Let $(A,\mathfrak{m})$ be a local ring. (I'd be ok assuming it's noetherian, if that's helpful, although I'd also be surprised if the answer really depended on this.) Let $A^{sh}$ be a strict henselization of $A$. My question is this:
Does the structure map $A\rightarrow A^{sh}$ satisfy Going-Up?
Intuitively, I think it should, for reasons I'll mention in a moment; on the other hand, in standard lists of basic properties of the strict henselization such as those at Stacks Project and EGA IV 18.8.12 and 18.8.13, I don't recall seeing this mentioned.
The (quite hand-wavy) reasons I believe that $A\rightarrow A^{sh}$ "ought" to satisfy Going-Up are as follows:
- Fixing a map $\phi: A\rightarrow k^{sep}$ from $A$ to a separable closure $k^{sep}$ of its residue field, $A\rightarrow A^{sh}$ is the limit over diagrams of the form $A \xrightarrow{f} A' \xrightarrow{\psi} k^{sep}$, where $f$ is étale and $\phi = \psi\circ f$. (This is the definition of $A^{sh}$ given in Milne's book on étale cohomology, see p. 38.) In general, étale extensions don't satisfy Going-Up. However, this is only because they allow a "certain amount of localization": by this I mean that any étale extension is, up to isomorphism, standard étale, i.e., of the form $A[x]_h/(g)$, where $g,h \in A[x]$ and $g$ is monic (there's also a condition how $g,h$ relate); it is a localization of the integral extension $A \rightarrow A[x]/(g)$, which does satisfy Going-Up. Going-Up can thus only fail if the "localization step" of inverting $h$ destroys the primes of $A[x]/(g)$ that lie over a target prime of $A$ in the right way. Furthermore, the étale extensions $A'$ used in constructing $A^{sh}$ are not arbitrary but precisely those having a maximal ideal that pulls back to $\mathfrak{m}$ in $A$; this is what makes it possible for them to fit into the diagram $A \xrightarrow{f} A' \xrightarrow{\psi} k^{sep}$. Thus $\mathfrak{m}$, and therefore any prime of $A$ (as they are all contained in $\mathfrak{m}$), "survives the localization step" in the construction of all the étale extensions used to build $A^{sh}$. Now it seems to me that, given a prime $\mathfrak{p}$ of $A$, some of the primes lying over it in the intermediate étale extensions $A'=A[x]_h/(g)$ will be contained in other maximals of $A[x]/(g)$ than the one that ends up being the kernel of $\psi:A'\rightarrow k^{sep}$, which could therefore get destroyed by the localization at $h$; so the intermediate $A'$'s probably won't satisfy Going-Up. However, any such primes will be destroyed by a later localization, since $A^{sh}$ is local. Thus it seems to me that in each intermediate $A'$, Going-Up is going to hold for the primes inside $\ker \psi$ (I mean if $\mathfrak{q}\subset \ker \psi$ pulls back to $\mathfrak{p}$ in $A$, and $\mathfrak{p}_1\subset A$ contains $\mathfrak{p}$, then there exists $\mathfrak{q}_1\subset\ker \psi$ pulling back to $\mathfrak{p}_1$); and then in $A^{sh}$, a similar situation will exist with respect to the kernel of $A^{sh}\rightarrow k^{sep}$, except that these will be all the primes. This train of thought is, of course, rather soft (or I wouldn't be asking the question).
- It is a standard property of $A\rightarrow A^{sh}$ that it is faithfully flat and of relative dimension zero. Thus Going-Down, Lying-Over, and Incomparability obtain for $A\rightarrow A^{sh}$. It is not the case that every faithfully flat map of relative dimension zero satisfies Going-Up (see here); still, it renders it at least plausible in this case?
- This is insanely vague, but: while I haven't seen Going-Up listed as a standard property of $A\rightarrow A^{sh}$, other standard properties do suggest that it "behaves well" with respect to chains of primes. E.g., $A^{sh}$ is universally catenary if and only if $A$ is universally catenary (EGA IV 18.8.17).
- This is perhaps even more insanely vague. I understand $\operatorname{Spec}A^{sh}$ geometrically as a kind of "universal cover of a very small neighborhood of [the closed point of] $\operatorname{Spec} A$". Ascending chains of primes in $A$ are, geometrically, descending flags of irreducible subschemes each of which hits the closed point of $\operatorname{Spec}A$. Thinking by analogy with covering space theory in, e.g., the category of smooth manifolds, any such flag ought to be able to be seen in a very small neighborhood, and ought to be able to be lifted to a universal cover. If I start with a lift of some irreducible subscheme of $\operatorname{Spec}A$ to $\operatorname{Spec}A^{sh}$, I ought (again purely by analogy with topological covering space theory) be able to lift any irreducible sub-subscheme downstairs to a sub-subscheme upstairs as well. This would be Going-Up if the analogy actually holds.
I'm looking forward to your thoughts. Thanks in advance.
