Unknown
≤ K ≤ E, where E/F is finite and separable, then
TE/F = TK/F
◦ TE/K and NE/F = NK/F
◦ NE/K.
Proof. Let σ1, . . . , σn be the distinct F-embeddings of K into L, and let τ1, . . . , τm be
the distinct K-embeddings of E into L, where L is the normal closure of E over F. Then
L/F is Galois, and each mapping σi and τj extends to an automorphism of L. Therefore
it makes sense to allow the mappings to be composed. By (2.1.6),
TK/F (TE/K(x)) =
n
i=1
σi(
m
j=1
τj(x)) =
n
i=1
m
j=1
σi(τj(x)).
Each σiτj = σi ◦ τj is an F-embedding of E into L, and the number of mappings is given
by mn = [E : K][K : F] = [E : F]. Furthermore, the σiτj are distinct when restricted
to E. For if σiτj = σkτl on E, then σi = σk on K, because τj and τk coincide with
the identity on K. Thus i = k, so that τj = τl on E. But then j = l. By (2.1.6),
TK/F (TE/K(x)) = TE/F (x). The norm is handled the same way, with sums replaced by
products. ♣
Here is another application of (2.1.6).
2.1.8 Proposition
If E/F is a finite separable extension, then the trace TE/F (x)cannot be 0 for every x ∈ E.
Proof. If T(x)= 0 for all x, then by (2.1.6),
n
i=1 σi(x)= 0 for all x. This contradicts
Dedekind’s lemma on linear independence of monomorphisms. ♣
2.1.9 Remark
A statement equivalent to (2.1.8)is that if E/F is finite and separable, then the trace
form, that is, the bilinear form (x, y) → TE/F (xy), is nondegenerate. In other words, if
T(xy)= 0 for all y, then x = 0. Going from (2.1.9)to (2.1.8)is immediate, so assume
T(xy)= 0 for all y, with x
= 0. Let x0 be an element with nonzero trace, as provided by
(2.1.8). Choose y so that xy = x0 to produce a contradiction.
2.1.10 Example
Let x = a + b
√
m be an element of the quadratic extension Q(
√
m)/Q, where m is a
square-free integer. We will find the trace and norm of x.
The Galois group of the extension consists of the identity and the automorphism
σ(a + b
√
m) = a − b
√
m. Thus by (2.1.6),
T(x) = x + σ(x) = 2a, and N(x) = xσ(x) = a2 − mb2.
2.2. THE BASIC SETUP FOR ALGEBRAIC NUMBER THEORY 5
Problems For Section 2.1
1. If E = Q(θ)where θ is a root of the irreducible cubic X3 −3X +1, find the norm and
trace of θ2.
2. Find the trace of the primitive 6th root of unity ω in the cyclotomic extension Q6 =
Q(ω).
3. Let θ be a root of X4 − 2 over Q. Find the trace over Q of θ, θ2, θ3 and
√
3θ.
4. Continuing Problem 3, show that
√
3 cannot belong to Q[θ].
2.2 The Basic Setup For Algebraic Number Theory
2.2.1 Assumptions
Let A be an integral domain with fraction field K, and let L be a finite separable extension
of K. Let B be the set of elements of L that are integral over A, that is, B is the integral
closure of A in L. The diagram below summarizes the information.
L B
K A
In the most important special case, A = Z, K = Q, L is a number field, that is, a finite
(necessarily separable)extension of Q, and B is the ring of algebraic integers of L. From
now on, we will refer to (2.2.1)as the AKLB setup.
2.2.2 Proposition
If x ∈ B, then the coefficients of charL/K(x)and min(x,K)are integral over A. In
particular, TL/K(x)and NL/K(x)are integral over A, by (2.1.3). If A is integrally closed,
then the coefficients belong to A.
Proof. The coefficients of min(x,K)are sums of products of the roots xi, so by (2.1.4),
it suffices to show that the xi are integral over A. Each xi is a conjugate of x over K, so
there is a K-isomorphism τi : K(x) → K(xi)suc h that τi(x) = xi. If we apply τi to an
equation of integral dependence for x over A, we get an equation of integral dependence
for xi over A. Since the coefficients belong to K [see (2.1.1)], they must belong to A if A
is integrally closed. ♣
2.2.3 Corollary
Assume A integrally closed, and let x ∈ L. Then x is integral over A, that is, x ∈ B, if
and only if the minimal polynomial of x over K has coefficients in A.
Proof. If min(x,K) ∈ A[X], then x is integral over A by definition of integrality. (See
(1.1.1); note also that A need not be integrally closed for this implication.)The converse
follows from (2.2.2). ♣
polynomial of a over Q is X − a, so by (2.2.3), a ∈ Z. ♣
2.2.5Quadratic Extensions of the Rationals
We will determine the algebraic integers of L = Q(
√
m), where m is a square-free integer
(a product of distinct primes). The restriction on m involves no loss of generality, for
example, Q(
√
12)= Q(
√
3).
