most recent 30 from http://mathoverflow.net 2013-05-21T07:15:07Z http://mathoverflow.net/feeds/question/47212 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47212/universal-homeomorphisms-and-the-etale-topology Laurent Moret-Bailly 2010-11-24T09:11:51Z 2010-11-24T22:21:44Z

<p>Let $f:X\to S$ be a universal homeomorphism of schemes. Assume $X(S')\neq\emptyset$ for some étale surjective $S'\to S$. Does $f$ have a section?</p> <p>The answer is yes if $S$ is reduced, by descent. Indeed, note that if $S_1$ is a reduced $S$-scheme then $X(S_1)$ has at most one element. Apply this to $S_1=S'\times_S S'$.</p> <p>Interesting special case: if $S$ has prime characteristic $p$, let $G$ be a finite locally free $S$-group scheme with connected (i.e. "infinitesimal") fibers, such as $\alpha_p$ or $\mu_p$. Is $H^1_{\mathrm{et}}(S,G)$ trivial?</p>

http://mathoverflow.net/questions/47212/universal-homeomorphisms-and-the-etale-topology/47277#47277 Angelo 2010-11-24T22:21:44Z 2010-11-24T22:21:44Z

<p>I think that the following might work. Let $X_0$ be a reduced scheme over a field $k$ of characteristic $p > 0$, and let $X$ be the product of $X_0$ with the ring of dual numbers $k[\epsilon]$. Then $\mathcal O_X = \mathcal O_{X_0} \oplus \epsilon\mathcal O_{X_0}$, and the $p^{\rm th}$ roots of 1 are those of the form $1 + \epsilon f$; hence the Zariski sheaf of $p^{\rm th}$ roots on 1 on $X$ is isomorphic to $\mathcal O_{X_0}$. Hence if $\mathrm H^1(X_0, \mathcal O_{X_0}) ≠ 0$ there is a non-trivial $\mu_p$-torsor on $X$ that is locally trivial in the Zariski topology, thus giving a counterexample.</p>