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I am curious about the analogue of this question, as stated in the title. Namely,

If $Z \subset X$ is a closed subscheme and $Y \to Z$ is faithfully flat (let's also say of finite presentation), can we find a map $Y' \to X$ which is flat, whose image contains $Z$ (and is thus a neighborhood of $Z$), and whose restriction to $Z$ is the given map?

This may be too strong; i.e. just as for étale maps, in the other question, this may be true only Zariski locally on $X$. That's fine too. In that case, in its most basic form it becomes a question of commutative algebra:

If $R$ is a ring, $I \subset R$ an ideal, and $S$ a faithfully flat $R/I$-algebra, is there a set of elements $f \in R$ such that $\sum (R/I) f = R/I$ and a corresponding set of faithfully flat $R_f$-algebras ${}_f\widetilde{S}$ such that ${}_f \widetilde{S}/I_f \cong S_f$ for each $f$?

Also, it would be nice to know that, in the event that this is true, it preserves finiteness hypotheses like "finitely presented".

For those who are curious about my motivations: I want to show that "$!$ pushforwards (resp. pullbacks) commute with $*$ pushforwards (resp. pullbacks)" under appropriate circumstances, those being when the $!$'s are along closed immersions and the $*$'s are along faithfully flat covers or vice-versa. At the very least I need to be able to switch the order in which maps of these types appear; that is, rewrite a cover of a closed immersion as a closed immersion into a cover, as my question asks.

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    $\begingroup$ I think the first $Y$ should not be the same object as the second $Y$. $\endgroup$
    – S. Carnahan ♦
    Commented Dec 26, 2011 at 4:34
  • $\begingroup$ You're right, of course. Fixed. $\endgroup$
    – Ryan Reich
    Commented Dec 26, 2011 at 5:32
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    $\begingroup$ I think this question is asking for something too strong; here's an example. Let R = Z_p, and let S_0 be a finite flat F_p-algebra that does not lift (as in mathoverflow.net/questions/63969/…). Since R is local, the present question asks: does there exist a faithfully flat R-algebra S lifting S_0? If there was such an S, then the p-adic completion of S would be a finite flat R-algebra lifting S_0 (by Nakayama, completeness, and finiteness modulo p), which cannot exist. $\endgroup$
    – Bhargav
    Commented Dec 26, 2011 at 17:34
  • $\begingroup$ @Bhargav: That is reasonably convincing. Can you think of an alternative, weaker formulation that excludes this counterexample? I am not sure what underlies this phenomenon: is it mixed characteristic? In any case, you should make your comment an answer. $\endgroup$
    – Ryan Reich
    Commented Dec 26, 2011 at 21:39
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    $\begingroup$ @Ryan: I am not sure if this is mixed characteristic specific. It seems reasonable to me that one can cook up similar examples over C with non-smooth points of Hilbert schemes of points, though I didn't try. As far as a fix is concerned, I don't have a solution, but a guess could be to allow fppf refinements on what you call Y, though my gut feeling is that even this is too strong. $\endgroup$
    – Bhargav
    Commented Dec 27, 2011 at 9:57

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(Essentially copied from the comments as requested.)

I think this question is asking for something too strong; here's an example. Let $R = \mathbf{Z}_p$, and let $S_0$ be a finite flat $\mathbf{F}_p$-algebra that does not lift to a finite flat $R$-algebra; an explicit example can be found here. Since $R$ is local, the present question asks: does there exist a faithfully flat $R$-algebra $S$ lifting $S_0$? If there was such an $S$, then the $p$-adic completion $\widehat{S}$ of S would be a finite flat R-algebra lifting $S_0$ (the flatness is clear, and the finiteness comes from Nakayama, completeness, and finiteness modulo $p$), which cannot exist. Hence, there is no such $S$ either.

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