5
$\begingroup$

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 $C$.
$\endgroup$
  • 1
    $\begingroup$ Perhaps this will help: stacks.math.columbia.edu/tag/0BSK $\endgroup$ – Will Chen Apr 26 '18 at 21:03
  • 1
    $\begingroup$ @WilliamChen: it seems that the main point is the adjective finite, as opposed to considering arbitrary étale algebras. $\endgroup$ – R. van Dobben de Bruyn Apr 27 '18 at 3:14
  • 3
    $\begingroup$ 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. $\endgroup$ – R. van Dobben de Bruyn Apr 27 '18 at 3:30
  • 3
    $\begingroup$ I guess that if $A$ is henselian, then $A^\text{sh, fin}=A^\text{sh}$. This might well be the only case. $\endgroup$ – Laurent Moret-Bailly Apr 27 '18 at 7:28
  • $\begingroup$ @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) $\endgroup$ – Will Chen Apr 27 '18 at 19:34
3
$\begingroup$

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 $\mathrm{Spec}(B)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.