In this post we shall develop the very basics of Galois cohomology, see two basic but important applications: Kummer Theory and the Weak Mordell-Weil Theorem, and remark on the general shape of both arguments. Hopefully at a later point I will post on Neukirch’s General’ Class Field Theory which is perhaps a more complex realisation of a similar philosophy. I guess the main theme is that even very basic Galois cohomology can be a powerful tool that allows you to get your money’s worth’ out of Galois theory without having to think too hard.

Firstly, let us set up the general framework of Galois cohomology. In general we will have a profinite group ${G}$ and want to study the category of ${G}$-modules, namely topological abelian groups ${A}$ with a continuous action by ${G}$. There is an obvious functor on this category ${A \mapsto A^G = \{a \in A| \sigma(a) = a \ \ \forall \sigma \in G\}}$, called taking ${G}$-invariants.’ In fact, for any subgroup (or even subset) ${H}$ of ${G}$ we have a similar functor ${A \mapsto A^H}$. In the case where ${H}$ is a closed subgroup, these will be of interest to us. Thus in this post, in its most general setting Galois cohomology will be about studying ${G}$-modules ${A}$ together with their submodules fixed by closed subgroups of ${H}$.

Example (the most basic reason for the name): If we consider ${G=Gal(\bar{k}/k)}$ the absolute Galois group of a field ${k}$, ${G}$ acts on ${\bar{k}}$ and by the fundamental theorem of Galois theory the closed subgroups of ${G}$ are in one to one correspondence with the subfields of ${\bar{k}}$ containing ${k}$, so the Galois cohomology of ${\bar{k}}$ is just the study of the field extension ${\bar{k}/k}$ and its subextensions. In fact, in all of the examples that follow we will take ${G}$ to be an absolute Galois group, but it is useful to remember that the general theory that follows is entirely group-theoretical.

So let us go back to our functor ${A \mapsto A^G}$. As a functor in a category of abelian groups it is reasonable to investigate its exactness properties. In fact, given a short exact sequence of ${G}$-modules

$\displaystyle 0\rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0$

it is true that

$\displaystyle 0 \rightarrow A'^G \rightarrow A^G \rightarrow A''^G$

is exact, but the final map need no longer be surjective. However, by general homological algebra nonsense (including that the category of ${G}$-modules has enough injectives) this is enough to get some right-derived functors going, and allow us to define Galois cohomology groups ${H^i(G, A)}$ with all the properties you would expect, but in particular such that ${H^0(G,A) = A^G}$ and such that the above short exact sequence gives rise to a long exact sequence of cohomology:

$\displaystyle 0 \rightarrow A'^G \rightarrow A^G \rightarrow A''^G \rightarrow H^1(G,A') \rightarrow \\ H^1(G,A) \rightarrow H^1(G,A'') \rightarrow H^2(G,A') \rightarrow ...$

In fact, for our examples we will only need to go up to the ${H^1}$ level, which already provides a powerful tool in virtue of the following very basic fact. If a ${G}$-module ${A}$ is in fact invariant under ${G}$, then ${H^1(G,A) \cong Hom(G,A)}$. I.e. it is just group homomorphisms from ${G}$ to ${A}$. In particular, if ${A'}$ is ${G}$-invariant, the ${H^1(G,A')}$ term of the above sequence has a concrete interpretation, and will be of crucial importance.

We can already start thinking about our first application. Given an arithmetically interesting field ${k}$, we want to study the relationship between the group ${k^* / k^{*n}}$ and its abelian extensions of exponent ${n}$ (i.e. every element of the Galois group has order dividing ${n}$). More specifically, we want to prove the following:

Theorem 1 (Kummer Theory) Let ${k}$ be a field containing all the ${n}$th roots of unity. Then every degree ${n}$ cyclic extension of ${k}$ is of the form ${k(\sqrt[n]{a})}$ for some ${a \in k}$. More generally, there is a bijective inclusion-preserving correspondence between abelian extensions of $k$ of exponent $n$ and subgroups of $k^*/k^{*n}$ given by ${L \mapsto \{l^n| l \in L^*\}}$, and ${H \mapsto k(\sqrt[n]{H})}$. There is a perfect pairing

