<< ÒÚ. 3(‚ÒÂ„Ó 4)—Œƒ≈–∆¿Õ»≈ >>
7.2.3 Lemma
Again assume [KL : Q] = mn. Let a1 , . . . , am and b1 , . . . , bn be integral bases for R and
S respectively. If Œ± ‚àà T , then
m n
cij
ai bj , cij ‚àà Z, r ‚àà Z
Œ±=
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 ‚àà R, wi ‚àà S.
basis for RS. (This is because RS consists of all Ô¬Ånite sums
Each vi is a linear combination of the ak with integer coeÔ¬Écients, and so on.) It follows
that Œ± is a linear combination of the ai bj with rational coeÔ¬Écients. 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 ‚äÜ d RS. Thus if d = 1, then T = RS.
1

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

cij cij (d/r)
= .
r d
In turn, it suÔ¬Éces 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 œÉ be an embedding of K in C. By (7.2.2), œÉ extends to an embedding (also called
œÉ) of KL in C such that œÉ is the identity on L. By (7.2.3), if Œ± ‚àà T we have
cij
œÉ(Œ±) = œÉ(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
œÉ(ai )xi = œÉ(Œ±)
i=1

where there is one equation for each of the m embeddings œÉ from K to C. Solving for xi
by Cramer‚Ä™s rule, we get xi = Œ≥i /Œ¥, where Œ¥ is the determinant formed from the œÉ(ai ) and
Œ≥i is the determinant obtained by replacing the ith column of Œ¥ with the œÉ(Œ±). Note that
by (2.3.3), Œ¥ 2 is the discriminant of R, call it e. Since all the œÉ(ai ) and œÉ(Œ±) are algebraic
integers, so are Œ¥ and all the Œ≥i . Now
Œ≥i Œ≥i Œ¥ Œ≥i Œ¥
xi = = 2=
Œ¥ Œ¥ e
so exi = Œ≥i Œ¥ is an algebraic integer. By deÔ¬Ånition 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 Œ∂ be a primitive nth root of unity, and denote the discriminant of {1, Œ∂, . . . , Œ∂ œ•(n)‚à’1 }
by disc(Œ∂). Then disc(Œ∂) divides nœ•(n) .
Proof. Let f (= Œ¦n , the nth cyclotomic polynomial) be the minimal polynomial of Œ∂
over Q. Since Œ∂ is a root of X n ‚à’ 1, we have X n ‚à’ 1 = f (X)g(X) for some g ‚àà Q[X].
But f ‚àà Z[X] (because Œ∂ is an algebraic integer), and f , hence g, is monic, so g ‚àà Z[X].
DiÔ¬Äerentiate both sides of the equation to get nX n‚à’1 = f (X)g (X) + f (X)g(X). Setting
X = Œ∂, which is a root of f , we have nŒ∂ n‚à’1 = f (Œ∂)g(Œ∂). But Œ∂ n‚à’1 = Œ∂ n /Œ∂ = 1/Œ∂, so

n = Œ∂f (Œ∂)g(Œ∂).

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

nœ•(n) = N (f (Œ∂))N (Œ∂g(Œ∂)).

But by (2.3.6), N (f (Œ∂)) = ¬±disc (Œ∂), and N (Œ∂g(Œ∂)) ‚àà Z by (2.2.2). The desired result
follows. ‚™£
7.2. AN INTEGRAL BASIS OF A CYCLOTOMIC FIELD 7

7.2.6 Theorem
If Œ∂ is a primitive nth root of unity, then the ring of algebraic integers of Q(Œ∂) is Z[Œ∂]. in
other words, the powers of Œ∂ form an integral basis.
Proof. We have proved this when Œ∂ is a prime power, so let n = m1 m2 where the mi are
relatively prime and greater than 1. Now

Œ∂ m1 = (ei2œÄ/n )m1 = ei2œÄm1 /n = ei2œÄ/m2 = Œ∂2 ,

a primitive (m2 )th root of unity, and similarly Œ∂ m2 = Œ∂1 , a primitive (m1 )th root of unity.
Thus Q(Œ∂1 ) and Q(Œ∂2 ) are contained in Q(Œ∂). On the other hand, since m1 and m2 are
relatively prime, there are integers r, s such that rm2 + sm1 = 1. Thus

Œ∂ = Œ∂ rm2 +sm1 = Œ∂1 Œ∂2 .
rs

It follows that Q(Œ∂) = Q(Œ∂1 )Q(Œ∂2 ), and we can apply (7.2.4). In that proposition, we
take K = Q(Œ∂1 ), L = Q(Œ∂2 ), KL = Q(Œ∂), R = Z[Œ∂1 ], S = Z[Œ∂2 ] (induction hypothesis),
T = RS. The hypothesis on the degree [KL : Q] is satisÔ¬Åed because œ•(n) = œ•(m1 )œ•(m2 ).
By (7.2.5), disc(Œ∂1 ) divides a power of m1 and disc(Œ∂2 ) 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 satisÔ¬Åed. The conclusion is that the ring T of algebraic integers of KL coincides with
RS. But the above argument that Q(Œ∂) = Q(Œ∂1 )Q(Œ∂2 ) may be repeated verbatim with Q
replaced by Z. We conclude that Z[Œ∂] = Z[Œ∂1 ]Z[Œ∂2 ] = RS = T . ‚™£

7.2.7 The Discriminant of a General Cyclotomic Extension
The Ô¬Åeld discriminant of Q(Œ∂), where Œ∂ is a primitive nth root of unity is given by

(‚à’1)œ•(n)/2 nœ•(n)
.
œ•(n)/(p‚à’1)
p|n p

A direct veriÔ¬Åcation, 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 Ô¬Åeld K, L is
a Ô¬Ånite 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 œÉ ‚àà G, then œÉ(B) = B. If Q is a prime ideal of B, then so is œÉ(Q). Moreover, if Q
lies above the nonzero prime ideal P of A, then so does œÉ(Q). Thus G acts on the set of
prime ideals lying above P .
Proof. If x ‚àà B, then œÉ(x) ‚àà B (apply œÉ to an equation of integral dependence). Thus
œÉ(B) ‚äÜ B. But œÉ ‚à’1 (B) is also contained in B, hence B = œÉœÉ ‚à’1 (B) ‚äÜ œÉ(B). If P B =
Qei , then apply œÉ to get P B = œÉ(Qi )ei . The œÉ(Qi ) must be prime ideals because œÉ
i
preserves all algebraic relations. Note also that œÉ is a K-automorphism, hence Ô¬Åxes every
element of A (and of P ). Therefore Q ‚à© A = P ‚á’ œÉ(Q) ‚à© 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 œÉ ‚àà G we have œÉ(Q) = Q1 .
Proof. If the assertion is false, then for each œÉ, the ideals Q1 and œÉ(Q) are maximal and
distinct, so Q1 ‚äÜ œÉ(Q). By the prime avoidance lemma (Section 3.1, exercises), there is
an element x ‚àà Q1 belonging to none of the œÉ(Q). Computing the norm of x relative to
L/K, we have N (x) = œÉ‚ààG œÉ(x) by (2.1.6). But one of the œÉ‚Ä™s is the identity, Q1 is an
ideal, and [by (8.1.1)] œÉ(x) ‚àà B for all œÉ. Consequently, N (x) ‚àà Q1 . But N (x) ‚àà A by

1
2 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

(2.2.2), so N (x) ‚àà Q1 ‚à© A = P = Q ‚à© A. Thus N (x) belongs to the prime ideal Q, and
therefore some œÉ ‚à’1 (x) belongs to Q as well. This gives x ‚àà œÉ(Q), a contradiction. ‚™£

8.1.3 Corollary
g
In the factorization P B = i=1 Piei of the nonzero prime ideal P , the ramiÔ¬Åcation indices
ei are the same for all i, as are the relative degrees fi . Thus the ram-rel identity simpliÔ¬Åes
to ef g = n, where n = [L : K] = |G|.
Proof. This follows from (8.1.2), along with the observation that an automorphism œÉ
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.

