Tipo V. Área especialmente ruidosa:
NORMAS GENERALES
We give the additive and multiplicative characters on Fq explicitly. We know that F×q is
cyclic; let ξ be a generator.
Theorem 1.9 (Multiplicative characters of Fq): The multiplicative characters of Fq are
given by
ψj(ξn) = e
2πijn q−1
for 0 ≤ j < q − 1.
Proof. By identifying ξ ∈ F×q with 1 ∈ Z/(q − 1)Z, this follows directly from Theorem 1.4.
Describing the additive characters takes slightly more creativity, since it is inconvenient to decompose F+q into cyclic groups.
Theorem 1.10 (Additive characters of Fq): Suppose q = pr with p prime. The additive
characters of Fq are given by
χa(g) = e
2πi
p Tr(ag) (12.5)
for a ∈ Fq where2
Tr(g) = g + gp+ · · · + gpr−1.
Proof. The automorphisms of Fq fixing Fp are generated by the Frobenius automorphism σ
sending g to gp. Since Tr(g) is fixed under this operation, it must be in the ground field F p.
This makes (12.5) well-defined since only the value of Tr(ag) modulo p matters in (12.5). The fact that χais a homomorphism comes directly from the fact that σ is a homomorphism.
Since χ1(ag) = χa(g), if χa = χb then χ1(ag) = χ1(bg) and χ1((a − b)g) = 0. However,
χ1 is not trivial (identically equal to 1) since there are at most pr−1 values of g such that
g + · · · + gpr−1
= 0. Thus a = b. This shows all characters in our list are distinct. Since we have found |G| characters we have found all of them.
Remark 1.11: In general, a n-dimensional complex representation of a group G is a ho- momorphism ρ from G into GLn(C), and the character χ of a representation is defined
by χ(g) = Tr(ρ(g)). This coincides with Definition 1.1 for abelian G, if we just consider 1-dimensional representations, since ρ is multiplication by a constant and χ is just that constant.
The general case of Corollary 1.5 is replaced by the following: every finite group has a number of irreducible characters equal to the number of conjugacy classes. The orthogonality relations hold when we consider just irreducible characters, and with |G| replaced by the size of the centralizer of g in the equation for column orthogonality.
Number Theory, §12.2.
§2 Gauss Sums
To relate additive characters to multiplicative characters, we need to evaluate sums in the form
G(ψ, χ) = X
y∈R×
ψ(y)χ(y). (12.6)
where ψ is a multiplicative character and χ is an additive character.
Suppose we wanted to write an additive character on Fq in terms of multiplicative char-
acters. By row orthogonality, q−11 P
ψ∈ cF×q
ψ(y)ψ(g) equals 1 if y = g and is 0 otherwise. This allows us to introduce multiplicative characters as follows: for y ∈ F×q,
χ(y) = 1 q − 1 X g∈F×q χ(g) X ψ∈ cF×q ψ(y)ψ(g) = 1 q − 1 X ψ∈ cF×q ψ(y)X g∈F×q ψ(g)χ(g) = 1 q − 1 X ψ∈ cF×q G(ψ, χ)ψ(y). (12.7)
The Gauss sums are the coefficients of the expansion of χ in terms of multiplicative charac- ters. The next theorem tells us how to calculate Gauss sums.
Theorem 2.1: Let ψ0 and χ0 denote the trivial multiplicative and additive characters on
Fq, respectively. Then for multiplicative and additive characters ψ and χ on Fq, we have
G(ψ, χ) = q − 1, ψ = ψ0, χ = χ0 −1, ψ = ψ0, χ 6= χ0 0, ψ 6= ψ0, χ = χ0 and |G(ψ, χ)| =√q, ψ 6= ψ0, χ 6= χ0.
If ψ is a nontrivial multiplicative character and χ is a primitive additive character on Z/N Z, then
|G(ψ, χ)| =√N . Proof. The first case is trivial. For the second case,
G(ψ0, χ) = X y∈F×q χ(y) = X y∈Fq χ(y) − 1 = −1
Now we consider the case case when ψ is nontrivial, and either χ 6= χ0 (in the case
R = Fq) or χ is primitive (in the case R = Z/NZ), respectively. We have
|G(ψ, χ)|2 = X g1,g2∈R× ψ(g1)ψ(g2)χ(g1)χ(g2) = X g1,g2∈R× ψ(g1−1g2)χ(g2− g1) = X h∈R× X g1∈R× ψ(h)χ(g1(h − 1)) setting h = g−11 g2 = X h∈R× ψ(h) X g1∈R χ(g1(h − 1)) ! − X y∈R\R× χ(y) = X h∈R× ψ(h) X g1∈R χ(g1(h − 1)) ! by Corollary 1.7 with ψ Now we note the following: when h = 1 all terms in the inner sum are 1, so it equals q or N , respectively. When h 6= 1, consider two cases.
