ω) = ω + ω = eiπ/3 + e−iπ/3 = 2 cos π/3 = 1.

3. We √ have min(θ, Q) = X 4 − 2, min(θ2 , Q) = X 2 − 2, min(θ3 , Q) = X 4 − 8, and

min( 3θ, Q) = X − 18. (To
4
√ compute the last two minimal polynomials, note that
(θ3 )4 = (θ4 )3 = 23√= 8 and ( 3θ)4 = 18.) Therefore all four traces are 0.
4. Suppose that 3 = a +√bθ + cθ2 + dθ3 . Take the trace of both sides to conclude
that
√ a = 0. 2(The 3trace √of 3 is20 because its minimal polynomial is X 2 − 3.) Thus
3
3 = bθ + cθ + dθ , so √ 3θ = bθ + cθ +
√ 2 Again3 take the trace of both sides to get
2d.
2
d =
√ 2 0. We now have 3 =
√ bθ + cθ , so 3θ = bθ + 2c. The minimal polynomial of
3θ is X 2 − 6,√because( 3θ2 )2 = 6. Once again taking the trace of both sides, we get
c = 0. Finally, 3 = bθ implies 9 = 2b4 , and we reach a contradiction.
 3

Section 2.2
√
1. By the quadratic formula, L = Q( b2 − 4c). Since b2 − 4c ∈ Q, we may write
b2 − 4c = s/t = st/t2 for relatively prime integers s and t. We also√have s = √ uy 2 and
√ square-free. Thus L = Q( uv) = Q( d).
2
t = vz , √
with u and
√ v relatively
√ prime and
2. If Q( d) = Q( e), then d =√a + b √e for rational numbers a and b. Squaring both
sides, we have d = a2 + b2 e + 2ab e, so e is rational, a contradiction (unless a = 0 and
b = 1). √ √ √ √
3. Any isomorphism of Q( d) and Q( e) must √ carry d into a+b e for rational numbers
a and b. Thus d is mapped to a2 + b2 + 2ab e. But a Q-isomorphism maps d to d, and
we reach a contradiction as in Problem 2.
2
4. Since ωn = ω2n , we have ωn ∈ Q(ω2n ), so Q(ωn ) ⊆ Q(ω2n ). If n is odd, then n+1 = 2r,
so

ω2n = −ω2n
2r
= −(ω2n
2 r
) = −ωnr .

√ Q(ω2n ) ⊆ Q(ωn ).
Therefore √
5. Q( −3) = Q(ω) where ω = − 12 + 12 −3 is a primitive cube root of unity.
6. If l(y) = 0, then (x, y) = 0 for all x. Since the bilinear form is nondegenerate, we must
have y = 0.
7. Since V and V ∗ have the same dimension, the map y → l(y) is surjective.
8. We have (xi , yj ) = l(yj )(xi ) = fj (xi ) = δij . Since the fj = l(yj ) form a basis, so do
the yj . n
9. Write xi = k=1 aik yk , and take the inner product of both sides with xj to conclude
that aij = (xi , yj ).

Section 2.3
1. The first statement follows because multiplication of each element of a group G by a
particular element g ∈ G permutes the elements of G. We can work in a Galois extension
of Q containing L, and each automorphism in the Galois group restricts to one of the σi
on L. Thus P + N and P N belong to the fixed field of the Galois group, which is Q.
2. Since the xj are algebraic integers, so are the σi (xj ), as in the proof of (2.2.2). Thus P
and N , hence P + N and P N , are algebraic integers. By (2.2.4), P + N and P N belong
to Z.
3. D = (P − N )2 = (P + N )2 − 4P N ≡ (P + N )2 mod 4. But any square is congruent
to 0 or 1 mod 4, and
n the result follows.
4. We have yi = j=1 aij xj with aij ∈ Z. By (2.3.2), D(y) = (det A)2 D(x). Since D(y)
is square-free, det A = ±1, so A has an inverse with coefficients in Z. Thus x = A−1 y, as
claimed.
5. Every algebraic integer can be expressed as a Z-linear combination of the xi , hence of
the yi by Problem 4. Since the yi form a basis for L over Q, they are linearly independent
and the result follows. √
6. No. For example, take L = Q( m), where m is a square-free integer with m ≡ 1
mod 4. By (2.3.11), the field discriminant is 4m, which is not square-free.
4