We say that the prime ideals œÉ(Q), œÉ ‚àà 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 œÉ ‚àà G such that œÉ(Q) = Q. (This does not mean that œÉ Ô¬Åxes
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.
‚à’1
For if œÉ1 D = œÉ2 D, then œÉ2 œÉ1 ‚àà D, so œÉ1 (Q) = œÉ2 (Q).
There is a particular subgroup of D that will be of interest. By (8.1.1), œÉ(B) = B for
every œÉ ‚àà G. If œÉ ‚àà D, then œÉ(Q) = Q. It follows that œÉ induces an automorphism œÉ of
B/Q. (Note that x ‚â° y mod Q iÔ¬Ä œÉx ‚â° œÉy mod Q.) Since œÉ is a K-automorphism, œÉ
is an A/P -automorphism. The mapping œÉ ‚Ü’ œÉ is a group homomorphism from D to the
group of A/P -automorphisms of B/Q.

8.1.5 DeÔ¬Ånition
The kernel I of the above homomorphism, that is, the set of all œÉ ‚àà D such that œÉ 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 = {œÉ ‚àà D : œÉ(x) + Q = x + Q ‚àÄx ‚àà B} = {œÉ ‚àà D : œÉ(x) ‚à’ x ‚àà Q ‚àÄx ‚àà B}.
8.1. DECOMPOSITION AND INERTIA GROUPS 3

We now introduce an intermediate Ô¬Åeld and ring into the basic AKLB setup, as follows.
L B

K A

Take KD to be the Ô¬Åxed Ô¬Åeld of D, and let AD = B ‚à© KD be the integral closure of A in
KD . Let PD be the prime ideal Q ‚à© AD . Note that Q is the only prime factor of PD B.
This is because all primes in the factorization are conjugate, and œÉ(Q) = Q for all œÉ ‚àà D,
by deÔ¬Ånition of D.

8.1.7 Lemma
Let PD B = Qe and f = [B/Q : AD /PD ]. Then e = e and f = f . Moreover,
=
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 ‚äÜ AD /PD ‚äÜ B/Q, so f ‚â¤ f . Also, P AD ‚äÜ PD , so PD
divides P AD , hence PD B divides P AD B = P B. Consequently, e ‚â¤ 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 subÔ¬Åeld of AD /PD , the proof is
complete. ‚™£

8.1.8 Theorem
The homomorphism œÉ ‚Ü’ œÉ of D to Gal[(B/Q)/(A/P )] introduced in (8.1.4) is surjective
with kernel I. Therefore Gal[(B/Q)/(A/P )] ‚àº D/I.
=
Proof. Let x be a primitive element of B/Q over A/P . Let x ‚àà B be a representative
of x. Let h(X) = X r + ar‚à’1 X r‚à’1 + ¬· + a0 be the minimal polynomial of x over KD ; the
coeÔ¬Écients ai belong to AD by (2.2.2). The roots of h are all of the form œÉ(x), œÉ ‚àà D.
(We are working in the extension L/KD , with Galois group D.) By (8.1.7), if we reduce
the coeÔ¬Écients of h mod PD , the resulting polynomial h(X) has coeÔ¬Écients in A/P . The
roots of h are of the form œÉ(x), œÉ ‚àà D (because x is a primitive element). Since œÉ ‚àà D
means that œÉ(Q) = Q, all conjugates of x over A/P lie in B/Q. By the basic theory of
splitting Ô¬Åelds, B/Q is a Galois extension of A/P .
To summarize, since every conjugate of x over A/P is of the form œÉ(x), every A/P -
automorphism of B/Q (necessarily determined by its action on x), is of the form œÉ where
œÉ ‚àà D. Since œÉ is trivial iÔ¬Ä œÉ ‚àà I, it follows that the map œÉ ‚Ü’ œÉ 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 deÔ¬Ånition 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
deÔ¬Ånition of stabilizer subgroup that D(œÉ(Q)) = œÉD(Q)œÉ ‚à’1 for every œÉ ‚àà G. Show that
the inertia subgroup also behaves in this manner, that is, I(œÉ(Q)) = œÉI(Q)œÉ ‚à’1 .
2. If L/K is an abelian extension (the Galois group G = Gal(L/K) is abelian), show that
the groups D(œÉ(Q)), œÉ ‚àà G, are all equal, as are the I(œÉ(Q)), œÉ ‚àà 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 Ô¬Åelds.

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 Ô¬Ånite extension of the Ô¬Ånite Ô¬Åeld A/P [see
(4.2.3)], so the Galois group is cyclic. Moreover, there is a canonical generator given by
x+Q ‚Ü’ xq +Q, x ‚àà B, where q = |A/P |. Thus we have identiÔ¬Åed a distinguished element
œÉ ‚àà 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 ‚àà B,

œÉ(x) ‚â° 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
œ„ ‚à’1 .
L/K L/K
If œ„ ‚àà G, then =œ„
œ„ (Q) Q

œ„ ‚à’1 x ‚â° (œ„ ‚à’1 x)q = œ„ ‚à’1 xq mod Q. Apply œ„ to both sides to
L/K
Proof. If x ‚àà B, then Q

conclude that œ„ L/K œ„ ‚à’1 satisÔ¬Åes the deÔ¬Åning equation for L/K
. Since the Frobenius
œ„ (Q)
Q
is determined by its deÔ¬Åning 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 œ„ (Q), œ„ ‚àà G, hence
by (8.1.2), for all prime ideals lying over P . ‚™£

8.2.4 Intermediate Fields
We now introduce an intermediate Ô¬Åeld between K and L, call it F . We can then lift P
to the ring of algebraic integers in F , namely B ‚à© F . A prime ideal lying over P has the
form Q ‚à© F , where Q is a prime ideal of P B. We will compare decomposition groups with
respect to the Ô¬Åelds L and F , with the aid of the identity

[B/Q : A/P ] = [B/Q : (B ‚à© F )/(Q ‚à© F )][(B ‚à© F )/(Q ‚à© 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 Ô¬Årst term on the
right is the order of the decomposition group of Q over Q ‚à© F . The second term on the
right is the relative degree of Q ‚à© F over P , call if f . Thus

|D(Q, Q ‚à© F )| = |D(Q, P )|/f

Since D = D(Q, P ) is cyclic and is generated by the Frobenius automorphism œÉ, the
unique subgroup of D with order |D|/f is generated by œÉ f . Note that D(Q, Q ‚à© 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 œÉ f .

