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 ’ jbi ∈ [0, 1).

r

2. We have x = x1 + i=1 bi ei with x1 ∈ H © T and the ei ∈ H © T . Since H © T is a

¬nite set, there are only ¬nitely many choices for x1 . Since there are only ¬nitely many

ei , H is ¬nitely generated.

3. There are only ¬nitely many distinct xj and in¬nitely 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 ¬nite number of elements that are linear

combinations of the ei with rational coe¬cients. If d is a common denominator of these

r

coe¬cients, 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 ∼ H, H is free, and since H ⊇ i=1 Zei , the rank of H is at least r, and

r

=

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 = 1 (1 + 5) and t = 1.

√ 2

m = 6 ’ 6 — 22 = 52 ’ 1, so u = 5 + 2√6 and t = 2.

m = 7 ’ 7 — 32 = 82 ’ 1, so u = 8 + 3 √ and t = 3.

7

m = 10 ’ 10 — 1 = 3 + 1, so u = 3 + 10 and t = 1.

2 2

√

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 = 1 (3 + 13) and t = 1.

√2

m = 14 ’ 14 — 4 = 15 ’ 1, so u = 15 √ 4 14 and t = 4.

2 2

+

m = 15 ’ 15 — 1 = 4 ’ 1, so u = 4 + 15 and t = 1. √

2 2

√

m = 17 ’ 17 — 22 = 82 + 4, so u = 1 (8 + 2 17) = 4 + 17 and t = 2.

2

2. Note that a/2 and b/2 are both integers, so u ∈ B0 .

√

3. With u = 1 (a + b m), we compute

2

√

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 = 2 (1 + 5), so 8u3 = 1 + 3 5 + (3 — 5) + 5 5. Thus

1

√ √ √

u3 = 2 + √ 5. Also, 4u2 = 6 +√ 5, so u2 = (3 + √

2 5)/2. When m = 13, we √ have

u = 2 (3 + 13), so 8u = 27 + 27 13 + (3 — 3 — 13) + 13 13. Therefore u = 18 + 5 13.

1 3 3

√ √

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

a2 ’ mb2 = ±1. For m = 5 we have 5 — 12 = 22 + 1, so a = 2, b = 1. For m = 13 we have

13 — 52 = 182 + 1, so a = 18, b = 5.

Section 7.1

1. The missing terms in the product de¬ning 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] = •(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 simpli¬cation. 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 i¬ r > 2.

Section 8.1

1. If „ ∈ I(Q) and x ∈ B, then

σ„ σ ’1 (x) ’ x = σ(„ σ ’1 (x) ’ σ ’1 (x)) ∈ σ(Q)

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

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

9

„ σ(x)’σ(x) ∈ σ(Q), so „ σ(x)’σ(x) = σ(y) for some y ∈ Q. Thus σ ’1 „ 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|b . Find a sequence of rational numbers s/t converging to b from above. Then

1

s/t s/t

|x|1 = |y|b < |y|1 , so |xt /y s |1 < 1. By hypothesis, |xt /y s |2 < 1, so |x|2 < |y|2 . Let

1

s/t ’ b to get |x|2 ¤ |y|b . But by taking a sequence of rationals converging to b from

2

below, we get |x|2 ≥ |y|b , hence |x|2 = |y|b . To summarize,

2 2

|x|1 = |y|b ’ |x|2 = |y|b .

1 2

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

1

Section 9.2

1. Let a = ± pri , hence a ∞ = pri . If p is one of the pi , then a p = p’ri , and

i i i

if p is not one of the pi , then a p = 1. Thus only ¬nitely many terms of the product

are unequal to 1, and the in¬nite prime cancels the e¬ect of the ¬nite 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

10

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 ¬nitely many conditions to be satis¬ed,

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 ’ 1

pm pm pm

+ 2 + ··· + m = 1 + p + ··· + p

m m’1

v[(p )!] = = .

p’1

p p p

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,

pm ’ 1

pm 1

pm

/(p )!] = p v(x) ’ v[(p )!] = ’

n m m m

v(x /n!) = v[x = .

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 ¬eld that extends the

p-adic absolute value.

11

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 i¬ m is a

quadratic residue mod p.

3. Successively ¬nd a0 , a1 , . . . , such that (a0 + a1 p + a2 p2 + · · · )2 = m in Zp . If we take

p = 5, m = 6, then the ¬rst four coe¬cients 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.

