Etale cohomology with coefficients in the integers - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:40:08Z http://mathoverflow.net/feeds/question/84414 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84414/etale-cohomology-with-coefficients-in-the-integers Etale cohomology with coefficients in the integers Benjamin Antieau 2011-12-27T21:24:56Z 2012-09-30T02:45:33Z <p>Here is a basic question. When does $H^1_{et}(X,\mathbb{Z})$ vanish? Using the exact sequence of constant etale sheaves $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}\rightarrow 0$, it is enough to show that $H^1_{et}(X,\mathbb{Q})$ vanishes. It is known, for instance by 2.1 of Deninger's 1988 JPAA paper, that $H^1_{et}(X,\mathbb{Q})$ vanishes when $X$ is normal.</p> <p>Note: there are two arguments I think are incorrect that claim to show $H^1_{et}(X,\mathbb{Z})$ always vanishes. The first is that $\mathbb{Z}$ is flasque in the etale topology. This is false. For instance, over the function field $\mathbb{C}(x,y)$, the long exact sequence in cohomology for $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/n\rightarrow 0$ shows that $H^2(\mathbb{C}(x,y),\mathbb{Z})$ is non-zero. So, $\mathbb{Z}$ cannot be flasque. The second argument is that $H^1_{et}(X,\mathbb{Z})=Hom_{cont}(\pi_1^{et}(X),\mathbb{Z})$, where $\pi_1^{et}(X)$ is the etale fundamental group of $X$, which is a profinite group. Since it is profinite, the $Hom$ group above vanishes. But, the claimed equality between $H^1_{et}(X,-)$ and $Hom_{cont}(\pi_1^{et}(X),-)$ only holds for torsion sheaves, as far as I have been able to determine.</p> <p>I am in fact interested in several things. First, either an example of $X$ such that $H^1_{et}(X,\mathbb{Z})$ is non-zero, or a proof that this always vanishes. Second, the same thing but where we only look at affine $X$. In particular, if it exists, I would love to see an example of a commutative ring $R$ where $H^1_{et}(Spec R,\mathbb{Z})$ is non-zero, if this is possible.</p> http://mathoverflow.net/questions/84414/etale-cohomology-with-coefficients-in-the-integers/84417#84417 Answer by Angelo for Etale cohomology with coefficients in the integers Angelo 2011-12-27T22:23:36Z 2011-12-27T22:23:36Z <p>The standard example is a copy of $\mathbb A^1_k$, where $k$ is an algebraically closed field, with two points glued. In algebraic terms, $X = \mathop{\rm Spec}R$, where $R := k[x,y]/(y^2 - x^3 + x^2)$. Consider the finite morphism $\pi\colon \mathbb A^1 \to X$, which yields an exact sequence $$0 \to \mathbb Z_X \to \pi_*\mathbb Z_{\mathbb A^1} \to i_*\mathbb Z_p \to 0,$$ where $p$ is the singular point of $X$. Since both $\pi_*\mathbb Z_{\mathbb A^1}$ and $i_*\mathbb Z_p$ have trivial étale cohomology , by taking global sections we see that $\mathrm H^1(X, \mathbb Z) = \mathbb Z$.</p> http://mathoverflow.net/questions/84414/etale-cohomology-with-coefficients-in-the-integers/108435#108435 Answer by Thomas Geisser for Etale cohomology with coefficients in the integers Thomas Geisser 2012-09-30T02:45:33Z 2012-09-30T02:45:33Z <p>More generally, if $X$ is proper over an algebraically closed field, then $H^1(X,\mathbb Z)$ is isomorphic to the cocharacter module of the maximal torus of the Picard variety $Hom(\mathbb G_m,Pic^0)$.</p>