1. R = Fq: As g1 ranges over Fq, g1(h − 1) ranges over Fq.
2. R = Z/NZ: As g1 ranges over Z/NZ, g1(h − 1) ranges over a subgroup H ⊆ Z/NZ,
hitting each element |H|N times. Since χ is primitive, χ|H is nontrivial.
In either case, Corollary 1.7 gives the inner sum to be 0. Hence |G(ψ, χ)|2 evaluates to
ψ(1)q = q or ψ(1)N = N , respectively. We will need the following fact later on.
Proposition 2.2: Let R = Fq or Z/NZ. For a ∈ R× and b ∈ R,
G(ψ, χab) = ψ(a)G(ψ, χb).
Proof. Using the fact that χc(g) = χ1(cg),
G(ψ, χab) = X y∈R× ψ(y)χab(y) = X y∈R× ψ(y)χb(ay) = X y∈R×
ψ(a−1y)χb(y) replacing y → a−1y
= ψ(a)−1 X
y∈R×
ψ(y)χb(y)
Number Theory, §12.3.
§3 Enumerating Solutions
We return to our original problem. Rather than just work with sums of dth powers, we work with diagonal equations
a1yd11 + · · · + anyndn = b (12.8)
where ai ∈ F×q and di ∈ N. First, note that because of the following lemma, we can restrict
to case where di|q − 1.
Lemma 3.1: The multisets {yd|y ∈ F
q} and {ygcd(d,q−1)|y ∈ Fq} are equal.
Proof. Let ξ be a generator for F×q, and write d = k gcd(d, q − 1) where gcd(k, q − 1) = 1.
Then removing the one occurrence of 0 in the two sets, we get {ξjd|0 ≤ j < q − 1} and
{ξj gcd(d,q−1)|0 ≤ j < q − 1}. The lemma follows from the fact that as multisets,
{jd (mod q − 1)|0 ≤ j < q − 1} = {j gcd(d, q − 1) (mod q − 1)|0 ≤ j < q − 1}. Indeed, each multiple of gcd(d, q − 1) appears gcd(d,q−1)q−1 times on both sides.
As (12.8) always has the trivial solution when b = 0, we just need to estimate the number of solutions to (12.8) when b 6= 0.
Theorem 3.2: [?, 6.37] Fix b 6= 0, di|q − 1 and let N be the number of solutions to (12.8)
when b 6= 0 is fixed. Then |N − qn−1| ≤ [(d 1− 1) · · · (dn− 1) − (1 − q− 1 2)M (d 1, . . . , dn)]q n−1 2
where M (d1, . . . , dn) is the number of n-tuples in the set
S := ( (j1, . . . , jn) ∈ Zn|1 ≤ ji ≤ di− 1 and n X i=1 ji di ∈ Z ) .
Note that we would expect N to be close to qn−1, because there are qn possible choices
for (y1, . . . , yn) and q possible values for their sum.
Proof. We use the idea mentioned in the introduction. We have N = 1 q X y1,...,yn∈Fq, χ∈ cF+q χ(a1y1d1 + · · · + anydnn)χ(b) = 1 q X y1,...,yn∈Fq, χ∈ cF+q χ(a1y1d1) · · · χ(anyndn)χ(b)
since by row orthogonality the inner sum is 1 if a1y1d1+ · · · + anyndn = b and 0 otherwise. Note
that χ0 contributes qn to the sum. Taking it out and factoring the remaining terms gives
N = qn−1+ 1 q X χ∈ cF+q,χ6=χ0 χ(b) n Y j=1 X yj∈Fq χ(ajy dj j ) (12.9)
We write the sums of additive characters as sums of multiplicative characters using the following lemma.
Lemma 3.3: Let χ be a nontrivial additive character and λ a multiplicative character of order d dividing q − 1. Then
X y∈Fq χ(ayd) = d−1 X j=1 λ(a)jG(λj, χ).
