Again assume [KL : Q] = mn. Let a1 , . . . , am and b1 , . . . , bn be integral bases for R and

S respectively. If ± ∈ T , then

m n

cij

ai bj , cij ∈ Z, r ∈ Z

±=

r

i=1 j=1

with r having no factor (except ±1) in common with all the cij .

Proof. The assumption that [KL : Q] = mn implies that the ai bj form a basis for KL/Q.

[See the process of constructing KL discussed in (7.2.1).] In fact the ai bj form an integral

vi wi , vi ∈ R, wi ∈ S.

basis for RS. (This is because RS consists of all ¬nite sums

Each vi is a linear combination of the ak with integer coe¬cients, and so on.) It follows

that ± is a linear combination of the ai bj with rational coe¬cients. Form a common

denominator and eliminate common factors to obtain the desired result. ™

7.2.4 Proposition

We are still assuming that [KL : Q] = mn. If d is the greatest common divisor of the

discriminant of R and the discriminant of S, then T ⊆ d RS. Thus if d = 1, then T = RS.

1

Proof. It su¬ces to show that in (7.2.3), r divides d. To see this, write

cij cij (d/r)

= .

r d

In turn, it su¬ces to show that r divides the discriminant of R. Then by symmetry, r

will also divide the discriminant of S, and therefore divide d.

Let σ be an embedding of K in C. By (7.2.2), σ extends to an embedding (also called

σ) of KL in C such that σ is the identity on L. By (7.2.3), if ± ∈ T we have

cij

σ(±) = σ(ai )bj .

r

i,j

6 CHAPTER 7. CYCLOTOMIC EXTENSIONS

If we set

n

cij

xi = bj ,

r

j=1

we have the system of linear equations

m

σ(ai )xi = σ(±)

i=1

where there is one equation for each of the m embeddings σ from K to C. Solving for xi

by Cramer™s rule, we get xi = γi /δ, where δ is the determinant formed from the σ(ai ) and

γi is the determinant obtained by replacing the ith column of δ with the σ(±). Note that

by (2.3.3), δ 2 is the discriminant of R, call it e. Since all the σ(ai ) and σ(±) are algebraic

integers, so are δ and all the γi . Now

γi γi δ γi δ

xi = = 2=

δ δ e

so exi = γi δ is an algebraic integer. By de¬nition of xi ,

n

ecij

exi = bj ,

r

j=1

an algebraic integer in RS. But e is a Z-linear combination of the ai , and the ai bj are

an integral basis for RS, so ecij /r is an integer. Thus r divides every ecij . By (7.2.3), r

has no factor (except the trivial ±1) in common with every cij . Consequently, r divides

e, the discriminant of R. ™

We need one more preliminary result.

7.2.5 Lemma

Let ζ be a primitive nth root of unity, and denote the discriminant of {1, ζ, . . . , ζ •(n)’1 }

by disc(ζ). Then disc(ζ) divides n•(n) .

Proof. Let f (= ¦n , the nth cyclotomic polynomial) be the minimal polynomial of ζ

over Q. Since ζ is a root of X n ’ 1, we have X n ’ 1 = f (X)g(X) for some g ∈ Q[X].

But f ∈ Z[X] (because ζ is an algebraic integer), and f , hence g, is monic, so g ∈ Z[X].

Di¬erentiate both sides of the equation to get nX n’1 = f (X)g (X) + f (X)g(X). Setting

X = ζ, which is a root of f , we have nζ n’1 = f (ζ)g(ζ). But ζ n’1 = ζ n /ζ = 1/ζ, so

n = ζf (ζ)g(ζ).

Now [Q(ζ) : Q] = •(n), so taking the norm of each side yields

n•(n) = N (f (ζ))N (ζg(ζ)).

But by (2.3.6), N (f (ζ)) = ±disc (ζ), and N (ζg(ζ)) ∈ Z by (2.2.2). The desired result

follows. ™

7.2. AN INTEGRAL BASIS OF A CYCLOTOMIC FIELD 7

7.2.6 Theorem

If ζ is a primitive nth root of unity, then the ring of algebraic integers of Q(ζ) is Z[ζ]. in

other words, the powers of ζ form an integral basis.

Proof. We have proved this when ζ is a prime power, so let n = m1 m2 where the mi are

relatively prime and greater than 1. Now

ζ m1 = (ei2π/n )m1 = ei2πm1 /n = ei2π/m2 = ζ2 ,

a primitive (m2 )th root of unity, and similarly ζ m2 = ζ1 , a primitive (m1 )th root of unity.

Thus Q(ζ1 ) and Q(ζ2 ) are contained in Q(ζ). On the other hand, since m1 and m2 are

relatively prime, there are integers r, s such that rm2 + sm1 = 1. Thus

ζ = ζ rm2 +sm1 = ζ1 ζ2 .

rs

It follows that Q(ζ) = Q(ζ1 )Q(ζ2 ), and we can apply (7.2.4). In that proposition, we

take K = Q(ζ1 ), L = Q(ζ2 ), KL = Q(ζ), R = Z[ζ1 ], S = Z[ζ2 ] (induction hypothesis),

T = RS. The hypothesis on the degree [KL : Q] is satis¬ed because •(n) = •(m1 )•(m2 ).

By (7.2.5), disc(ζ1 ) divides a power of m1 and disc(ζ2 ) divides a power of m2 . Thus the

greatest common divisor of disc(R) and disc(S) is 1, and again the hypothesis of (7.2.4)

is satis¬ed. The conclusion is that the ring T of algebraic integers of KL coincides with

RS. But the above argument that Q(ζ) = Q(ζ1 )Q(ζ2 ) may be repeated verbatim with Q

replaced by Z. We conclude that Z[ζ] = Z[ζ1 ]Z[ζ2 ] = RS = T . ™

7.2.7 The Discriminant of a General Cyclotomic Extension

The ¬eld discriminant of Q(ζ), where ζ is a primitive nth root of unity is given by

(’1)•(n)/2 n•(n)

.

•(n)/(p’1)

p|n p

A direct veri¬cation, with the aid of (7.1.7) and Problem 3 of Section 7.1, shows that

the formula is correct when n = pr . The general case is handled by induction, but the

computation is very messy.

In the next chapter, we will study factorization of primes in Galois extensions. The

results will apply, in particular, to cyclotomic extensions.

Chapter 8

Factoring of Prime Ideals in

Galois Extensions

8.1 Decomposition and Inertia Groups

We return to the general AKLB setup: A is a Dedekind domain with fraction ¬eld K, L is

a ¬nite separable extension of K, and B is the integral closure of A in L. But now we add

the condition that the extension L/K is normal, hence Galois. We will see shortly that

the Galois assumption imposes a severe constraint on the numbers ei and fi in the ram-rel

identity (4.1.6). Throughout this chapter, G will denote the Galois group Gal(L/K).

8.1.1 Proposition

If σ ∈ G, then σ(B) = B. If Q is a prime ideal of B, then so is σ(Q). Moreover, if Q

lies above the nonzero prime ideal P of A, then so does σ(Q). Thus G acts on the set of

prime ideals lying above P .

