# étale cohomology with values in an abelian scheme is torsion?

Let $A/X$ be an abelian scheme. Is $H^n(X,A)$ torsion for $n > 0$?

Perhaps this can be proved analogously as Proposition IV.2.7 of Milne's Étale cohomology (where it is proved that the cohomological Brauer group of a quasi-compact scheme is torsion) exploiting the Kummer sequence. For integral affine schemes and the multiplicative group one can also prove this using that the Brauer group injects into the Brauer group of the fraction field, which is torsion as a Galois cohomology group.

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Just added the "algebraic geometry" tag. –  Laurent Moret-Bailly Nov 28 '11 at 15:48
If by "cohomological Brauer group" you mean $\mathrm{H}^2(X, \mathbb{G}_{\mathrm{m}})$, then this is not what Milne's proposition says. Milne is using $\mathrm{Br}(X)$ to mean the group of equivalence classes of Azumaya algebras, which is indeed torsion, but in general the cohomology group need not be torsion: see Milne's Remark 2.8 on the same page. (The counterexample by Mumford referred to there is not too outlandish - something like a cone over an elliptic curve, IIRC.) Likewise, in your last sentence you need the scheme to be regular. –  Martin Bright Nov 28 '11 at 17:19

For $H^1$ there are many results in chapter XIII of Raynaud's thesis (LNM 119).

First (XIII.3.1) it is easy to construct of $A$-torsors which have infinite order: Let $k$ be a field, $A_0$ an abelian variety over $k$ having a point $a$ of infinite order. Pick two rational points $x$ and $y$ (e.g. $0$ and $1$) on the affine line $L$, and let $X$ be obtained from $L$ by identifying $x$ and $y$. Consider the trivial $A_0$-torsor $P:=A_0\times_k L$ over $L$. Identifying $P_x$ and $P_y$ via translation by $a$ (which is an isomorphism of $A_0$-torsors) you get $Q_a\to X$ which is a torsor over $X$ under $A=X\times_k A_0$. It cannot be trivial: if $s$ were a section, it would give rise to a $k$-morphism $s':L\to A_0$, which must be constant, but this contradicts the requirement $s'(y)=s'(x)+a$. Clearly, if $n\in\mathbb{Z}$, the class $nQ_a\in H^1(X,A)$ is just $Q_{na}$ which is also nontrivial unless $n=0$.

There are even counterexamples over a normal two-dimensional base, but the construction is harder (XIII.3.2).

In general, if $c\in H^1(X,A)$, the property that $c$ is torsion is related to the representability or projectivity of the corresponding torsor: see XIII.2.3 and XIII.2.6.

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OK, so are there also counterexamples for $n > 1$? –  Timo Keller Nov 28 '11 at 13:46
Of course that's a good question; I don't know the answer. –  Laurent Moret-Bailly Nov 28 '11 at 14:09

Let me point out that the analogous question on the analytic topology is false in all positive degrees. Take an abelian scheme $A/X$ over a complex variety, and let $\mathcal{A}$ denote the sheaf of holomorphic section on $X_{an}$. We have an exact sequence $$0\to L\to V\to \mathcal{A}\to 0$$ where $L$ is locally constant, and $V$ is a vector bundle (cf Deligne, Hodge II, p 50). If we assume that $X$ is affine, then $X_{an}$ is Stein, so by Cartan B, we have an isomorphism $$H^n(X,\mathcal{A})\cong H^{n+1}(X,L)$$ for $n>0$. It is easy to find examples, where the right side is nontorsion for any given degree. In fact, we may as well take $A$ to be a product of $X$ with an abelian variety. Then we just need to choose $X$ so that (n+1)st Betti number is nonzero.

While this doesn't directly address your actual question for etale cohomology, my guess would be no for this as well.

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In [Milne, Arithmetic Duality Theorems], II.5, it is proved that $H^r(U,\mathcal{A})$ is torsion for $U$ the ring of integers of a number field or a complete smooth curve over a finite field, respectively.

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