1
$\begingroup$

I need the reference to a detailed proof the following fact.

Let $g\geq 2$ be an integer. Let $\mathcal M^{ss}_g: Sch\rightarrow Groupoids$ be the prestack which sends a scheme $T$ to the groupoid of families of semistable curves of genus $g$ over $T$. Then $\mathcal M^{ss}_g$ is a Artin algebraic stack in the Etale topology.

(Also "curves" are schemes, not algebraic spaces.)

$\endgroup$
9
  • 1
    $\begingroup$ I am not sure what you mean by "Artin algebraic stack in the Etale topology." Anyway, I recommend that you look at "Gromov-Witten invariants in algebraic geometry" by Kai Behrend: personal.math.ubc.ca/~behrend/gwag.pdf (published in Invent. math.). Although that Artin stack is not the main point of the article, Behrend does work with that stack in the article. $\endgroup$ Commented Sep 4, 2022 at 21:08
  • 1
    $\begingroup$ Of course my article with Johan de Jong and Xuhua He proves that the stack of all curves is an Artin stack. (This is a folk result; we just wrote down one proof.) $\endgroup$ Commented Sep 4, 2022 at 21:10
  • 1
    $\begingroup$ OK, the last parenthetical sentence of the question suggests that @S.D. wants to only work with schemes. Then in the prestable case it doesn't work. I don't know about the semistable case. $\endgroup$
    – Johan
    Commented Sep 4, 2022 at 22:51
  • 1
    $\begingroup$ Well... that is the thing that is wrong for the prestable curves: there exists a scheme $T$ and a prestable curve $X$ over $T$ such that $X$ is not a scheme. So the morphism from the universal curve to the moduli stack is not representable by schemes (for the prestable case). $\endgroup$
    – Johan
    Commented Sep 5, 2022 at 0:26
  • 2
    $\begingroup$ OK, I think you can make a semistable genus 2 example by gluing 2 elliptic tails to the genus $0$ Example 2.3 in Fulghesu's "The stack of rational curves". It is a bit delicate. Good luck! Also Jason meant "etale local" and not "zariski locally on the target" above. $\endgroup$
    – Johan
    Commented Sep 6, 2022 at 15:39

0

You must log in to answer this question.