I've stumbled upon the statement that the morphism $\pi$ from a root stack of the form $\sqrt[r]{\mathscr{L}/\mathscr{Y}}$ (i.e. the "generic" version, not the one concentrated along a divisor) to its underlying stack $\mathscr{Y}$ is "étale".

Now, I know what "étale" means for representable morphisms (which the above $\pi$ is not). If $\boldsymbol{\mathrm{p}}$ is a property of morphisms of schemes (usually required to be stable under base change), then $f:\mathscr{X}\to\mathscr{Y}$ is $\boldsymbol{\mathrm{p}}$ if every base change $\mathscr{X}\times_\mathscr{Y} U\to U$ is $\boldsymbol{\mathrm{p}}$.

What about the case of a non representable $f:\mathscr{X}\to\mathscr{Y}$ when we have atlases $\alpha:X\to\mathscr{X}$, $\beta:Y\to \mathscr{Y}$ (with $X$ and $Y$,say, schemes)? I had a look to the stacks project but wasn't able to find it.

At least when $\alpha$ and $\beta$ are étale atlases (i.e. when $\mathscr{X}$ and $\mathscr{Y}$ are Deligne-Mumford), I think it would make sense to say that $f$ is $\boldsymbol{\mathrm{p}}$ if $f'':=f'\circ \alpha'$ is $\boldsymbol{\mathrm{p}}$, where $f':\mathscr{X}':=\mathscr{X}\times_{\mathscr{Y}}Y\to Y$ and $\alpha':\mathscr{X}'\times_{\mathscr{X}}X\to\mathscr{X}'$. Edit: This is equivalent to requiring that $X_Y:=X\times_{f\circ\alpha, \mathscr{Y},\beta}Y\to Y$ is $\boldsymbol{\mathrm{p}}$.

Is this one the right definition people use? Is it correct also for Artin stacks?

Edit: Also, how does "my" definition compare with others in the literature?

Edit: So far, it seems that for DM stacks all the definitions agree and work for $\boldsymbol{\mathrm{p}}=$"étale". On page 10 here Alper says "finite morphisms of stacks are necessarily representable": why?

  • $\begingroup$ I think $X\to Y$ is smooth if for all schemes $U$ and all (representable) smooth $U\to Y$, the (representable) morphism $U\to Y$ is smooth. Similar definition should go for etale. I read this in Toric Stacks II, p. 8 arxiv.org/pdf/1107.1907.pdf $\endgroup$ – Ariyan Javanpeykar Nov 20 '15 at 18:13
  • $\begingroup$ @Ariyan: thanks for the pointer, I've just read that page. So, you actually meant that X→Y is smooth if for every smooth cover U→X the composite U→Y is smooth, right? What about other properties such as "étale", or "finite": same criterion? $\endgroup$ – Qfwfq Nov 20 '15 at 19:05
  • $\begingroup$ Sorry for that typo. My guess is that the same definition should go for etale and finite. (I guess one should show that if you have $X\to Y \to Z$ such that $X\to Y$ is P and $X\to Z$ is P then $Y\to Z$ is P. If that's the case, then I think this definition makes sense.) $\endgroup$ – Ariyan Javanpeykar Nov 20 '15 at 19:07
  • 3
    $\begingroup$ @Ariyan Javanpeykar : for finite morphisms, I doubt it. For instance they are defined as quasi-finite and proper in the reference text : Vistoli, Angelo Intersection theory on algebraic stacks and on their moduli spaces.ams.org/mathscinet-getitem?mr=1005008 definition 1.8 $\endgroup$ – Niels Nov 20 '15 at 21:33
  • 1
    $\begingroup$ @pro: I think that taking K\"ahler differentials everywhere doesn't give a cartesian sheaf, since the $\Omega$ is not preserved by pullback along smooth maps. $\endgroup$ – Mattia Talpo Nov 22 '15 at 4:56

For non-representable morphisms of Artin (i.e. algebraic) stacks, different properties are defined in different ways, depending on their particular nature.

E.g. for properties which are smooth local on source and target, see here in the Stacks Project.

The property of being etale is not smooth local on source and target, so this approach doesn't work in that case. (If we restrict to DM stacks, we can use the etale analogue of this definition, and so in that case we get the correct definition of etale; see David Carchedi's answer.)

This doesn't mean that we can't define the notion of etale morphisms, though; it just means that this particular framework doesn't apply.

One approach is via infinitesimal lifting properties: using these one can define what it means for a morphism of stacks (representable or not) to be formally smooth, formally etale, or formally unramified. By imposing locally fin. pres. in the first two cases, or locally fin. type in the third case, one then gets notions of smooth, etale, and unramified.

Alternatively, one can define a morphism to be unramified iff it is loc. finite type and has etale diagonal. (Since diagonal morphisms are always representable, we know what etale means for the diagonal.) We can then define a a morphism to be etale if it is flat, loc. fin. pres., and unramified.

Note that one can also define a smooth morphism using the framework of smooth-local-on-source-and-target discussed above, because being smooth is smooth local.

