5
$\begingroup$

Let$\newcommand{\mM}{\mathcal{M}}$ $\mM_{1,1}$ be the moduli stack of elliptic curves. Let $R$ be a Dedekind domain, say $\mathbb{Z}[1/N]$ for simplicity, and suppose we have a finite etale cover:

$$\mM\rightarrow\mM_{1,1}[1/N]$$ Must the coarse moduli scheme $M$ of $\mM$ be smooth over $\mathbb{Z}[1/N]$? This is certainly true if $\mM$ is representable, since $\mM_{1,1}$ is smooth over $\mathbb{Z}$. Furthermore, for a geometric point Spec $k\rightarrow$ Spec $R$ of characteristic not 2 or 3, $M_k$ is the coarse moduli scheme of $\mM_k$, which is normal since it's the quotient of the scheme $\mM\times_{\mM_{1,1}[1/N]} \mM(N^2)$ by $GL_2(\mathbb{Z}/N)$, where $\mM(N^2)$ is the representable stack over $\mathbb{Z}[1/N]$ classifying elliptic curves with full level $N$ structure. Since $M_k$ is a normal curve, it's smooth, and thus since $R$ is a Dedekind domain, we find that $M[1/6]$ is smooth over $\mathbb{Z}[1/6N]$.

The same argument doesn't work at $p = $ 2 or 3, since for a residue field $k$ of $R$ of characteristic 2 or 3, $M_k$ is not necessarily the coarse moduli scheme of $\mM_k$ (the stack is not tame over 2 and 3). Certainly the only potential singularities are the preimages of $j = 0\equiv 1728\mod p$ in $M_k$.

Is this a real obstruction? Of course if $\mM$ is representable then this is a non-issue, and maybe for special problems like $\Gamma_0(N)$ one can use division polynomials, but that's kind of ad hoc. I'd like to understand better what can go wrong generally speaking. If this is a real problem, then can we at least say it's flat?

$\endgroup$
5
$\begingroup$

Yes, $M$ is smooth.

In proving this we may focus on a fixed residue characteristic, so we loose no generality by assuming that $N$ is divisible by a prime $\ell \ge 3$. The morphism $\mathcal{Y}(\ell)[1/N] \rightarrow \mathcal{M}_{1, 1}[1/N]$ is a $\mathrm{GL}_2(\mathbb{Z}/\ell\mathbb{Z})$-torsor, and so is its base change $\mathcal{Y}(\ell)[1/N]_{\mathcal{M}} \rightarrow \mathcal{M}$. Moreover, $\mathcal{Y}(\ell)[1/N]_{\mathcal{M}}$ is a smooth $\mathbb{Z}[1/N]$-curve (say, by IV.2.5 in Deligne--Rapoport) and $M$ is the categorical (ring of invariants) quotient $\mathcal{Y}(\ell)[1/N]_{\mathcal{M}}/\mathrm{GL}_2(\mathbb{Z}/\ell\mathbb{Z})$. In order to conclude it remains to recall the following useful theorem from the appendix of Katz--Mazur (see "Notes on Chapters 8 and 10").

Theorem. If $S$ is a Noetherian regular scheme and $X \rightarrow S$ is a relative curve that is affine over $S$ and that is equipped with an $S$-linear action of a finite group $G$, then the categorical quotient $X/G$ is a smooth $S$-curve.

The argument is the same one that also proves smoothness of coarse moduli spaces in the case of congruence level modular curves away from the level (see, for instance, the footnote on 10.10.3 (5) in Katz--Mazur) and works over any Dedekind domain $R$ as in your question.

$\endgroup$
  • $\begingroup$ WHAAAAAAAT. Okay what's weird is after that theorem in the appendix they seem to prove something that seems to imply that coarse base change holds in general, which is certainly false... $\endgroup$ – Will Chen Oct 31 '15 at 3:55
  • $\begingroup$ @oxeimon: They don't prove coarse base change "in general" there: they assume that $X$ is $S$-smooth, which is a strong assumption. For instance, using the result there you won't be able to get coarse base change for, say, $X_1(N)$ in characteristics dividing $N$. $\endgroup$ – Kestutis Cesnavicius Oct 31 '15 at 4:12
  • $\begingroup$ Sorry I don't mean in general, I mean away from the "level". It's slightly weird that you should be able to get coarse base change even on characteristics 2 or 3 (where the stack is not tame), but I guess it also makes key use of the assumption that the coarse moduli scheme is a (relative) curve. $\endgroup$ – Will Chen Oct 31 '15 at 4:21
  • $\begingroup$ @oxeimon: I agree that their result is surprising, but it is, of course, special to curves. Tame stacks give general guidelines concerning the behavior of coarse spaces under base change, but, for instance, they wouldn't even predict that the formation of the coarse space of $\mathcal{M}_{1, 1}$ itself commutes with arbitrary base change. $\endgroup$ – Kestutis Cesnavicius Oct 31 '15 at 4:30

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.