$\displaystyle k^* / k^{*n} \stackrel{\cong}{\rightarrow} Hom(G_k, \mu_n(k)).$

How do we prove this theorem? Let us note several striking features of its statement. Firstly, for it to be nontrivial, the field involved has to actually have some abelian extensions of exponent ${n}$. In particular, if ${k}$ is an algebraically closed field nothing interesting happens (and the theorem tells us that this is because every element of ${k}$ has an ${n}$th root which is already in ${k}$). Secondly, it seems that to get things going, ${k}$ does need to make a small arithmetic sacrifice: the ${n}$th roots of unity ${\mu_n(\bar{k})}$ must already be present in ${k}$. If we want to use Galois cohomology, there really is only one short exact sequence we could think of writing down. Maybe try to guess what it is before reading it.

$\displaystyle 0 \rightarrow \mu_n(\bar{k}) \rightarrow \bar{k}^* \stackrel{(-)^n}{\rightarrow} \bar{k}^{*} \rightarrow 0.$

What happens when we take invariants of the absolute Galois group ${G_k = Gal(\bar{k}/k)}$? Galois cohomology tells us that we have a long exact sequence

$\displaystyle 0 \rightarrow \mu_n(k) \rightarrow k^* \stackrel{(-)^n}{\rightarrow} k^* \rightarrow H^1(G_k, \mu_n(\bar{k})) \\ \rightarrow H^1(G_k, \bar{k}^*) \stackrel{(-)^n}{\rightarrow} H^1(G_k, \bar{k}^*) \rightarrow ...$

So in particular, if we truncate the exact sequence at the ${H^1(G_k, \mu_n(k))}$ term, we get

$\displaystyle 0 \rightarrow k^*/k^{*n} \rightarrow H^1(G_k, \mu_n(\bar{k})) \rightarrow H^1(G_k, \bar{k}^*).$

This looks like we have got somewhere, provided we can compute or interpret the first cohomology groups. And here is where our assumption that ${k}$ contains all the ${n}$th roots of unity comes into play. Given this, ${\mu_n(\bar{k})}$ is ${G_k}$-invariant, so we have ${H^1(G_k, \mu_n(\bar{k}))}$ is in fact isomorphic as an abelian group to ${Hom(G_k, \mu_n(k))}$.

Thus, certainly to prove the Kummer isomorphism it will suffice to prove that the next map in the exact sequence is the zero map, and in fact what we need is precisely Hilbert’s Theorem 90, which says that ${H^1(G_k, \bar{k}^*) = 0}$.

Finally, the inclusion-preserving correspondence and the claim that cyclic extensions of exponent ${n}$ are generated by an ${n}$th root follow after a little more work from the Kummer isomorphism. Indeed, a homomorphism ${G_k \rightarrow \mu_n(k)}$ gives a cyclic quotient of ${G_k}$ with exponent ${n}$ and unravelling the definitions we see that it is generated by the ${n}$th root of the corresponding element in ${k^*/k^{*n}}$. Furthermore, a collection of such homomorphisms gives rise to a map into a product of some ${\mu_n(k)}$ whose kernel corresponds to a field of exponent ${n}$, etc.

So we are done, having quoted facts about Galois cohomology and Hilbert’s Theorem 90 (and suppressed several other less significant details). There are two questions we should ask ourselves. Firstly, what did the Galois cohomology do for us? It seems to me that it provided the key tool for passing from a slightly hard-to-handle object directly connected to the arithmetic of fields (${k^*/k^{*n}}$) to some more abstract and more group-theoretical objects (the ${H^1}$ groups) which turned out to be easier to compute. It did this by lifting all attention away from ${k^*}$, instead replacing it with the pair ${(\bar{k}^*, G_k)}$. The first object is in some sense arithmetically trivial’, being algebraically closed, and the second is a group which we might hope is easier to manipulate than the complicated field ${k}$ and its algebraic extensions. Secondly, what didn’t it do for us? Most notably, it didn’t actually do the necessary computations involving ${\bar{k}}$ and ${G_k}$. It just told us which ${H^1}$s to compute and we had to go away and do that. It was clever of \text{us} (or Kummer) to realise we needed the hypothesis that ${k}$ contained the ${n}$th roots of unity and this gave us a handle on one of the groups, and to control the other group we should really have proved Hilbert 90, a nontrivial theorem.

