<< стр. 4(всего 4)СОДЕРЖАНИЕ
9. By Problem 8,
ПЂ 3ПЂ ПЂ 3ПЂ 3ПЂ
log |d| в‰Ґ log + (n в€’ 1) log = log в€’ 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 в€€ T because jbi в€’ jbi в€€ [0, 1).
r
2. We have x = x1 + i=1 bi ei with x1 в€€ H в€© T and the ei в€€ H в€© T . Since H в€© T is a
п¬Ѓnite set, there are only п¬Ѓnitely many choices for x1 . Since there are only п¬Ѓnitely many
ei , H is п¬Ѓnitely generated.
3. There are only п¬Ѓnitely many distinct xj and inп¬Ѓnitely many integers, so xj = xk for
some j = k. By linear dependence of the ei , we have (j в€’ k)bi = jbi в€’ kbi for all i,
and the result follows.
4. By the previous problems, H is generated by a п¬Ѓnite number of elements that are linear
combinations of the ei with rational coeп¬ѓcients. If d is a common denominator of these
r
coeп¬ѓcients, then d = 0 and dH вЉ† 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 в€ј H, H is free, and since H вЉ‡ i=1 Zei , the rank of H is at least r, and
r
=
hence exactly r.

Section 6.3
в€љ
1. m = 2 в‡’ 2 Г— 12 = 12 + 1, so the fundamental unit u is 1 + 2 and we stop at step
t = 1. в€љ
m = 3 в‡’ 3 Г— 12 = 22 в€’ 1, so u = 2 + 3 and t = 1. в€љ
m = 5 в‰Ў 1 mod 4 в‡’ 5 Г— 12 = 12 + 4, so u = 1 (1 + 5) and t = 1.
в€љ 2
m = 6 в‡’ 6 Г— 22 = 52 в€’ 1, so u = 5 + 2в€љ6 and t = 2.
m = 7 в‡’ 7 Г— 32 = 82 в€’ 1, so u = 8 + 3 в€љ and t = 3.
7
m = 10 в‡’ 10 Г— 1 = 3 + 1, so u = 3 + 10 and t = 1.
2 2
в€љ
m = 11 в‡’ 11 Г— 32 = 102 в€’ 1, so u = 10 + 3 11 and t = 3.
8
в€љ
m = 13 в‰Ў 1 mod 4 в‡’ 13 Г— 12 = 32 + 4, so u = 1 (3 + 13) and t = 1.
в€љ2
m = 14 в‡’ 14 Г— 4 = 15 в€’ 1, so u = 15 в€љ 4 14 and t = 4.
2 2
+
m = 15 в‡’ 15 Г— 1 = 4 в€’ 1, so u = 4 + 15 and t = 1. в€љ
2 2
в€љ
m = 17 в‡’ 17 Г— 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 в€€ B0 .
в€љ
3. With u = 1 (a + b m), we compute
2
в€љ
8u3 = a(a2 + 3b2 m) + b(3a2 + b2 m) m.

Now a2 в€’ b2 m = В±4, and if we add 4b2 m to both sides, we get
a2 + 3b2 m = 4b2 m В± 4 = 4(b2 m В± 1). Since m в‰Ў 1 mod 4, m must be odd, and since b is
also odd, b2 m В± 1 is even, so 4(b2 m В± 1) is divisible by 8. Similarly,
3a2 + b2 m = 4a2 в€’ (a2 в€’ b2 m) = 4a2 В± 4, which is also divisible by 8 because a is odd. It
follows that u3 в€€ B0 .
4. If u2 в€€ B0 , then u2 is a positive unit in B0в€љ hence so is (u2 )в€’1 = uв€’2 . Therefore
,
3 в€’2
u = u u в€€ B0 . But a and b are odd, so u в€€ Z[ m], a contradiction.
в€љ/ в€љ в€љ
5. When m = 5, we have u = 2 (1 + 5), so 8u3 = 1 + 3 5 + (3 Г— 5) + 5 5. Thus
1
в€љ в€љ в€љ
u3 = 2 + в€љ 5. Also, 4u2 = 6 +в€љ 5, so u2 = (3 + в€љ
2 5)/2. When m = 13, we в€љ have
u = 2 (3 + 13), so 8u = 27 + 27 13 + (3 Г— 3 Г— 13) + 13 13. Therefore u = 18 + 5 13.
1 3 3
в€љ в€љ
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 в€’ mb2 = В±1. For m = 5 we have 5 Г— 12 = 22 + 1, so a = 2, b = 1. For m = 13 we have
13 Г— 52 = 182 + 1, so a = 18, b = 5.

