Let $R$ be a ring and let $\mathcal{M}_{1, R}$ be the algebraic $R$-stack of elliptic curves (over $R$-schemes as bases). One knows that the coarse moduli space of $\mathcal{M}_{1, R}$ is supposed to be $\mathcal{M}_{1, R} \rightarrow \mathbb{A}^1_R$, with the map being given by the $j$-invariant. Is there a reference for this fact for an arbitrary $R$? I know that Deligne and Rapoport treat the case $R = \mathbb{Z}$ in Theorem VI.1.1 of their article, but they note that even this case is slightly painful to prove directly. The general case does not seem to follow formally because the formation of a coarse moduli space does not commute with non-flat base change.

For completeness, I recall the definition of the coarse moduli space $M$ of an algebraic $R$-stack $\mathcal{M}$ as is given in Definition I.8.1 of Deligne-Rapoport: $M$ is an $R$-algebraic space equipped with a morphism $\mathcal{M} \rightarrow M$ that induces a bijection on (isomorphism classes of objects of) $\overline{k}$-points for every algebraically closed $R$-field $\overline{k}$ and that is initial among morphisms from $\mathcal{M}$ to $R$-algebraic spaces.

  • $\begingroup$ I don't have it with me, so I'm not sure if it's in there, but have you tried looking in Katz, Mazur - Arithmetic Moduli of Elliptic Curves ? $\endgroup$ Mar 4, 2015 at 9:07
  • $\begingroup$ Yes, I have looked into that. Unfortunately, they don't prove it there. They say in (8.2.1) that this is "well-known" and they give a reference to some paper of Igusa's (unfortunately, I may need a time machine to properly understand the latter, but I guess that the property justified there is bijectivity on $\overline{k}$-points). Katz and Mazur do, however, base their constructions on the $j$-line, but perhaps the case $R = \mathbb{Z}$ is the most important one (where Deligne-Rapoport applies), or perhaps they get around it with their ad hoc definition of a coarse moduli space in 8.1.1. $\endgroup$ Mar 4, 2015 at 16:22
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    $\begingroup$ For the coordinate ring $A$ of $Y(p)$ over $\mathbf{Z}[1/p]$ and prime $p\ge 3$ and $G={\rm{GL}}_2(\mathbf{F}_p)$, we want $R\otimes A^G \rightarrow (R\otimes A)^G$ is an equality. We know $A^G=\mathbf{Z}[j][1/p]$ by D-R, so $A^G\rightarrow A$ is finite flat by regularity stuff, so injectivity holds for any $R$. But $R$ matters through its underlying abelian group, so (via direct limits) for surjectivity we reduce to $R=\mathbf{Z}/n\mathbf{Z}$ and then $\mathbf{F}_{\ell}$ for prime $\ell\ne p$. But the finite flat $j$-map on $Y(p)_{\mathbf{F}_{\ell}}$ has degree $\#G$, so we are done. QED $\endgroup$
    – user74230
    Mar 5, 2015 at 4:05
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    $\begingroup$ @QuestionMark: The only method which works is to use Raynaud's construction via formal schemes as developed in Deligne-Rapoport (and so it goes through generalized elliptic curves!). Of course, one might want to then prove that Raynaud's construction actually coincides with the usual Tate curve, another fact you'll never find proved in the literature (but is a nice exercise with formal GAGA and inspection of Tate's classical computations), but actually this agreement is not so important since the explicit equation is basically useless (though has psychological value). $\endgroup$
    – user74230
    Mar 9, 2015 at 7:00
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    $\begingroup$ Section 3.3. of arxiv.org/pdf/1511.07475.pdf is a reference in the compactified case. $\endgroup$ Jul 3, 2016 at 18:12

2 Answers 2


The following strategy should work although I do not claim that it is the most elegant.

Claim 1: The coarse moduli stack of elliptic curves $\mathcal{M}_{1,R}$ is affine.

Proof: It is enough to show this if $2$ or $3$ is invertible in $R$; we can glue in the general case.

If $2$ or $3$ is invertible, there is a finite etale cover $X \to \mathcal{M}_{1,R}$ by an affine scheme $X$. This $X$ can be chosen to be the moduli stack of elliptic curves with some high enough level structure. Then we can proceed a la Section 3 of Brian Conrad's http://math.stanford.edu/~conrad/papers/coarsespace.pdf to get that $\mathcal{M}_{1,R}$ has an affine coarse moduli space.