Proof. Note that λ exists since the group of multiplicative characters is isomorphic to Z/(q − 1)Z by Theorem 1.4. Suppose χ = χc. We write χ as a sum of multiplicative characters
using (12.7), get the Gauss sum to be independent of a by using Proposition 2.2, and take out the exponent as we were hoping to do:
X y∈Fq χ(ayd) = X y∈Fq χac(yd) = 1 + X y∈F×q χac(yd) = 1 + 1 q − 1 X ψ∈ cF×q X y∈F×q G(ψ, χac)ψ(yd) = 1 + 1 q − 1 X ψ∈ cF×q ψ(a)G(ψ, χc) X y∈Fq ψ(y)d (12.10) = 1 + d−1 X j=0 λ(a)jG(λj, χ) (12.11) = d−1 X j=1 λ(a)jG(λj, χ) (12.12)
Note (12.11) follows since by Corollary1.7,P
y∈F×q ψ(y)
d= 0 unless ψdis the trivial character,
which is true iff ψ is a power of λ. In that case, the inner sum in (12.10) is q − 1. In (12.12) we used G(ψ0, χ) = −1 (Theorem 2.1).
Using Lemma 3.3 and letting λj be the multiplicative character with λj(ξt) = e
2πit dj we rewrite (12.9) as N − qn−1 = 1 q X χ∈ cF+q,χ6=χ0 χ(b) n Y j=1 d−1 X k=1 λj(aj)kG(λkj, χ) ! = 1 q X χ∈ cF+q,χ6=χ0 X (k1,...,kn),1≤ki≤di−1 χ(b)λ1 k1 (a1) · · · λn kn (an)G(λ1k1, χ) · · · G(λnkn, χ) = 1 q X c∈F×q X (k1,...,kn),1≤ki≤di−1 χc(b)λ1 k1 (a1) · · · λn kn (an)G(λ1k1, χc) · · · G(λnkn, χc) = 1 q X (k1,...,kn),1≤ki≤di−1 G(λ1k1, χa1) · · · G(λn kn, χ an) X c∈F×q χb(c)λ1 k1 (c) · · · λn kn (c) (12.13)
Number Theory, §12.4. = 1 q X (k1,...,kn),1≤ki≤di−1 G(λ1k1, χa1) · · · G(λn kn, χ an)G(λ1 k1 · · · λn kn , χb) (12.14)
where in (12.13) we used Proposition 2.2 twice, to get λj kj (aj)G(λjkj, χc) = λj kj (c)λj kj (aj)G(λjkj, χ1) = λj kj (c)G(λjkj, χaj). Now we apply Theorem 2.1 to get that |G(λki
i , χai)| = √ q. Note (λ1 k1 · · · λn kn )(ξt) = e(2πi) k1 d1+···+ kn dn t
is the trivial character iff (k1, . . . , kn) ∈ S. Hence |G(λ1 k1
· · · λn kn
, χb)| = 1 if (k1, . . . , kn) ∈ S
and √q otherwise. Using this and the triangle inequality, (12.14) becomes |N − qn−1| ≤ 1 q[q n 2|S| + q n+1 2 ((d 1− 1) · · · (dn− 1) − |S|)],
proving the theorem.
§4 Applications to Waring’s Problem
Now we derive Small’s bound for Waring’s constant g(d, q), the minimum n such that (12.8) has a solution with d1 = · · · = dn= d for all b. By Lemma 3.1, g(d, q) = g(gcd(d, q − 1), q),
so it suffices to consider the case d|q − 1.
First, note that sufficient condition for Waring’s constant to exist is that the set {yd|y ∈
Fq} is not contained in a proper subfield of Fq. Since this set is generated multiplicatively
by ξd, and any subfield is multiplicatively generated by ξ pr −1
pk −1 for some k|d, writing q = pr with p prime we need
pr− 1
pk− 1 - d for every proper divisor k of r. (12.15)
Apply Theorem 3.2 (dropping the term with M (d1, . . . , dn)) to get
N ≥ qn−1− (d − 1)nqn−12 (12.16)
This is positive when
qn−12 > (d − 1)n ⇐⇒ n
2(ln q − 2 ln(d − 1)) > ln q
2 (12.17)
Thus we obtain the following bound for g(d, q):
Theorem 4.1: Suppose d|q − 1 and q > (d − 1)2. Then
g(d, q) ≤ ln q ln q − 2 ln(d − 1) + 1 .