Proof. If x ∈ B, then σ(x) ∈ B (apply σ to an equation of integral dependence). Thus

σ(B) ⊆ B. But σ ’1 (B) is also contained in B, hence B = σσ ’1 (B) ⊆ σ(B). If P B =

Qei , then apply σ to get P B = σ(Qi )ei . The σ(Qi ) must be prime ideals because σ

i

preserves all algebraic relations. Note also that σ is a K-automorphism, hence ¬xes every

element of A (and of P ). Therefore Q © A = P ’ σ(Q) © A = P . ™

We now show that the action of G is transitive.

8.1.2 Theorem

Let Q and Q1 be prime ideals lying above P . Then for some σ ∈ G we have σ(Q) = Q1 .

Proof. If the assertion is false, then for each σ, the ideals Q1 and σ(Q) are maximal and

distinct, so Q1 ⊆ σ(Q). By the prime avoidance lemma (Section 3.1, exercises), there is

an element x ∈ Q1 belonging to none of the σ(Q). Computing the norm of x relative to

L/K, we have N (x) = σ∈G σ(x) by (2.1.6). But one of the σ™s is the identity, Q1 is an

ideal, and [by (8.1.1)] σ(x) ∈ B for all σ. Consequently, N (x) ∈ Q1 . But N (x) ∈ A by

1

2 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

(2.2.2), so N (x) ∈ Q1 © A = P = Q © A. Thus N (x) belongs to the prime ideal Q, and

therefore some σ ’1 (x) belongs to Q as well. This gives x ∈ σ(Q), a contradiction. ™

8.1.3 Corollary

g

In the factorization P B = i=1 Piei of the nonzero prime ideal P , the rami¬cation indices

ei are the same for all i, as are the relative degrees fi . Thus the ram-rel identity simpli¬es

to ef g = n, where n = [L : K] = |G|.

Proof. This follows from (8.1.2), along with the observation that an automorphism σ

preserves all algebraic relations. ™

Since we have a group G acting on the prime factors of P B, it is natural to consider

the stabilizer subgroup of each prime factor Q.

8.1.4 De¬nitions and Comments

We say that the prime ideals σ(Q), σ ∈ G, are the conjugates of Q. Thus (8.1.2) says that

all prime factors of P B are conjugate. The decomposition group of Q is the subgroup D

of G consisting of those σ ∈ G such that σ(Q) = Q. (This does not mean that σ ¬xes

every element of Q.) By the orbit-stabilizer theorem, the size of the orbit of Q is the

index of the stabilizer subgroup D. Since there is only one orbit, of size g,

g = [G : D] = |G|/|D|, hence |D| = n/g = ef g/g = ef,

independent of Q. Note also that distinct conjugates of Q determine distinct cosets of D.

’1

For if σ1 D = σ2 D, then σ2 σ1 ∈ D, so σ1 (Q) = σ2 (Q).

There is a particular subgroup of D that will be of interest. By (8.1.1), σ(B) = B for

every σ ∈ G. If σ ∈ D, then σ(Q) = Q. It follows that σ induces an automorphism σ of

B/Q. (Note that x ≡ y mod Q i¬ σx ≡ σy mod Q.) Since σ is a K-automorphism, σ

is an A/P -automorphism. The mapping σ ’ σ is a group homomorphism from D to the

group of A/P -automorphisms of B/Q.

8.1.5 De¬nition

The kernel I of the above homomorphism, that is, the set of all σ ∈ D such that σ is

trivial, is called the inertia group of Q.

8.1.6 Remarks

The inertia group is a normal subgroup of the decomposition group, as it is the kernel of

a homomorphism. It is given explicitly by

I = {σ ∈ D : σ(x) + Q = x + Q ∀x ∈ B} = {σ ∈ D : σ(x) ’ x ∈ Q ∀x ∈ B}.

8.1. DECOMPOSITION AND INERTIA GROUPS 3

We now introduce an intermediate ¬eld and ring into the basic AKLB setup, as follows.

L B

KD AD

K A

Take KD to be the ¬xed ¬eld of D, and let AD = B © KD be the integral closure of A in

KD . Let PD be the prime ideal Q © AD . Note that Q is the only prime factor of PD B.

This is because all primes in the factorization are conjugate, and σ(Q) = Q for all σ ∈ D,

by de¬nition of D.

8.1.7 Lemma

Let PD B = Qe and f = [B/Q : AD /PD ]. Then e = e and f = f . Moreover,

A/P ∼ AD /PD .

=

Proof. First, observe that by the ram-rel identity [see (8.1.3)], e f = [L : KD ], which is

|D| by the fundamental theorem of Galois theory. But |D| = ef by (8.1.4), so e f = ef .

Now as in (4.1.3)-(4.1.5), A/P ⊆ AD /PD ⊆ B/Q, so f ¤ f . Also, P AD ⊆ PD , so PD

divides P AD , hence PD B divides P AD B = P B. Consequently, e ¤ e, and this forces

e = e and f = f . Thus the dimension of B/Q over AD /PD is the same as the dimension

of B/Q over A/P . Since A/P can be regarded as a sub¬eld of AD /PD , the proof is

complete. ™

8.1.8 Theorem

The homomorphism σ ’ σ of D to Gal[(B/Q)/(A/P )] introduced in (8.1.4) is surjective

with kernel I. Therefore Gal[(B/Q)/(A/P )] ∼ D/I.

=

Proof. Let x be a primitive element of B/Q over A/P . Let x ∈ B be a representative

of x. Let h(X) = X r + ar’1 X r’1 + · + a0 be the minimal polynomial of x over KD ; the

coe¬cients ai belong to AD by (2.2.2). The roots of h are all of the form σ(x), σ ∈ D.

(We are working in the extension L/KD , with Galois group D.) By (8.1.7), if we reduce

the coe¬cients of h mod PD , the resulting polynomial h(X) has coe¬cients in A/P . The

roots of h are of the form σ(x), σ ∈ D (because x is a primitive element). Since σ ∈ D

means that σ(Q) = Q, all conjugates of x over A/P lie in B/Q. By the basic theory of

splitting ¬elds, B/Q is a Galois extension of A/P .

To summarize, since every conjugate of x over A/P is of the form σ(x), every A/P -

automorphism of B/Q (necessarily determined by its action on x), is of the form σ where

σ ∈ D. Since σ is trivial i¬ σ ∈ I, it follows that the map σ ’ σ is surjective and has

kernel I. ™

4 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

8.1.9 Corollary

The order of I is e. Thus the prime ideal P does not ramify if and only if the inertia

group of every prime ideal Q lying over P is trivial.

Proof. By de¬nition of relative degree, the order of Gal[(B/Q)/(A/P )] is f . By (8.1.4),

the order of D is ef . Thus by (8.1.8), the order of I must be e. ™

Problems For Section 8.1

1. Let D(Q) be the decomposition group of the prime ideal Q. It follows from the

de¬nition of stabilizer subgroup that D(σ(Q)) = σD(Q)σ ’1 for every σ ∈ G. Show that

the inertia subgroup also behaves in this manner, that is, I(σ(Q)) = σI(Q)σ ’1 .

2. If L/K is an abelian extension (the Galois group G = Gal(L/K) is abelian), show that

the groups D(σ(Q)), σ ∈ G, are all equal, as are the I(σ(Q)), σ ∈ G. Show also that the

groups depend only on the prime ideal P of A.

8.2 The Frobenius Automorphism

In the basic AKLB setup, with L/K a Galois extension, we now assume that K and L

are number ¬elds.

