$\DeclareMathOperator\sh{sh}$This question is cross-posted from Math.SE where it has gone unanswered for a week -- perhaps it is harder than I guessed. My question is this:

Let $A$ be a local commutative, unital ring and let $A^{\sh}$ be a strict henselization of $A$. Does the structure map $A\rightarrow A^{\sh}$ satisfy Going-Up? (If the answer is no, are there broad natural conditions on $A$ that would make it yes, such as being noetherian, or excellent?)

The Going-Up property does not appear in standard lists of properties of the strict henselization such as those at the Stacks Project and in EGA IV 18.8.12 and 18.8.13. I have some vague reasons to think the strict henselization "should" satisfy Going-Up, and some vague reasons to think it's plausible that it does. But one of these reasons would also seem to apply to the completion, which does not always satisfy Going-Up, see here. In any case, here they are:

  • The most concrete reason I have is that the strict henselization is a colimit of a directed system of étale $A$-algebras, which themselves are localizations of finite $A$-algebras, with the directed system strung together in a way that guarantees every prime of $A$ (has a lift that) survives all the localizations. Finite $A$-algebras do satisfy Going-Up; the localization step can interfere with this by destroying primes, but the structure of the directed system guarantees that (i) $A$'s maximal $\mathfrak{m}$ lifts to the maximal of $A^{\sh}$, so that $A\rightarrow A^{\sh}$ is faithfully flat, thus all primes of $A$ do have a lift in $A^{\sh}$, and (ii) all lifts of any of $A$'s primes in any intermediate $A'$, that are not in the kernel of $A'\rightarrow k^{sep}$, will eventually be destroyed. This does not add up to a proof, of course (or I wouldn't be asking the question), but it does mean that if $A\rightarrow A^{\sh}$ fails to have Going-Up, it does so in kind of a funny way: it seems to me it would have to look like a chain $\mathfrak{p}\subset\mathfrak{q}$ in $A$, and a prime $\mathfrak{P}$ of $A^{\sh}$ lifting $\mathfrak{p}$, and some intermediate étale algebra $A'=B_h$ with $B$ finite over $A$ and $h\in B$, such that none of the lifts of $\mathfrak{q}$ in $B$ containing the pullback of $\mathfrak{P}$ in $B$ (which exist, because $A\rightarrow B$ satisfies going up) are contained in the kernel of $B\rightarrow B_h \rightarrow k^{sep}$, even though this kernel does contain at least one prime that pulls back to $\mathfrak{q}$, and any such prime does contain a prime that pulls back to $\mathfrak{p}$ (this is all known by pulling back to $B$ a chain in $A^{\sh}$ meeting these requirements, which must exist because $A\rightarrow A^{\sh}$ is faithfully flat and thus satisfies Lying-Over and Going-Down). Does this really happen?
  • On a more metaphorical and handwavy level, I think of $A^{\sh}$ as a kind of "universal cover of a very small neighborhood of (the closed point of) $\operatorname{Spec}A$." If I reason by analogy with covering space theory in topology, Going-Up translates to the claim that if I have a flag of irreducible subvarieties all going through the same point, and I have a lift of the highest-dimension of these to the universal cover, then I can extend it to a lift of the whole flag. This is true in covering space theory. (But the imprecision of this reasoning is made clear by considering that it's not obvious why something similar wouldn't also apply in the completion, where Going-Up does not always hold.)
  • On a perhaps even less precise level, there do exist theorems about the strict henselization that seem to hew to the motto, "chains of primes are respected by strict henselization"; for example, $A$ is universally catenary iff $A^{\sh}$ is universally catenary (EGA IV 18.8.17). This is not true for the completion, which is always universally catenary (at least in the noetherian case), whether or not $A$ is.

I'm looking forward to your thoughts. Thanks in advance.


1 Answer 1


Here is a counterexample.

Let $S$ be a smooth surface (irreducible). Let $C_1, C_2 \subset S$ be two disjoint smooth curves in $S$ which happen to be isomorphic. Let $X$ be the result of glueing $C_1$ and $C_2$ by this isomorphism. The singular locus of $X$ is a curve $C$ which is mapped onto isomorphically by $C_1$ and $C_2$ via the morphism $S \to X$. On the other hand, let $Y$ be the result of glueing two copies $S_1, S_2$ of $S$ where $C_1 \subset S_1$ is glued to $C_2 \subset S_2$ via the given isomorphism of $C_1$ with $C_2$. Again the singular locus of $Y$ is a curve $C' \subset Y$ which is mapped onto isomorphically by $C_1 \subset S_1$ and $C_2 \subset S_2$ via the morphism $S_1 \amalg S_2 \to Y$. We may and do think of $S_1$ and $S_2$ as closed subschemes of $Y$ (in fact $S_1$ and $S_2$ are the irreducible components of $Y$). There is a morphism $$ f : Y \longrightarrow X $$ compatible with the obvious morphism $S_1 \amalg S_2 \to S$ and the morphisms $S \to X$ and $S_1 \amalg S_2 \to Y$ mentioned above. For any closed point $c' \in C'$ corresponding to $c = f(c') \in C$ the morphism $f$ is etale at $c'$.

OK, now we are going to choose a curve $D \subset X$ which is smooth, meets $C \subset X$ exactly at $c$. Just take a general smooth curve on $S$ passing through one of the two points above $c$ and take the image. As primes in $\mathcal{O}_{X, c}$ we will use $(0)$ and the prime ideal cutting out $D$. We can do this because $X$ and $D$ are irreducible. Then $f^{-1}(D)$ will have two irreducible components $D_1$ and $D_2$ with $D_1 \subset S_1$ and $D_2 \subset S_2$. But only one of these will pass through $c'$ by our choice of D (to see this I suggest drawing a picture). Say $c' \in D_1$. Then we pick our prime ideal in $\mathcal{O}_{Y, c'}$ to be the prime ideal $\mathfrak P$ cutting out $S_2$ which does indeed lie over $(0) \subset \mathcal{O}_{X, c}$ as $S_2 \to X$ is dominant. But there is no prime containing $\mathfrak P$ in $\mathcal{O}_{Y, c'}$ lying over the prime ideal cutting out $D$ because $D_2$ does not pass through $c'$.


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