Note that in particular, (12.17) for n = 2 allows us to make the “inverse” statement that if q > (d − 1)4, then the equation yd
1 + y2d= b has a solution for any b ∈ Fq. That is, for any
Part III
Chapter 13
Rings of integers
When we have a field extension L of Q, we would like to define a ring of integers for L, with properties similar to the ring Z ⊆ Q. We will define this ring of integers in a slightly more general context.
§1 Integrality
Definition 1.1: Let A be an integral domain and L a field containing A. An element of x ∈ L is integral over A if it is the zero of a monic polynomial with coefficients in A:
xn+ an−1xn−1+ · · · + a1x + a0 = 0, n ≥ 1, a0, . . . , an−1 ∈ A.
The integral closure of A in L is the set of elements of L integral over A.
Example 1.2: The integral closure of Z in Q is simply Z itself (we see this more generally in Proposition 1.8). Thus, integral closure generalizes the notion of what it means to be an “integer” in other number fields. As we will see in Example4.7, for d squarefree, the integral closure of Q(√d) is Z[√d] when d ≡ 3 (mod 4) and Zh1+
√ d 2
i
when d ≡ 1 (mod 4). Algebra is much nicer in integral extensions—which is why, for instance, we would study Zh1+
√ −3 2
i
rather than just Z[√−3].
Theorem 1.3: Let L be a field containing the ring A. Then the elements of L integral over A form a ring.
Proof. We give two proofs. We need to show that if a, b are algebraic over A then so are a + b and ab.
Proof 1: Let p, q be the minimal polynomials of a, b, let a1, . . . , akbe the conjugates of a and
b1, . . . , bl be the conjugates of b. The coefficients of
Y 1 ≤ i ≤ m 1 ≤ j ≤ n (x − (ai+ bj)), Y 1 ≤ i ≤ m 1 ≤ j ≤ n (x − (aibj))
are symmetric in the ai and symmetric in the bj so by the Fundamental Theorem of Sym-
ai and in the bj, with coefficients in A. By Vieta’s Theorem these are expressible in terms
of the coefficients of p, q, which are in A. Hence these polynomials have coefficients in A. They have a + b, ab as roots, as desired.
Proof 2: We use the following lemma.
Lemma 1.4 (Criterion for integrality): An element α ∈ L is integral over A if and only if there exists a nonzero finitely generated A-submodule of L such that αM ⊆ M . If so, then we can take M = A[α].
Example 1.5: For example, √1
2 fails this criterion over Z—multiplying by it has the effect
of making M “finer.” √2, however, is integral.
In the case A = Z and B = Q, a ∈ Q is integral over Z iff a ∈ Z. Indeed, a ∈ Z satisfies x − a, and if a 6∈ Z, then powers of a contain arbitrarily large denominators so Z[α] is not finitely generated.
Proof. ⇒: If α satisfies a monic polynomial of degree n, then A[α] is generated by 1, α, . . . , αn−1.
⇐: Suppose M is generated by v1, . . . , vn. Then we can find a matrix T with coefficients
in A such that α v1 .. . vn = T v1 .. . vn .
Since v1, . . . , vn 6= 0, αI − T is singular, and det(αI − T ) = 0. This gives a monic polynomial
equation satisfied by α.
Now for α, β ∈ L and let M = A[α] and N = A[β]. Note
1. if M, N are finitely generated by {αi} and {βj}, then M N is finitely generated by
{αiβj}.
2. αβM N ⊆ M N and (α + β)M N ⊆ M N .
Hence αβ and α + β are integral over A by Lemma 1.4 as needed.
For the rest of this chapter, A is an integral domain, K is its fraction field, L is an extension of K, and B is the integral closure of A in L.
L B
K A
(13.1)
Definition 1.6: A is integrally closed or normal if its integral closure in K = Frac(A) is itself.
Proposition 1.7: If L is algebraic over K then every element of L can be written as ab where b ∈ B and a ∈ A. Thus L = Frac(B). In particular, for any extension L/Q, Frac(OL) = L.
Number Theory, §13.1.
