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Let $X$ be a variety over a field $K$ and let $L/K$ be a field extension. Let $\ell$ be a prime number.

Do we always have that $ H^m(X,\mathbb{Q}_{\ell}) \cong H^m(X_L,\mathbb{Q}_\ell)^{\Gamma(L/K)}$?

If so, could someone sketch the argument?

If not, what additional assumptions on $X$, $K$, $L$ and/or $\ell$ would make this true?

For example, what happens if we assume that $\ell$ and $[L : K]$ (as as supernatural number) are coprime?

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2 Answers 2

up vote 6 down vote accepted

To precise the answer of Timo. Yes, the spectral sequence of Timo is the way to answer your question in general. Of course, it depends on computing Galois cohomology, which may not be easy, and is very dependent of the type of fields $K$ you are considering. As Keerthy remarked, if $l$ and $[L:K]$ are coprime, your isomorphism is correct.

If you want interesting conditions for this to be true, you need to be more specific on your situation. In general, one situation that has been studied (for arithmetic purposes) are the cases where $K$ is a local or a global field, and $L$ an algebraic closure of $K$ (or, when $K$ is global, the maximal algebraic extension unramified outside a specified, often finite, set of places). In this case, general theorems tells you that you don't have to worry about the Galois cohomology group $H^i$ with $i>2$, which simplifies a lot your spectral sequence, and you have deep conjectures, sone of them theorems, which tells you that some of those Galois cohomology groups vanish, which in many cases can tell you that your isomorphism holds. These conjectures are called the Jannsen's conjectures, and they are related (but not in anyway equivalent) to the Bloch-Kato conjectures on the rank of Selmer groups. You can read about them (and the applications to your question, which is explicitly mentioned in the introduction as a motivation) in Jannsen's paper in the book "Galois groups over $\mathbb{Q}$, edited by Ribet, Serre, Ihara.

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Thanks for the answer! – Wanderer Jun 26 '12 at 20:57

The Hochschild-Serre spectral sequence gives for $L/K$ finite Galois $H^p(G(L/K),H^q(X_L,\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F})$, and you are asking for the edge morphism $E^m \to E_2^{0,m}$ to be an isomorphism

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So, in other words, he needs $H^p(G(L/K), H^q(X_L,\mathcal F))=0$ for all $p>0$. – Will Sawin Jun 25 '12 at 14:21
Isn't the higher group cohomology killed by the order of $G(L/K)$? This would mean that it shouldn't make a difference for $\mathbb{Q}_{\ell}$-cohomology. – Keerthi Madapusi Pera Jun 25 '12 at 14:21

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