Section 3.1
1. We may assume that I is not contained in the union of any collection of s − 1 of the
Pi ’s. (If so, we can simply replace s by s − 1.) It follows that elements of the desired form
exist.
2. Assume that I ⊆ P1 and I ⊆ P2 . We have a1 ∈ P1 , a2 ∈ / P1 , so a1 + a2 ∈
/ P1 . Similarly,
a1 ∈/ P2 , a2 ∈ P2 , so a1 + a2 ∈/ P2 . Thus a1 + a2 ∈/ I ⊆ P1 ∪ P2 , contradicting a1 , a2 ∈ I.
3. For all i = 1, . . . , s − 1 we have ai ∈
/ Ps , hence a1 · · · as−1 ∈ / Ps because Ps is prime.
But as ∈ Ps , so a cannot be in Ps . Thus a ∈ I and a ∈ / P 1 ∪ · · · ∪ Ps .

Section 3.2
1. The product of ideals is always contained in the intersection. If I and J are relatively
prime, then 1 = x + y with x ∈ I and y ∈ J. If z ∈ I ∩ J, then z = z1 = zx + zy ∈ IJ.
The general result follows by induction, along with the computation

R = (I1 + I3 )(I2 + I3 ) ⊆ I1 I2 + I3 .

Thus I1 I2 and I3 are relatively prime. Continue in this manner with

R = (I1 I2 + I4 )(I3 + I4 ) ⊆ I1 I2 I3 + I4

and so on.
2. We have R = Rr = (P1 + P2 )r ⊆ P1r + P2 . Thus P1r and P2 are relatively prime for all
r ≥ 1. Assuming inductively that P1r and P2s are relatively prime, it follows that

P2s = P2s R = P2s (P1r + P2 ) ⊆ P1r + P2s+1

so

R = P1r + P2s ⊆ P1r + (P1r + P2s+1 ) = P1r + P2s+1

completing the induction.

3. Let r be a nonzero element of R such that rK ⊆ R, hence K ⊆ r−1 R ⊆ K. Thus
K = r−1 R. Since r−2 ∈ K we have r−2 = r−1 s for some s ∈ R. But then r−1 = s ∈ R,
so K ⊆ R and consequently K = R.

Section 3.3
√
1. By (2.1.10), the norms are 6,6,4 and 9. Now if x = a + b −5 and x = yz, then
N (x) = a2 + 5b2 = N (y)N (z). The only algebraic integers of norm 1 are ±1, and there
are no √algebraic integers of norm 2 or 3. Thus there cannot be a nontrivial factorization
of 1 ± −5, √ 2 or 3. √
2. If (a + b −5)(c + d −5) = 1, take norms to get (a + 5b2 )(c2 + 5d2 ) = 1, so b = d = 0,
a = ±1, c = ±1.
3. By Problem 2, if two factors are associates, then the quotient of the factors is ±1,
which is impossible. √ √
4. This is done as in Problems 1-3, using the factorization 18 = (1 + −17)(1 − −17) =
 5

2 × 32 . √
5. By (2.2.6)√or (2.3.11), the algebraic integers are of the form a + b −3, a, b ∈ Z, or
(u/2) + (v/2) −3 with u and v odd integers. If we require that the norm be 1, we only
get ±1 in the first case. But in the second case, we have u2 + 3v 2 = 4, so u = ±1, v = ±1.
Thus if ω = eiπ/3 , then the algebraic integers of norm 1 are ±1, ±ω, and ±ω 2 .