8.2.1 De¬nitions and Comments

Let P be a prime ideal of A that does not ramify in B, and let Q be a prime lying over P .

By (8.1.9), the inertia group I(Q) is trivial, so by (8.1.8), Gal[(B/Q)/(A/P )] is isomorphic

to the decomposition group D(Q). But B/Q is a ¬nite extension of the ¬nite ¬eld A/P [see

(4.2.3)], so the Galois group is cyclic. Moreover, there is a canonical generator given by

x+Q ’ xq +Q, x ∈ B, where q = |A/P |. Thus we have identi¬ed a distinguished element

σ ∈ D(Q), called the Frobenius automorphism, or simply the Frobenius, of Q, relative to

the extension L/K. The Frobenius automorphism is determined by the requirement that

for every x ∈ B,

σ(x) ≡ xq mod Q.

We use the notation L/K for the Frobenius automorphism. The behavior of the Frobe-

Q

nius under conjugation is similar to the behavior of the decomposition group as a whole

(see the exercises in Section 8.1).

8.2.2 Proposition

„ ’1 .

L/K L/K

If „ ∈ G, then =„

„ (Q) Q

„ ’1 x ≡ („ ’1 x)q = „ ’1 xq mod Q. Apply „ to both sides to

L/K

Proof. If x ∈ B, then Q

conclude that „ L/K „ ’1 satis¬es the de¬ning equation for L/K

. Since the Frobenius

„ (Q)

Q

is determined by its de¬ning equation, the result follows. ™

8.2. THE FROBENIUS AUTOMORPHISM 5

8.2.3 Corollary

L/K

If L/K is abelian, then depends only on P , and we write the Frobenius automor-

Q

L/K

phism as , and sometimes call it the Artin symbol.

P

Proof. By (8.2.2), the Frobenius is the same for all conjugate ideals „ (Q), „ ∈ G, hence

by (8.1.2), for all prime ideals lying over P . ™

8.2.4 Intermediate Fields

We now introduce an intermediate ¬eld between K and L, call it F . We can then lift P

to the ring of algebraic integers in F , namely B © F . A prime ideal lying over P has the

form Q © F , where Q is a prime ideal of P B. We will compare decomposition groups with

respect to the ¬elds L and F , with the aid of the identity

[B/Q : A/P ] = [B/Q : (B © F )/(Q © F )][(B © F )/(Q © F ) : A/P ].

The term on the left is the order of the decomposition group of Q over P , denoted by

D(Q, P ). (We are assuming that P does not ramify, so e = 1.) The ¬rst term on the

right is the order of the decomposition group of Q over Q © F . The second term on the

right is the relative degree of Q © F over P , call if f . Thus

|D(Q, Q © F )| = |D(Q, P )|/f

Since D = D(Q, P ) is cyclic and is generated by the Frobenius automorphism σ, the

unique subgroup of D with order |D|/f is generated by σ f . Note that D(Q, Q © F ) is

a subgroup of D(Q, P ), because Gal(L/F ) is a subgroup of Gal(L/K). It is natural to

expect that the Frobenius automorphism of Q, relative to the extension L/F , is σ f .

8.2.5 Proposition

f

L/F L/K

= .

Q Q

L/K

. Then σ ∈ D, so σ(Q) = Q; also σ(x) ≡ xq mod Q, x ∈ B, where

Proof. Let σ = Q

f

q = |A/P |. Thus σ f (Q) = Q and σ f (x) ≡ xq . Since q f is the cardinality of the ¬eld

(B © F )/(Q © F ), the result follows. ™

8.2.6 Proposition

L/K F/K

If the extension F/K is Galois, then the restriction of σ = to F is .

Q Q©F

Proof. Let σ1 be the restriction of σ to F . Since σ(Q) = Q, it follows that σ1 (Q © F ) =

Q © F . (Note that F/K is normal, so σ1 is an automorphism of F .) Thus σ1 belongs

to D(Q © F, P ). Since σ(x) ≡ xq mod Q, we have σ1 (x) ≡ xq mod (Q © F ), where

F/K

q = |A/P |. Consequently, σ1 = .™

Q©F

6 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

8.2.7 De¬nitions and Comments

We may view the lifting from the base ¬eld K to the extension ¬eld L as occurring in

three distinct steps. Let FD be the decomposition ¬eld of the extension, that is, the ¬xed

¬eld of the decomposition group D, and let FI be the inertia ¬eld, the ¬xed ¬eld of the

inertia group I. We have the following diagram:

L

e=|I|

FI

f =|D|/e

FD

g=n/ef

K

All rami¬cation takes place at the top (call it level 3), and all splitting at the bottom

(level 1). There is inertia in the middle (level 2). Alternatively, the results can be

expressed in tabular form:

e f g

Level 1 1 1 g

2 1 f 1

3 e 1 1

As we move up the diagram, we multiply the rami¬cation indices and relative degrees.

This is often expressed by saying that e and f are multiplicative in towers. The basic

point is that if Q = Qe1 · · · and Q1 = Qe2 · · · , then Q = Qe1 e2 · · · . The multiplicativity

1 2 2

of f follows because f is a vector space dimension.

8.3 Applications

8.3.1 Cyclotomic Fields

Let ζ be a primitive mth root of unity, and let L = Q(ζ) be the corresponding cyclotomic

¬eld. (We are in the AKLB setup with A = Z and K = Q.) Assume that p is a rational

prime that does not divide m. Then by (7.2.5) and the exercises for Section 4.2, p is

unrami¬ed. Thus (p) factors in B as Q1 · · · Qg , where the Qi are distinct prime ideals.

Moreover, the relative degree f is the same for all Qi , because the extension L/Q is

Galois. In order to say more about f , we ¬nd the Frobenius automorphism σ explicitly.

The de¬ning equation is σ(x) ≡ xp mod Qi for all i, and consequently

σ(ζ) = ζ p .

8.3. APPLICATIONS 7

(The idea is that the roots of unity remain distinct when reduced mod Qi , because the

polynomial X n ’ 1 is separable over Fp .)

Now the order of σ is the size of the decomposition group D, which is f . Thus f is

the smallest positive integer such that σ f (ζ) = ζ. Since ζ is a primitive mth root of unity,

we conclude that

f is the smallest positive integer such that pf ≡ 1 mod m.

Once we know f , we can ¬nd the number of prime factors g = n/f , where n = •(m).

(We already know that e = 1 because p is unrami¬ed.)

When p divides m, the analysis is more complicated, and we will only state the result.

Say m = pa m1 , where p does not divide m1 . Then f is the smallest positive integer such

that pf ≡ 1 mod m1 . The factorization is (p) = (Q1 · · · Qg )e , with e = •(pa ). The Qi are

distinct prime ideals, each with relative degree f . The number of distinct prime factors

is g = •(m1 )/f .

We will now give a proof of Gauss™ law of quadratic reciprocity.

8.3.2 Proposition

Let q be an odd prime, and let L = Q(ζq ) be the cyclotomic ¬eld generated by a primitive

q th root of unity. Then L has a unique quadratic sub¬eld F . Explicitly, if q ≡ 1 mod 4,

√

√

then the quadratic sub¬eld is Q( q), and if q ≡ 3 mod 4, it is Q( ’q). More compactly,

√

F = Q( q — ), where q — = (’1)q’1)/2 q.

