EDIT: Thanks to Harry Gindi and Anonymous below for insightful comments, I've refined the definitions here. Recall that a formally etale morphism of schemes $Y \to X$ is a morphism which has the unique right lifting property with respect to all nilpotent thickenings $Z \to W$. One salient feature of nilpotent thickenings is that they are universal homeomorphisms.

Definition: Say that a morphism of schemes $Y \to X$ is strongly formally etale if it has the unique right lifting property with respect to all universal homeomorphisms $Z \to W$. That is, for every commutative square as below, there exists a unique diagonal filler $W \to Y$, as indicated, making the two triangles commute.

$$\require{AMScd} \begin{CD} Z @>>> Y \\ @VVV \nearrow @VVV\\ W @>>> X \end{CD}$$

By definition, then, if $Y \to X$ is strongly formally etale, then $Y \to X$ is formally etale. The converse presumably does not hold. However, etale morphisms have an additional finiteness condition (etale = formally etale + locally of finite presentation) which makes me hope for an affirmative answer to the first question below:


  1. Let $Y \to X$ be an etale morphism. Then is $Y \to X$ strongly formally etale?

  2. Does there exist standard terminology for "strongly formally etale"?

  3. Is there a characterization of the class of morphisms which have the unique left lifting property with respect to all etale morphisms? How about (strongly) formally etale?

  • $\begingroup$ @downvoter I'd welcome any criticism you may have. $\endgroup$ – Tim Campion Apr 16 '20 at 14:59
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    $\begingroup$ I'm not the downvoter, but the notion of 'weakly nilpotent' that you have is too weak. You need the condition to be pullback stable. As soon as you do that, you get the notion of universal homeomorphism, and I think this becomes related to a theorem of Grothendieck and also to some open problems in anabelian geometry. $\endgroup$ – Harry Gindi Apr 16 '20 at 15:00
  • $\begingroup$ @HarryGindi Thanks! What does the theorem of Grothendieck say? $\endgroup$ – Tim Campion Apr 16 '20 at 15:02
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    $\begingroup$ A universal homeomorphism induces an equivalence of étale topoi. Barwick has some notes about this whole question: maths.ed.ac.uk/~cbarwick/papers/trig1.pdf $\endgroup$ – Harry Gindi Apr 16 '20 at 15:02
  • $\begingroup$ Being "weakly nilpotent" is a really weak condition: any field extension gives such a map. Corresponding, being "strongly formally etale" is a ridiculously strong condition. For example, given a field $k$, a $k$-algebra domain $A$ is strongly formally etale over $k$ only when $A=k$. $\endgroup$ – Anonymous Apr 16 '20 at 15:16

This follows from the well-known "topological invariance of the étale site"; see e.g. Tag 04DZ.

Indeed, write $S' \to S$ for $Z \to W$, and consider the pullbacks of the étale map $Y \to X$ to étale maps $T \to S$ and $T' \to S$. We need to show that a section to $T' \to S'$ uniquely arises by pullback from a section of $T \to S$. This is a special case of Tag 0BTY.

Presumably the same should hold for weakly étale affine morphisms, since these are very close to ind-étale; see Tag 097Y. The only "topological invariance of the pro-étale site" I have seen is Lemma 5.4.2 is Bhatt–Scholze, which is only on the level of sheaves (rather than an actual equivalence of sites).