Section 3.4
√ √ √ √
1. 1 − −5 = 2 − (1 + −5) ∈ P2 , so (1 + −5)(1 − −5) = 6 ∈ P22 .
√ that 4 ∈ P2 ,√so by Problem
2. Since 2√∈ P2 , it follows 2
√ 1,2 2 = 6 − 4 √ ∈ P22 . √
3. (2, 1 + −5)(2, 1 + −5) = (4, 2(1 + −5), (1 + −5) ), and (1 + −5)2 = −4 + 2 −5.
Therefore each of the generators of the ideal P22 is divisible by 2, hence belongs to (2).
Thus P22 ⊆ (2). √
4. x2 +5 ≡ (x+1)(x−1)
√ mod 3, which suggests that (3) = P3 P3 , where P3 = (3, 1+ −5)
and P3 = (3, 1 − −5).
√ √
5. P3 P3 = (3, 3(1+ −5), 3(1− −5), 6) ⊆ (3), because each generator of P3 P3 is divisible
by 3. But 3 ∈ P3 ∩ P3 , hence 9 ∈ P3 P3 , and therefore 9 − 6 = 3 ∈ P3 P3 . Thus (3) ⊆ P3 P3 ,
and the result follows.

Section 4.1
1. The kernel is {a ∈ A : a/1 ∈ MS −1 A} = A ∩ (MS −1 A) = M by (1.2.6).
2. By hypothesis, M ∩ S = ∅, so s ∈/ M. By maximality of M we have M + As = A, so
y + bs = 1 for some y ∈ M, b ∈ A. Thus bs ≡ 1 mod M.
3. Since 1 − bs ∈ M, (a/s) − ab = (a/s)(1 − bs) ∈ MS −1 A. Therefore (a/s) + MS −1 A =
ab + MS −1 A = h(ab).

Section 4.2

1. By the Chinese remainder theorem, B/(p) ∼ = i B/Piei . If p does not ramify, then
ei = 1 for all i, so B/(p) is a product of fields, hence has no nonzero nilpotents. On the
other hand, suppose that e = ei > 1, with P = Pi . Choose x ∈ P e−1 \ P e and observe
that (x + P e )e is a nonzero nilpotent in B/P e .
2. The minimal polynomial of a nilpotent element is a power of X, and the result follows
from (2.1.5).
n
3. let β = i=1 bi ωi with bi ∈ Z. Then, with T denoting trace,

n 
n
T (A(βωj )) = T ( bi A(ωi ωj )) = bi T (ωi ωj ) ≡ 0 mod p.
i=1 i=1

If β ∈
/ (p), then not all the bi can be 0 mod p, so the determinant of the matrix (T (ωi ωj )),
which is the discriminant D by (2.3.1), is 0 mod p. Therefore, p divides d. ♣
4. This follows from the Chinese remainder theorem, as in Problem 1. The fields Fi all
have characteristic p because p annihilates B/(p). 
5. The Ti are nondegenerate by separability, and i Ti is nondegenerate by orthogonality,
that is, πi (x)πj (y) = 0 for i = j.
6

6. Since Fi /Fp is a finite extension of a finite field, it is a Galois extension, so all em-
beddings are actually automorphisms. Thus for any z ∈ Fi , the endomorphism given by
multiplication by z has trace TFi /Fp (z) = Ti (z). Since B/(p) is, in particular, a direct
sum of the Fi , the result follows.

Section 4.3
√
1. Factoring (2) is covered by case (c1) of (4.3.2), and we have (2) = (2, 1 + −5)2 .
√ case (a1), and x + 5 ≡ (x + 1)(x − 1) mod 3. Therefore
2
Factoring (3)√is covered by
(3) = (3, 1 + −5) (3, √1 − −5).
2. We have (5) = (5, −5)2 , as in case
√ (b). To factor
2
√ (7), note that x + 5 factors mod 7
as (x + 3) (x − 3), so (7) = (7, 3 + −5) (7, 3 − −5), as in case (a1). Since -5 is not a
quadratic residue mod 11, we are in case (a2) and 11 remains prime.
3. Mod 5 we have x3 − 2 ≡ x3 − 27 = x3 − 33 = (x − 3)(x2 + 3x + 9) = (x + 2)(x2 + 3x − 1).
Thus

(5) = (5, α + 2)(5, α2 + 3α − 1)

√
3
where α = 2.

