# Are completions stalks under some Grothendieck topology?

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 $R_{\mathfrak{p}}$).

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?

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The stalk of a sheaf $F$ at a point $p$ is always some sort of colimit of sets of sections $F(U)$ -- this is true for any topology with enough points. The completion looks like an (inverse) limit. So the naive answer to your question is no. – Charles Rezk Dec 19 '10 at 3:07
That sounds about right. I wonder how Grothendieck thought of these things. Surely it occurred to him that these notions don't work together, and yet that they are analogous... – James D. Taylor Dec 19 '10 at 5:42
I use to heuristically see $\mathrm{Spec}(\mathcal{O}_{X,x})$ as an "intersection of open neighbourhoods of $x$", while $\mathrm{Spec}(\widehat{\mathcal{O}}_{X,x})$ as an "increasing union of closed subsets containing $x$". – Qfwfq Dec 23 '10 at 22:10

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 $x$.
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 $X$.