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:

- This is true if $A$ is a field.
- 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 $C$.

finite, as opposed to considering arbitrary étale algebras. $\endgroup$ – R. van Dobben de Bruyn Apr 27 '18 at 3:14finiteétale 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. $\endgroup$ – R. van Dobben de Bruyn Apr 27 '18 at 3:30etaleextension which doesn't split) $\endgroup$ – Will Chen Apr 27 '18 at 19:34