Are completions stalks under some Grothendieck topology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:43:26Z http://mathoverflow.net/feeds/question/49843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49843/are-completions-stalks-under-some-grothendieck-topology Are completions stalks under some Grothendieck topology? James D. Taylor 2010-12-19T01:47:39Z 2010-12-19T09:56:44Z <p>Let $R$ be a ring, and $\mathfrak{p}$ be a prime ideal. The stalk at $\mathfrak{p}$ with respect to the etale topology is $(R_{\mathfrak{p}})^{sh}$ (the strict henselization of $R_{\mathfrak{p}}$). The stalk at $\mathfrak{p}$ with respect to the Nisnevich topology is $(R_{\mathfrak{p}})^h$ (the henselization of $R_{\mathfrak{p}}$).</p> <p>Grothendieck also spoke of formal neighborhoods, and I wonder if this fits into the pattern above. To be precise: is there some Grothendieck topology for which the stalk at $\mathfrak{p}$ would be the completion of $R_{\mathfrak{p}}$ with respect to $\mathfrak{p}$? If so, what is it?</p> http://mathoverflow.net/questions/49843/are-completions-stalks-under-some-grothendieck-topology/49852#49852 Answer by Emerton for Are completions stalks under some Grothendieck topology? Emerton 2010-12-19T06:21:16Z 2010-12-19T09:56:44Z <p>You may want to look at the Artin approximation theorem. Roughly, it says (in the context of varieties, say) that any phenomenon you observe at the level of completions is already achieved etale locally. </p> <p>E.g. suppose that $X$ is an algebraic curve over $k$ (an algebraically closed field) and $x$ is a closed point of $X$ such that completed local ring $\hat{\mathcal O}_{X,x}$ has two irreducible components in its spectrum. Then Artin's theorem says that there is an etale n.h. of $x$ which is the union of two branches passing through $x$.</p> <p>This has the practical consequence that notions such as ``$x$ is a node'' can be defined either by a condition on the completion $\hat{\mathcal O}_{X,x}$ or by an etale local condition. The former is more elementary, and usually easier to check; the latter is often more powerful in proofs, because it is more tightly connected to the geometry of the whole curve $X$.</p>