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Algebraic Number Theory – Lecture 10
Sandro Bettin
“You know that I write slowly. This is chiefly because I am never satisfied until I have said as much as possible in a few words, and writing briefly takes far more time than writing at length.”
The goal of the lecture is to prove Gauss’s reciprocity law, using methods related to what we have done in the past. Firstly, let’s introduce the Legendre symbol.
Definition. For every integer a coprime to p, the Legendre symbol
a p
is defined by ( a p
1 if x^2 ≡ a (mod p) has a solution −1 otherwise.
Elementary properties of group theory show us that the Legendre symbol is multiplicative and that one has (^) ( a p
≡ a
p− 1 (^2) (mod p)
and this gives (^) ( − 1 p
p− 2 1 ,
p
p^28 − 1
for any odd prime p. Gauss’s reciprocity law gives us what’s left for a full understanding of the Legendre symbol.
Theorem (Gauss’s reciprocity law). For two distinct odd prime numbers p and q, the following identity holds: (^) ( p q
q p
= (−1)(p−1)(q−1)/^4.
To prove this we need to state some other new results from ramification theory.
Proposition 1. Let L/K be a Galois extension, B an ideal of OL with ramification index e and inertia degree f , and ZB the decomposition field of B over K. Let BZ = B ∩ ZB be the prime ideal of ZB below B. Then
i) BZ is nonsplit in L, i.e. B is the only prime ideal of L above BZ. ii) B over ZB has ramification index e and inertia degree f. iii) The ramification index and the inertia degree of BZ over K both equal 1.
Definition. Let K ⊆ L = K(θ) be number fields with θ ∈ OL and let p(x) ∈ OK [x] be the minimal polynomial of θ. The conductor of the ring OK [θ] is the biggest ideal F of OL which is contained in OK [θ], i.e. F = {α ∈ OL | αOL ⊆ OK [θ]}. 1
2
Proposition 2. Let p ⊆ OK be a prime ideal, coprime with the conductor F of OK [θ], and let
p(x) = p 1 (x)e^1 · · · pr (x)er
be the factorization of the polynomial p(x) = p(x) (mod p) into irreducibles pi(x) = pi(x) (mod p) over the residue class field Ok/p, with all pi ∈ OK [x]. Then
Bi = pOL + pi(θ)OL, i = 1,... , r,
are the different prime ideals of OL above p. The inertia degree fi of Bi is the degree of pi(x), and one has p = Be 11 · · · Be r r.
Proof. See Neukirch.
Corollary 2.1. For squarefree a and (p, 2 a) = 1, we have that
a p
= 1 if and only if p is totally
split in Q(
a).
Proof.
a p
= 1 signifies that
x^2 − a ≡ (x − α)(x + α) (mod p)
for some α ∈ Z. Since OQ(√a) ⊆ Z[
√a 2 ], the conductor of^ Z[
a] is a divisor of 2 and so we can apply the previous proposition.
Now we need to now something about the factorization of primes in the cyclotomic fields.
Proposition 3. Let ζn be a primitive nth root of unity with n =
p p νp (^) the prime factorization
of n. Moreover, let fp be the smallest positive integer such that
pfp^ ≡ 1 (mod n/pνp^ ).
Then one has in Q(ζn) the factorization
p = (p 1 · · · pr )φ(p
νp (^) ) ,
where p 1 ,... , pr are distinct prime ideals, all of degree fp.
Proof. See Neukirch (it’s a consequence of Proposition 2).
Proposition 4. Let q and p be odd primes, q∗^ = (−1) q− 2 1 q and ζq a primitive qth root of unity. Then p is totally split in Q(
q∗) if and only if p splits in Q(ζq ) into an even number of prime ideals.
Proof. A little computation shows that q∗^ = τ 2 ,
where τ is the Gauss sum
τ =
a∈F (^) q∗
a q
ζq a.
Therefore Q(
q∗) ⊆ Q(ζq ). Let’s assume p is totally split in Q(
q∗), say p = p 1 p 2. Since all the extensions Q ⊆ Q(
q∗) ⊆ Q(ζq ) are Galois extension, the number of ideals above p 1 is equal to the number of ideals above p 2 and so p splits in Q(ζq ) into an even number of prime ideals. By the previous proposition these ideals must be distinct since for n = q we have νp = 0.
Conversely, assume pOQ(ζq ) = p 1 · · · pr ,