Proof. The Galois group of the extension is cyclic of even order q ’ 1, hence has a unique

subgroup of index 2. Therefore L has a unique quadratic sub¬eld. By (7.1.7) √ the and

∈ Q. But d ∈ Q,

(q’1)/2 q’2

exercises to Section 7.1, the ¬eld discriminant is d = (’1) q /

because d has an odd number of factors of q. If q ≡ 1 mod 4, then the sign of d is

√ √

positive and Q( d) = Q( q). Similarly, if q ≡ 3 mod 4, then the sign of d is negative

√ √

and Q( d) = √ ’q). [Note that the roots of the cyclotomic polynomial belong to L,

Q(

hence so does d; see (2.3.5).] ™

8.3.3 Remarks

Let σp be the Frobenius automorphism F/Q , where F is the unique quadratic sub¬eld

p

of L, and p is an odd prime unequal to q. By (4.3.2), case (a1), if q — is a quadratic residue

mod p, then p splits, so g = 2 and therefore f = 1. Thus the decomposition group D

is trivial, and since σp generates D, σp is the identity. If q — is not a quadratic residue

mod p, then by (4.3.2), case (a2), p is inert, so g = 1, f = 2, and σp is nontrivial. Since

the Galois group of F/Q has only two elements, it may be identi¬ed with {1, ’1} under

—

multiplication, and we may write (using the standard Legendre symbol) σp = ( qp ). On

L/Q

the other hand, σp is the restriction of σ = to F , by (8.2.6). Thus σp is the identity

p

on F i¬ σ belongs to H, the unique subgroup of Gal(L/Q) of index 2. This will happen i¬

σ is a square. Now the Frobenius may be viewed as a lifting of the map x ’ xp mod q.

p

[As in (8.3.1), σ(ζq ) = ζq .] Thus σ will belong to H i¬ p is a quadratic residue mod q. In

other words, σp = ( p ).

q

8 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

8.3.4 Quadratic Reciprocity

If p and q are distinct odd primes, then

p q

= (’1)(p’1)(q’1)/4 .

q p

Proof. By (8.3.3),

q—

(q’1)/2

’1

(’1)(q’1)/2

p q q

= = = .

q p p p p p

But by elementary number theory, or by the discussion in the introduction to Chapter 1,

’1

= (’1)(p’1)/2 ,

p

and the result follows. ™

8.3.5 Remark

Let L = Q(ζ), where ζ is a primitive pth root of unity, p prime. As usual, B is the ring

of algebraic integers of L. In this case, we can factor (p) in B explicitly. By (7.1.3) and

(7.1.5),

(p) = (1 ’ ζ)p’1 .

Thus the rami¬cation index e = p ’ 1 coincides with the degree of the extension. We say

that p is totally rami¬ed.

Chapter 9

Local Fields

The de¬nition of global ¬eld varies in the literature, but all de¬nitions include our primary

source of examples, number ¬elds. The other ¬elds that are of interest in algebraic number

theory are the local ¬elds, which are complete with respect to a discrete valuation. This

terminology will be explained as we go along.

9.1 Absolute Values and Discrete Valuations

9.1.1 De¬nitions and Comments

An absolute value on a ¬eld k is a mapping x ’ |x| from k to the real numbers, such that

for every x, y ∈ k,

1. |x| ≥ 0, with equality if and only if x = 0;

2. |xy| = |x| |y|;

3. |x + y| ¤ |x| + |y|.

The absolute value is nonarchimedean if the third condition is replaced by a stronger

version:

3 . |x + y| ¤ max(|x|, |y|).

As expected, archimedean means not nonarchimedean.

The familiar absolute values on the reals and the complex numbers are archimedean.

However, our interest will be in nonarchimedean absolute values. Here is where most of

them come from.

A discrete valuation on k is a surjective map v : k ’ Z∪{∞}, such that for every x, y ∈ k,

(a) v(x) = ∞ if and only if x = 0;

(b) v(xy) = v(x) + v(y);

(c) v(x + y) ≥ min(v(x), v(y)).

A discrete valuation induces a nonarchimedean absolute value via |x| = cv(x) , where c

is a constant with 0 < c < 1.

1

2 CHAPTER 9. LOCAL FIELDS

9.1.2 Example

Let A be a Dedekind domain with fraction ¬eld K, and let P be a nonzero prime ideal of A.

Then (see page 3 of Chapter 4) the localized ring AP is a discrete valuation ring (DVR)

with unique maximal ideal (equivalently, unique nonzero prime ideal) P AP . Choose a

generator π of this ideal; this is possible because a DVR is, in particular, a PID. Now if

x ∈ K — , the set of nonzero elements of K, then by factoring the principal fractional ideal

(x)AP , we ¬nd that x = uπ n , where n ∈ Z and u is a unit in AP . We de¬ne vP (x) = n,

with vP (0) = ∞. We can check that vP is a discrete valuation, called the P -adic valuation

on K. Surjectivity and conditions (a) and (b) follow directly from the de¬nition. To verify

(c), let x = uπ m , y = vπ n with m ≥ n. Then x + y = (v ’1 uπ m’n + 1)vπ n , and since

the term in parentheses belongs to AP , the exponent in its prime factorization will be

nonnegative. Therefore vP (x + y) ≥ n = min(vP (x), vP (y)).

Now consider the special case A = Z, K = Q, P = (p). If x is rational and x = pr a/b

where neither a nor b is divisible by p, then we get the p-adic valuation on the rationals,

given by vp (pr a/b) = r.

Here are some of the basic properties of nonarchimedean absolute values. It is often

convenient to exclude the trivial absolute value, given by |x| = 1 for x = 0, and |0| = 0.

Note also that for any absolute value, |1| = | ’ 1| = 1, | ’ x| = |x|, and |x’1 | = 1/|x| for

x = 0. (Observe that 1 — 1 = (’1) — (’1) = x — x’1 = 1.)

9.1.3 Proposition

Let | | be a nonarchimedean absolute value on the ¬eld K. Let A be the corresponding

valuation ring, de¬ned as {x ∈ K : |x| ¤ 1}, and P the valuation ideal {x ∈ K : |x| < 1}.

Then A is a local ring with unique maximal ideal P and fraction ¬eld K. If u ∈ K, then

u is a unit of A if and only if |u| = 1. If the trivial absolute value is excluded, then A is

not a ¬eld.

Proof.

1. A is a ring, because it is closed under addition, subtraction and multiplication, and

contains the identity.

2. K is the fraction ¬eld of A, because if z is a nonzero element of K, then either z or its

inverse belongs to A.

3. A is a local ring with unique maximal ideal P . It follows from the de¬nition that P

is a proper ideal. If Q is any proper ideal of A, then Q ⊆ P , because A \ P ⊆ A \ Q.

(If x ∈ A \ P , then |x| = 1, hence |x’1 | = 1, so x’1 ∈ A. Thus x ∈ Q implies that

xx’1 = 1 ∈ Q, a contradiction.)

4. If u ∈ K, then u is a unit of A i¬ |u| = 1. For if u and v belong to A and uv = 1, then

|u| |v| = 1. But both |u| and |v| are at most 1, hence they must equal 1. Conversely, if

|u| = 1, then |u’1 | = 1. But then both u and its inverse belong to A, so u is a unit of A.

