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A Brief Introduction to Classical and Adelic Algebraic ... - William Stein

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68CHAPTER 10. DISCRIMANNTS, NORMS, AND FINITENESS OF THE CLASS GROUP<br />

Before proving Theorem 10.4.2, we prove a few lemmas. The strategy of the<br />

proof will be <strong>to</strong> start with any nonzero ideal I, <strong>and</strong> prove that there is some nonzero<br />

a ∈ K, with very small norm, such that aI is an integral ideal. Then Norm(aI) =<br />

Norm K/Q(a)Norm(I) will be small, since Norm K/Q(a) is small. The trick is <strong>to</strong><br />

determine precisely how small an a we can choose subject <strong>to</strong> the condition that aI<br />

be an integral ideal, i.e., that a ∈ I −1 .<br />

Let S be a subset of V = R n . Then S is convex if whenever x, y ∈ S then the<br />

line connecting x <strong>and</strong> y lies entirely in S. We say that S is symmetric about the<br />

origin if whenever x ∈ S then −x ∈ S also. If L is a lattice in V , then the volume<br />

of V/L is the volume of the compact real manifold V/L, which is the same thing as<br />

the absolute value of the determinant of any matrix whose rows form a basis for L.<br />

Lemma 10.4.3 (Blichfeld). Let L be a lattice in V = R n , <strong>and</strong> let S be a bounded<br />

closed convex subset of V that is symmetric about the origin. Assume that Vol(S) ≥<br />

2 n Vol(V/L). Then S contains a nonzero element of L.<br />

Proof. First assume that Vol(S) > 2n · Vol(V/L). If the map π : 1<br />

2S → V/L is<br />

injective, then<br />

<br />

1<br />

1<br />

Vol(S) = Vol<br />

2n 2 S<br />

<br />

≤ Vol(V/L),<br />

a contradiction. Thus π is not injective, so there exist P1 = P2 ∈ 1<br />

2S such that<br />

P1 − P2 ∈ L. By symmetry −P2 ∈ 1<br />

1<br />

2S. By convexity, the average 2 (P1 − P2) of P1<br />

<strong>and</strong> −P2 is also in 1<br />

2S. Thus 0 = P1 − P2 ∈ S ∩ L, as claimed.<br />

Next assume that Vol(S) = 2n · Vol(V/L). Then for all ε > 0 there is 0 = Qε ∈<br />

L ∩ (1 + ε)S, since Vol((1 + ε)S) > Vol(S) = 2n · Vol(V/L). If ε < 1 then the Qε<br />

are all in L ∩ 2S, which is finite since 2S is bounded <strong>and</strong> L is discrete. Hence there<br />

exists Q = Qε ∈ L∩(1+ε)S for arbitrarily small ε. Since S is closed, Q ∈ L∩S.<br />

Lemma 10.4.4. If L1 <strong>and</strong> L2 are lattices in V , then<br />

Vol(V/L2) = Vol(V/L1) · [L1 : L2].<br />

Proof. Let A be an au<strong>to</strong>morphism of V such that A(L1) = L2. Then A defines an<br />

isomorphism of real manifolds V/L1 → V/L2 that changes volume by a fac<strong>to</strong>r of<br />

| Det(A)| = [L1 : L2]. The claimed formula then follows.<br />

Fix a number field K with ring of integers OK. Let σ : K → V = R n be the<br />

embedding<br />

σ(x) = σ1(x), σ2(x), . . .,σr(x),<br />

Re(σr+1(x)), . . . ,Re(σr+s(x)), Im(σr+1(x)), . . . ,Im(σr+s(x)) ,<br />

where σ1, . . .,σr are the real embeddings of K <strong>and</strong> σr+1, . . .,σr+s are half the complex<br />

embeddings of K, with one representative of each pair of complex conjugate<br />

embeddings. Note that this σ is not exactly the same as the one at the beginning<br />

of Section 10.2.

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