Section 7.1
1. The missing terms in the product deп¬Ѓning 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 в€’ c)2 is a negative real number, so each pair of complex embeddings
contributes a negative sign.
3. We have 2r2 = [Q(О¶) : Q] = П•(pr ) = prв€’1 (p в€’ 1), so the sign is (в€’1)s , where, assuming
pr > 2, s = prв€’1 (p в€’ 1)/2. To show that there are no real embeddings, note that if О¶
is mapped to -1, then в€’О¶ 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 simpliп¬Ѓcation. If p is odd, the sign
will be positive if and only if p в‰Ў 1 mod 4. If p = 2, the sign will be positive iп¬Ђ r > 2.

Section 8.1
1. If П„ в€€ I(Q) and x в€€ B, then

ПѓП„ Пѓ в€’1 (x) в€’ x = Пѓ(П„ Пѓ в€’1 (x) в€’ Пѓ в€’1 (x)) в€€ Пѓ(Q)

so ПѓI(Q)Пѓ в€’1 вЉ† I(Пѓ(Q)). Conversely, let П„ в€€ I(Пѓ(Q)), x в€€ B. Then П„ = Пѓ(Пѓ в€’1 П„ Пѓ)Пѓ в€’1 ,
so we must show that Пѓ в€’1 П„ Пѓ в€€ I(Q), in other words, Пѓ в€’1 П„ Пѓ(x) в€’ x в€€ Q. Now we have
9

П„ Пѓ(x)в€’Пѓ(x) в€€ Пѓ(Q), so П„ Пѓ(x)в€’Пѓ(x) = Пѓ(y) for some y в€€ Q. Thus Пѓ в€’1 П„ Пѓ(x)в€’x = y в€€ Q,
the desired result.
2. Since G is abelian, ПѓD(Q)Пѓ в€’1 = ПѓПѓ в€’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 в€’ О¶ is the product of the conjugates by (2.1.6), and the result follows
from (7.1.6).
3. The ideals (1 в€’ О¶)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 в†’ b to get |x|2 в‰¤ |y|b . But by taking a sequence of rationals converging to b from
2
below, we get |x|2 в‰Ґ |y|b , hence |x|2 = |y|b . To summarize,
2 2

|x|1 = |y|b в‡’ |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 = В± pri , hence a в€ћ = pri . If p is one of the pi , then a p = pв€’ri , and
i i i
if p is not one of the pi , then a p = 1. Thus only п¬Ѓnitely many terms of the product
are unequal to 1, and the inп¬Ѓnite prime cancels the eп¬Ђect of the п¬Ѓnite primes. The result
follows.

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

|yi |i в‰¤ |yi в€’ a|i + |a|i < + |a|i
10

so |a|i > |yi |i в€’ , and we need 0 < в‰¤ |yi |i в€’ 1. On the other hand, if i > r, then
|a|i в‰¤ |a в€’ zi |i + |zi |i < + |zi |i
so we need 0 < в‰¤ 1 в€’ |zi |i . Since there are only п¬Ѓnitely many conditions to be satisп¬Ѓed,
a single can be chosen, and the result follows.

