Relationship between Galois cohomology and etale cohomology.

Why is étale cohomology a natural generalization of Galois cohomology ?

I would like to inform you that I have a few quite sufficient prerequisites Galois cohomology and its application to solve the $90$ - th problem of Hilbert . So I can understand what you're going to tell me eventually.

However , in theory , I take a little well how to define étale cohomology which is based on the sheaves theory and sheaf cohomology , but so far, I have not had the opportunity to apply the etale cohomology theory to a calculation that proves practical . So it can pose obstacles thereafter, but I count on you anyway. If I do not well captured something that repulses me the way, I point you.

Thank you in advance for your help.

1 Answer

The technical fact is the following:

Theorem: There is an equivalence of categories between $$\mathsf{Ab}\left((\mathrm{Spec}(k))_\mathrm{\acute{e}t})\right)$$ and the category $$\mathsf{DiscMod}_{G_k}$$ of discrete $$G_k$$-modules preserving cohomology.

Here $$\mathsf{Ab}$$ denotes the category of abelian sheaves on the site $$(\mathrm{Spec}(k))_\mathrm{\acute{e}t}$$ and discrete $$G_k$$-modules are abelian groups $$M$$ with an additive action of $$G_k$$ such that $$G_k\times M\to M$$ is abelian.

Moreover, what I mean by preserving cohomology is that if $$\mathcal{F}$$ is associated to $$M$$ then

$$H^1_\mathrm{cont.}(G_k,M)=H^1((\mathrm{Spec}(k))_\mathrm{\acute{e}t},\mathcal{F})$$

where the left hand side is Galois cohomology and the right hand side is étale cohomology.

This equivalence is actually pretty easy to see. Namely, whatever $$(\mathrm{Spec}(k)))_\mathrm{\acute{e}t}$$ is rigorously, one can think about it, after fixing a separable closure $$\overline{k}$$, as the lattice of separable finite subextensions of $$\overline{k}$$. Then, an abelian sheaf $$\mathcal{F}$$ on $$(\mathrm{Spec}(k))_\mathrm{\acute{e}t}$$ is just an association of an abelian group $$\mathcal{F}(L)$$ for every finite separable extension such that an inclusion $$L\subseteq L'$$ gives a group map $$\mathcal{F}(L)\to\mathcal{F}(L')$$ (satisfying the obvious compatibility conditions) and subject to the condition that

$$\mathcal{F}(L)=\mathcal{F}(L')^{\mathrm{Gal}(L'/L)}$$

whenever $$L'/L$$ is Galois, and where the Galois group acts by the functoriality I mentioned above.

One can then check that if one defines

$$\mathcal{F}_{\overline{k}}:=\varinjlim_L \mathcal{F}(L)$$

where $$L$$ travels over finite separable (or Galois—they're cofinal) extensions, then $$\mathcal{F}_{\overline{k}}$$ defines a discrete $$G_k$$-module. This gives us the association

$$\mathsf{Ab}\left((\mathrm{Spec}(k))_\mathrm{\acute{e}t}\right)\to \mathsf{DiscMod}_{G_k}$$ and the association the other way takes a discrete $$G_k$$-module $$M$$ and associates to a Galois $$L/K$$ the set $$M^{\mathrm{Gal}(\overline{k}/L)}$$. This defines the equivalence of categories.

Note that under this equivalence, the 'global sections functor' $$\mathcal{F}\mapsto \mathcal{F}(k)'$$ gets mapped to the $$G_k$$-invariants functor $$M\mapsto M^{G_k}$$. Thus, the cohomologies of these functors should agree. But, they are just the étale cohomology and the Galois cohomology respectively.

Now, how is this helpful in discussing Hilbert's theorem 90? Well, we can use geometric intuition then. Namely, we think about the étale site as being essentially the categories of 'generalized opens' of a scheme, trying to capture something better approximating the 'classical topology' of schemes over $$\mathbb{C}$$. In particular, it's natural to think about vector bundles on this generalized type of open set. But, in any reasonable category of open sets, the set of vector bundles is classified by the cohomology group $$H^1(X,\mathrm{GL}_n)$$ (i.e. vector bundles should always correspond to principal $$\mathrm{GL}_n$$-bundles). In particular, vector bundles on this category of generalized open sets $$X_\mathrm{\acute{e}t}$$ should be classified by the cohomology group $$H^1(X_{\mathrm{\acute{e}t}},\mathrm{GL}_n)$$.

But, the following miracle occurs:

$$H^1(X_{\mathrm{\acute{e}t}},\mathrm{GL}_n)=H^1(X_{\mathrm{Zar}},\mathrm{GL}_n)$$

which is not actually difficult to check (it comes down to finitely presented and flat can be checked étale locally). Thus, the vector bundles on this larger class of generalized opens on $$X$$ agree precisely the vector bundles on the far more restrictive 'usual opens' on $$X$$.

In particular, for $$X=\mathrm{Spec}(k)$$, we see that $$H^1((\mathrm{Spec}(k)_{\mathrm{\acute{e}t}},\mathrm{GL}_n)=0$$ since there are no non-trivial vector bundles on a literal point! In particular, we see that $$H^1((\mathrm{Spec}(k))_\mathrm{\acute{e}t},\mathbf{G}_m)=0$$ since there are no non-trivial line bundles on a point.

But, $$\mathbf{G}_m$$ is an abelian sheaf on $$(\mathrm{Spec}(k))_\mathrm{\acute{e}t}$$ associating to any $$L$$ the group $$L^\times$$. Thus, under our equivalence from above, we see that

$$H^1((\mathrm{Spec}(k))_\mathrm{\acute{e}t},\mathbf{G}_m)=H^1(G_k,\mathbf{G}_m(\overline{k}))=H^1(G_k,\overline{k}^\times)$$

and thus we recover Hilbert's theorem 90.

Summing this up, once one has the zen geometric view of étale cohomology, one sees Hilbert's theorem 90 as the statement that there are no non-trivial line bundles on a point.

• Thank you SIr. Your answear is extremely clear. :-) – Albert007 Sep 6 '15 at 13:10
• @Albert007 No problem! Glad I could help. – Alex Youcis Sep 6 '15 at 16:07
• @peterag Thanks--fixed. – Alex Youcis Nov 30 at 19:23