8.2.5 Proposition
f
L/F L/K
= .
Q Q

L/K
. Then œÉ ‚àà D, so œÉ(Q) = Q; also œÉ(x) ‚â° xq mod Q, x ‚àà B, where
Proof. Let œÉ = Q
f
q = |A/P |. Thus œÉ f (Q) = Q and œÉ f (x) ‚â° xq . Since q f is the cardinality of the Ô¬Åeld

8.2.6 Proposition
L/K F/K
If the extension F/K is Galois, then the restriction of œÉ = to F is .

Proof. Let œÉ1 be the restriction of œÉ to F . Since œÉ(Q) = Q, it follows that œÉ1 (Q ‚à© F ) =
Q ‚à© F . (Note that F/K is normal, so œÉ1 is an automorphism of F .) Thus œÉ1 belongs
to D(Q ‚à© F, P ). Since œÉ(x) ‚â° xq mod Q, we have œÉ1 (x) ‚â° xq mod (Q ‚à© F ), where
F/K
q = |A/P |. Consequently, œÉ1 = .‚™£
6 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

We may view the lifting from the base Ô¬Åeld K to the extension Ô¬Åeld L as occurring in
three distinct steps. Let FD be the decomposition Ô¬Åeld of the extension, that is, the Ô¬Åxed
Ô¬Åeld of the decomposition group D, and let FI be the inertia Ô¬Åeld, the Ô¬Åxed Ô¬Åeld of the
inertia group I. We have the following diagram:

L
e=|I|

FI
f =|D|/e

FD
g=n/ef

K

All ramiÔ¬Åcation 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 ramiÔ¬Åcation 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 Œ∂ be a primitive mth root of unity, and let L = Q(Œ∂) be the corresponding cyclotomic
Ô¬Åeld. (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
unramiÔ¬Åed. 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 Ô¬Ånd the Frobenius automorphism œÉ explicitly.
The deÔ¬Åning equation is œÉ(x) ‚â° xp mod Qi for all i, and consequently

œÉ(Œ∂) = Œ∂ p .
8.3. APPLICATIONS 7

(The idea is that the roots of unity remain distinct when reduced mod Qi , because the
polynomial X n ‚à’ 1 is separable over Fp .)
Now the order of œÉ is the size of the decomposition group D, which is f . Thus f is
the smallest positive integer such that œÉ f (Œ∂) = Œ∂. Since Œ∂ is a primitive mth root of unity,
we conclude that
f is the smallest positive integer such that pf ‚â° 1 mod m.
Once we know f , we can Ô¬Ånd the number of prime factors g = n/f , where n = œ•(m).
(We already know that e = 1 because p is unramiÔ¬Åed.)
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 ‚â° 1 mod m1 . The factorization is (p) = (Q1 ¬· ¬· ¬· Qg )e , with e = œ•(pa ). The Qi are
distinct prime ideals, each with relative degree f . The number of distinct prime factors
is g = œ•(m1 )/f .
We will now give a proof of Gauss‚Ä™ law of quadratic reciprocity.

8.3.2 Proposition
Let q be an odd prime, and let L = Q(Œ∂q ) be the cyclotomic Ô¬Åeld generated by a primitive
q th root of unity. Then L has a unique quadratic subÔ¬Åeld F . Explicitly, if q ‚â° 1 mod 4,
‚àö
‚àö
then the quadratic subÔ¬Åeld is Q( q), and if q ‚â° 3 mod 4, it is Q( ‚à’q). More compactly,
‚àö
F = Q( q ‚à— ), where q ‚à— = (‚à’1)q‚à’1)/2 q.
Proof. The Galois group of the extension is cyclic of even order q ‚à’ 1, hence has a unique
subgroup of index 2. Therefore L has a unique quadratic subÔ¬Åeld. By (7.1.7) ‚àö the and
‚àà Q. But d ‚àà Q,
(q‚à’1)/2 q‚à’2
exercises to Section 7.1, the Ô¬Åeld discriminant is d = (‚à’1) q /
because d has an odd number of factors of q. If q ‚â° 1 mod 4, then the sign of d is
‚àö ‚àö
positive and Q( d) = Q( q). Similarly, if q ‚â° 3 mod 4, then the sign of d is negative
‚àö ‚àö
and Q( d) = ‚àö ‚à’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 œÉp be the Frobenius automorphism F/Q , where F is the unique quadratic subÔ¬Åeld
p
of L, and p is an odd prime unequal to q. By (4.3.2), case (a1), if q ‚à— 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 œÉp generates D, œÉp is the identity. If q ‚à— is not a quadratic residue
mod p, then by (4.3.2), case (a2), p is inert, so g = 1, f = 2, and œÉp is nontrivial. Since
the Galois group of F/Q has only two elements, it may be identiÔ¬Åed with {1, ‚à’1} under
‚à—
multiplication, and we may write (using the standard Legendre symbol) œÉp = ( qp ). On
L/Q
the other hand, œÉp is the restriction of œÉ = to F , by (8.2.6). Thus œÉp is the identity
p
on F iÔ¬Ä œÉ belongs to H, the unique subgroup of Gal(L/Q) of index 2. This will happen iÔ¬Ä
œÉ is a square. Now the Frobenius may be viewed as a lifting of the map x ‚Ü’ xp mod q.
p
[As in (8.3.1), œÉ(Œ∂q ) = Œ∂q .] Thus œÉ will belong to H iÔ¬Ä p is a quadratic residue mod q. In
other words, œÉp = ( p ).
q
8 CHAPTER 8. FACTORING OF PRIME IDEALS IN GALOIS EXTENSIONS

If p and q are distinct odd primes, then

p q
= (‚à’1)(p‚à’1)(q‚à’1)/4 .
q p

Proof. By (8.3.3),

q‚à—
(q‚à’1)/2
‚à’1
(‚à’1)(q‚à’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,

‚à’1
= (‚à’1)(p‚à’1)/2 ,
p

and the result follows. ‚™£

8.3.5 Remark
Let L = Q(Œ∂), where Œ∂ 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 ‚à’ Œ∂)p‚à’1 .

Thus the ramiÔ¬Åcation index e = p ‚à’ 1 coincides with the degree of the extension. We say
that p is totally ramiÔ¬Åed.
Chapter 9

Local Fields

The deÔ¬Ånition of global Ô¬Åeld varies in the literature, but all deÔ¬Ånitions include our primary
source of examples, number Ô¬Åelds. The other Ô¬Åelds that are of interest in algebraic number
theory are the local Ô¬Åelds, 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
An absolute value on a Ô¬Åeld k is a mapping x ‚Ü’ |x| from k to the real numbers, such that
for every x, y ‚àà k,
1. |x| ‚â• 0, with equality if and only if x = 0;
2. |xy| = |x| |y|;
3. |x + y| ‚â¤ |x| + |y|.
The absolute value is nonarchimedean if the third condition is replaced by a stronger
version:
3 . |x + y| ‚â¤ 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 ‚Ü’ Z‚à™{‚àû}, such that for every x, y ‚àà k,
(a) v(x) = ‚àû if and only if x = 0;
(b) v(xy) = v(x) + v(y);
(c) v(x + y) ‚â• 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 Ô¬Åeld 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 œÄ of this ideal; this is possible because a DVR is, in particular, a PID. Now if
x ‚àà K ‚à— , the set of nonzero elements of K, then by factoring the principal fractional ideal
(x)AP , we Ô¬Ånd that x = uœÄ n , where n ‚àà Z and u is a unit in AP . We deÔ¬Åne vP (x) = n,
with vP (0) = ‚àû. 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 deÔ¬Ånition. To verify
(c), let x = uœÄ m , y = vœÄ n with m ‚â• n. Then x + y = (v ‚à’1 uœÄ m‚à’n + 1)vœÄ n , and since
the term in parentheses belongs to AP , the exponent in its prime factorization will be
nonnegative. Therefore vP (x + y) ‚â• 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| = | ‚à’ 1| = 1, | ‚à’ x| = |x|, and |x‚à’1 | = 1/|x| for
x = 0. (Observe that 1 √— 1 = (‚à’1) √— (‚à’1) = x √— x‚à’1 = 1.)

