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 and there is a unique element satisfying two conditions: , and for all and .

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 it is just an element of A. is the set of all n-cochains, and can be made into a group by the operation .

We can also get from to (which is what people who know about cohomology were hoping for), by the function :

.

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 (i.e. we have a chain complex).

Now if we label them

and

. Then we form . We call the elements of the n-cocycles and the elements of the n-coboundaries.

So if you don’t like that, we can scratch it now, since in HT90 we only care about , so let’s take a closer look at that. We can completely classify what the elements of look like. For any and any we need . Which is to say that .

Now let’s classify what looks like. Well, if , then for some . So for some . Well, I think you might be able to see the previous formulation of the theorem coming from unravelling these definitions.

Theorem statement: If is a finite Galois extension and , then is trivial.

Proof: Let a be a cocycle. Then let by . As in the last post, is not 0 by linear independence. So let such that and set .

Then for any we have

. Now we use that is a cocycle (in the kernel) to continue the equality as

.

Aha, so 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).

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?

I was reading the blog topological musings, and the category talk reminded me of this neat thing I posted on AoPS awhile ago.

By using the Mayer-Vietoris sequence in cohomology, we determine whether or not can be written as the union of two open connected sets, U and V, such that is disconnected.

Well, it seems that we are concerned with stuff, since that tells us the number of connected components.

The sequence gives , a nice short exact sequence in which we mostly know they are connected, so

, basically the question reduces to: Is it possible for the second to last term to have dimension greater than 1, but the sequence to remain exact?

Well, no, since the exactness tells us the dimension of is the dimension of .

So we have a nifty result: the plane cannot be broken into two connected parts in which the intersection of these two parts is disconnected.

Challenge for the readers: This seems extremely simple and obvious. Find a proof that does not require advanced techniques such as cohomology.