5. If | | is nontrivial, then A is not a ¬eld. For if x = 0 and |x| = 1, then either |x| < 1

and |x’1 | > 1, or |x| > 1 and |x’1 | < 1. Either way, we have an element of A whose

inverse lies outside of A. ™

9.1. ABSOLUTE VALUES AND DISCRETE VALUATIONS 3

9.1.4 Proposition

If the nonarchimedean and nontrivial absolute value | | on K is induced by the discrete

valuation v, then the valuation ring A is a DVR.

Proof. In view of (9.1.3), we need only show that A is a PID. Choose an element π ∈ A

such that v(π) = 1. If x is a nonzero element of A and v(x) = n ∈ Z, then v(xπ ’n ) = 0,

so xπ ’n has absolute value 1 and is therefore a unit u by (9.1.3). Thus x = uπ n . Now if I

is any proper ideal of A, then I will contain an element uπ n with |n| as small as possible,

say |n| = n0 . Either π n0 or π ’n0 will be a generator of I (but not both since I is proper).

We conclude that every ideal of A is principal. ™

The proof of (9.1.4) shows that A has exactly one nonzero prime ideal, namely (π).

9.1.5 Proposition

If | | is a nonarchimedean absolute value , then |x| = |y| implies |x + y| = max(|x|, |y|).

Hence by induction, if |x1 | > |xi | for all i = 2, . . . , n, then |x1 + · · · + xn | = |x1 |.

Proof. We may assume without loss of generality that |x| > |y|. Then

|x| = |x + y ’ y| ¤ max(|x + y|, |y|) = |x + y|,

otherwise max(|x + y|, |y|) = |y| < |x|, a contradiction. Since |x + y| ¤ max(|x|, |y|) = |x|,

the result follows. ™

9.1.6 Corollary

With respect to the metric induced by a nonarchimedean absolute value, all triangles are

isosceles.

Proof. Let the vertices of the triangle be x, y and z. Then |x ’ y| = |(x ’ z) + (z ’ y)|.

If |x ’ z| = |z ’ y|, then two side lengths are equal. If |x ’ z| = |z ’ y|, then by (9.1.5),

|x ’ y| = max(|x ’ z|, |z ’ y|), and again two side lengths are equal. ™

9.1.7 Proposition

The absolute value | | is nonarchimedean if and only if |n| ¤ 1 for every integer n =

1 ± · · · ± 1, equivalently if and only if the set {|n| : n ∈ Z} is bounded.

Proof. If the absolute value is nonarchimedean, then |n| ¤ 1 by repeated application of

condition 3 of (9.1.1). Conversely, if every integer has absolute value at most 1, then it

su¬ces to show that |x + 1| ¤ max(|x|, 1) for every x. (Apply this result to x/y, y = 0.)

By the binomial theorem,

n n

n n

|x + 1| = x¤ |x|r .

n r

r r

r=0 r=0

By hypothesis, the integer n has absolute value at most 1. If |x| > 1, then |x|r ¤ |x|n

r

for all r = 0, 1, . . . , n. If |x| ¤ 1, then |x|r ¤ 1. Consequently,

|x + 1|n ¤ (n + 1) max(|x|n , 1).

4 CHAPTER 9. LOCAL FIELDS

Take nth roots and let n ’ ∞ to get |x + 1| ¤ max(|x|, 1). Finally,to show that bounded-

ness of the set of integers is an equivalent condition, note that if |n| > 1, then |n|j ’ ∞

as j ’ ∞ ™

Problems For Section 9.1

1. Show that every absolute value on a ¬nite ¬eld is trivial.

2. Show that a ¬eld that has an archimedean absolute value must have characteristic 0.

3. Two nontrivial absolute values | |1 and | |2 on the same ¬eld are said to be equivalent

if for every x, |x|1 < 1 if and only if |x|2 < 1. [Equally well, |x|1 > 1 if and only if

|x|2 > 1; just replace x by 1/x if x = 0.] This says that the absolute values induce the

same topology (because they have the same sequences that converge to 0). Show that two

nontrivial absolute values are equivalent if and only if for some real number a, we have

|x|a = |x|2 for all x.

1

9.2 Absolute Values on the Rationals

In (9.1.2), we discussed the p-adic absolute value on the rationals (induced by the p-adic

valuation, with p prime), and we are familiar with the usual absolute value. In this section,

we will prove that up to equivalence (see Problem 3 of Section 9.1), there are no other

nontrivial absolute values on Q.

9.2.1 Preliminary Calculations

Fix an absolute value | | on Q. If m and n are positive integers greater than 1, expand

m to the base n. Then m = a0 + a1 n + · · · + ar nr , 0 ¤ ai ¤ n ’ 1, ar = 0.

(1) r ¤ log m/ log n.

This follows because nr ¤ m.

(2) For every positive integer l we have |l| ¤ l, hence in the above base n expansion,

|ai | ¤ ai < n.

This can be done by induction: |1| = 1, |1 + 1| ¤ |1| + |1|, and so on.

There are 1 + r terms in the expansion of m, each bounded by n[max(1, |n|)]r . [We

must allow for the possibility that |n| < 1, so that |n|i decreases as i increases. In this

case, we will not be able to claim that |a0 | ¤ n(|n|r ).] With the aid of (1), we have

(3) |m| ¤ (1 + log m/ log n)n[max(1, |n|)]log m/ log n .

Replace m by mt and take the tth root of both sides. The result is

(4) |m| ¤ (1 + t log m/ log n)1/t n1/t [max(1, |n|)]log m/ log n .

Let t ’ ∞ to obtain our key formula:

(5) |m| ¤ [max(1, |n|)]log m/ log n .

9.3. ARTIN-WHAPLES APPROXIMATION THEOREM 5

9.2.2 The Archimedean Case

Suppose that |n| > 1 for every n > 1. Then by (5), |m| ¤ |n|log m/ log n , and therefore

log |m| ¤ (log m/ log n) log |n|. Interchanging m and n gives the reverse inequality, so

log |m| = (log m/ log n) log |n|. It follows that log |n|/ log n is a constant a, so |n| = na .

Since 1 < |n| ¤ n [see (2)], we have 0 < a ¤ 1. Thus our absolute value is equivalent to

the usual one.

9.2.3 The Nonarchimedean Case

Suppose that for some n > 1 we have |n| ¤ 1. By (5), |m| ¤ 1 for all m > 1, so

|n| ¤ 1 for all n ≥ 1, and the absolute value is nonarchimedean by (9.1.7). Excluding

the trivial absolute value, we have |n| < 1 for some n > 1. (If every nonzero integer

has absolute value 1, then every nonzero rational number has absolute value 1.) Let

P = {n ∈ Z : |n| < 1}. Then P is a prime ideal (p). (Note that if ab has absolute value

less than 1, so does either a or b.) Let c = |p|, so 0 < c < 1.

Now let r be the exact power of p dividing n, so that pr divides n but pr+1 does not.

Then n/pr ∈ P , so |n|/cr = 1, |n| = cr . Note that n/pr+1 also fails to belong to P , but

/

this causes no di¬culty because n/pr+1 is not an integer.

