# Is the strict henselization isomorphic to the filtered colimit of finite etale algebras?

Let $(A,\mathfrak{m})$ be a local ring, and let $A^{\mathrm{sh}}$ be the strict henselization of $A$ at $\mathfrak{m}$. Let me denote $A^{\mathrm{sh},\mathrm{fin}}$ for the filtered colimit of finite etale $A$-algebras (with a fixed map to the separable closure of $A/\mathfrak{m}$). There is a canonical map \begin{align} \varphi : A^{\mathrm{sh},\mathrm{fin}} \to A^{\mathrm{sh}} \end{align} of $A$-algebras. When is $\varphi$ an isomorphism?

Remarks/thoughts:

1. This is true if $A$ is a field.
2. We can consider using Zariski's Main Theorem (say this version) in some way. If $A^{\mathrm{sh},\mathrm{fin}} \to B$ is an etale ring map, there exists a factorization $A^{\mathrm{sh},\mathrm{fin}} \to C \to B$ where $A^{\mathrm{sh},\mathrm{fin}} \to C$ is finite and $C \to B$ induces an open immersion $\operatorname{Spec} B \to \operatorname{Spec} C$, but I don't know whether $A^{\mathrm{sh},\mathrm{fin}} \to C$ is etale (or whether it can be made etale after a refinement of $B$). It seems we can make $A^{\mathrm{sh},\mathrm{fin}} \to C$ finite flat, namely using the structure theorem for etale ring maps (e.g. 00UE) which says we can take $C = A^{\mathrm{sh},\mathrm{fin}}[t]/(f(t))$ for some monic polynomial $f(t) \in A^{\mathrm{sh},\mathrm{fin}}[t]$ and $B$ to be a principal localization of $C$.
• Perhaps this will help: stacks.math.columbia.edu/tag/0BSK Commented Apr 26, 2018 at 21:03
• @WilliamChen: it seems that the main point is the adjective finite, as opposed to considering arbitrary étale algebras. Commented Apr 27, 2018 at 3:14
• One of the reasons étale algebras (as opposed to finite étale algebras) take a central role in the theory is that a finite étale $A$-algebra is only semi-local. The proof that $A^{\operatorname{sh}}$ is local relies crucially on the ability to localise away any unwanted primes, which is not available in the finite étale case. It seems that your $A^{\operatorname{sh,fin}}$ is not necessarily a local ring. Commented Apr 27, 2018 at 3:30
• I guess that if $A$ is henselian, then $A^\text{sh, fin}=A^\text{sh}$. This might well be the only case. Commented Apr 27, 2018 at 7:28
• @R.vanDobbendeBruyn Ah good point I missed that. I guess it boils down to whether or not you can have a non-henselian local ring for which all finite etale extensions split into products of local rings. (Non-henselianness guarantees that there exist finite extensions which don't split, but it doesn't seem to imply that we can always find a finite etale extension which doesn't split) Commented Apr 27, 2018 at 19:34

If your proposed description is correct, then the strict henselization of $A$ would be integral over $A$, and hence the same holds true for any subring of the strict henselization. But this is essentially never true (unless $A$ is henselian).
For an explicit example, take $A = \mathbf{C}[x]_{(x)}$. Choose a map $f:X \to \mathbf{A}^1$ of smooth affine curves of degree $3$ such that $f^{-1}(0) = \{x,y\}$ with $x$ unramified and $y$ ramified. Take $B = \mathcal{O}_{X,x}$, so $\mathrm{Spec}(B)$ is the affine open subscheme of $X \times_{\mathbf{A}^1} \mathrm{Spec}(A)$ obtained by removing $y$. Then $A \to B$ is a local \'etale map of local domains with the same residue field, so $B$ occurs as a subring of the strict henselization of $A$. But $A \to B$ is not integral as it fails the valuative criterion of properness: there is a valuation on $\mathrm{Frac}(B) = K(X)$ corresponding to the point $y \in X$ that has a center on $\mathrm{Spec}(A)$ but not one on $\mathrm{Spec}(B)$.