Generalized HT90

I officially promise this is my last post on Hilbert’s Theorem 90, but because of that it is going to go really fast for those who have not seen group cohomology (it is really cool, so I couldn’t pass it up).

An abelian group is a G-module (G a group) if for all \sigma\in G and a\in A there is a unique element \sigma(a)\in A satisfying two conditions: \sigma(a+b)=\sigma(a)+\sigma(b), and (\sigma\tau)(a)=\sigma(\tau(a)) for all \sigma, \tau\in G and a,b\in A.

Just check any algebra text or here for more information on modules.

Now define an n-cochain of G over A to be a a function of n variables from G into A. If n=0 it is just an element of A. C^n(G, A) is the set of all n-cochains, and can be made into a group by the operation (f+g)(\sigma_1, \ldots, \sigma_n)=f(\sigma_1, \ldots, \sigma_n)+g(\sigma_1, \ldots, \sigma_n).

We can also get from C^n(G, A) to C^{n+1}(G, A) (which is what people who know about cohomology were hoping for), by the function \delta:

(\delta f)(\sigma_1, \ldots, \sigma_{n+1})=\sigma_1(f(\sigma_2, \ldots, \sigma_{n+1}))+

\sum_{i=1}^n(-1)^i f(\sigma_1, \ldots, \sigma_i\sigma_{i+1}, \ldots, \sigma_{n+1})+(-1)^{n+1}f(\sigma_1, \ldots, \sigma_n).

OK. That looks bad, but really in some sense it is the natural choice. I’ll leave it to you to check that this is both a homomorphism and that \delta\delta f=0 (i.e. we have a chain complex).

Now if we label them \delta_0: A\to C^1(G, A)

\delta_1: C^1(G, A)\to C^2(G, A) and

\delta_n: C^n(G, A)\to C^{n+1}(G, A). Then we form \displaystyle H^n(G, A)=\frac{\ker\delta_n}{im\delta_{n-1}}=\frac{Z^n(G, A)}{B^n(G, A)}. We call the elements of Z^n(G, A) the n-cocycles and the elements of B^n(G, A) the n-coboundaries.

So if you don’t like that, we can scratch it now, since in HT90 we only care about H^1(G, A), so let’s take a closer look at that. We can completely classify what the elements of Z^1(G, A) look like. For any f\in Z^1(G, A) and any \sigma, \tau\in G we need (\delta f)(\sigma, \tau)=\sigma(f(\tau))-f(\sigma\tau)+f(\sigma)=0. Which is to say that f(\sigma\tau)=\sigma(f(\tau))+f(\sigma).

Now let’s classify what B^1(G,A) looks like. Well, if g\in B^1(G, A), then g=\delta h for some h\in A=C^0(G,A). So g(\sigma)=(\delta h)(\sigma)=\sigma(a)-a for some a\in A. Well, I think you might be able to see the previous formulation of the theorem coming from unravelling these definitions.

Theorem statement: If K/F is a finite Galois extension and G=Gal(K/F), then H^1(G, K^\times) is trivial.

Proof: Let a be a cocycle. Then let \alpha: K\to K by c\mapsto \sum_{\sigma\in G}a(\sigma)\sigma(c). As in the last post, \alpha is not 0 by linear independence. So let c\in K such that \alpha(c)\neq 0 and set b=\alpha(c).

Then for any \tau\in G we have

\displaystyle\tau(b)=\sum_{\sigma\in G}\tau(a(\sigma)\sigma(c))

\displaystyle =\sum_{\sigma\in G}\tau(a\sigma)(\tau\sigma)(c)

\displaystyle =\sum_{\sigma\in G}a(\tau)^{-1}a(\tau\sigma)(\tau\sigma)(c). Now we use that a is a cocycle (in the kernel) to continue the equality as

\displaystyle = a(\tau)^{-1}\sum_{\sigma\in G}a(\tau\sigma)(\tau\sigma)(c)


Aha, so a(\tau)=b\tau(b)^{-1} is a coboundary! Thus every cocycle is a coboundary, so the quotient is trivial.

Test your understanding by now trying to prove the other formulation as a corollary to this (remember you assume that G is cyclic in that version and have to relate it back to the norm).


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