To summarize, our absolute value agrees, up to equivalence, with the p-adic absolute

value on the positive integers, hence on all rational numbers. (In going from a discrete

valuation to an absolute value, we are free to choose any constant in (0,1). A di¬erent

constant will yield an equivalent absolute value.)

Problems For Section 9.2

If vp is the p-adic valuation on Q, let p be the associated absolute value with the

’r

r

particular choice c = 1/p. Thus p p = p . Denote the usual absolute value by ∞.

1. Establish the product formula: If a is a nonzero rational number, then

a =1

p

p

where p ranges over all primes, including the “in¬nite prime” p = ∞.

9.3 Artin-Whaples Approximation Theorem

The Chinese remainder theorem states that if I1 , . . . In are ideals in a ring R that are

relatively prime in pairs, and ai ∈ Ii , i = 1, . . . , n, then there exists a ∈ R such that

a ≡ ai mod Ii for all i. We are going to prove a result about mutually equivalent absolute

values that is in a sense analogous. The condition a ≡ ai mod Ii will be replaced by

the statement that a is close to ai with respect to the ith absolute value. First, some

computations.

6 CHAPTER 9. LOCAL FIELDS

9.3.1 Lemma

Let | | be an arbitrary absolute value. Then

(1) |a| < 1 ’ an ’ 0;

(2) |a| < 1 ’ an /(1 + an ) ’ 0;

(3) |a| > 1 ’ an /(1 + an ) ’ 1.

Proof. The ¬rst statement follows from |an | = |a|n . To prove (2), use the triangle

inequality and the observation that 1 + an = 1 ’ (’an ) to get

1 ’ |a|n ¤ |1 + an | ¤ 1 + |a|n ,

so by (1), |1 + an | ’ 1. Since |±/β| = |±|/|β|, another application of (1) gives the desired

result. To prove (3), write

a’n

an 1

1’ ’ 0 by (2). ™

= =

1 + a’n

1 + an 1 + an

Here is the key step in the development.

9.3.2 Proposition

Let | |1 , . . . , | |n be nontrivial, mutually inequivalent absolute values on the same ¬eld.

Then there is an element a such that |a|1 > 1 and |a|i < 1 for i = 2, . . . , n.

Proof. First consider the case n = 2. Since | |1 and | |2 are inequivalent, there are

elements b and c such that |b|1 < 1, |b|2 ≥ 1, |c|1 ≥ 1, |c|2 < 1. If a = c/b, then |a|1 > 1

and |a|2 < 1.

Now if the result holds for n ’ 1, we can choose an element b such that |b|1 > 1, |b|2 <

1, . . . , |b|n’1 < 1. By the n = 2 case, we can choose c such that |c|1 > 1 and |c|n < 1.

Case 1. Suppose |b|n ¤ 1. Take ar = cbr , r ≥ 1. Then |ar |1 > 1, |ar |n < 1, and |ar |i ’ 0

as r ’ ∞ for i = 2, . . . , n ’ 1. Thus we can take a = ar for su¬ciently large r.

Case 2. Suppose |b|n > 1. Take ar = cbr /(1 + br ). By (3) of (9.3.1), |ar |1 ’ |c|1 > 1 and

|ar |n ’ |c|n < 1 as r ’ ∞. If 2 ¤ i ¤ n ’ 1, then |b|i < 1, so by (2) of (9.3.1), |ar |i ’ 0

as r ’ ∞. Again we can take a = ar for su¬ciently large r. ™

9.3.3 Approximation Theorem

Let | |1 , . . . , | |n be nontrivial mutually inequivalent absolute values on the ¬eld k.

Given arbitrary elements x1 , . . . , xn ∈ k and any positive real number , there is an

element x ∈ k such that |x ’ xi |i < for all i = 1, . . . , n.

Proof. By (9.3.2), ∀i ∃yi ∈ k such that |yi |i > 1 and |yi |j < 1 for j = i. Take zi =

r r

yi /(1 + yi ). Given δ > 0, it follows from (2) and (3) of (9.3.1) that for r su¬ciently large,

|zi ’ 1|i < δ and |zj | < δ, j = i.

Our candidate is

x = x1 z1 + · · · xn zn .

9.4. COMPLETIONS 7

To show that x works, note that x ’ xi = xj zj + xi (zi ’ 1). Thus

j=i

n

|x ’ xi |i ¤ δ |xj |i + δ|xi |i = δ |xj |i .

j=1

j=i

Choose δ so that the right side is less than , and the result follows. ™

Problems For Section 9.3

1. Let | |1 , . . . , | |n be nontrivial mutually inequivalent absolute values on the ¬eld k.

Fix r with 0 ¤ r ¤ n. Show that there is an element a ∈ k such that |a|1 > 1, . . . , |a|r > 1

and |a|r+1 , . . . , |a|n < 1.

2. There is a gap in the ¬rst paragraph of the proof of (9.3.2), which can be repaired by

showing that the implication |a|1 < 1 ’ |a|2 < 1 is su¬cient for equivalence. Prove this.

9.4 Completions

You have probably seen the construction of the real numbers from the rationals, and

the general process of completing a metric space using equivalence classes of Cauchy

sequences. If the metric is induced by an absolute value on a ¬eld, then we have some

additional structure that we can exploit to simplify the development. If we complete the

rationals with respect to the p-adic rather than the usual absolute value, we get the p-adic

numbers, the most popular example of a local ¬eld.

9.4.1 De¬nitions and Comments

Let K be a ¬eld with an absolute value | |, and let C be the set of Cauchy sequences

with elements in K. Then C is a ring under componentwise addition and multiplication.

Let N be the set of null sequences (sequences converging to 0). Then N is an ideal of C

(because every Cauchy sequence is bounded). In fact N is a maximal ideal, because every

Cauchy sequence not in N is eventually bounded away from 0, hence is a unit in C. The

ˆ

completion of K with respect to the given absolute value is the ¬eld K = C/N . We can

ˆ

embed K in K via c ’ {c, c, . . . } + N .

ˆ ˆ

We now extend the absolute value on K to K. If (cn )+N ∈ K, then (|cn |) is a Cauchy

sequence of real numbers, because by the triangle inequality, |cn | ’ |cm | has (ordinary)

absolute value at most |cn ’ cm | ’ 0 as n, m ’ ∞. Thus |cn | converges to a limit, which

we take as the absolute value of (cn ) + N . Since the original absolute value satis¬es the

de¬ning conditions in (9.1.1), so does the extension.

ˆ

To simplify the notation, we will denote the element (cn ) + N of K by (cn ). If

cn = c ∈ K for all n, we will write the element as c.

9.4.2 Theorem

ˆ ˆ

K is dense in K and K is complete.

