étale cohomology with values in an abelian scheme is torsion? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:41:37Z http://mathoverflow.net/feeds/question/82076 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82076/etale-cohomology-with-values-in-an-abelian-scheme-is-torsion étale cohomology with values in an abelian scheme is torsion? norondion 2011-11-28T12:54:01Z 2012-01-16T20:57:08Z <p>Let $A/X$ be an abelian scheme. Is $H^n(X,A)$ torsion for $n > 0$?</p> <p>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.</p> http://mathoverflow.net/questions/82076/etale-cohomology-with-values-in-an-abelian-scheme-is-torsion/82082#82082 Answer by Laurent Moret-Bailly for étale cohomology with values in an abelian scheme is torsion? Laurent Moret-Bailly 2011-11-28T13:41:59Z 2011-11-28T14:07:36Z <p>For $H^1$ there are many results in chapter XIII of Raynaud's thesis (LNM 119). </p> <p>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 <code>$P:=A_0\times_k L$</code> 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 <code>$Q_{na}$</code> which is also nontrivial unless $n=0$.</p> <p>There are even counterexamples over a normal two-dimensional base, but the construction is harder (XIII.3.2).</p> <p>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.</p> http://mathoverflow.net/questions/82076/etale-cohomology-with-values-in-an-abelian-scheme-is-torsion/82105#82105 Answer by Donu Arapura for étale cohomology with values in an abelian scheme is torsion? Donu Arapura 2011-11-28T18:22:39Z 2011-11-28T18:22:39Z <p>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.</p> <p>While this doesn't directly address your actual question for etale cohomology, my guess would be no for this as well.</p> http://mathoverflow.net/questions/82076/etale-cohomology-with-values-in-an-abelian-scheme-is-torsion/85843#85843 Answer by Timo Keller for étale cohomology with values in an abelian scheme is torsion? Timo Keller 2012-01-16T20:57:08Z 2012-01-16T20:57:08Z <p>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.</p>