See Appendix B of this paper by David Rydh for a discussion of these various definitions and their equivalences. (Note though that the affineness condition on the schemes involved in the infinitesimal lifting properties seems to have been accidentally omitted.)

Note also that in the non-representable context, etale is stronger than smooth and quasi-finite, or smooth and relative dimension zero. (Since smooth implies flat, etale is equivalent to smooth and unramified.)

E.g. if G is a positive dimensional smooth alg. group over Spec $k$, then $$BG := [ \mathrm{Spec} \, k / G] \to \mathrm{Spec} \, k$$ is smooth and quasi-finite, but not etale. It is of negative relative dimension (equal to $- \dim G$.
The morphism $$ [\mathbb A^1/ \mathbb G_m] \to \mathrm{Spec }\, k$$ is smooth and of relative dimension zero, but it is not etale.

(An etale morphism is unramified, thus has etale diagonal, thus has unramified diagonal, and thus is a DM morphism. In particular, if the target is a scheme, then the source is a DM stack, which $BG$ (for positive dimensional $G$) and $[\mathbb A^1/\mathbb G_m]$ are not.)

Added later: here is a correct characterization of etale morphisms of Artin stacks as certain kinds of smooth morphisms (akin to the fact that etale morphisms of schemes are smooth morphisms that are locally quasi-finite).

Recall that a morphism of Artin stacks is called Deligne--Mumford if it has unramified diagonal, or equivalently (but non-obviously) if it's base-change over any scheme yields a Deligne--Mumford stack (see here in the Stacks Project, especially footnote 1). (So this is a weakening of representability --- which is equivalent to the diagonal being a monomorphism --- and is automatic for morphisms between DM stacks.)

Then a morphism $X \to Y$ is etale iff it is smooth, Deligne--Mumford, and locally quasi-finite.

(For the proof: note that an unramified morphism has (by definition) an etale, and so unramified, diagonal, hence is Deligne--Mumford. Using this, one can also check that an unramified morphism is locally quasi-finite, because this reduces to checking that a Deligne--Mumford stack which is unramified over the Spec of a field is locally quasi-finite --- which is easy. Since etale morphisms are in particular smooth, we get the only if direction.

For the if direction, by pulling back over a chart of $Y$ we may assume that $Y$ is a scheme, and hence that $X$ is a DM stack. Now we have to check that a smooth and loc. quasi-fin. morphism from a DM stack to a scheme is etale, which is again easy.)

Yet another formulation of the same notion, adopting the view-point expressed by user t3suji in a comment on another answer:

A morphism $X \to Y$ of Artin stacks is etale if it is DM, and if for any morphism $V \to Y$ whose source is a scheme, the base-changed morphism $$X \times_Y V \to V$$ (which is now a morphism from a DM stack to a scheme) is etale (in the sense of morphisms of DM stacks, where we already know what etale means: choose an etale chart $U \to X\times_Y V$, and require that the composite $U \to V$ be etale).

(The equivalence with the various preceding definitions is easily checked.)

| cite | improve this answer | |
  • $\begingroup$ Yes, indeed, in Definition B.5 of my paper "The canonical embedding of an unramified morphism in an étale morphism" the assumption that $T$ is affine is missing (this is crucial in the definition of formal smoothness). Thanks for pointing this out. Also see this question. $\endgroup$ – David Rydh Jun 3 at 10:27
  • $\begingroup$ Let me also add that I think that the "correct" definition of quasi-finite for non-representable maps should impose that the diagonal is (locally) quasi-finite so that stabilizer groups are discrete. This would rule out $BG\to \operatorname{Spec} k$ as quasi-finite when $\dim G\geq 1$. But "smooth + quasi-finite $\implies$ étale" is still not true in characteristic $p$ since $B\pmb{\mu}_p \to \operatorname{Spec} k$ is smooth and quasi-finite but not étale. $\endgroup$ – David Rydh Jun 3 at 10:30

I've been trying to find a good reference for a while, it doesn't seem there are many.

In Simplicial Methods for Operads and Algebraic Geometry, Moerdijk and Toën define an étale $n$-algebraic morphisms of (derived, I'll omit this adjective from now on) stacks $f\colon \mathscr{X}\to \mathscr{Y}$ by requiring that for every scheme $U\to \mathscr{Y}$, the fibered product $U\times_\mathscr{Y}^h \mathscr{X}$ admits a smooth $n$-atlas $V\to U\times_\mathscr{Y}^h \mathscr{X}$ (i.e. a smooth $(n-1)$-algebraic epimorphism from a scheme, $n=0$ means representable) such that $V\to U$ is étale.

If we ignore the "derived" everywhere (I don't know if this is bad in any way) and take $n=1$, this would say that a morphism $\mathscr{X}\to \mathscr{Y}$ between algebraic stacks is étale if for every scheme with a map $U\to \mathscr{Y}$ there exists a smooth atlas $V\to U\times_\mathscr{Y}\mathscr{X}$ such that $V\to U$ is étale.

