Hilbert's Theorem 90...the math

December 11, 2008 by hilbertthm90 | 1 Comment

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 cyclic, say of order n. If , then if and only if for some .

Let’s parse this a little before the proof. First off let’s define the notation . Notice that we can view K as a vector sapce over F, so by is a linear transformation. So we want a norm, so we define . 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, . This is left as an (not required) exercise to familiarize yourself with the terms before moving on. Note that we can determine similarly to be the trace of .

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 for some . Then let’s simply use the result (that you proved!) to see that since has order n.

Now for the forwards direction. Suppose . Again using a standard linear algebra argument we know that 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, from K to K is not 0. By the thing you proved the coefficient on is which is 1. More importantly, there is some such that , so let’s define .

But . Thus . 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 , then 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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