Proof. Given α ∈ L, suppose that it satisfies the equation
P (x) := anxn+ an−1xn−1+ · · · + a0 = 0
with a0, . . . , an ∈ K and an 6= 0. Since Frac(A) = K, by multiplying by an element of A as
necessary we may assume a0, . . . , an ∈ A. Then
an−1n P x d
:= xn+ an−1xn−1+ anan−2xn−2+ · · · + an−1n a0.
Hence anα is integral over A, i.e. anα ∈ B. This shows α is in the desired form.
For the last part, take K = Q and A = Z.
For short we call (13.1) the “AKLB” setup if we further assume A is integrally closed in K. In the usual case, A is the integral closure of Z in K. in this case, we write A = OK.1
When F = Q, the algebraic closure of Q, a ∈ Q is called an algebraic number and a ∈ OQ
is an algebraic integer.
Theorem 1.8 (Rational Roots Theorem): A UFD is integrally closed.
Proof. Suppose R is a UFD with field of fractions K. Let x ∈ K be integral over R; suppose x satisfies
xn+ an−1xn−1+ · · · + a0 = 0
where a0, . . . , an−1 ∈ R. Write x = pq where p, q ∈ R are relatively prime. Then multiplying
the above by qn gives
pn+ an−1pn−1q + · · · + a1pqn−1+ a0qn = 0
q(an−1pn−1+ · · · + a0qn−1) = −pn
Thus q | p, possible only if q = 1. This shows x ∈ R.
Note that in the definition of integrality, an element is integral if it is the zero of any monic polynomial in A[x]. However, it suffices to check that its minimal polynomial is in A[x].
Proposition 1.9: Let L be an algebraic extension of K and A be integrally closed. Then α ∈ L is integral over A iff its minimal polynomial f over K has coefficients in A.
Proof. The reverse direction is clear. For the forward direction, note all zeros of f are integral over K since they satisfy the same polynomial equation that α satisfies. The coefficients of f are polynomial expressions in the roots so are integral over A, and hence in A (since they are already in K).
Proposition 1.10 (Finite generation):
1
Later on, when we take K to be an extension of the p-adic field Qp, we will use OKto denote the integral
1. Let A ⊆ B ⊆ C be rings. If B is finitely generated as an A-module and C is finitely generated as a B-module, then C is finitely generated as an A-module.
2. If B is integral over A and finitely generated as an A-algebra, then it is finitely generated as an A-module.
Proof.
1. Take products of generators.
2. Let algebra generators be β1, . . . , βm. Then
A ⊆ A[β1] ⊆ · · · ⊆ A[β1, . . . , βm]
is a chain of integral extensions, so item 2 follows from 1. Combining this proposition with Lemma 1.4 we get the following:
Proposition 1.11 (Transitivity of integrality): Let A ⊆ B ⊆ C be integral domains and K, L, M be their fraction fields.
1. If B is integral over A and C is integral over B, then C is integral over A.
2. Let A0 is the integral closure of A over B and A00 be the integral closure of A0 over C. Let A000 be the integral closure of A in C.
3. The integral closure of A is integrally closed. Proof.
1. For γ ∈ C, let bi be the coefficients of the minimal polynomial of C over B. Then γ
is integral over A[b0, . . . , bm], so by Proposition 1.10, item 2, A[b0, . . . , bm, γ] is finitely
generated over A. Since γA[b0, . . . , bm, γ] ⊆ A[b1, . . . , bm, γ], by Lemma1.4, γ is integral
over A.
2. By item 1 applied to A ⊆ A0 ⊆ A00, A00 is integral over A so A00 ⊆ A000. Conversely, any
element a ∈ A000 is integral over A so a fortiori integral over A00; thus A000 ⊆ A00.
3. Follows from item 2 applied to A = B = C.
§2 Norms and Traces
Let B be a free A-module of rank n. Then any element β ∈ B defines an A-linear map mβ
(or [β]), multiplication by β. It is helpful to think of β as a linear map because then we can apply results from linear algebra.
Definition 2.1: The trace, determinant, and characteristic polynomial of mβ are called the
Number Theory, §13.2.
These are computed by choosing any basis of e1, . . . , en for B over A, and then computing
the action of β on this basis.
