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Algebraic Number Theory – Lecture 4
Sandro Bettin
“What the world needs is more geniuses with humility, there are so few of us left.”
- Discriminants
Definition 1. Let K be a number field and α = {α 1 ,... , αn} be a basis for K. The discriminant of α is ∆[α] = (det(σi(αj )))^2
where σi are the embeddings K ↪→ C.
Remark. ∆[α] ∈ Q since
σρ(∆[α]) = (det(σρσi(αj )))^2 = (± det(σi(αj )))^2 = ∆[α].
If α ⊂ OK then ∆[α] ∈ OK ∩ Q = Z.
Theorem S1. There exists an integral basis α = {α 1... , αn} with n = [K : Q].
Sketch proof. Take a Q-basis α ⊂ OK of K with ∆[α] minimal. Then suppose that α is not an integral basis, so there exists ω ∈ OK with, say,
ω = θ 1 α 1 +... + θnαn
where θ 1 6 ∈ Z, i.e. θ 1 = θ + r for some 0 < r < 1. Then α′^ = {ω − θα 1 , α 2 ,... , αn} is given by
α′^ =
θ 1 − θ θ 2 θ 3 · · · θn 0 1 0 · · · 0 0 0 1 0 .. .
α.
So ∆[α′] = r^2 ∆[α] < ∆[α]
contradicting the minimality of ∆[α].
Corollary. All integral bases of a given number field have the same discriminant up to sign, say |∆|.
Corollary. If α is a Q-basis of K and α ⊂ OK , and if ∆[α] is square free then α is an integral basis.
1
Proof. If α′^ is an integral basis then α = (ci,j )α′^ for some matrix (ci,j ) ∈ Zn,n. So ∆[α] = (det(ci,j ))^2 ∆, whence det(ci,j ) = ± 1 as ∆[α] is square free. So (ci,j ) ∈ Gln(Z) and so α′^ = (ci,j )−^1 α, thus α is an integral basis as well.
Theorem S2. If K = Q(θ) is a number field of degree n, then
∆[1, θ,... , θn−^1 ] = (−1)n(n−1)/^2 N (Dp(θ))
where p is the minimal polynomial of θ and D is the formal derivative.
- Quadratic fields
K is a quadratic field if [K : Q] = 2. If K = Q(θ) is a quadratic field then
θ =
−a ±
a^2 − 4 b 2
i.e. θ is a root of t^2 + at + b. Writing
a^2 − 4 b = r
d with d ∈ Z square free, then clearly K = Q(
d).
Theorem S3. Let d ∈ Z be square free and K = Q(
d). Then
- if d 6 ≡ 1 (mod 4) then OK = Z[
d] and ∆ = 4d;
- if d ≡ 1 (mod 4) then OK = Z[(1 +
d)/2] and ∆ = d.
Proof. Let α ∈ K, so α =
a + b
d c
with hcf(a, b, c) = 1. Claim that α ∈ OK if and
only if (^) (
t − a + b
d c
t − a − b
d c
∈ Z[t].
So if and only if
2 a c
(1) ∈ Z, and
a^2 − b^2 d c^2
(2) ∈ Z.
Let q = hcf(a, c). From (2), q^2 | a^2 − b^2 d. But q^2 | a^2 and d is square free, so q | b. But hcf (a, b, c) = 1 so q = 1. From (1), then, c = 1 or 2. If c = 1 then α ∈ OK anyway.
If c = 2 then a^2 − b^2 d ≡ 0 (mod 4) by (2). But a is odd as q = 1 and so b must be odd too, whence a^2 ≡ b^2 ≡ 1 (mod 4). Hence 1 − d ≡ 0 (mod 4).
- Factorisation into irreducibles
Definition 2. Given a ring R,
(1) we say x ∈ R is irreducible if and only if x = mn implies m or n is a unit; (2) we say p ∈ R is prime if and only if p is not a unit or zero, and p | mn implies p | m or p | n.
Every prime is irreducible, but not necessarily vice versa. We often denote the units of a ring R by U (R) or, if the ring is clear from the context, then just U.
Definition 3. An integral domain D is called noetherian if one of the following holds:
(1) every ideal in D is finitely generated; (2) (the ascending chain condition) if I 0 ⊆ I 1 ⊆ I 2 ⊆... are all ideals then there exists N ∈ N such that In = IN for every n > N ; (3) (maximality condition) every nonempty set of ideals of D has a maximal element by inclusion.
Theorem S6. If D is noetherian then every nonzero element can be written as a product of irreducible elements.
Proof. Exercise. Hints: proceed by contradiction and let
X = {x ∈ D\U | x cannot be expressed as a product of irreducible elements} ⊂ D.
Consider the ideals (x) with x ∈ X, and choose the maximal one – which we can do since D is noetherian. Note that x is not irreducible since it is in X so write x = yz for non-units y and z and consider the ideals (y) and (z). Show these aren’t in X and hence derive a contradiction to x ∈ X.
Theorem S7. For any number field K the ring OK is noetherian.
Proof. Let I ⊆ OK be an ideal. As an additive group OK is free abelian of rank n = [K : Q], so the subgroup (I, +) is free abelian of rank s 6 n. If {x 1 ,... , xs} is a Z-basis for (I, +) then I = (x 1 ,... , xs), so I is finitely generated and hence OK is noetherian.
Corollary. Factorisation into irreducibles is possible in OK for any number field K.
