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Geometry of Numbers

3 Unit theorem

To study the units of OK, we will consider the

log embedding OKRr+s,x[Fi:R]log(|σi(x)|)=li(x), where Fi is the field σi is an embedding into.

  • Lemma 3.1. The kernel of the log embedding consists of roots of unity.

  • Proof. Certainly they are in the kernel. Conversely, note that anything in the kernel has bounded coefficients of the characteristic polynomial, so that there are finitely many possibilities. But its powers are in the kernel, so it must be a root of unity.

  • Lemma 3.2. The image of the units via the log embedding is discrete.

  • Proof. Any thing whose image is near zero has bounded coefficients of the characteristic polynomial, so there are only finitely many such things.

The units lie in the hyperplane li=0. It turns out that they are full rank, and hence free of rank r+s1.

  • Theorem 3.3. The units form a lattice in the hyperplane li=0.

  • Proof. Suppose that we have a nonzero linear form on icili=0 on li=0. We will show there is a unit such that the form on that unit is nonzero. We can assume WLOG that cn=0,c1>0. Consider the region in Rr×Cs defined by |σi|bi. For fixed bi, r+s>i, we want to choose br+s large enough such that the region has area contains a lattice point. This will happen when c=2rπs|disc(K)|biR×(biC)2. Now choose any values of b1,,br+s1, and a sufficiently large br+s will make this an equality. Now if x is a nonzero lattice point in the region, then 1|N(x)|ibiR×i(biC)2. On the other hand, for i<r+s, |σi|=N(σi)jiσjbic. Thus li is between [Fi:R]log(bi)log(c) and [Fi:R]log(bi). Thus we can manufacture each li to be in any interval of our choosing of a fixed size, so we can produce infinitely many x such with norm at most c such that the icili takes distinct values on each. Since there are finitely many ideals of a given norm, two must differ by a unit on which the linear form doesn’t vanish.