Section 9.4
1. The condition stated is equivalent to v(a/b) в‰Ґ 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 в€’1 = (p в€’ 1) в€’ p = (p в€’ 1) + [(p в€’ 1) в€’ p]p = (p в€’ 1) + (p в€’ 1)p в€’ p2 . Continuing
inductively, we get
в€’1 = (p в€’ 1) + (p в€’ 1)p + (p в€’ 1)p2 + В· В· В· .
The result can also be obtained by multiplying by -1 on each side of the equation
1 = (1 в€’ p)(1 + p + p2 + В· В· В· ).
4. Since n! = 1В·2 В· В· В· p В· В· В· 2p В· В· В· 3p В· В· В· , it follows that if rp в‰¤ n < (r +1)p, then |n!| = 1/pr .
Thus |n!| в†’ 0 as n в†’ в€ћ.
5. No. Although |pr | = 1/pr в†’ 0 as r в†’ в€ћ, 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) в‰¤ n, so v(n) в‰¤ log n/ log p and consequently v(n)/n в†’ 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.
в€ћ
7. Since n/pi в‰¤ n/pi and i=1 1/pi = (1/p)/(1 в€’ 1/p) = 1/(1 в€’ p), the result follows.
8. By Problem 7,
pm в€’ 1
pm pm pm
+ 2 + В·В·В· + m = 1 + p + В·В·В· + p
m mв€’1
v[(p )!] = = .
pв€’1
p p p
9. We have 1/|n!| = pv(n!) в‰¤ pn/(pв€’1) by Problem 7. Thus |an |1/n в‰¤ p1/(pв€’1) . Thus
the radius of convergence is at least pв€’1/(pв€’1) . Now let |x| = pв€’1/(pв€’1) = (1/p)v(x) , so
v(x) = 1/(p в€’ 1). Taking n = pm , we have, using Problem 8,
pm в€’ 1
pm 1
pm
/(p )!] = p v(x) в€’ v[(p )!] = в€’
n m m m
v(x /n!) = v[x = .
pв€’1 pв€’1 pв€’1
Since 1/(p в€’ 1) is a constant independent of m, xn /n! does not converge to 0, so the series
diverges.
Note that 0 < 1/(p в€’ 1) < 1, and since v is a discrete valuation, there is no x в€€ Qp
such that v(x) = 1/(p в€’ 1). Thus |x| < pв€’1/(pв€’1) is equivalent to |x| < 1. But the sharper
bound is useful in situations where Qp is embedded in a larger п¬Ѓeld that extends the
11

Section 9.5
1. Take F (X) = X pв€’1 в€’ 1, which has p в€’ 1 distinct roots mod p. (The multiplicative
group of nonzero elements of Z/pZ is cyclic.) All roots are simple (because deg F = pв€’1).
By (9.5.3), the roots lift to distinct roots of unity in Zp .
2. Take F (X) = X 2 в€’ 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 iп¬Ђ m is a
quadratic residue mod p.
3. Successively п¬Ѓnd a0 , a1 , . . . , such that (a0 + a1 p + a2 p2 + В· В· В· )2 = m in Zp . If we take
p = 5, m = 6, then the п¬Ѓrst four coeп¬ѓcients are a0 = 1, a1 = 3, a2 = 0, a3 = 4. There is
a second solution, the negative of this one. When computing, donвЂ™t forget the carry. For
example, (1 + 3 Г— 51 + a2 Г— 52 + В· В· В· )2 = 1 + 1 Г— 51 yields a term 6 Г— 51 = 1 Г— 51 + 1 Г— 52 ,
so the equation for a2 is 2a2 + 10 (not 9) в‰Ў 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 п¬Ѓeld 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 п¬Ѓeld, 8-6
group, 8-2
Dedekind domain, 3-1
DedekindвЂ™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
п¬Ѓeld 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, 8-1п¬Ђ.
global п¬Ѓeld, 9-1
greatest common divisor of ideals, 3-6
HenselвЂ™s lemma, 9-10
ideal class group, 3-8
п¬Ѓniteness of, 5-6
inert prime, 4-8
inertia п¬Ѓeld, 8-6
group, 8-2
inertial degree, see relative degree
inп¬Ѓnite prime, 9-5
integral basis, 2-10, 2-11
of a cyclotomic п¬Ѓeld, 7-4п¬Ђ.
integral closure, 1-3
integral element, extension, 1-2п¬Ђ.
integral ideal, 3-3
integrally closed, 1-3
isosceles triangle, 9-3
KummerвЂ™s theorem, 4-7
lattice, 5-1
least common multiple of ideals, 3-6
lifting of prime ideals, 4-1
local п¬Ѓeld, 9-1, 9-8
local ring, 1-7
localization, 1-5п¬Ђ.
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
MinkowskiвЂ™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 п¬Ѓeld, 2-5
number ring, 4-4
p-adic logarithm and exponential, 9-9
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
ram-rel identity, 4-2
ramiп¬Ѓcation, 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 п¬Ѓeld, 9-10
ring of fractions, 1-5
splitting of a prime, 4-8
stabilizing a module, 1-2
StickelbergerвЂ™s theorem, 2-12
totally ramiп¬Ѓed, 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

 << стр. 4(всего 4)СОДЕРЖАНИЕ