Proposition 2.2 (Elementary properties): The following hold (a ∈ A; β, β0 ∈ B): 1. Tr(β + β0) = Tr(β) + Tr(β0)
2. Tr(aβ) = aTr(β) 3. Tr(a) = na
4. Nm(ββ0) = Nm(β) · Nm(β0) 5. Nm(a) = an
Proposition 2.3 (Behavior with respect to field extensions): Suppose L/K is a degree n field extension, M is a finite extension of L, and β ∈ L.
1. (Relationship with roots of minimal polynomial) If f (X) is the minimal polynomial of β over K and β1, . . . , βm are the roots of f (X) = 0 in a Galois closure of K, then
letting r = [L : K(β)] = mn, (a) charL/K(β) = f (X)r
(b) TrL/K(β) = r(β1+ · · · + βm)
(c) NmL/K(β) = (β1· · · βm)r
2. (Relationship with embeddings) Suppose L is separable over K, M is a Galois extension of K, and σ1, . . . , σn are the set of distinct embeddings L → M fixing K. Then
(a) TrL/K(β) = σ1(β) + · · · + σn(β)
(b) NmL/K(β) = σ1(β) · · · σn(β)
In particular, this is true when L = M is a Galois extension of K, and we can think of the σk as simply the elements of G(L/K).
3. (Transitivity of trace and norm) Suppose β ∈ M and M/K is separable.2 Then
(a) TrM/K(β) = TrL/K(TrM/L(β))
(b) NmM/K(β) = NmL/K(NmM/L(β))
4. (Integrality) Assume AKLB. If β ∈ B, then the coefficients of charL/K(β), and hence
TrL/K(β) and NmL/K(β), are integral over A. In particular, if A is integrally closed in
L then they are in A. Proof.
1. If r = 1, i.e. K[β] = L, then by the Cayley-Hamilton Theorem, f (mβ) = 0. Since
f (X) is irreducible, f (X) | charL/K(β). However, these are monic polynomials of the
same degree so they are equal.
In the general case, take a basis xi of K[β] over K and a basis yj of L over K[β]. Then
xiyj form a basis of L over K, and the matrix of mβ with respect to this basis is n
copies of A. This proves (a), which implies the rest of the statements.
2. Let β1, . . . , βm be the conjugates of β. There are m distinct imbeddings K(β) → M ;
they each take β to a different βk. Each of these imbeddings extend to r := [L :
K(β)] = mn imbeddings L → M . Now use item 1.
3. Note that for any finite extensions K ⊆ L ⊆ N with N Galois, an imbedding L ,→ N fixing K can be extended to a K-automorphism on N , and so be considered an element of the set G(N/K)/G(N/L).3
Let N be a Galois extension containing M . By item 2, TrM/K(β) = X σ∈G(N/K)/G(N/M ) σ(β) TrL/K(TrM/L(β)) = TrL/K X σ∈G(N/L)/G(N/M ) σ(β) = X τ ∈G(N/K)/G(N/L) X σ∈G(N/L)/G(N/M ) τ (σ(β))
where in the second sum we take arbitrary representatives τ ∈ G(N/K) and σ ∈ G(N/L). These are equal because for any choice of these representatives,
{σ ∈ G(N/K)/G(N/M )} = {τ σ | τ ∈ G(N/K)/G(N/L), σ ∈ G(N/L)/G(N/M )} when considered in G(N/K)/G(N/M ) (i.e. as imbeddings M ,→ N fixing K). The same is true of the norm.
4. The minimal polynomial of α has coefficients in A, by Proposition 1.9. Hence the result follows from item 1.
§3 Discriminant
Definition 3.1: If B is a ring and free A-module of rank m, and β1, . . . , βm ∈ B, then their
discriminant is
D(β1, . . . , βm) = det[TrB/A(βiβj)]1≤i,j≤m.
3Using the primitive element theorem, write L = K(β). The imbeddings L → N are those taking β to a
conjugate; there are [L : K] imbeddings. But we know G(N/K)/G(N/L) = [L : K], so all of the imbeddings must be extendable. We also use this fact (in addition to a counting argument) in the proof of 2.
Number Theory, §13.3.
Proposition 3.2: If the change of basis matrix from γi to βi is T , then
D(γ1, . . . , γm) = det(T )2· D(β1, . . . , βm).