  • $\begingroup$ Is it necessary for the weakly étale morphisms to be affine? It seems to me that when a weakly étale morphism $Y\rightarrow X$ gets tested against a universal homeomorphism $Z \rightarrow W$, by shrinking neighborhoods around chosen points repeatedly, one can make all the schemes involved affine (while maintaining that the restriction $Z'\rightarrow W'$ is a universal homeomorphism, using that univ. homeo's are affine). After obtaining the lifts in the affine situation, uniqueness of the lifts should allow them to glue to a global one, no? $\endgroup$ – Pavel Čoupek Apr 17 '20 at 2:50
  • $\begingroup$ @PavelČoupek even in the étale case I'm not sure how to do this. When you choose an affine in $X$, you want its preimage in $Y$ to still be affine, or at least to contain an affine containing the image of $Z$. I see no a priori reason why this should be true. (Like in my answer, you may assume $W = X$ if you want, but I don't see how that helps.) $\endgroup$ – R. van Dobben de Bruyn Apr 17 '20 at 2:54
  • $\begingroup$ I thought that after choosing affine $X'$ in $X$, you just choose some affine $Y'$ of $Y$ in the preimage of $X'$. Then $Y' \rightarrow X'$ should be still w. étale since w. étale is local on the source and target. After that, some yoga on the LHS: take similarly $W'$ affine in $W$ in preimage of $X'$, then intersect the preimages of $Y', W'$ in $Z$ and choose affine open $Z'$ inside this intersection. Image of $Z'$ in $W$ is open in $W'$, so after possibly takings even smaller affine $W''$ inside, taking preimage $Z''$ of $W''$; then $Z'' \rightarrow W''$ is univ. homeo. Is this incorrect? $\endgroup$ – Pavel Čoupek Apr 17 '20 at 3:15
  • $\begingroup$ @PavelČoupek: the thing that's unclear to me is how is lifting of this local version going to imply the global version. The hardest part is probably showing that $W$ can be covered by such $W''$. But you might be right that this just works; I just didn't see an obvious argument. $\endgroup$ – R. van Dobben de Bruyn Apr 17 '20 at 4:11
  • $\begingroup$ I think that is OK because $Z \rightarrow W$ is bijective. Given $w \in W$, let $z$ be the preimage of $w$ in $Z$, and denote by $y, x$ their images in $Y, X,$ resp. (in particular, $y \mapsto x$). Then replacing "affine open" by "affine neighbourhood of [the respective fixed point]" in the above, one produces $W''$ containing $w$. Of course, you might not cover $Y$ this way, but you cover at least an open subscheme containing the full (set-theoretic) image of $Z \rightarrow Y$, which agrees with the image of any possible lift $W \rightarrow Y$ (again due to $Z \rightarrow W$ bijective). $\endgroup$ – Pavel Čoupek Apr 17 '20 at 4:35

EDIT: I edited this answer rather substantially since I misinterpreted the question a bit at first. Sorry about that.

Formally étale is indeed very weak with respect to what you want:

For example, let me consider the case of a $k$-algebra $A=k[X^{p^{-\infty}}]$ and the map $A \rightarrow k$ given by killing all the variables. This is formally étale but obviously does not have the unique lifting property with respect e.g. to the thickening $A/(x) \rightarrow k,$ which is again given by killing all the variables. A version of this example (with details showing why this is formally étale) is given in this answer.

One candidate for what strongly formally étale might be is what was called ln-formally étale last year in M. Morrow's project for the Arizona winter school: Notes with project description, it is project C.1, towards the end of the notes. The "ln" refers to the fact that the lifting property considered there is with respect of killing locally nilpotent ideals. (You can also read there that "ln-formally étale" is an ad hoc term. So does not really answer the question about standard terminology.)

Another candidate is the notion of weakly étale morphism, from B. Bhatt's and P. Scholze's The pro-étale topology for schemes (arXiv). A morphism $X \rightarrow Y$ is weakly étale if both itself and its diagonal $\Delta_{X/Y}$ are flat.

It seems that your "strongly formally étale" sits somewhere in between the two (or agrees with one of them):

Clearly strongly formally étale morphism has to be ln-formally étale, since locally nilpotent thickenings are universal homeomorphisms. On the other hand, the argument used in the above linked AWS notes showing "weakly étale $\Rightarrow$ ln-formally étale" should verbatim show that a weakly étale morphism is strongly formally étale: In terms of algebras, if $A \rightarrow B$ is weakly étale, then there is a faithfully flat map $B \rightarrow C$ such that the composite $A \rightarrow C$ is a filtered colimit of étale $A$-algebras (Theorem 1.3 of the Bhatt-Scholze paper linked above). But such algebras have the desired lifting property since étale algebras do, and hence, by fpqc descent of morphisms, so does $A \rightarrow B$.

Unfortunately, it turns out that weakly étale is strictly stronger that ln-formally étale, so there is still some room there. An example demonstrating this is, more or less, the above mentioned example of formally étale algebra, on steroids. A writeup of the example is here.

