A Mind for Madness

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Hilbert’s Theorem 90…the math

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So now that I’ve presented the origins of Hilbert’s Theorem 90, I thought it might be good to actually present it mathematically with a proof.

Statement: Suppose K is a finite Galois extension of F with G=Gal(K/F) cyclic, say G=<\sigma > of order n. If a\in K, then N_{K/F}(a)=1 if and only if a=b/\sigma(b) for some b\in K.

Let’s parse this a little before the proof. First off let’s define the notation N_{K/F}(a). Notice that we can view K as a vector sapce over F, so t_a: K\to K by u\mapsto au is a linear transformation. So we want a norm, so we define N_{K/F}(a)=\det (t_a). i.e. the determinant of the matrix that represents that linear transformation. Thus, we get all our nice results from linear algebra like independence of basis choice. If we fight with definitions and linear algebra for a bit we also get that if the extension is Galois, \displaystyle N_{K/F}(a)=\prod_{\phi\in Gal(K/F)} \phi (a). This is left as an (not required) exercise to familiarize yourself with the terms before moving on. Note that we can determine Tr_{K/F}(a) similarly to be the trace of t_a.

Proof of the theorem: Let’s do the backwards way first (I’m going to suppress the norm notation since we know where it is happening). Suppose a=b/\sigma(b) for some b\in K. Then let’s simply use the result (that you proved!) to see that \displaystyle N(a)=\left(\frac{b}{\sigma(b)}\right)\left(\frac{\sigma(b)}{\sigma^2(b)}\right)\cdots \left(\frac{\sigma^{n-1}(b)}{\sigma^n(b)}\right)=1 since \sigma has order n.

Now for the forwards direction. Suppose N(a)=1. Again using a standard linear algebra argument we know that 1, \sigma, \ldots, \sigma^{n-1} are K-linearly independent (in fact, any set of field automorphisms of K are K-linearly independent over the vector space of all functions from K to K).

Thus, \displaystyle \phi = a\cdot 1+(a\sigma(a))\sigma + (a\sigma(a)\sigma^2(a))\sigma^2+\cdots + \left(\prod_{i=0}^{n-1}\sigma^i (a)\right)\sigma^{n-1} from K to K is not 0. By the thing you proved the coefficient on \sigma^{n-1} is N(a) which is 1. More importantly, there is some c\in K such that \phi(c)\neq 0, so let’s define b=\phi(c).

But \sigma(b)=\frac{1}{a}(b-ac)+\sigma^n(c)=\frac{1}{a}(b-ac)+c=\frac{b}{a}. Thus a=b/\sigma(b). And we are done.

Does this proof remind anyone of Lagrange resolvents? Eh, whatever, I won’t dig for a connection now.

I also mentioned that there is a generalization of this.

Statement: If K/F is a finite Galois extension and G=Gal(K/F), then H^1(G, K^\times) is trivial. I’m debating whether or not to parse and prove this version next time, or to just drop the Hilbert Theorem 90 posts. Suggestions?

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

One thought on “Hilbert’s Theorem 90…the math

  1. Pingback: The Giant’s Shoulders #6 « Rigorous Trivialities

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