Proof. Let M1 and M2 be the matrices of the bilinear form
(α, α0) = TrB/A(αα0)
with respect to the bases (β1, . . . , βm) and (γ1, . . . , γm), respectively. Then, using the change
of basis formula for bilinear forms,
D(β1, . . . , βm) = det(M1)
D(γ1, . . . , γm) = det(M2)
M2 = TtM1T
det(M2) = det(T )2· det(M1)
from which the result follows.
Consider the discriminant of an arbitrary basis of B over A. By the above fact, this is well-defined up to multiplication by the square of a unit. The residue in A/(A×)2 is called
the discriminant disc(B/A). The discriminant also refers to the ideal of A this element generates.
Note disc(B/A) can be thought of as the determinant of the matrix of the bilinear form (β, β0) = TrB/A(ββ0).
Proposition 3.3 (Criterion for integral basis): Let A ⊆ B be integral domains and B be a free A-module of rank m with disc(B/A) 6= 0. Then γ1, . . . , γm ∈ B form a basis for B as
an A-module iff
(D(γ1, . . . , γm)) = (disc(B/A))
as ideals.
Proof. Let βi be a basis. If the change of basis matrix from γi to βi is T , then by Proposi-
tion 3.2,
D(γ1, . . . , γm) = det(T )2· D(β1, . . . , βm) = det(T )2disc(B/A)
Now γi is basis iff T is invertible, iff det(T ) is a unit, iff (D(γ1, . . . , γm)) = (disc(B/A)).
Proposition 3.4 (Discriminants and Field Extensions):
1. (Relationship with embeddings) Let L be separable finite over K of degree m, and σ1, . . . , σm be the embeddings of L into a Galois extension M fixing K. Then for any
basis β1, . . . , βm of L over K,
D(β1, . . . , βm) = det(σiβj)2 6= 0.
2. (Nondegeneracy of trace pairing) If B is free of rank m over A (with fraction fields K, L as above), then the pairing
(β, β0) 7→ Tr(ββ0)
Here perfect means that the map a 7→ (b 7→ (a, b)) is an isomorphism L → L∗, and similarly for b 7→ (a 7→ (a, b)). This is equivalent to saying that the bilinear form is nondegenerate. Proof. Use Proposition 2.3(1b), and that σk, det are both multiplicative. Inequality follows
from independence of characters:
Let G be a group, F a field. Then the homomorphisms G → F×are linearly independent.
Thus for K of degree m over Q, we can talk of disc(OK/Z).
A closely related quantity to the discriminant is the different.
Definition 3.5: Assume AKLB, and suppose L/K is a finite separable extension. The codifferent of B with respect to A is
B∗ = {y ∈ L | Tr(xy) ∈ A for all x ∈ B}. The different of B with respect to A is
DB/A = (B∗)−1.
In other words, it is the largest B-submodule satisfying Tr(E) ⊆ A. Note that DB/A = (B∗)−1.
Remark 3.6: We will define the discriminant in general, when B is not necessarily a free A- module, in Chapter21. The relationship between the two definitions is the following: Let p be an ideal in A. Then Apis a principal ideal domain (in fact, a DVR). Let S = A−p; then S−1B
is free over S−1A by the structure theorem for modules. We have (disc(S−1B/S−1A)) = (pAp)m(p) for some m(p). Then
disc(B/A) =Y
p
pm(p).
§4 Integral bases
Proposition 4.1 (Finite generation of integral extensions): Let A be integrally closed and L separable of degree m over K. There are free finite A-submodules M and M0 of L such that M ⊆ B ⊆ M0. B is a finitely generated A-module if A is Noetherian, and free of rank m if A is a PID.4
Proof. Let {β1, . . . , βm} ⊆ B be a basis for L over K. Take a basis βi0 so that Tr(βiβj0) = δij.
Then
Aβ1+ · · · + Aβm ⊆ B ⊆ Aβ10 + · · · + Aβ 0 m.
The second inclusion follows because if β ∈ B, then writing β = P
jbjβ 0
j, we have that
bi = Tr(ββi) ∈ A. (In other words, the βi0 form a basis for the codifferent B
∗, which contains
B.)
Number Theory, §13.4.
If A is Noetherian, then M0 is finitely generated, so its submodule B is finitely generated over A. If A is a PID, then by the Structure Theorem for Modules (over PIDs), M is a direct sum of cyclic modules and a free module. Since it is contained in a free module of rank m and contains a free module of rank m, it must be free of rank m.