Index

m-n means chapter m, page n

absolute value, 9-1

on the rationals, 9-4

AKLB setup, 2-5

algebraic integer, 1-2

approximation theorem, 9-5, 9-6

archimedean absolute value, 9-1

Artin symbol, 8-5

Artin-Whaples, see approximation theorem

Cauchy sequence, 9-7

characteristic polynomial, 2-1

class number, 5-7

coherent sequence, 9-8

completion of a ¬eld with an absolute value, 9-7

conjugates of an element, 2-3

of a prime ideal, 8-2

contraction of an ideal, 4-1

cyclotomic extension, 2-5, 2-7, 6-5, 7-1, 8-6

polynomial, 7-1

decomposition ¬eld, 8-6

group, 8-2

Dedekind domain, 3-1

Dedekind™s lemma, 2-4

denominator of a fractional ideal, 3-3

Dirichlet unit theorem, 6-1, 6-3, 6-4

discrete valuation, 9-1

discrete valuation ring, 4-3, 9-2, 9-3

discriminant, 2-8, 7-3, 7-4

divides means contains, 3-6

DVR, see discrete valuation ring

embedding, canonical, 5-4

complex, 5-4

logarithmic, 6-1

real, 5-4

equation of integral dependence, 1-2

1

2

equivalent absolute values, 9-4

extension of an ideal, 4-1

factoring of prime ideals in extensions, 4-1

¬eld discriminant, 2-10

fractional ideal, 3-2, 3-3

Frobenius automorphism, 8-4

fundamental domain, 5-1

fundamental system of units, 6-5

fundamental unit, 6-6

Galois extensions, 8-1¬.

global ¬eld, 9-1

greatest common divisor of ideals, 3-6

Hensel™s lemma, 9-10

ideal class group, 3-8

¬niteness of, 5-6

inert prime, 4-8

inertia ¬eld, 8-6

group, 8-2

inertial degree, see relative degree

in¬nite prime, 9-5

integral basis, 2-10, 2-11

of a cyclotomic ¬eld, 7-4¬.

integral closure, 1-3

integral element, extension, 1-2¬.

integral ideal, 3-3

integrally closed, 1-3

isosceles triangle, 9-3

Kummer™s theorem, 4-7

lattice, 5-1

least common multiple of ideals, 3-6

lifting of prime ideals, 4-1

local ¬eld, 9-1, 9-8

local ring, 1-7

localization, 1-5¬.

functor, 1-8

of modules, 1-7

localized ring, 1-5

lying over, 4-1

minimal polynomial, 2-2

Minkowski bound on element norms, 5-5

on ideal norms, 5-6

Minkowski™s convex body theorem, 5-2

multiplicative property of norms, 2-2, 4-4, 4-5

multiplicative set, 1-5

nonarchimedean absolute value, 9-1

nondegenerate bilinear form, 2-4

3

norm, 1-1, 2-1

norm of an ideal, 4-4

null sequence, 9-7

number ¬eld, 2-5

number ring, 4-4

p-adic logarithm and exponential, 9-9

p-adic integers, 9-9

p-adic numbers, 9-9

P -adic (and p-adic) valuation, 9-2

power series, 9-8

prime avoidance lemma, 3-2

prime element, 9-8

principal fractional ideal, 3-8

product formula, 9-5

quadratic extension, 2-4, 2-6, 2-7, 4-8, 6-6, 6-7

quadratic reciprocity, 8-8

ram-rel identity, 4-2

rami¬cation, 4-2

and the discriminant, 4-6

index, 4-2

of a prime, 4-8

rational integers, 2-6, 2-11

relative degree, 4-2

residue class degree, see relative degree

residue ¬eld, 9-10

ring of fractions, 1-5

splitting of a prime, 4-8

stabilizing a module, 1-2

Stickelberger™s theorem, 2-12

totally rami¬ed, 8-8

trace, 2-1

form, 2-4

transitivity of integral extensions, 1-3

of trace and norm, 2-4

trivial absolute value, 9-2

uniformizer, 9-8

unimodular matrix, 2-11, 5-1

unique factorization of ideals, 3-5

unit theorem, see Dirichlet unit theorem

valuation ideal, 9-2

valuation ring, 9-2

Vandermonde determinant, 2-9