A remark on notation: To make sure there is no confusion between algebraic integers
and ordinary integers, we will often use the term “rational integer” for a member of Z.
Now by direct verification or by (2.1.10)and (2.1.3), the minimal polynomial over Q
of the element a+b
√
m ∈ L (with a, b ∈ Q)is X2 −2aX +a2 −mb2. By (2.2.3), a+b
√
m
is an algebraic integer if and only if 2a and a2 − mb2 are rational integers. In this case,
we also have 2b ∈ Z. For we have (2a)2 − m(2b)2 = 4(a2 − mb2) ∈ Z, so m(2b)2 ∈ Z. If
2b is not a rational integer, its denominator would include a prime factor p, which would
appear as p2 in the denominator of (2b)2. Multiplication of (2b)2 by m cannot cancel the
p2 because m is square-free, and the result follows.
Here is a more convenient way to characterize the algebraic integers of a quadratic
field.
2.2.6 Proposition
The set B of algebraic integers of Q(
√
m), m square-free, can be described as follows.
(i)If m
≡ 1 mod 4, then B consists of all a + b
√
m, a, b ∈ Z;
(ii)If m ≡ 1 mod 4, then B consists of all (u/2)+ (v/2)
√
m, u, v ∈ Z, where u and v
have the same parity (both even or both odd).
[Note that since m is square-free, it is not divisible by 4, so the condition in (i)can be
written as m ≡ 2 or 3 mod 4.]
Proof. By (2.2.5), the algebraic integers are of the form (u/2)+ (v/2)
√
m, where u, v ∈ Z
and (u2 − mv2)/4 ∈ Z, that is, u2 − mv2 ≡ 0 mod 4. It follows that u and v have the
same parity. [The square of an even number is congruent to 0 mod 4, and the square of
an odd number is congruent to 1 mod 4.] Moreover, the “both odd” case can only occur
when m ≡ 1 mod 4. The “both even” case is equivalent to u/2, v/2 ∈ Z, and we have
the desired result. ♣
When we introduce integral bases in the next section, we will have an even more
convenient way to describe the algebraic integers of Q(
√
m).
If [L : K] = n, then a basis for L/K consists of n elements of L that are linearly
independent over K. In fact we can assemble a basis using only elements of B.
2.2.7 Proposition
There is a basis for L/K consisting entirely of elements of B.
Proof. Let x1, . . . , xn be a basis for L over K. Each xi is algebraic over K, and therefore
satisfies a polynomial equation of the form
amxmi
+ · · · + a1xi + a0 = 0
with am
= 0 and the ai ∈ A. (Initially, we only have ai ∈ K, but then ai is the ratio of
two elements of A, and we can form a common denominator.)Multiply the equation by
am−1
m to obtain an equation of integral dependence for yi = amxi over A. The yi form
the desired basis. ♣
2.2.8 Corollary of the Proof
If x ∈ L, then there is a nonzero element a ∈ A and an element y ∈ B such that x = y/a.
In particular, L is the fraction field of B.
Proof. In the proof of (2.2.7), take xi = x, am = a, and yi = y. ♣
In Section 2.3, we will need a standard result from linear algebra. We state the result
now, and an outline of the proof is given in the exercises.
2.2.9 Theorem
Suppose we have a nondegenerate symmetric bilinear form on an n-dimensional vector
space V , written for convenience using inner product notation (x, y) . If x1, . . . , xn is any
basis for V , then there is a basis y1, . . . , yn for V , called the dual basis referred to V , such
that
(xi, yj) = δij =
1, i= j
0, i
= j.
Problems For Section 2.2
1. Let L = Q(α), where α is a root of the irreducible quadratic X2+bX+c, with b, c ∈ Q.
Show that L = Q(
√
m)for some square-free integer m. Thus the analysis of this section
covers all possible quadratic extensions of Q.
2. Show that the quadratic extensions Q(
√
m), m square-free, are all distinct.
3. Continuing Problem 2, show that in fact no two distinct quadratic extensions of Q are
Q-isomorphic.
Cyclotomic fields do not exhibit the same behavior. Let ωn = ei2π/n, a primitive nth
root of unity. By a direct computation, we have ω2
2n = ωn and
−ωn+1
2n = −eiπ(n+1)/n = eiπeiπeiπ/n = ω2n.
4. Show that if n is odd, then Q(ωn) = Q(ω2n).
5. Give an example of a quadratic extension of Q that is also a cyclotomic extension.
We now indicate how to prove (2.2.9).
6. For any y in the finite-dimensional vector space V , the mapping x → (x, y)is a linear
form l(y)on V , that is, a linear map from V to the field of scalars. Show that the linear
transformation y → l(y)from V to V ∗ (the space of all linear forms on V )is injective.
