Recall that if K is an algebraic number field of degree n = r + 2s, then the unit group of K is isomorphic to a W * Z^(r+s-1) where W is a finite abelian group. As in previous posts, r is the number of ring homomorphisms from K to R, and s is the number of complex conjugate pairs of ring homomorphisms from K to C whose images aren’t contained in R. This is Dirichlet’s theorem on units.
The basic way of proving Dirichlet’s theorem is by constructing a group homomorphism from the unit group of K to R^n. First, we denote the n homomorphisms from K to C by w1, w2, …, w(r+2s), where the first r homomorphisms are real, and the next 2s are such that w(r+k) and w(r+s+k) are complex conjugates (i.e. w(r+k) = w(r+s+k)). We send a unit u to (log|w1(u)|, log|w2(u)|, …, log|w(r+2s)(u)|), where the logs can be taken to any base, as long as all are taken to the same base. Since log(xy) = log(x) + log(y), this is a group homomorphism from K* to the additive group of R^n.
In fact, only the first r+s-1 coordinates matter. This is because complex conjugates have the same absolute value, so that log|w(r+s+k)(u)| = log|w(r+k)(u)|; this means that the last s coordinates depend on the first r+s. Also, |w1(u)|*|w2(u)|*…*|w(r+2s)(u)| = 1 since u is a unit, so log|w1(u)| + log|w2(u)| + … + log|w(r+2s)(u)| = 0 since log1 = 0; then if s = 0 we have log|w(r)(u)| = -log|w1(u)| – log|w2(u)| – … – log|w(r-1)(u)|, and if s > 0 we have 2log|w(r+s)(u)| = -log|w1(u)| – log|w2(u)| – … – log|w(r)(u)| – 2log|w(r+1)(u)| – … – 2log|w(r+s-1)(u)|.
The main trick is to show that the image in R^(r+s-1) is then Z^(r+s-1), which I’ll prove later. Now I’ll only show that a unit is mapped to 0 iff it’s a root of unity (i.e. the kernel of the map in K* is the group of roots of unity). The if direction is almost trivial: if u^m = 1 for some m, then for each i, |w(i)(u)|^m = 1, so |w(i)(u)| = 1 and log|w(i)(u)| = 0.
For the only if direction, I’ll first prove that for a given positive real constant c, there are only finitely many elements such that |w(i)(a)| <= c for every i. The minimal polynomial of such an element over Q is f(x) = (x–w1)(x–w2)…(x–w(n)) = x^n – (w1 + w2 + … + w(n))x^(n-1) + (w1w2 + w1w3 + … + w(n-1)w(n))x^(n-2) – … + (-1)^n(w1w2…w(n)), where each coefficient is an integer. There are at most nc choices for the n-1 coefficient, n(n-1)c^2 for the n-2 coefficient… and c^n for the constant. So there are only finitely many polynomials whose roots satisfy the condition, hence only finitely many elements.
If u is a unit such that |w(i)(u)| <= 1 for every i, then u must be a root of unity. This is because it’s easy to see that 1, u^2, u^3, u^4… all satisfy the condition. There are only finitely many elements satisfying the condition, so for some distinct m1 and m2, say with m1 < m2, we have u^m1 = u^m2, so u^(m2-m1) = 1 and u is a root of unity.
Now, let’s return to the group homomorphism from K* to R^n. If u maps to 0, then log|w(i)(u)| = 0 for every i, so that |w(i)(u)| = 1 for every i, and u is a root of unity.
Abstract Nonsense 