Section 5.3

1. We have r2 = 1 and n = 2, so the bound is (4/π)(2/4) |d| = (2/π) |d|. The
discriminant may be calculated from (2.3.11). We have d = 4m for m = −1, −2, √ and
d = m for m = −3, −7. The largest |d| is 8, and the corresponding bound is 4 2/π,
which is about 1.80. Thus all the class numbers are 1.
2. We have r2 = 0 and n = 2, so the bound is |d|/2. We have d = 4m for √ m = 2, 3, and
d = m for m = 5, 13. The largest |d| is 13, and the corresponding bound is 13/2, which
is about 1.803. Thus all the class numbers are 1. √
3. The discriminant is -20 and the Minkowski bound is 2 20/π, which is about 2.85.
Since 2 ramifies [see (4.3.2), case (c1)], there√is only one ideal of norm 2. Thus class
number is at most 2. But we know that Q( −5) is not a UFD, by the exercises for
Section 3.3. Therefore the class number is 2. √ √
4. The discriminant is 24 and the bound is 24/2 = 6, which is about 2.45. Since √ 2
ramifies [see (4.3.2), case (b)],√the argument
√ proceeds as in Problem
√ 3. Note
√ that Q( 6) is
not a UFD because
√ −2 =√ (2+ 6)(2−√ 6). Note also that 2+ 6 and √ 2− 6 are√associates,
because (2 + 6)/(2 − 6) = −5 − 2 6, which √ is a unit [(−5 − 2 6)(−5 + 2 6) = 1].
5. The discriminant is 17 and the bound is 17/2, which is about 2.06. Since 2 splits
[(4.3.2), case (c2)], there are 2 ideals of norm
√ 2. In fact these
√ ideals are principal, as can
be seen from the factorization −2 = [(3 + 17)/2] [(3 − 17)/2]. Thus every ideal class
contains a principal ideal, so the ideal class group
√ is trivial.
√
6. The discriminant is 56 and the bound is 56/2 = 14, which is about 3.74. Since 3
remains prime [(4.3.2), case (a2)], there are no ideals of norm 3. (The norm of the principal
ideal (3) is 32 = 9.) Since 2 ramifies [(4.3.2), case (b)], there is only one
√ ideal of√norm 2.
This ideal is principal, as can be seen from the factorization 2 = (4 + 14)(4 − 14). As
in Problem 5, the class number is 1.
 7

7. This follows from the Minkowski bound (5.3.5) if we observe that N (I) ≥ 1 and
2r2 ≤ n.
8. By a direct computation, we get a2 and

an+1 π (n + 1)2n+2 1 π 1
= = (1 + )2n .
an 4 n2n (n + 1)2 4 n

By the binomial theorem, an+1 /an = (π/4)(1 + 2 + positive terms) ≥ 3π/4. Thus

a3 an π2
|d| ≥ a2 ··· ≥ (3π/4)n−2 ,
a2 an−1 4

and we can verify by canceling common factors that (π 2 /4)(3π/4)n−2 ≥ (π/3)(3π/4)n−1 .

9. By Problem 8,
π 3π π 3π 3π
log |d| ≥ log + (n − 1) log = log − log + n log
3 4 3 4 4
and the result follows.
10. This follows from the bound given in Problem 8.

Section 6.1
1. Since x, hence jx, as well as ei , hence bi ei , all belong to H, so does xj . We have
xj ∈ T because jbi − jb
ri  ∈ [0, 1).
2. We have x = x1 + i=1 bi ei with x1 ∈ H ∩ T and the ei ∈ H ∩ T . Since H ∩ T is a
finite set, there are only finitely many choices for x1 . Since there are only finitely many
ei , H is finitely generated.
3. There are only finitely many distinct xj and infinitely many integers, so xj = xk for
some j = k. By linear dependence of the ei , we have (j − k)bi = jbi  − kbi  for all i,
and the result follows.
4. By the previous problems, H is generated by a finite number of elements that are linear
combinations of the ei with rational coefficients. If d is a common denominator of these
r
coefficients, then d = 0 and dH ⊆ i=1 Zei . Thus dH is a subgroup of a free abelian
group of rank r, hence is free of rank at most r.
5. Since dH ∼
r
= H, H is free, and since H ⊇ i=1 Zei , the rank of H is at least r, and
hence exactly r.