7. Show that any linear form on V is l(y)for some y.
8. Let f1, . . . , fn be the dual basis corresponding to x1, . . . , xn. Thus each fj belongs to
V ∗ (not V )and fj(xi) = δij. If fj = l(yj ), show that y1, . . . , yn is the required dual basis
referred to V .
9. Show that xi =
n
j=1(xi, xj)yj . Thus in order to compute the dual basis referred to
V , we must invert the matrix ((xi, xj )).
2.3 The Discriminant
The discriminant of a polynomial is familiar from basic algebra, and there is also a discriminant
in algebraic number theory. The two concepts are unrelated at first glance, but
there is a connection between them. We assume the basic AKLB setup of (2.2.1), with
n = [L : K].
2.3.1 Definition
If n = [L : K], the discriminant of the n-tuple x = (x1, . . . , xn)of elements of L is
D(x)= det(TL/K(xixj)).
Thus we form a matrix whose ij entry is the trace of xixj , and take the determinant of
the matrix; by (2.1.1), D(x) ∈ K. If the xi ∈ B, then by (2.2.2), D(x)is integral over A,
that is, D(x) ∈ B. Thus if A is integrally closed and the xi ∈ B, then D(x)b elongs to A.
The discriminant behaves quite reasonably under linear transformation.
2.3.2 Lemma
If y = Cx, where C is an n by n matrix over K and x and y are n-tuples written as
column vectors, then D(y)= (detC)2D(x).
Proof. The trace of yrys is
T(
i,j
cricsjxixj) =
i,j
criT(xixj)csj
hence
(T(yrys)) = C(T(xixj))C
where C is the transpose of C. The result follows upon taking determinants. ♣
Here is an alternative expression for the discriminant.
2.3.3 Lemma
Let σ1, . . . , σn be the distinct K-embeddings of L into an algebraic closure of L, as in
(2.1.6). Then D(x)= [det(σi(xj ))]2.
determinants. ♣
The discriminant “discriminates” between bases and non-bases, as follows.
2.3.4 Proposition
If x = (x1, . . . , xn), then the xi form a basis for L over K if and only if D(x)
= 0.
Proof. If
j cjxj = 0, with the cj ∈ K and not all 0, then
j cjσi(xj)= 0 for all i, so
the columns of the matrix H = (σi(xj)) are linearly dependent. Thus linear dependence
of the xi implies that D(x)= 0. Conversely, assume that the xi are linearly independent,
and therefore a basis because n = [L : K]. If D(x)= 0, then the rows of H are linearly
dependent, so for some ci ∈ K, not all 0, we have
i ciσi(xj)= 0 for all j. Since the xj
form a basis, it follows that
i ciσi(u)= 0 for all u ∈ L, so the monomorphisms σi are
linearly dependent. This contradicts Dedekind’s lemma. ♣
We now make the connection between the discriminant defined above and the discriminant
of a polynomial.
2.3.5Prop osition
Assume that L = K(x), and let f be the minimal polynomial of x over K. Let D be the
discriminant of the basis 1, x, x2, . . . ,xn−1 over K, and let x1, . . . , xn be the roots of f
in a splitting field, with x1 = x. Then D coincides with
i<j(xi −xj)2, the discriminant
of the polynomial f.
Proof. Let σi be the K-embedding that takes x to xi, i = 1, . . . , n. Then σi(xj) =
xj
i , 0 ≤ j ≤ n − 1. By (2.3.3), D is the square of the determinant of the matrix
V =
1 x1 x21
· · · xn−1
1
1 x2 x22
· · · xn−1
2
...
...
...
. . .
...
1 xn x2
n
· · · xn−1
n
.
But det V is a Vandermonde determinant, whose value is
i<j(xj − xi), and the result
follows. ♣
Proposition 2.3.5 yields a formula that is often useful in computing the discriminant.
2.3.6 Corollary
Under the hypothesis of (2.3.5),
D = (−1)(n2
)NL/K(f
(x))
where f is the derivative of f.
Proof. Let c = (−1)(n2
). By (2.3.5),
D =
i<j
(xi − xj)2 = c
i =j
(xi − xj) = c
i
j =i
(xi − xj).
But f(X) = (X − x1) · · · (X − xn) , so
f
(xi) =
k
j =k
(X − xj)
with X replaced by xi. When the substitution X = xi is carried out, only the k = i term
is nonzero, hence
f
(xi) =
j =i
(xi − xj).
Consequently,
D = c
n
i=1
f
(xi).
But
f
(xi) = f
(σi(x)) = σi(f
(x))
so by (2.1.6),
D = cNL/K(f
(x)). ♣
2.3.7 Definitions and Comments
In the AKLB setup with [L : K] = n, suppose that B turns out to be a free A-module
of rank n. A basis for this module is said to be an integral basis of B (or of L) . An
integral basis is, in particular, a basis for L over K, because linear independence over A
is equivalent to linear independence over the fraction field K. We will see shortly that an
integral basis always exists when L is a number field. In this case, the discriminant is the
same for all integral bases. It is called the field discriminant.
