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A theorem of Giraud says that gerbes over a scheme $X$ bounded by a sheaf of Abelian groups $A$ are classified by elements of the etale cohomology group $H^2(X,A)$. Similar statements hold in other categories (differential, topological) as well.

One of the points of view on gerbes (emphasized, for example, by Hitchin in "Lectures on special Lagrangian manifolds") is that a gerbe can be glued from trivial gerbes just like a vector bundle can be glued from trivial vector bundles, but the gluing data is not transition functions but A-torsors. I am trying to see if this philosophy can be applied to gerbes over a spectrum of a (perfect) field $k$. Giraud's theorem tells in this case that isomorphism classes of gerbes are in bijective correspondence with elements of the Galois cohomology group $H^2(k, A)$.

Let $k'/k$ be Galois with Galois group $G$ and let $A$ be an Abelian algebraic group defined over $k$ (we use $A$ to denote the corresponding Galois module). The Hochshild-Serre spectral sequence yields the following exact sequence: $$ \ldots \to H^2(G,A^{G_{k'}}) \to \mathrm{Ker}(H^2(k,A) \to H^2(k',A)) \to H^1(G,H^1(k',A)) \to H^3(G,A^{G_{k'}})\ldots $$ where $G_{k'}$ is the absolute Galois group of $k'$. Suppose we have a gerbe over $\mathrm{Spec}\ k$ that trivialises after base change to $\mathrm{Spec}\ k'$. An element of $H^1(G,H^1(k',A))$ is the gluing data that has been mentioned above (indeed, $H^1(k',A)$ classifies $A$-torsors defined over $k'$). If also the second and third cohomology of $G$ with coeffecients in $A^{G_{k'}}$ vanished then we would have an isomorphism between two middle terms of the sequence which means that any gerbe that trivialises over $\mathrm{Spec}\ k'$ can be described in terms of this gluing data.

I have the following question: is it true that for any gerbe over $k$ one can always find a Galois extension $k'$ with Galois group $G$ such that the gerbe trivialises after base change to $k'$ and $H^2(G,A)$ and $H^3(G,A)$ vanish?

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I don't think so. I think a gerbe bound by $A$ over the spectrum of a field $k$ gives a cohomology class $\eta\in H^2(k,A)$, and the gerbe trivializes over an extension $k'/k$ if and only if this cohomology class trivializes. Now take $k=\mathbb{R}$, $A=\mathbf{G}_{m,\mathbb{R}}$, then $H^2(k, A)={\rm Br}(\mathbb{R})=\frac{1}{2}\mathbb{Z}/\mathbb{Z}$. Let $\eta\in{\rm Br}(\mathbb{R})$ be the nontrivial element (corresponding to Hamilton's quaternions). Then $\eta$ trivializes over $\mathbb{C}$, but not over $\mathbb{R}$, so $k'=\mathbb{C}$, $G={\rm Gal}(\mathbb{C}/\mathbb{R})$, and $H^2(G,A)=H^2(\mathbb{C}/\mathbb{R},\mathbf{G}_m)={\rm Br}(\mathbb{R})\neq 0$, so $H^2(G,A)$ does not vanish.

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indeed, the statement I hoped for is false. Do you think it might still be true for some classes of fields? For example, fields finitely generated over an algebraically closed field? –  Dima Sustretov Apr 24 '13 at 7:43

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