9.1.3 Proposition
Let | | be a nonarchimedean absolute value on the Ô¬Åeld K. Let A be the corresponding
valuation ring, deÔ¬Åned as {x ‚àà K : |x| ‚â¤ 1}, and P the valuation ideal {x ‚àà K : |x| < 1}.
Then A is a local ring with unique maximal ideal P and fraction Ô¬Åeld K. If u ‚àà 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 Ô¬Åeld.
Proof.
1. A is a ring, because it is closed under addition, subtraction and multiplication, and
contains the identity.
2. K is the fraction Ô¬Åeld 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 deÔ¬Ånition that P
is a proper ideal. If Q is any proper ideal of A, then Q ‚äÜ P , because A \ P ‚äÜ A \ Q.
(If x ‚àà A \ P , then |x| = 1, hence |x‚à’1 | = 1, so x‚à’1 ‚àà A. Thus x ‚àà Q implies that
xx‚à’1 = 1 ‚àà Q, a contradiction.)
4. If u ‚àà K, then u is a unit of A iÔ¬Ä |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‚à’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 Ô¬Åeld. For if x = 0 and |x| = 1, then either |x| < 1
and |x‚à’1 | > 1, or |x| > 1 and |x‚à’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 œÄ ‚àà A
such that v(œÄ) = 1. If x is a nonzero element of A and v(x) = n ‚àà Z, then v(xœÄ ‚à’n ) = 0,
so xœÄ ‚à’n has absolute value 1 and is therefore a unit u by (9.1.3). Thus x = uœÄ n . Now if I
is any proper ideal of A, then I will contain an element uœÄ n with |n| as small as possible,
say |n| = n0 . Either œÄ n0 or œÄ ‚à’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 (œÄ).

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 ‚à’ y| ‚â¤ max(|x + y|, |y|) = |x + y|,
otherwise max(|x + y|, |y|) = |y| < |x|, a contradiction. Since |x + y| ‚â¤ 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 ‚à’ y| = |(x ‚à’ z) + (z ‚à’ y)|.
If |x ‚à’ z| = |z ‚à’ y|, then two side lengths are equal. If |x ‚à’ z| = |z ‚à’ y|, then by (9.1.5),
|x ‚à’ y| = max(|x ‚à’ z|, |z ‚à’ y|), and again two side lengths are equal. ‚™£

9.1.7 Proposition
The absolute value | | is nonarchimedean if and only if |n| ‚â¤ 1 for every integer n =
1 ¬± ¬· ¬· ¬· ¬± 1, equivalently if and only if the set {|n| : n ‚àà Z} is bounded.
Proof. If the absolute value is nonarchimedean, then |n| ‚â¤ 1 by repeated application of
condition 3 of (9.1.1). Conversely, if every integer has absolute value at most 1, then it
suÔ¬Éces to show that |x + 1| ‚â¤ 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‚â¤ |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 ‚â¤ |x|n
r
for all r = 0, 1, . . . , n. If |x| ‚â¤ 1, then |x|r ‚â¤ 1. Consequently,
|x + 1|n ‚â¤ (n + 1) max(|x|n , 1).
4 CHAPTER 9. LOCAL FIELDS

Take nth roots and let n ‚Ü’ ‚àû to get |x + 1| ‚â¤ 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 ‚Ü’ ‚àû
as j ‚Ü’ ‚àû ‚™£

Problems For Section 9.1
1. Show that every absolute value on a Ô¬Ånite Ô¬Åeld is trivial.
2. Show that a Ô¬Åeld that has an archimedean absolute value must have characteristic 0.
3. Two nontrivial absolute values | |1 and | |2 on the same Ô¬Åeld 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 ‚â¤ ai ‚â¤ n ‚à’ 1, ar = 0.
(1) r ‚â¤ log m/ log n.
This follows because nr ‚â¤ m.
(2) For every positive integer l we have |l| ‚â¤ l, hence in the above base n expansion,
|ai | ‚â¤ ai < n.
This can be done by induction: |1| = 1, |1 + 1| ‚â¤ |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 | ‚â¤ n(|n|r ).] With the aid of (1), we have
(3) |m| ‚â¤ (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| ‚â¤ (1 + t log m/ log n)1/t n1/t [max(1, |n|)]log m/ log n .
Let t ‚Ü’ ‚àû to obtain our key formula:

(5) |m| ‚â¤ [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| ‚â¤ |n|log m/ log n , and therefore
log |m| ‚â¤ (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| ‚â¤ n [see (2)], we have 0 < a ‚â¤ 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| ‚â¤ 1. By (5), |m| ‚â¤ 1 for all m > 1, so
|n| ‚â¤ 1 for all n ‚â• 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 ‚àà 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 ‚àà P , so |n|/cr = 1, |n| = cr . Note that n/pr+1 also fails to belong to P , but
/
this causes no diÔ¬Éculty 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 diÔ¬Äerent
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
‚à’r
r
particular choice c = 1/p. Thus p p = p . Denote the usual absolute value by ‚àû.

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 ‚ÄúinÔ¬Ånite prime‚Äù p = ‚àû.

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 ‚àà Ii , i = 1, . . . , n, then there exists a ‚àà R such that
a ‚â° 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 ‚â° 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 ‚á’ an ‚Ü’ 0;
(2) |a| < 1 ‚á’ an /(1 + an ) ‚Ü’ 0;
(3) |a| > 1 ‚á’ an /(1 + an ) ‚Ü’ 1.
Proof. The Ô¬Årst statement follows from |an | = |a|n . To prove (2), use the triangle
inequality and the observation that 1 + an = 1 ‚à’ (‚à’an ) to get

1 ‚à’ |a|n ‚â¤ |1 + an | ‚â¤ 1 + |a|n ,

so by (1), |1 + an | ‚Ü’ 1. Since |Œ±/Œ≤| = |Œ±|/|Œ≤|, another application of (1) gives the desired
result. To prove (3), write

a‚à’n
an 1
1‚à’ ‚Ü’ 0 by (2). ‚™£
= =
1 + a‚à’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 Ô¬Åeld.
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 ‚â• 1, |c|1 ‚â• 1, |c|2 < 1. If a = c/b, then |a|1 > 1
and |a|2 < 1.
Now if the result holds for n ‚à’ 1, we can choose an element b such that |b|1 > 1, |b|2 <
1, . . . , |b|n‚à’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 ‚â¤ 1. Take ar = cbr , r ‚â• 1. Then |ar |1 > 1, |ar |n < 1, and |ar |i ‚Ü’ 0
as r ‚Ü’ ‚àû for i = 2, . . . , n ‚à’ 1. Thus we can take a = ar for suÔ¬Éciently large r.
Case 2. Suppose |b|n > 1. Take ar = cbr /(1 + br ). By (3) of (9.3.1), |ar |1 ‚Ü’ |c|1 > 1 and
|ar |n ‚Ü’ |c|n < 1 as r ‚Ü’ ‚àû. If 2 ‚â¤ i ‚â¤ n ‚à’ 1, then |b|i < 1, so by (2) of (9.3.1), |ar |i ‚Ü’ 0
as r ‚Ü’ ‚àû. Again we can take a = ar for suÔ¬Éciently large r. ‚™£

9.3.3 Approximation Theorem
Let | |1 , . . . , | |n be nontrivial mutually inequivalent absolute values on the Ô¬Åeld k.
Given arbitrary elements x1 , . . . , xn ‚àà k and any positive real number , there is an
element x ‚àà k such that |x ‚à’ xi |i < for all i = 1, . . . , n.
Proof. By (9.3.2), ‚àÄi ‚àÉyi ‚àà k such that |yi |i > 1 and |yi |j < 1 for j = i. Take zi =
r r
yi /(1 + yi ). Given Œ¥ > 0, it follows from (2) and (3) of (9.3.1) that for r suÔ¬Éciently large,

|zi ‚à’ 1|i < Œ¥ and |zj | < Œ¥, j = i.