Section 6.3
√
1. m = 2 ⇒ 2 × 12 = 12 + 1, so the fundamental unit u is 1 + 2 and we stop at step
t = 1. √
m = 3 ⇒ 3 × 12 = 22 − 1, so u = 2 + 3 and t = 1. √
m = 5 ≡ 1 mod 4 ⇒ 5 × 12 = 12 + 4, so u = 12 (1 + 5) and t = 1.
√
m = 6 ⇒ 6 × 22 = 52 − 1, so u = 5 + 2√6 and t = 2.
m = 7 ⇒ 7 × 32 = 82 − 1, so u = 8 + 3 √7 and t = 3.
m = 10 ⇒ 10 × 12 = 32 + 1, so u = 3 + 10√and t = 1.
m = 11 ⇒ 11 × 32 = 102 − 1, so u = 10 + 3 11 and t = 3.
8
√
m = 13 ≡ 1 mod 4 ⇒ 13 × 12 = 32 + 4, so u = 12 (3 + 13) and t = 1.
√
m = 14 ⇒ 14 × 42 = 152 − 1, so u = 15 √ + 4 14 and t = 4.
m = 15 ⇒ 15 × 12 = 42 − 1, so u = 4 + 15√and t = 1. √
m = 17 ⇒ 17 × 22 = 82 + 4, so u = 12 (8 + 2 17) = 4 + 17 and t = 2.
2. Note that a/2 and√b/2 are both integers, so u ∈ B0 .
3. With u = 12 (a + b m), we compute
√
8u3 = a(a2 + 3b2 m) + b(3a2 + b2 m) m.

Now a2 − b2 m = ±4, and if we add 4b2 m to both sides, we get

a2 + 3b2 m = 4b2 m ± 4 = 4(b2 m ± 1). Since m ≡ 1 mod 4, m must be odd, and since b is
also odd, b2 m ± 1 is even, so 4(b2 m ± 1) is divisible by 8. Similarly,
3a2 + b2 m = 4a2 − (a2 − b2 m) = 4a2 ± 4, which is also divisible by 8 because a is odd. It
follows that u3 ∈ B0 .
4. If u2 ∈ B0 , then u2 is a positive unit in B0√ , hence so is (u2 )−1 = u−2 . Therefore
3 −2
u = u u ∈ B0 . But a and b are odd, so √u∈ / Z[ m], a contradiction.
√ √
5. When m = 5, we have u = 12 (1 + 5), so 8u3 = 1 + 3 5 + (3 × 5) + 5 5. Thus
√ √ √
u3 = 2 + √ 5. Also, 4u2 = 6 +√2 5, so u2 = (3 + √ 5)/2. When m = 13, we √ have
u = 12 (3 + 13), so 8u3 = 27 + 27 13 + (3 × 3 × 13) + 13 13. Therefore u3 = 18 + 5 13.
√ √
Also, 4u2 = 22 + 6 13 = (11 + 3 13)/2.
Note that the results for u3 in Problem 5 are exactly what we would get by solving
a − mb2 = ±1. For m = 5 we have 5 × 12 = 22 + 1, so a = 2, b = 1. For m = 13 we have
2

13 × 52 = 182 + 1, so a = 18, b = 5.