The following is immediate:
Theorem 4.2: If K is finite over Q (i.e. a number field), then OK is a finitely generated
Z-module. It is the largest subring that is finitely generated over Z.
Definition 4.3: A basis for OK as a Z-module is called an integral basis.
Proposition 4.4: Suppose K has characteristic 0 (so L separable over K), L = K[β], and f is the minimal polynomial of β over K. Let f (X) =Q(X − βi) in the Galois closure of L.
Then
D(1, β, . . . , βm−1) = Y
1≤i<j≤m
(βi− βj)2 = (−1)m(m−1)/2· NmL/K(f0(β)).
This is called the discriminant of f .5
Proof. Note the βi are conjugates of β; assume β = β1.
By Proposition 3.4, we have D(1, β, . . . , βm−1) = 1 β1 · · · β1m−1 1 β2 · · · β2m−1 .. . ... . .. ... 1 βm · · · βmm−1 2 = Y 1≤i<j≤m (βi− βj)2,
where the last statement follows by evaluating the Vandermonde determinant. For the second equality, note by Proposition 2.3(1c) that
NmL/K(f0(β)) = NmL/K((β1− β2) · · · (β1− βm)) = Y 1≤i≤m Y 1≤j≤m, j6=i (βi− βj) = (−1)m(m−1)2 Y 1≤i<j≤m (βi − βj)2.
Proposition 4.5: If K = Q[α], α ∈ OK, and D(1, α, . . . , αm−1) = disc(O/Z) then {1, α, . . . , αm−1}
is an integral basis.
Proof. Using change-of-basis and the correspondence between index and determinant, D(1, α, . . . , αm−1) = disc(OK/Z) · [OK : Z[α]]2.
Now disc(OK/Z) ∈ Z so [OK : Z[α]] = 1.
Theorem 4.6 (Stickelberger’s Theorem):
1. Let s is the number of complex (nonreal) embeddings K → C. Then sign[disc(K/Q)] = (−1)s/2.
2. disc(OK/Z) ≡ 0 or 1 (mod 4).
Proof. 1. Write K = Q[α] by the Primitive Element Theorem and α1, . . . , αr be the real
conjugates and β1, β1, . . . , βs, βs be the complex conjugates. By Proposition 4.4,
sign(D(1, α, . . . , αm−1)) = sign Y 1≤j≤s (βj − βj)2 ! = Y 1≤j≤s i2 = (−1)s/2.
2. Let α1, . . . , αm be an integral basis. Let P and −N be the sum of the terms in the
expansion of det(σiαj) corresponding to even and odd permutations, respectively:
P = X even π∈Sm m Y i=1 σiαπ(i) N = X odd π∈Sm m Y i=1 σiαπ(i). Then disc(OK/Z) = det(σiαj)2 = (P − N )2 = (P + N )2− 4P N. Take σ ∈ G(Kgal/Q). Note composition by σ permutes the σ
i, say by ν. Then P = X even π∈Sm m Y i=1 σiαν−1π(i) N = X odd π∈Sm m Y i=1 σiαν−1π(i)
and hence σ permutes {P, N }. Hence σ fixes P + N, P N and they are rational. Since they are integral over Z they are integers. Thus the above is congruent to 0 or 1 modulo 4.
Example 4.7 (Quadratic extensions): Any quadratic extension of Q is in the form Q(√m) for some squarefree integer m. We find the ring of integers of Q(√m). Consider two cases.
Number Theory, §13.4.
1. m ≡ 2, 3 (mod 4): The minimal polynomial of √m is X2− m, so
disc(1,√m) = (√m − (−√m))2 = 4m. Note disc(1,
√ m)
disc(Q(√m)/Q) must be a square by Proposition 3.2 so disc(Q(
√
m)/Q) equals m or 4m. However, by Stickelberger’s Theorem, disc(Q(√m)/Q) ≡ 0, 1 (mod 4). Hence disc(Q(√m)/Q) 6= m and disc(Q(√m)/Q) = 4m. By Proposition 3.3, 1,√m is an integral basis.
2. m ≡ 1 (mod 4): Note 1+
√ m
2 is integral with minimal polynomial X
2− X − m−1 4 , so disc 1,1 + √ m 2 = 1 + √ m 2 − 1 −√m 2 2 = m.
Since m is squarefree, disc(Q(√m)/Q) = m and Proposition 3.3 says 1,1+