Our candidate is

x = x1 z1 + ¬· ¬· ¬· xn zn .
9.4. COMPLETIONS 7

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

n
|x ‚à’ xi |i ‚â¤ Œ¥ |xj |i + Œ¥|xi |i = Œ¥ |xj |i .
j=1
j=i

Choose Œ¥ 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 Ô¬Åeld k.
Fix r with 0 ‚â¤ r ‚â¤ n. Show that there is an element a ‚àà k such that |a|1 > 1, . . . , |a|r > 1
and |a|r+1 , . . . , |a|n < 1.
2. There is a gap in the Ô¬Årst paragraph of the proof of (9.3.2), which can be repaired by
showing that the implication |a|1 < 1 ‚á’ |a|2 < 1 is suÔ¬Écient 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 Ô¬Åeld, 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 Ô¬Åeld.

Let K be a Ô¬Åeld 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
ÀÜ
completion of K with respect to the given absolute value is the Ô¬Åeld K = C/N . We can
ÀÜ
embed K in K via c ‚Ü’ {c, c, . . . } + N .
ÀÜ ÀÜ
We now extend the absolute value on K to K. If (cn )+N ‚àà K, then (|cn |) is a Cauchy
sequence of real numbers, because by the triangle inequality, |cn | ‚à’ |cm | has (ordinary)
absolute value at most |cn ‚à’ cm | ‚Ü’ 0 as n, m ‚Ü’ ‚àû. Thus |cn | converges to a limit, which
we take as the absolute value of (cn ) + N . Since the original absolute value satisÔ¬Åes the
deÔ¬Åning conditions in (9.1.1), so does the extension.
ÀÜ
To simplify the notation, we will denote the element (cn ) + N of K by (cn ). If
cn = c ‚àà K for all n, we will write the element as c.

9.4.2 Theorem
ÀÜ ÀÜ
K is dense in K and K is complete.
8 CHAPTER 9. LOCAL FIELDS

ÀÜ
Proof. Let Œ± = (cn ) ‚àà K, with Œ±n = cn . Then
|Œ± ‚à’ Œ±n | = lim |cm ‚à’ cn | ‚Ü’ 0 as n ‚Ü’ ‚àû,
m‚Ü’‚àû

ÀÜ ÀÜ
proving that K is dense in K. To prove completeness of K, let (Œ±n ) be a Cauchy sequence
ÀÜ
in K. Since K is dense, for every positive integer n there exists cn ‚àà K such that
ÀÜ
|Œ±n ‚à’cn | < 1/n. But then (cn ) is a Cauchy sequence in K, hence in K, and we are assured
ÀÜ
that Œ± = (cn ) is a legal element of K. Moreover, |Œ±n ‚à’ Œ±| ‚Ü’ 0, proving completeness. ‚™£

9.4.3 Uniqueness of the Completion
Suppose K is isomorphic to a dense subÔ¬Åeld of the complete Ô¬Åeld L, where the absolute
ÀÜ
value on L extends that of (the isomorphic copy of) K. If x ‚àà K, then there is a sequence
xn ‚àà K such that xn ‚Ü’ x. But the sequence (xn ) is also Cauchy in L, hence converges
to an element y ‚àà L. If we deÔ¬Åne f (x) = y, then f is a well-deÔ¬Åned homomorphism of
Ô¬Åelds, necessarily injective. If y ‚àà L, then y is the limit of a Cauchy sequence in K, which
ÀÜ ÀÜ
converges to some x ‚àà 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 deÔ¬Åne a local Ô¬Åeld 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 Œ± ‚àà K,
then by (9.1.4) we can write Œ± = uœÄ r with r ‚àà Z, u a unit in A and œÄ an element of
A such that v(œÄ) = 1. Often, œÄ is called a prime element or a uniformizer. Note that
A = {Œ± ‚àà K : v(Œ±) ‚â• 0} and P = {Œ± ‚àà K : v(Œ±) ‚â• 1} = AœÄ.
Let S be a Ô¬Åxed set of representatives of the cosets of A/P . We will show that each
Œ± ‚àà K has a Laurent series expansion
Œ± = a‚à’m œÄ ‚à’m + ¬· ¬· ¬· + a‚à’1 œÄ ‚à’1 + a0 + a1 œÄ + a2 œÄ 2 + ¬· ¬· ¬· , ai ‚àà S,
and if ar is the Ô¬Årst nonzero coeÔ¬Écient (r may be negative), then v(Œ±) = r.
The idea is to expand the unit u in a power series involving only nonnegative powers of
œÄ. For some a0 ‚àà S we have u ‚à’ a0 ‚àà P . But then v(u ‚à’ a0 ) ‚â• 1, hence v((u ‚à’ a0 )/œÄ) ‚â• 0,
so (u ‚à’ a0 )/œÄ ‚àà A. Then for some a1 ‚àà S we have [(u ‚à’ a0 )/œÄ] ‚à’ a1 ‚àà P , in other words,
u ‚à’ a0 ‚à’ a1 œÄ
‚àà P.
œÄ
Repeating the above argument, we get
u ‚à’ a0 ‚à’ a1 œÄ
‚àà A.
œÄ2
Continue inductively to obtain the desired series expansion. Note that by deÔ¬Ånition of S,
the coeÔ¬Écients ai are unique. Thus an expansion of Œ± that begins with a term of degree
r in œÄ corresponds to a representation Œ± = uœÄ r and a valuation v(Œ±) = r. Also, since
|œÄ| < 1, high positive powers of œÄ are small with respect to the given absolute value. The
partial sums sn of the series form a coherent sequence, that is, sn ‚â° sn‚à’1 mod (œÄ)n .
9.4. COMPLETIONS 9

9.4.5 Proposition
Let an be any series of elements in a local Ô¬Åeld. Then the series converges if and only
if an ‚Ü’ 0.
Proof. If the series converges, then an ‚Ü’ 0 by the standard calculus argument, so assume
that an ‚Ü’ 0. Since the absolute value is nonarchimedean, n ‚â¤ m implies that
m
| ai | ‚â¤ max(an , . . . , am ) ‚Ü’ 0 as n ‚Ü’ ‚àû. ‚™£
i=n