2.3.8 Theorem
If A is integrally closed, then B is a submodule of a free A-module of rank n. If A is a
PID, then B itself is free of rank n over A, so B has an integral basis.
Proof. By (2.1.9), the trace is a nondegenerate symmetric bilinear form defined on the
n-dimensional vector space L over K. By (2.2.2), the trace of any element of B belongs to
A. Now let x1, . . . , xn be any basis for L over K consisting of elements of B [see (2.2.7)],
and let y1, . . . , yn be the dual basis referred to L [see (2.2.9)]. If z ∈ B, then we can write
z =
j=1 ajyj with the aj ∈ K. We know that the trace of xiz belongs to A, and we
also have
T(xiz) = T(
n
j=1
ajxiyj) =
n
j=1
ajT(xiyj) =
n
j=1
ajδij = ai.
Thus each ai belongs to A, so that B is an A-submodule of the free A-module ⊕nj
=1Ayj .
Moreover, B contains the free A-module ⊕nj
=1Axj . Consequently, if A is a principal ideal
domain, then B is free over A of rank exactly n. ♣
2.3.9 Corollary
The set B of algebraic integers in any number field L is a free Z-module of rank n = [L : Q].
Therefore B has an integral basis. The discriminant is the same for every integral basis.
Proof. Take A = Z in (2.3.8)to show that B has an integral basis. The transformation
matrix C between two integral bases [see (2.3.2)] is invertible, and both C and C−1 have
rational integer coefficients. Take determinants in the equation CC−1 = I to conclude
that detC is a unit in Z. Therefore detC = ±1, so by (2.3.2), all integral bases have the
same discriminant. ♣
2.3.10 Remark
An invertible matrix C with coefficients in Z is said to be unimodular if C−1 also has
coefficients in Z. We have just seen that a unimodular matrix has determinant ±1.
Conversely, a matrix over Z with determinant ±1 is unimodular, by Cramer’s rule.
2.3.11 Theorem
Let B be the algebraic integers of Q(
√
m), where m is a square-free integer.
(i)If m
≡ 1 mod 4, then 1 and
√
m form an integral basis, and the field discriminant is
d = 4m.
(ii)If m ≡ 1 mod 4, then 1 and (1+
√
m)/2 form an integral basis, and the field discriminant
is d = m.
Proof.
(i)By (2.2.6), 1 and
√
m span B over Z, and they are linearly independent because
√
m
is irrational. By (2.1.10), the trace of a+b
√
m is 2a, so by (2.3.1), the field discriminant
is
2 0
0 2m
= 4m.
(ii)By (2.2.6), 1 and (1 +
√
m)/2 are algebraic integers. To show that they span B,
consider (u + v
√
m)/2, where u and v have the same parity. Then
1
2
(u + v
√
m) = (u − v
2
)(1) + v [
1
2
(1 +
√
m)]
with (u − v)/2 and v in Z. To prove linear independence, assume that a, b ∈ Z and
a + b [
1
2
(1 +
√
m) ] = 0.
Then 2a + b + b
√
m = 0, which forces a = b = 0. Finally, by (2.1.10), (2.3.1), and the
computation [(1 +
√
m)/2]2 = (1+m)/4 +
√
m/2, the field discriminant is
2 1
1 (1+m)/2
= m. ♣
Problems For Section 2.3
Problems 1-3 outline the proof of Stickelberger’s theorem, which states that the discriminant
of any n-tuple in a number field is congruent to 0 or 1 mod 4.
1. Let x1, . . . ,xn be arbitrary algebraic integers in a number field, and consider the
determinant of the matrix (σi(xj )), as in (2.3.3). The direct expansion of the determinant
has n! terms. let P be the sum of those terms in the expansion that have plus signs in front
of them, and N the sum of those terms prefixed by minus signs. Thus the discriminant D
of (x1, . . . , xn)is (P − N)2. Show that P + N and PN are fixed by each σi, and deduce
that P + N and PN are rational numbers.
2. Show that P + N and PN are rational integers.
3. Show that D ≡ 0 or 1 mod 4.
4. Let L be a number field of degree n over Q, and let y1, . . . , yn be a basis for L over
Q consisting of algebraic integers. Let x1, . . . ,xn be an integral basis. Show that if
the discriminant D(y1, . . . , yn)is square-free, then each xi can be expressed as a linear
combination of the yj with integer coefficients.
5. Continuing Problem 4, show that if D(y1, . . . , yn)is square-free, then y1, . . . , yn is an
integral basis.
6. Is the converse of the result of problem 5 true?
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