Section 7.1
1. The missing terms in the product defining the discriminant are either squares of
real numbers or occur as a complex number and its conjugate. Thus the missing terms
contribute a positive real number, which cannot change the overall sign.
2. Observe that (c − c)2 is a negative real number, so each pair of complex embeddings
contributes a negative sign.
3. We have 2r2 = [Q(ζ) : Q] = ϕ(pr ) = pr−1 (p − 1), so the sign is (−1)s , where, assuming
pr > 2, s = pr−1 (p − 1)/2. To show that there are no real embeddings, note that if ζ
is mapped to -1, then −ζ is mapped to 1. But 1 is also mapped to 1, and (assuming a
nontrivial extension), we reach a contradiction.
Examination of the formula for s allows further simplification. If p is odd, the sign
will be positive if and only if p ≡ 1 mod 4. If p = 2, the sign will be positive iff r > 2.

Section 8.1
1. If τ ∈ I(Q) and x ∈ B, then

στ σ −1 (x) − x = σ(τ σ −1 (x) − σ −1 (x)) ∈ σ(Q)

so σI(Q)σ −1 ⊆ I(σ(Q)). Conversely, let τ ∈ I(σ(Q)), x ∈ B. Then τ = σ(σ −1 τ σ)σ −1 ,

so we must show that σ −1 τ σ ∈ I(Q), in other words, σ −1 τ σ(x) − x ∈ Q. Now we have
 9

τ σ(x)−σ(x) ∈ σ(Q), so τ σ(x)−σ(x) = σ(y) for some y ∈ Q. Thus σ −1 τ σ(x)−x = y ∈ Q,

the desired result.
2. Since G is abelian, σD(Q)σ −1 = σσ −1 D(Q) = D(Q), so by Problem 1 and (8.1.2),
all the decomposition groups are the same. The decomposition groups depend only on P
because P determines the unique factorization of P B into prime ideals of B. The analysis
is the same for the inertia groups.

Section 8.3
1. This follows from (7.1.6), along with (4.2.6) and (4.2.8).
2. The norm of 1 − ζ is the product of the conjugates by (2.1.6), and the result follows
from (7.1.6).
3. The ideals (1 − ζ)r are all equal by (7.1.2).

Section 9.1
1. This follows from (6.1.5) and the observation that a root of unity must have absolute
value 1.
2. If the characteristic is p = 0, then there are only p integers, and the result follows from
(9.1.7).
3. Assume the absolute values equivalent. By nontriviality, there is an element y with
|y|1 > 1. Take a = log |y|2 / log |y|1 . For every x there is a real number b such that
|x|1 = |y|b1 . Find a sequence of rational numbers s/t converging to b from above. Then
s/t s/t
|x|1 = |y|b1 < |y|1 , so |xt /y s |1 < 1. By hypothesis, |xt /y s |2 < 1, so |x|2 < |y|2 . Let
s/t → b to get |x|2 ≤ |y|b2 . But by taking a sequence of rationals converging to b from
below, we get |x|2 ≥ |y|b2 , hence |x|2 = |y|b2 . To summarize,

|x|1 = |y|b1 ⇒ |x|2 = |y|b2 .

Taking logarithms (if x = 0), we have log |x|2 / log |x|1 = a, hence |x|a1 = |x|2 .

Section 9.2
 
1. Let a = ± pri i , hence a∞ = pri i . If p is one of the pi , then ap = p−r
i
i
, and
if p is not one of the pi , then ap = 1. Thus only finitely many terms of the product
are unequal to 1, and the infinite prime cancels the effect of the finite primes. The result
follows.

Section 9.3
1. For each i = 1, . . . , n, choose yi , zi ∈ k such that |yi |i > 1 and |zi |i < 1. This is
possible by (9.3.2). Take xi = yi if i ≤ r, and xi = zi if i > r. By (9.3.3), there is an
element a ∈ k such that |a − xi |i < < for all i. (We will specify < in a moment.) If i ≤ r,
then

|yi |i ≤ |yi − a|i + |a|i < < + |a|i

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so |a|i > |yi |i − <, and we need 0 < < ≤ |yi |i − 1. On the other hand, if i > r, then
|a|i ≤ |a − zi |i + |zi |i < < + |zi |i
so we need 0 < < ≤ 1 − |zi |i . Since there are only finitely many conditions to be satisfied,
a single < can be chosen, and the result follows.