The completion of the rationals with respect to the p-adic valuation is called the Ô¬Åeld of
p-adic numbers, denoted by Qp . The valuation ring A = {Œ± : v(Œ±) ‚â• 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 œÄ = p. If in addition, there is no constant term, we get the
valuation ideal P = {Œ± : v(Œ±) ‚â• 1}. The set S of coset representatives may be chosen to
be {0, 1, . . . , p ‚à’ 1}. (Note that if a = b and a ‚â° b mod p, then a ‚à’ b ‚àà 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 ‚Äúcarry‚Äù. 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‚à’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 inÔ¬Ånite 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?
‚àû
6. Show that the p-adic power series for log(1 + x), namely n=1 (‚à’1)n+1 xn /n, converges
in Qp for |x| < 1 and diverges elsewhere. This allows a deÔ¬Ånition of a p-adic logarithm:
logp (x) = log[1 + (x ‚à’ 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
‚àû 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 ‚à’ 1).
8. Show that the p-adic valuation of (pm )! is (pm ‚à’ 1)/(p ‚à’ 1).
‚àû
9. Show that the exponential series n=0 xn /n! converges for |x| < p‚à’1/(p‚à’1) and diverges
elsewhere.
10 CHAPTER 9. LOCAL FIELDS

9.5 Hensel‚Ä™s Lemma
9.5.1 The Setup
Let K be a local Ô¬Åeld 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 Ô¬Åeld k = A/P is called the
residue Ô¬Åeld of A or of K. If a ‚àà A, then the coset a + P ‚àà k will be denoted by a. If f is
a polynomial in A[X], then reduction of the coeÔ¬Écients of f mod P yields a polynomial
f in k[X]. Thus

d d
ai X ‚àà A[X], f (X) = ai X i ‚àà k[X].
i
f (X) =
i=0 i=0

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

9.5.2 Hensel‚Ä™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 ‚à’ r. We will inductively construct
gn , hn ‚àà A[X], n = 1, 2, . . . , such that deg gn = r, deg hn = d ‚à’ r, g n = G, hn = H, and

f (X) ‚à’ gn (X)hn (X) ‚àà P n [X].

Thus the coeÔ¬Écients of f ‚à’ gn hn belong to P n .
The basis step: Let n = 1. Choose monic g1 , h1 ‚àà A[X] such that g 1 = G and h1 = H.
Then deg g1 = r and deg h1 = d ‚à’ r. Since f = g 1 h1 , we have f ‚à’ g1 h1 ‚àà P [X].
The inductive step: Assume that gn and hn have been constructed. Let f (X)‚à’gn (X)hn (X) =
d
i=0 ci X with the ci ‚àà 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 ‚à’ r, and similarly the degree of wi is
at most r. Moreover,

X i ‚à’ vi (X)gn (X) ‚à’ wi (X)hn (X) ‚àà P [X]. (1)

We deÔ¬Åne
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‚Ä™S LEMMA 11

Since the ci belong to P n ‚äÜ 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 ‚à’ r. To check the remaining condition,

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

= (f ‚à’ gn hn ‚à’ ci (X i ‚à’ gn vi ‚à’ hn wi ) ‚à’
ci X i ) + ci cj wi vj .
i i i,j

By the induction hypothesis, the Ô¬Årst 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
Ô¬Ånal term belongs to P 2n [X] ‚äÜ P n+1 [X], completing the induction.
Finishing the proof. By deÔ¬Ånition of gn+1 , we have gn+1 ‚à’ gn ‚àà P n [X], so for any
Ô¬Åxed i, the sequence of coeÔ¬Écients of X i in gn (X) is Cauchy and therefore converges.
To simplify the notation we write gn (X) ‚Ü’ g(X), and similarly hn (X) ‚Ü’ h(X), with
g(X), h(X) ‚àà A[X]. By construction, f ‚à’ gn hn ‚àà P n [X], and we may let n ‚Ü’ ‚àû 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)‚à’1 X d‚à’r respectively, with a ‚àà P . (Note that 1 + a must reduce to 1 mod P .) By
replacing g and h by (1 + a)‚à’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 Œ· ‚àà k. Then f has a simple root a ‚àà A such that a = Œ·.
Proof. We may write f (X) = (X ‚à’ Œ·)H(X) where X ‚à’ Œ· and H(X) are relatively prime
in k[X]. By Hensel‚Ä™s lemma, we may lift the factorization to f (X) = (X ‚à’ a)h(X) with
h ‚àà A[X], a ‚àà A and a = Œ·. If a is a multiple root of f , then Œ· is a multiple root of f ,

Problems For Section 9.5
1. Show that for any prime p, there are p ‚à’ 1 distinct (p ‚à’ 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 eÔ¬Äective procedure for doing this.
3. In Problem 2, suppose that we ‚àö only want to decide if m is a square in Zp , but
not
to Ô¬Ånd 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‚à’1 to get

a‚à’1 = ‚à’(cn‚à’1 + ¬· ¬· ¬· + c1 an‚à’2 + c0 an‚à’1 ) ‚àà 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 Ô¬Ånite-dimensional vector space is injective iÔ¬Ä it is
surjective. Thus if b ‚àà B and b = 0, there is an element c ‚àà A[b] ‚äÜ B such that bc = 1.
Therefore B is a Ô¬Åeld.
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 ‚Ü’ a + Q is a well-deÔ¬Åned injection of A/P into B/Q, since P = Q ‚à© A.
Thus A/P can be viewed as a subring of B/Q.
5. If b + Q ‚àà B/Q, then b satisÔ¬Åes an equation of the form

xn + an‚à’1 xn‚à’1 + ¬· ¬· ¬· + a1 x + a0 = 0, ai ‚àà A.

By Problem 4, b + Q satisÔ¬Åes 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 Ô¬Åeld if and only if B is a Ô¬Åeld, 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 ‚àà M, then by maximality of M, the ideal generated by M and x is R. Thus there
/
exists y ‚àà M and z ‚àà 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 ‚à’1 (g ‚—¦ f ) takes m/s to g(f (m))/s, as does (S ‚à’1 g) ‚—¦ (S ‚à’1 f ). If f is the identity on
M , then S ‚à’1 f is the identity on S ‚à’1 M .
5. By hypothesis, g ‚—¦f = 0, so (S ‚à’1 g)‚—¦(S ‚à’1 f ) = S ‚à’1 (g ‚—¦f ) = S ‚à’1 0 = 0. Thus im S ‚à’1 f ‚äÜ
ker S ‚à’1 g. Conversely, let y ‚àà N, s ‚àà S, with y/s ‚àà ker S ‚à’1 g. Then g(y)/s = 0/1, so for
some t ‚àà S we have tg(y) = g(ty) = 0. Therefore ty ‚àà ker g = im f , so ty = f (x) for some
x ‚àà M . We now have y/s = ty/st = f (x)/st = (S ‚à’1 f )(x/st) ‚àà im S ‚à’1 f .
6. The sequence 0 ‚Ü’ N ‚Ü’ M ‚Ü’ M/N ‚Ü’ 0 is exact, so by Problem 5, the sequence
0 ‚Ü’ NS ‚Ü’ MS ‚Ü’ (M/N )S ‚Ü’ 0 is exact. (If f is one of the maps of the Ô¬Årst sequence,
the corresponding map in the second sequence is S ‚à’1 f .) It follows from the deÔ¬Ånition of
localization of a module that NS ‚â¤ MS , and by exactness of the second sequence we have
(M/N )S ‚àº MS /NS .
=

Section 2.1
1. A basis for E/Q is 1, Œ∏, Œ∏2 , and

Œ∏2 1 = Œ∏2 , Œ∏2 Œ∏ = Œ∏3 = 3Œ∏ ‚à’ 1, Œ∏2 Œ∏2 = Œ∏4 = Œ∏Œ∏3 = 3Œ∏2 ‚à’ Œ∏.