Now for your case, the base change of $\sqrt[r]{\mathscr{L}/\mathscr{Y}}$ along $T\to \mathscr{Y}$ is $\sqrt[r]{L/T}$ where $L$ is the pullback of $\mathscr{L}$, and since $\sqrt[r]{L/T}$ is a $\mu_r$-gerbe over $T$, if $r$ is invertible you have an atlas $U\to\sqrt[r]{L/T}$ such that $U\to T$ is étale.

EDIT: I'm adding a couple of things in response to comments.

If by "your" definition you mean the one with atlases, that has the problem that I outlined in my first comment to your question, that being étale is not smooth-local in source and target. In the definition I wrote down above, you require something for all maps of schemes to the target, and the existence of an atlas of the pullback that does what you want.

In the DM case this is easier, because being étale is étale local on source and target, and you can just ask it for one pair atlases and it will be true for any.

About affine morphisms, the only definition I've ever seen implies representability. You might be interested in cohomologically affine morphisms (see http://arxiv.org/pdf/0804.2242v3.pdf )

About finite, you can either say "representable finite" (and I think this is the standard) or quasi-finite (that is a condition on geometric points) and proper. Proper is separated (the diagonal, which is representable, is proper), finite type and universally closed (that you define using the associated topological spaces). Alternatively, there is a valuative criterion for properness.

Maybe you've already seen those, but let me also point out this discussion https://math.stackexchange.com/questions/1104790/this-property-is-local-on-properties-of-morphisms-of-s-schemes and this http://stacks.math.columbia.edu/tag/04QW and the neighboring sections of the stacks project.

| cite | improve this answer | |
  • $\begingroup$ Just read your answer. So, if I understand correctly, Moerdijk--Toën's definition is more or less equivalent to mine (assuming "derived" can be dropped)... $\endgroup$ – Qfwfq Nov 21 '15 at 13:28
  • $\begingroup$ [in place of $U\times_\mathscr{Y}^h \mathscr{Y}$ I assumed you meant $U\times_\mathscr{Y}^h \mathscr{X}$] $\endgroup$ – Qfwfq Nov 21 '15 at 13:35
  • 1
    $\begingroup$ nitpick: in the lectures mentioned by Mattia, Moerdijk and Toen write two separate articles (the former on dendroidal sets, the latter is a survey of dag/hag stuff, which is joint with Vezzosi). So it should be the definition of Toen or Toen-Vezzosi. $\endgroup$ – pro Nov 21 '15 at 15:54
  • 1
    $\begingroup$ @Mattia: though the definition formally seems to works also in the Artin case, there is one feature which is not clear to me. The map $BG\to *$, for $G$ positive dimensional, would be étale; do we want this? $\endgroup$ – Qfwfq Nov 21 '15 at 19:09
  • 1
    $\begingroup$ I would assume that an etale morphism of stacks $X\to Y$ is required to be `relative DM'; that is, that any base change from $Y$ to a scheme gives a DM stack. (Obviously, this is my opinion, but I don't see any other well-behaved definition.) $\endgroup$ – t3suji Nov 23 '15 at 16:33

The way I have always used the word étale in reference to a possibly-not-representable morphism of Deligne-Mumford stacks $f:\mathscr{X} \to \mathscr{Y}$ is that for any étale morphism $X \to \mathscr{X}$ from a scheme, the composite $$X \to \mathscr{Y}$$ is étale (both diagonals are representable so this makes sense). This is equivalent to the induced geometric morphism

$$St\left(\mathscr{X}_{et}\right)\to St\left(\mathscr{Y}_{et}\right)$$

being an étale geometric morphism of small étale 2-topoi (i.e. $f:\mathscr{X} \to \mathscr{Y}$ is a stack on the small étale site of $\mathscr{Y}$). Whether or not this is standard, I don't know, but it should be. This definition, at least at first glance, seems reasonable for Artin stacks as well (without the bit on topoi).

| cite | improve this answer | |
  • 6
    $\begingroup$ The problem with this definition is that if $\mathscr{X}$ is not DM, then there's no surjective étale morphism from a scheme. So étale maps from schemes are not going to cover the source, and you're not imposing anything on the map over the `pieces' that are not covered... $\endgroup$ – Mattia Talpo Nov 21 '15 at 3:33
  • 1
    $\begingroup$ Good point. I did say at first glance. At second glance, good point ;-). What would be reasonable for the case of Artin stacks? $\endgroup$ – David Carchedi Nov 21 '15 at 3:54
  • 1
    $\begingroup$ Ok, but at least for DM stacks it works fine, right? How does it compare with "my" definition? And with Moerdijk--Toën's (which, at a first glance, I would think is equivalent to mine)? $\endgroup$ – Qfwfq Nov 21 '15 at 13:32
  • 1
    $\begingroup$ @Qfwfq: It's equivalent to yours for Deligne-Mumford stacks. It's a standard trick that you don't need to check for all etale maps from a scheme, but only on atlases in this case, and then it boils down to exactly what you wrote. $\endgroup$ – David Carchedi Nov 21 '15 at 18:44

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.