The general shape of this argument was rather pleasant however, and seems to be quite common in number theory. We have an operation on some geometrical object which is surjective over an algebraically closed field because inverting it involves solving algebraic equations (like taking ${n}$th roots), but not necessarily surjective over arbitrary fields. Such a failure to be surjective reflects interesting arithmetic behaviour, and the cokernels involved can be important arithmetic invariants but difficult to get our hands on. Galois cohomology allows us to translate this difficult arithmetic-geometric behaviour of geometrical spaces over a field like ${{\mathbb Q}}$ into group-theoretical behaviour of the absolute Galois group together with the much simpler geometric behaviour of our space considered over an algebraically closed field.

Our second striking example is the Weak Mordell-Weil theorem on elliptic curves.

Theorem 2 (Weak Mordell-Weil) Let ${K}$ be a number field, ${E/K}$ an elliptic curve with full ${n}$-torsion: ${E(K)[n] = E(\bar{K})[n]}$. Then ${E(K)/nE(K)}$ is finite.

This might not superficially resemble Kummer Theory, but look at the data and the structure of the situation. We have an algebraic operation, the multiplication-by-${n}$ map ${E \rightarrow E}$, and in particular it is surjective over ${\bar{K}}$ but probably not surjective over ${K}$. And, perhaps the biggest give-away of all, we have been told that the kernel of this map, the ${n}$-torsion, is all ${G_K}$ invariant. This time, at least if you’re happy with elliptic curves and everything I’ve said so far, you should have no trouble working out which exact sequence to write down. Refer to the one above if it helps.

We consider

$\displaystyle 0 \rightarrow E(\bar{K})[n] \rightarrow E(\bar{K}) \stackrel{[n]}{\rightarrow} E(\bar{K}) \rightarrow 0.$

Taking ${G_K}$-invariants, we get the usual long exact sequence:

$\displaystyle 0 \rightarrow E(K)[n] \rightarrow E(K) \stackrel{[n]}{\rightarrow} E(K) \rightarrow \\ H^1(G_K, E(K)[n]) \rightarrow H^1(G_K, E(\bar{K})) \stackrel{[n]}{\rightarrow} H^1(G_K, E(\bar{K})).$

and since ${E(K)[n] = E(\bar{K})[n]}$, truncating at the usual place we can form the short exact sequence:

$\displaystyle 0 \rightarrow E(K)/nE(K) \rightarrow Hom(G_K, E(K)[n]) \rightarrow H^1(G_K, E(\bar{K}))[n] \rightarrow 0.$

Sadly, the group ${Hom(G_K, E(K)[n])}$ fails to be finite, even using the hypothesis that ${K}$ is a number field, so we do need slightly more arithmetic input, and it turns out that (certainly when ${\mu_n \subset K}$) we can use arithmetic arguments to prove that any homomorphism in the image of ${E(K)/nE(K)}$ factors through ${Gal(F/K)}$ where ${F}$ is the compositum of all degree ${n}$ cyclic extensions of ${K}$ which are unramified outside a finite set of primes (${2}$ and those primes dividing the discriminant of ${E}$). By Hermite’s theorem, ${F/K}$ is finite, so ${Hom(Gal(F/K), E(K)[n])}$ is finite, and hence we get to the theorem we wanted.

In fact, by doing slightly more work, one is able to use techniques known as ${n}$-descent to get good bounds on the size of this image. I should mention that I only feel comfortable when ${\mu_n \subset K}$, in which case Kummer theory works nicely to allow us to describe these homomorphisms explicitly via

$\displaystyle Hom(Gal(F/K), E(K)[n]) \cong Hom(Gal(F/K), {\mathbb Z}/n{\mathbb Z})^2 \subset (K^*/K^{*n})^2,$

but my suspicion is that things are in fact more general.

Again, as with the previous application, it is in some sense difficult to pin down exactly what Galois cohomology did for us. It translated our problem about the mysterious group ${E(K)/nE(K)}$ into one about an also slightly mysterious subgroup of the (at least superficially not too mysterious) ${Hom(G_K, E(K)[n])}$, which then has to be attacked by further arithmetic arguments. However, it seems that in enacting this transformation, Galois cohomology redefines the problem in a way that does make it genuinely much easier to solve and I know of no proof that isn’t on one level or another via this very basic level of Galois cohomology.