ADDED: Let me show how to prove, assuming the unique lifting property for weakly étale maps between affine spaces discussed above, that all weakly étale maps have the unique lifting property:

As in the question, assume the commutative square $$\require{AMScd}(*)\;\;\;\; \begin{CD} Z @>>> Y \\ @VVV \nearrow @VVV\\ W @>>> X \end{CD}$$ where $W \rightarrow Z$ is a universal homeomorphism and $Y \rightarrow X$ is weakly étale. We want to show that the indicated diagonal map, making the whole thing commutative, uniquely exists. (Clearly it uniquely exists as a map of topological spaces since $Z \rightarrow W$ is a homeomorphism - call this map of topological spaces $f$.)

Consider the collection of all commutative squares $$\require{AMScd}(*')\;\;\;\; \begin{CD} Z' @>>> Y' \\ @VVV @VVV\\ W' @>>> X' \end{CD}$$ such that

1) all the schemes $W'. X', Y', Z'$ are affine open subschemes of $W. X, Y, Z,$ resp., and the maps in the square are obtained by restriction from $(*)$,

2) $Z'$ is the preimage of $W'$, or equivalently, $Z' \rightarrow W'$ is still a universal homeomorphism.

Since the property of $Y \rightarrow X$ being weakly étale is local on the source and the target, it follows that the map $Y' \rightarrow X'$ is weakly étale automatically; thus, assuming (1) and (2), a diagonal map $f': W' \rightarrow Y'$ fitting commutatively into the diagram $(*')$ uniquely exist (and, forgetting the scheme structure, it is clear that this lift as a topological map is obtained by restriction of $f$). So one just needs to glue these maps to a one map $W \rightarrow Z$. To show that this works, one needs the following two claims:

Claim 1: Each point $w \in W$ is contained in $W'$ for some square $(*')$ as described above.

Claim 2: Given any $w \in W$ and squares $(*'),(*'')$ as above such that $w \in W' \cap W'', f(w) \in Y' \cap Y'',$ there exists another square $(*''')$ as above with $w \in W''' \subseteq W' \cap W'', f(w) \in Y''' \subseteq Y' \cap Y'' $.

Assuming both claims, the lifts $f': W' \rightarrow Y' \hookrightarrow Y$ necessarily glue to the desired map $f: W \rightarrow Y$ (as a map of schemes): Claim 1 makes sure that the map is everywhere defined while Claim 2 implies that the lifts agree on overlaps, because of uniqueness of the lifts (this can always be checked on a basis, in our case basis formed by the affine opens $W'''$ as in the claim). Commutativity of the upper triangle then can be checked again locally, so noting that Claim 1 also implies that the full space $Z$ gets covered (thanks to $Z \rightarrow W$ and all $Z' \rightarrow W'$ being homeomorphisms), the upper of the two triangles commutes, and similarly for the lower triangle. Uniquness follows from uniqueness locally.

So what remains is to check the two claims. For the first one, consider $w \in W$. Since $Z \rightarrow W$ is a homeomorphism, there is a unique $z \in Z$ mapping to $w$. Denote by $y$ the image of $z$ in $Y$ and by $x$ the image of $w$ in $X$ (so by commutativity of $(*),$ $y \mapsto x$). First choose $x \in X' \subseteq X$ an affine open neighborhood, then $y \in Y' \subseteq Y$ an affine open neighborhood contained in the preimage of $X'$. Next, let $W_1$ denote the preimage of $X'$ in $W$, and let $Z_1$ be the intersection of the preimages of $Y'$ and $W_1$ in $Y$. Since the homeomorphism $Z \rightarrow W$ is open, the image $W_2$ of $Z_1$ in W is an open neighborhood of $w$ contained in $W_1$. Finally, take $x \in W'\subseteq W_2$ an affine open neighborhood, and $Z'$ the preimage of $W'$. Universal homeomorphisms are affine, so $Z'$ is also affine, and hence $W', X', Y', Z'$ assemble to the desired square $(*')$.

The proof of Claim 2 is similar: fixing the same notation for the four points $x, y, z, w$ as above, one sees that $f(w)=y.$ So one can perform the same yoga as for the proof of claim 1, except with the extra condition that $Y'''$ (what was denoted by $Y'$ in the above paragraph) should also be contained in the prescribed intersection $Y' \cap Y''$, and similarly when choosing $W'''$, one has to choose it sufficiently small (inside $W' \cap W''$).


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