Thus
Ô£® Ô£π
0 ‚à’1 0
Ô£°0 3 ‚à’1Ô£»
2
m(Œ∏ ) =
10 3

and we have T (Œ∏2 ) = 6, N (Œ∏2 ) = 1. Note that if we had already computed the norm of Œ∏
(the matrix of Œ∏ is
Ô£® Ô£π
‚à’1
0 0
m(Œ∏) = Ô£°1 3Ô£»
0
0 1 0

and T (Œ∏) = 0, N (Œ∏) = ‚à’1), it would be easier to calculate N (Œ∏2 ) as [N (Œ∏)]2 = (‚à’1)2 = 1.
2. The cyclotomic polynomial Œ®6 has only two roots, œâ and its complex conjugate œâ. By
(2.1.5),

T (œâ) = œâ + œâ = eiœÄ/3 + e‚à’iœÄ/3 = 2 cos œÄ/3 = 1.

3. We have min(Œ∏, Q) = X 4 ‚à’ 2, min(Œ∏2 , Q) = X 2 ‚à’ 2, min(Œ∏3 , Q) = X 4 ‚à’ 8, and
‚àö
min( 3Œ∏, Q) = X 4 ‚à’ 18. (To compute the last two minimal polynomials, note that
‚àö
(Œ∏3 )4 = (Œ∏4 )3 = 23‚àö 8 and ( 3Œ∏)4 = 18.) Therefore all four traces are 0.
=
4. Suppose that 3 = a +‚àö + cŒ∏2 + dŒ∏3 . Take the trace of both sides to conclude
bŒ∏
of 3 is 0 because its minimal polynomial is X 2 ‚à’ 3.) Thus
that a = 0. (The trace ‚àö
‚àö
3 = bŒ∏ + cŒ∏2 + dŒ∏3 , so 3Œ∏ = bŒ∏2 + cŒ∏3 + 2d. Again take the trace of both sides to get
‚àö ‚àö
d = 0. We now have 3 ‚àö bŒ∏ + cŒ∏2 , so 3Œ∏2 = bŒ∏3 + 2c. The minimal polynomial of
=
‚àö2
3Œ∏ is X 2 ‚à’ 6, because( 3Œ∏2 )2 = 6. Once again taking the trace of both sides, we get
‚àö
c = 0. Finally, 3 = bŒ∏ implies 9 = 2b4 , and we reach a contradiction.
3

Section 2.2
‚àö
1. By the quadratic formula, L = Q( b2 ‚à’ 4c). Since b2 ‚à’ 4c ‚àà Q, we may write
b2 ‚à’ 4c = s/t = st/t2 for relatively prime integers s and t. We also have s = ‚àö 2 and
uy
‚àö
with u and v relatively prime and square-free. Thus L = Q( uv) = Q( d).
2
t = vz , ‚àö ‚àö
‚àö ‚àö
2. If Q( d) = Q( e), then d = a + b ‚àöe for rational numbers a and b. Squaring both
‚àö
sides, we have d = a2 + b2 e + 2ab e, so e is rational, a contradiction (unless a = 0 and
b = 1). ‚àö ‚àö
‚àö ‚àö
3. Any isomorphism of Q( d) and Q( e) must carry d into a+b e for rational numbers
‚àö
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 œân = œâ2n , we have œân ‚àà Q(œâ2n ), so Q(œân ) ‚äÜ Q(œâ2n ). If n is odd, then n+1 = 2r,
2

so

œâ2n = ‚à’œâ2n = ‚à’(œâ2n )r = ‚à’œân .
2r 2 r

Therefore Q(œâ2n ) ‚äÜ Q(œân ).
‚àö ‚àö
5. Q( ‚à’3) = Q(œâ) where œâ = ‚à’ 1 + 1 ‚à’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 ‚à— have the same dimension, the map y ‚Ü’ l(y) is surjective.
8. We have (xi , yj ) = l(yj )(xi ) = fj (xi ) = Œ¥ij . 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 Ô¬Årst statement follows because multiplication of each element of a group G by a
particular element g ‚àà 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 œÉi
on L. Thus P + N and P N belong to the Ô¬Åxed Ô¬Åeld of the Galois group, which is Q.
2. Since the xj are algebraic integers, so are the œÉi (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 ‚à’ N )2 = (P + N )2 ‚à’ 4P N ‚â° (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 ‚àà 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 coeÔ¬Écients in Z. Thus x = A‚à’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. ‚àö
6. No. For example, take L = Q( m), where m is a square-free integer with m ‚â° 1
mod 4. By (2.3.11), the Ô¬Åeld 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 ‚à’ 1 of the
Pi ‚Ä™s. (If so, we can simply replace s by s ‚à’ 1.) It follows that elements of the desired form
exist.
2. Assume that I ‚äÜ P1 and I ‚äÜ P2 . We have a1 ‚àà P1 , a2 ‚àà P1 , so a1 + a2 ‚àà P1 . Similarly,
/ /
a1 ‚àà P2 , a2 ‚àà P2 , so a1 + a2 ‚àà P2 . Thus a1 + a2 ‚àà I ‚äÜ P1 ‚à™ P2 , contradicting a1 , a2 ‚àà I.
/ / /
3. For all i = 1, . . . , s ‚à’ 1 we have ai ‚àà Ps , hence a1 ¬· ¬· ¬· as‚à’1 ‚àà Ps because Ps is prime.
/ /
But as ‚àà Ps , so a cannot be in Ps . Thus a ‚àà I and a ‚àà P1 ‚à™ ¬· ¬· ¬· ‚à™ 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 ‚àà I and y ‚àà J. If z ‚àà I ‚à© J, then z = z1 = zx + zy ‚àà IJ.
The general result follows by induction, along with the computation

R = (I1 + I3 )(I2 + I3 ) ‚äÜ I1 I2 + I3 .

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

R = (I1 I2 + I4 )(I3 + I4 ) ‚äÜ I1 I2 I3 + I4

and so on.
2. We have R = Rr = (P1 + P2 )r ‚äÜ P1 + P2 . Thus P1 and P2 are relatively prime for all
r r

r ‚â• 1. Assuming inductively that P1 and P2 are relatively prime, it follows that
r s

P2 = P2 R = P2 (P1 + P2 ) ‚äÜ P1 + P2
s+1
s s s r r

so

R = P1 + P2 ‚äÜ 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 ‚äÜ R, hence K ‚äÜ r‚à’1 R ‚äÜ K. Thus
K = r‚à’1 R. Since r‚à’2 ‚àà K we have r‚à’2 = r‚à’1 s for some s ‚àà R. But then r‚à’1 = s ‚àà R,
so K ‚äÜ R and consequently K = R.

Section 3.3
‚àö
1. By (2.1.10), the norms are 6,6,4 and 9. Now if x = a + b ‚à’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 ‚àöalgebraic integers of norm 2 or 3. Thus there cannot be a nontrivial factorization
of 1 ¬± ‚à’5, 2 or 3. ‚àö
‚àö
2. If (a + b ‚à’5)(c + d ‚à’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. ‚àö ‚àö
4. This is done as in Problems 1-3, using the factorization 18 = (1 + ‚à’17)(1 ‚à’ ‚à’17) =
5

2 √— 32 . ‚àö
5. By (2.2.6)‚àö (2.3.11), the algebraic integers are of the form a + b ‚à’3, a, b ‚àà Z, or
or
(u/2) + (v/2) ‚à’3 with u and v odd integers. If we require that the norm be 1, we only
get ¬±1 in the Ô¬Årst case. But in the second case, we have u2 + 3v 2 = 4, so u = ¬±1, v = ¬±1.
Thus if œâ = eiœÄ/3 , then the algebraic integers of norm 1 are ¬±1, ¬±œâ, and ¬±œâ 2 .