Section 9.4
1. The condition stated is equivalent to v(a/b) ≥ 0.
2. The product is 4+2p+4p2 +p3 +p4 . But 4 = 1+3 = 1+p and 4p2 = p2 +3p2 = p2 +p3 .
Thus we have 1 + 3p + p2 + 2p3 + p4 = 1 + 2p2 + 2p3 + p4 .
3. We have −1 = (p − 1) − p = (p − 1) + [(p − 1) − p]p = (p − 1) + (p − 1)p − p2 . Continuing
inductively, we get
−1 = (p − 1) + (p − 1)p + (p − 1)p2 + · · · .
The result can also be obtained by multiplying by -1 on each side of the equation
1 = (1 − p)(1 + p + p2 + · · · ).
4. Since n! = 1·2 · · · p · · · 2p · · · 3p · · · , it follows that if rp ≤ n < (r +1)p, then |n!| = 1/pr .
Thus |n!| → 0 as n → ∞.
5. No. Although |pr | = 1/pr → 0 as r → ∞, all integers n such that rp < n < (r + 1)p
have absolute value 1. Thus the sequence of absolute values |n| cannot converge, hence
the sequence itself cannot converge.
6. We have |an | = |1/n| = pv(n) , where v(n) is the highest power of p dividing n. Thus
pv(n) ≤ n, so v(n) ≤ log n/ log p and consequently v(n)/n → 0. We can apply the root
test to get lim sup |an |1/n = lim pv(n)/n = 1. The radius of convergence is the reciprocal
of the lim sup, namely 1. Thus the series converges for |x| < 1 and diverges for |x| > 1.
The series also diverges at |x| = 1 because |1/n| does not converge to 0.
∞
7. Since n/pi  ≤ n/pi and i=1 1/pi = (1/p)/(1 − 1/p) = 1/(1 − p), the result follows.
8. By Problem 7,
pm pm pm pm − 1
v[(pm )!] = + 2 + · · · + m = 1 + p + · · · + pm−1 = .
p p p p−1
9. We have 1/|n!| = pv(n!) ≤ pn/(p−1) by Problem 7. Thus |an |1/n ≤ p1/(p−1) . Thus
the radius of convergence is at least p−1/(p−1) . Now let |x| = p−1/(p−1) = (1/p)v(x) , so
v(x) = 1/(p − 1). Taking n = pm , we have, using Problem 8,
m pm pm − 1 1
v(xn /n!) = v[xp /(pm )!] = pm v(x) − v[(pm )!] = − = .
p−1 p−1 p−1
Since 1/(p − 1) is a constant independent of m, xn /n! does not converge to 0, so the series
diverges.
Note that 0 < 1/(p − 1) < 1, and since v is a discrete valuation, there is no x ∈ Qp
such that v(x) = 1/(p − 1). Thus |x| < p−1/(p−1) is equivalent to |x| < 1. But the sharper
bound is useful in situations where Qp is embedded in a larger field that extends the
p-adic absolute value.
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Section 9.5
1. Take F (X) = X p−1 − 1, which has p − 1 distinct roots mod p. (The multiplicative
group of nonzero elements of Z/pZ is cyclic.) All roots are simple (because deg F = p−1).
By (9.5.3), the roots lift to distinct roots of unity in Zp .
2. Take F (X) = X 2 − m. Since p does not divide m and p = 2, F and its derivative are
relatively prime, so there are no multiple roots. By (9.5.3), m is a square in Zp iff m is a
quadratic residue mod p.
3. Successively find a0 , a1 , . . . , such that (a0 + a1 p + a2 p2 + · · · )2 = m in Zp . If we take
p = 5, m = 6, then the first four coefficients are a0 = 1, a1 = 3, a2 = 0, a3 = 4. There is
a second solution, the negative of this one. When computing, don’t forget the carry. For
example, (1 + 3 × 51 + a2 × 52 + · · · )2 = 1 + 1 × 51 yields a term 6 × 51 = 1 × 51 + 1 × 52 ,
so the equation for a2 is 2a2 + 10 (not 9) ≡ 0 mod 5, so a2 = 0.