8 CHAPTER 9. LOCAL FIELDS

ˆ

Proof. Let ± = (cn ) ∈ K, with ±n = cn . Then

|± ’ ±n | = lim |cm ’ cn | ’ 0 as n ’ ∞,

m’∞

ˆ ˆ

proving that K is dense in K. To prove completeness of K, let (±n ) be a Cauchy sequence

ˆ

in K. Since K is dense, for every positive integer n there exists cn ∈ K such that

ˆ

|±n ’cn | < 1/n. But then (cn ) is a Cauchy sequence in K, hence in K, and we are assured

ˆ

that ± = (cn ) is a legal element of K. Moreover, |±n ’ ±| ’ 0, proving completeness. ™

9.4.3 Uniqueness of the Completion

Suppose K is isomorphic to a dense sub¬eld of the complete ¬eld L, where the absolute

ˆ

value on L extends that of (the isomorphic copy of) K. If x ∈ K, then there is a sequence

xn ∈ K such that xn ’ x. But the sequence (xn ) is also Cauchy in L, hence converges

to an element y ∈ L. If we de¬ne f (x) = y, then f is a well-de¬ned homomorphism of

¬elds, necessarily injective. If y ∈ L, then y is the limit of a Cauchy sequence in K, which

ˆ ˆ

converges to some x ∈ K. Consequently, f (x) = y. Thus f is an isomorphism of K and

L, and f preserves the absolute value.

9.4.4 Power Series Representation

We de¬ne a local ¬eld K as follows. There is an absolute value on K induced by a discrete

valuation v, and with respect to this absolute value, K is complete. For short, we say

that K is complete with respect to the discrete valuation v. Let A be the valuation ring

(a DVR), and P the valuation ideal; see (9.1.3) and (9.1.4) for terminology. If ± ∈ K,

then by (9.1.4) we can write ± = uπ r with r ∈ Z, u a unit in A and π an element of

A such that v(π) = 1. Often, π is called a prime element or a uniformizer. Note that

A = {± ∈ K : v(±) ≥ 0} and P = {± ∈ K : v(±) ≥ 1} = Aπ.

Let S be a ¬xed set of representatives of the cosets of A/P . We will show that each

± ∈ K has a Laurent series expansion

± = a’m π ’m + · · · + a’1 π ’1 + a0 + a1 π + a2 π 2 + · · · , ai ∈ S,

and if ar is the ¬rst nonzero coe¬cient (r may be negative), then v(±) = r.

The idea is to expand the unit u in a power series involving only nonnegative powers of

π. For some a0 ∈ S we have u ’ a0 ∈ P . But then v(u ’ a0 ) ≥ 1, hence v((u ’ a0 )/π) ≥ 0,

so (u ’ a0 )/π ∈ A. Then for some a1 ∈ S we have [(u ’ a0 )/π] ’ a1 ∈ P , in other words,

u ’ a0 ’ a1 π

∈ P.

π

Repeating the above argument, we get

u ’ a0 ’ a1 π

∈ A.

π2

Continue inductively to obtain the desired series expansion. Note that by de¬nition of S,

the coe¬cients ai are unique. Thus an expansion of ± that begins with a term of degree

r in π corresponds to a representation ± = uπ r and a valuation v(±) = r. Also, since

|π| < 1, high positive powers of π are small with respect to the given absolute value. The

partial sums sn of the series form a coherent sequence, that is, sn ≡ sn’1 mod (π)n .

9.4. COMPLETIONS 9

9.4.5 Proposition

Let an be any series of elements in a local ¬eld. Then the series converges if and only

if an ’ 0.

Proof. If the series converges, then an ’ 0 by the standard calculus argument, so assume

that an ’ 0. Since the absolute value is nonarchimedean, n ¤ m implies that

m

| ai | ¤ max(an , . . . , am ) ’ 0 as n ’ ∞. ™

i=n

9.4.6 De¬nitions and Comments

The completion of the rationals with respect to the p-adic valuation is called the ¬eld of

p-adic numbers, denoted by Qp . The valuation ring A = {± : v(±) ≥ 0} is called the ring

of p-adic integers, denoted by Zp . The series representation of a p-adic integer contains

only nonnegative powers of π = p. If in addition, there is no constant term, we get the

valuation ideal P = {± : v(±) ≥ 1}. The set S of coset representatives may be chosen to

be {0, 1, . . . , p ’ 1}. (Note that if a = b and a ≡ b mod p, then a ’ b ∈ P , so a and b

cannot both belong to S. Also, a rational number can always be replaced by an integer

with the same valuation.) Arithmetic is carried out via polynomial multiplication, except

that there is a “carry”. For example, if p = 7, then 3 + 6 = 9 = 2 + p. For some practice,

see the exercises.

We adopt the convention that in going from the p-adic valuation to the associated

absolute value |x| = cv(x) , 0 < c < 1, we take c = 1/p. Thus |pr | = p’r .

Problems For Section 9.4

1. Show that a rational number a/b (in lowest terms) is a p-adic integer if and only if p

does not divide b.

2. With p = 3, express the product of (2 + p + p2 ) and (2 + p2 ) as a p-adic integer.

3. Express the p-adic integer -1 as an in¬nite series.

4. Show that the sequence an = n! of p-adic integers converges to 0.

5. Does the sequence an = n of p-adic integers converge?

∞

6. Show that the p-adic power series for log(1 + x), namely n=1 (’1)n+1 xn /n, converges

in Qp for |x| < 1 and diverges elsewhere. This allows a de¬nition of a p-adic logarithm:

logp (x) = log[1 + (x ’ 1)].

In Problems 7-9, we consider the p-adic exponential function.

7. Recall from elementary number theory that the highest power of p dividing n! is

∞ i

i=1 n/p . (As an example, let n = 15 and p = 2. Calculate the number of multiples

of 2, 4,and 8 in the integers 1-15.) Use this result to show that the p-adic valuation of n!

is at most n/(p ’ 1).

8. Show that the p-adic valuation of (pm )! is (pm ’ 1)/(p ’ 1).

∞

9. Show that the exponential series n=0 xn /n! converges for |x| < p’1/(p’1) and diverges

elsewhere.

10 CHAPTER 9. LOCAL FIELDS

9.5 Hensel™s Lemma

9.5.1 The Setup

Let K be a local ¬eld with valuation ring A and valuation ideal P . By (9.1.3) and (9.1.4),

A is a local ring, in fact a DVR, with maximal ideal P . The ¬eld k = A/P is called the

residue ¬eld of A or of K. If a ∈ A, then the coset a + P ∈ k will be denoted by a. If f is

a polynomial in A[X], then reduction of the coe¬cients of f mod P yields a polynomial

f in k[X]. Thus

d d

ai X ∈ A[X], f (X) = ai X i ∈ k[X].

i

f (X) =

i=0 i=0

Hensel™s lemma is about lifting a factorization of f from k[X] to A[X]. Here is the precise

statement.

9.5.2 Hensel™s Lemma

Assume that f is a monic polynomial of degree d in A[X], and that the corresponding

polynomial F = f factors as the product of relatively prime monic polynomials G and H

in k[X]. Then there are monic polynomials g and h in A[X] such that g = G, h = H and

f = gh.

Proof. Let r be the degree of G, so that deg H = d ’ r. We will inductively construct

gn , hn ∈ A[X], n = 1, 2, . . . , such that deg gn = r, deg hn = d ’ r, g n = G, hn = H, and

f (X) ’ gn (X)hn (X) ∈ P n [X].

Thus the coe¬cients of f ’ gn hn belong to P n .

The basis step: Let n = 1. Choose monic g1 , h1 ∈ A[X] such that g 1 = G and h1 = H.

Then deg g1 = r and deg h1 = d ’ r. Since f = g 1 h1 , we have f ’ g1 h1 ∈ P [X].