Section 3.4
‚àö ‚àö ‚àö ‚àö
1. 1 ‚à’ ‚à’5 = 2 ‚à’ (1 + ‚à’5) ‚àà P2 , so (1 + ‚à’5)(1 ‚à’ ‚à’5) = 6 ‚àà P2 . 2

2. Since 2 ‚àà P2 , it follows that 4 ‚àà P2 ,‚àö by Problem 1, 2 = 6 ‚à’ 4 ‚àö P2 .
‚àà2
2
so
‚àö ‚àö ‚àö ‚àö
3. (2, 1 + ‚à’5)(2, 1 + ‚à’5) = (4, 2(1 + ‚à’5), (1 + ‚à’5)2 ), and (1 + ‚à’5)2 = ‚à’4 + 2 ‚à’5.
2
Therefore each of the generators of the ideal P2 is divisible by 2, hence belongs to (2).
Thus P2 ‚äÜ (2).
2
‚àö
4. x2 +5 ‚â° (x+1)(x‚à’1) mod 3, which suggests that (3) = P3 P3 , where P3 = (3, 1+ ‚à’5)
‚àö
and P3 = (3, 1 ‚à’ ‚à’5).‚àö ‚àö
5. P3 P3 = (3, 3(1+ ‚à’5), 3(1‚à’ ‚à’5), 6) ‚äÜ (3), because each generator of P3 P3 is divisible
by 3. But 3 ‚àà P3 ‚à© P3 , hence 9 ‚àà P3 P3 , and therefore 9 ‚à’ 6 = 3 ‚àà P3 P3 . Thus (3) ‚äÜ P3 P3 ,
and the result follows.

Section 4.1
1. The kernel is {a ‚àà A : a/1 ‚àà MS ‚à’1 A} = A ‚à© (MS ‚à’1 A) = M by (1.2.6).
2. By hypothesis, M ‚à© S = ‚à…, so s ‚àà M. By maximality of M we have M + As = A, so
/
y + bs = 1 for some y ‚àà M, b ‚àà A. Thus bs ‚â° 1 mod M.
3. Since 1 ‚à’ bs ‚àà M, (a/s) ‚à’ ab = (a/s)(1 ‚à’ bs) ‚àà MS ‚à’1 A. Therefore (a/s) + MS ‚à’1 A =
ab + MS ‚à’1 A = h(ab).

Section 4.2
1. By the Chinese remainder theorem, B/(p) ‚àº i B/Piei . If p does not ramify, then
=
ei = 1 for all i, so B/(p) is a product of Ô¬Åelds, hence has no nonzero nilpotents. On the
other hand, suppose that e = ei > 1, with P = Pi . Choose x ‚àà P e‚à’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 Œ≤ = i=1 bi œâi with bi ‚àà Z. Then, with T denoting trace,
n n
bi T (œâi œâj ) ‚â° 0 mod p.
T (A(Œ≤œâj )) = T ( bi A(œâi œâj )) =
i=1 i=1

If Œ≤ ‚àà (p), then not all the bi can be 0 mod p, so the determinant of the matrix (T (œâi œâj )),
/
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 Ô¬Åelds 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, œÄi (x)œÄj (y) = 0 for i = j.
6

6. Since Fi /Fp is a Ô¬Ånite extension of a Ô¬Ånite Ô¬Åeld, it is a Galois extension, so all em-
beddings are actually automorphisms. Thus for any z ‚àà 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
‚àö
1. Factoring (2) is covered by case (c1) of (4.3.2), and we have (2) = (2, 1 + ‚à’5)2 .
Factoring (3)‚àö covered by case (a1), and x2 + 5 ‚â° (x + 1)(x ‚à’ 1) mod 3. Therefore
is ‚àö
(3) = (3, 1 + ‚à’5) (3, 1 ‚à’ ‚à’5).
‚àö
2. We have (5) = (5, ‚à’5)2 , as in case (b). To factor (7), note that x2 + 5 factors mod 7
‚àö ‚àö
as (x + 3) (x ‚à’ 3), so (7) = (7, 3 + ‚à’5) (7, 3 ‚à’ ‚à’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 ‚à’ 2 ‚â° x3 ‚à’ 27 = x3 ‚à’ 33 = (x ‚à’ 3)(x2 + 3x + 9) = (x + 2)(x2 + 3x ‚à’ 1).
Thus

(5) = (5, Œ± + 2)(5, Œ±2 + 3Œ± ‚à’ 1)
‚àö
3
where Œ± = 2.

Section 5.3
1. We have r2 = 1 and n = 2, so the bound is (4/œÄ)(2/4) |d| = (2/œÄ) |d|. The
discriminant may be calculated from (2.3.11). We have d = 4m for m = ‚à’1, ‚à’2, and ‚àö
d = m for m = ‚à’3, ‚à’7. The largest |d| is 8, and the corresponding bound is 4 2/œÄ,
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 ‚àö = 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. ‚àö
3. The discriminant is -20 and the Minkowski bound is 2 20/œÄ, which is about 2.85.
Since 2 ramiÔ¬Åes [see (4.3.2), case (c1)], there‚àö only one ideal of norm 2. Thus class
is
number is at most 2. But we know that Q( ‚à’5) is not a UFD, by the exercises for
Section 3.3. Therefore the class number is 2. ‚àö ‚àö
4. The discriminant is 24 and the bound is 24/2 = 6, which is about 2.45. Since 2 ‚àö
ramiÔ¬Åes [see (4.3.2), case (b)],‚àö argument proceeds as in Problem 3. Note that Q( 6) is
the ‚àö ‚àö ‚àö
not a UFD because ‚à’2 = (2+ 6)(2‚à’‚àö 6). Note also that 2+ 6 and 2‚à’ 6 are associates,
‚àö ‚àö ‚àö ‚àö
because (2 + 6)/(2 ‚à’ 6) = ‚à’5 ‚à’ 2 6, which is a unit [(‚à’5 ‚à’ 2 6)(‚à’5 + 2 6) = 1].
‚àö
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
‚àö ‚àö
be seen from the factorization ‚à’2 = [(3 + 17)/2] [(3 ‚à’ 17)/2]. Thus every ideal class
contains a principal ideal, so the ideal class group is trivial.
‚àö ‚àö
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 ramiÔ¬Åes [(4.3.2), case (b)], there is only one ideal of norm 2.
‚àö ‚àö
This ideal is principal, as can be seen from the factorization 2 = (4 + 14)(4 ‚à’ 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) ‚â• 1 and
2r2 ‚â¤ n.
8. By a direct computation, we get a2 and

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

By the binomial theorem, an+1 /an = (œÄ/4)(1 + 2 + positive terms) ‚â• 3œÄ/4. Thus

œÄ2
a3 an
|d| ‚â• a2 ¬·¬·¬· ‚â• (3œÄ/4)n‚à’2 ,
a2 an‚à’1 4

and we can verify by canceling common factors that (œÄ 2 /4)(3œÄ/4)n‚à’2 ‚â• (œÄ/3)(3œÄ/4)n‚à’1 .
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