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The answer to my question might very well be standard, but I have had trouble finding the right keywords to search for it, so I apologize if this is something well-known to the experts.

I am learning the basics of the theory of algebraic stacks and, as probably many entering the field, I was perplexed by the fact that in the definition one merely demands the diagonal to be representable by algebraic spaces rather than (like in the definition of algebraic spaces themselves) representable by schemes. However, the example presented in this answer has greatly clarified why it is this notion that one may want to consider: the (suitably defined) stack $\mathscr M_1$ of genus 1 curves admits pairs of objects $E,X$ such that the isomorphism object between them is an algebraic space but not a scheme, which means precisely that the diagonal is not representable by schemes. The reason for this is the fact that $X$ is an $E$-torsor which is not a scheme, and so $X$ embeds into ${\underline{\mathrm {Isom}}}_{\mathscr{M}_1}(E,X)$.

However, this example, while conceptually reasonable, is relatively difficult to construct. I have recently found out about another example of a moduli stack which has objects which are not schemes: the stack of nodal curves (see Example 2.3 here for an exposition). This example, based on Hironaka's non-analytic complex 3-folds, I found much easier to understand. However it doesn't seem to lead to another example of non-representability of the diagonal by schemes - as there are no torsors in sight, this object doesn't seem to embed into any isomorphism object. The first question I have is whether this is the case or not:

Question 1: Let $\mathscr{M}_\mathrm{nod}$ be the moduli stack of rational nodal curves as defined in the above paper. Is it true that the diagonal $\mathscr{M}_\mathrm{nod}\to\mathscr{M}_\mathrm{nod}\times\mathscr{M}_\mathrm{nod}$ is not representable by schemes?

More generally I am interested in the following. I don't expect the answer to be always positive, but I suspect there should be some guiding principles in this direction, of which the above would be the special case:

Question 1.1: Consider a moduli stack $\mathscr{M}$ which classifies some kind of objects, and suppose that there are objects in it which are etale- (or fppf-, or fpqc-)locally schemes, but are not themselves schemes. Should one then expect the diagonal $\mathscr{M}\to\mathscr{M}\times\mathscr{M}$ to not be representable by schemes? What about the converse?

In the converse direction, the only stacks for which I know the behavior of the diagonal is for stacks of algebraic curves $\mathscr{M}_{g,n}$, and those are representable by schemes thanks to some projectivity results and representability of the Hilbert scheme.

Loosely related question concerns the situation for Deligne-Mumford stacks. I am certain some DM stacks which have diagonals not representable by schemes, although I don't think any of the moduli of (stable) curves provide an example.

Question 2: Are there some "natural" moduli problems which are representable by a DM stack whose diagonal is not representable by schemes?

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  • $\begingroup$ I am probably misunderstanding Q2. But isn't the diagonal of a DM stack by definition unramified, and thus representable? $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 10, 2021 at 14:42
  • $\begingroup$ @AriyanJavanpeykar Apologies if this is a sign of my ignorance, but I'm not aware how to conclude from diagonal being unramified that it is representable by schemes. $\endgroup$
    – Wojowu
    Commented Mar 10, 2021 at 15:39
  • $\begingroup$ @Wojuwu Let $X\to S$ be a finitely presented DM (algebraic) stack (with separated diagonal). Let $\Delta$ be its diagonal. Let $T\to X\times_S X$ be a morphism with $T$ a scheme. Since $\Delta$ is unramified, this means that the pull-back of $\Delta$ along $T\to X\times_S X$ is an algebraic space, let's call it $I$, which is unramified over $T$. Now, since $I\to T$ is unramified separated and finitely presented, it is quasi-finite separated. An algebraic space which is quasi-finite separated over a scheme is a scheme. So, $I$ is a scheme, as required.... $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 11, 2021 at 9:33
  • $\begingroup$ ....I did add some additional hypotheses of finite presentation and separated diagonal. Were you trying to avoid these? $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 11, 2021 at 9:34
  • $\begingroup$ @AriyanJavanpeykar I'm not able to fully follow this argument, but I find it believable. I don't mind the extra hypotheses you took - they are (at least in practice, I imagine) reasonably mild niceness properties. Feel free to post your argument as an answer. $\endgroup$
    – Wojowu
    Commented Mar 11, 2021 at 21:30

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