The inductive step: Assume that gn and hn have been constructed. Let f (X)’gn (X)hn (X) =

d

i=0 ci X with the ci ∈ P . Since G = g n and H = hn are relatively prime, for each

i n

i = 0, . . . , d there are polynomials v i and wi in k[X] such that

X i = v i (X)g n (X) + wi (X)hn (X).

Since g n has degree r, the degree of v i is at most d ’ r, and similarly the degree of wi is

at most r. Moreover,

X i ’ vi (X)gn (X) ’ wi (X)hn (X) ∈ P [X]. (1)

We de¬ne

d d

gn+1 (X) = gn (X) + ci wi (X), hn+1 (X) = hn (X) + ci vi (X).

i=0 i=0

9.5. HENSEL™S LEMMA 11

Since the ci belong to P n ⊆ P , it follows that g n+1 = g n = G and hn+1 = hn = H. Since

the degree of gn+1 is at most r, it must be exactly r, and similarly the degree of hn+1 is

d ’ r. To check the remaining condition,

f ’ gn+1 hn+1 = f ’ (gn + ci wi )(hn + ci vi )

i i

= (f ’ gn hn ’ ci (X i ’ gn vi ’ hn wi ) ’

ci X i ) + ci cj wi vj .

i i i,j

By the induction hypothesis, the ¬rst grouped term on the right is zero, and, with the

aid of Equation (1) above, the second grouped term belongs to P n P [X] = P n+1 [X]. The

¬nal term belongs to P 2n [X] ⊆ P n+1 [X], completing the induction.

Finishing the proof. By de¬nition of gn+1 , we have gn+1 ’ gn ∈ P n [X], so for any

¬xed i, the sequence of coe¬cients of X i in gn (X) is Cauchy and therefore converges.

To simplify the notation we write gn (X) ’ g(X), and similarly hn (X) ’ h(X), with

g(X), h(X) ∈ A[X]. By construction, f ’ gn hn ∈ P n [X], and we may let n ’ ∞ to get

f = gh. Since g n = G and hn = H for all n, we must have g = G and h = H. Since

f, G and H are monic, the highest degree terms of g and h are of the form (1 + a)X r and

(1 + a)’1 X d’r respectively, with a ∈ P . (Note that 1 + a must reduce to 1 mod P .) By

replacing g and h by (1 + a)’1 g and (1 + a)h, respectively, we can make g and h monic

without disturbing the other conditions. The proof is complete. ™

9.5.3 Corollary

With notation as in (9.5.1), let f be a monic polynomial in A[X] such that f has a simple

root · ∈ k. Then f has a simple root a ∈ A such that a = ·.

Proof. We may write f (X) = (X ’ ·)H(X) where X ’ · and H(X) are relatively prime

in k[X]. By Hensel™s lemma, we may lift the factorization to f (X) = (X ’ a)h(X) with

h ∈ A[X], a ∈ A and a = ·. If a is a multiple root of f , then · is a multiple root of f ,

which is a contradiction. ™

Problems For Section 9.5

1. Show that for any prime p, there are p ’ 1 distinct (p ’ 1)th roots of unity in Zp .

2. Let p be an odd prime not dividing the integer m. We wish to determine whether m

is a square in Zp . Describe an e¬ective procedure for doing this.

3. In Problem 2, suppose that we √ only want to decide if m is a square in Zp , but

not

to ¬nd the series representation of m explicitly. Indicate how to do this, and illustrate

with an example.

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 ¬nite-dimensional vector space is injective i¬ 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 ¬eld.

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-de¬ned 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 satis¬es an equation of the form

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

By Problem 4, b + Q satis¬es 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 ¬eld if and only if B is a ¬eld, 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 ¬rst sequence,

the corresponding map in the second sequence is S ’1 f .) It follows from the de¬nition 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

°0 3 ’1»

2

m(θ ) =

10 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

®

’1

0 0

m(θ) = °1 3»

0

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 4 ’ 18. (To 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 +√ + cθ2 + dθ3 . Take the trace of both sides to conclude

bθ

of 3 is 0 because its minimal polynomial is X 2 ’ 3.) Thus

that a = 0. (The trace √

√

3 = bθ + cθ2 + dθ3 , so 3θ = bθ2 + cθ3 + 2d. Again take the trace of both sides to get

√ √

d = 0. We now have 3 √ bθ + cθ2 , so 3θ2 = bθ3 + 2c. The minimal polynomial of

=

√2

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 = √ 2 and

uy

√

with u and v relatively prime and square-free. Thus L = Q( uv) = Q( d).

2

t = vz , √ √

√ √

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.

4. Since ωn = ω2n , we have ωn ∈ Q(ω2n ), so Q(ωn ) ⊆ Q(ω2n ). If n is odd, then n+1 = 2r,

2

so

ω2n = ’ω2n = ’(ω2n )r = ’ωn .

2r 2 r

Therefore Q(ω2n ) ⊆ Q(ωn ).

√ √

5. Q( ’3) = Q(ω) where ω = ’ 1 + 1 ’3 is a primitive cube root of unity.

2 2

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 ¬rst 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 ¬xed ¬eld 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 the result follows.

n

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 coe¬cients 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 ¬eld 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 ∈ P1 ∪ · · · ∪ 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 ⊆ P1 + P2 . Thus P1 and P2 are relatively prime for all

r r

r ≥ 1. Assuming inductively that P1 and P2 are relatively prime, it follows that

r s

P2 = P2 R = P2 (P1 + P2 ) ⊆ P1 + P2

s+1

s s s r r

so

R = P1 + P2 ⊆ P1 + (P1 + P2 ) = P1 + P2

s+1 s+1

r s r r r

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)√ (2.3.11), the algebraic integers are of the form a + b ’3, a, b ∈ Z, or

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 ¬rst 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 ∈ P2 . 2

2. Since 2 ∈ P2 , it follows that 4 ∈ P2 ,√ by Problem 1, 2 = 6 ’ 4 √ P2 .

∈2

2

so

√ √ √ √

3. (2, 1 + ’5)(2, 1 + ’5) = (4, 2(1 + ’5), (1 + ’5)2 ), and (1 + ’5)2 = ’4 + 2 ’5.

2

Therefore each of the generators of the ideal P2 is divisible by 2, hence belongs to (2).

Thus P2 ⊆ (2).

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 ¬elds, 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

bi T (ωi ωj ) ≡ 0 mod p.

T (A(βωj )) = T ( bi A(ωi ωj )) =

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 ¬elds 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 ¬nite extension of a ¬nite ¬eld, 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 .

Factoring (3)√ covered by case (a1), and x2 + 5 ≡ (x + 1)(x ’ 1) mod 3. Therefore

is √

(3) = (3, 1 + ’5) (3, 1 ’ ’5).

√

2. We have (5) = (5, ’5)2 , as in case (b). To factor (7), note that x2 + 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, ± + 2)(5, ±2 + 3± ’ 1)

√

3

where ± = 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 √ = 2, 3, and

m

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 rami¬es [see (4.3.2), case (c1)], there√ only one ideal of norm 2. Thus class

is

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 √

rami¬es [see (4.3.2), case (b)],√ argument proceeds as in Problem 3. Note that Q( 6) is

the √ √ √

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 rami¬es [(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

π (n + 1)2n+2

an+1 1 π 1

= (1 + )2n .

=

n2n (n + 1)2

an 4 4 n

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

π2

a3 an

|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 .