To see that moduli of elliptic curves with high enough level structure are affine schemes, we can for example proceed like follows:

By easy explicit computations, we hvae the following two equivalences:

$$\mathcal{M}_{1}(3)_R \simeq Spec \mathbb{R}[a_1,a_3,\Delta^{-1}]/\mathbb{G}_m \text{ if }2 \text{ is invertible}$$ $$\mathcal{M}_{1}(2)_R \simeq Spec \mathbb{R}[b_2,b_4,\Delta^{-1}]/\mathbb{G}_m \text{ if }3 \text{ is invertible}$$

Here, the / denote stack quotients. The necessary arguments can be found, e.g., in http://www.math.uiuc.edu/~rezk/tmf3-paper-final.pdf Section 3 and http://www3.nd.edu/~mbehren1/papers/K2S.pdf Section 1.3.

If you want to get rid of the $/\mathbb{G}_m$, observe first that $Spec A/\mathbb{G}_m$ has coarse moduli space $Spec A_0$ if $A_\bullet$ denotes the grading corresponding to the $\mathbb{G}_m$-action. Note furthermore that $\mathcal{M}(3)_R \to \mathcal{M}_1(3)_R$ is finite and thus also $\mathcal{M}(3)_R \to U$, where $U$ is the coarse moduli space of $\mathcal{M}_1(3)_R$. Likewise for $\mathcal{M}(4)_R \to \mathcal{M}_1(2)_R$. The stacks $\mathcal{M}(3)_R$ and $\mathcal{M}(4)_R$ are algebraic spaces by rigidity (see Katz-Mazur) and thus they are actually affine schemes. QED

Alternatively you could also give defining equations for $\mathcal{M}(3)$ and $\mathcal{M}(4)$ directly, but this is slightly more complicated. One could also have used $\mathcal{M}(2)_R \simeq R[x_2,y_2, \Delta^{-1}]/\mathbb{G}_m$ if $2$ is invertible, which is also quite easy.

Claim 2: If a stack $\mathcal{X}$ has affine coarse moduli space, it is given by $Spec\, \Gamma(\mathcal{X},\mathcal{O}_\mathcal{X})$.

Proof: Global sections are morphisms to $Spec\, \mathbb{Z}$. QED

Claim 3: $\Gamma(\mathcal{M}_{1,R},\mathcal{O}) \cong R[j]$

Edit: Using the suggestion of Tyler and the correction by Question Mark, this should read like this:

Say, we already know this result for $R=\mathbb{Z}$. In general, we have to show that the (ring) homomorphism $$\Gamma(\mathcal{M}, \mathcal{O}) \otimes_{\mathbb{Z}}R \to \Gamma(\mathcal{M}_{1,R},\mathcal{O}) \cong \Gamma(\mathcal{M},\mathcal{O}\otimes_{\mathbb{Z}} R)$$ is an isomorphism. We can actually show this for every abelian group $R$. It is clear if $R$ is a free abelian group. As $H^1(\mathcal{M},\mathcal{O}) = 0$ and $\mathcal{O}$ is flat over $\mathbb{Z}$, the sequence $0 \to \mathbb{Z}^? \to \mathbb{Z}^? \to R\to 0$ induces short exact sequences of source and target. This implies the result for every $R$.

The result $H^1(\mathcal{M},\mathcal{O}) = 0$ is only formal after inverting $2$ and $3$ -- integrally it can essentially be found in Sections 5 and 7 of http://arxiv.org/pdf/math/0311328.pdf (only that Bauer uses the language of Hopf algebroids; a translation into stack language can, for example, be found in Section 4.1 of http://arxiv.org/pdf/1307.8310.pdf). The global sections result over $\mathbb{Z}$ can also be deduced from Bauer's paper if you want as he computes the whole cohomology of the moduli stack.

I have not actually checked it, but this should be rather doable in a number of ways. We only have to check it for $R = \mathbb{Z}$ and $R= \mathbb{F}_p$ as everything else is flat over these. We get it actually for free if $p \geq 5$ because then $$\Gamma(\mathcal{M}_{1,\mathbb{F}_p},\mathcal{O}) \cong \Gamma(\mathcal{M}_{1,\mathbb{Z}},\mathcal{O})/p$$ as $H^1$ vanishes. Say you want to find out the case $\mathbb{F}_3$, then you just have to compute the fixed points $R[x_2,y_2, \Delta^{-1}]^{GL_2(\mathbb{Z}/2)}_0$, where $0$ stands for degree $0$ -- the group action is completely explicit. Or you can also use directly the Weierstrass equations and that all transformations between them are known...

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    $\begingroup$ I believe that Claim 3 boils down to the computation that $H^1({\cal M}_1, {\cal O})$ is flat (it actually vanishes) -- this isn't formal, and it fails for the moduli of {elliptic curves + nowhere-vanishing 1-forms}, where base-change is false. $\endgroup$ Mar 4, 2015 at 21:26
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    $\begingroup$ I don't understand the "everything else is flat over these" part. E.g., $\mathbb{Z}/p^2\mathbb{Z}$ is not flat over a $\mathbb{Z}$ and it is not a ring over $\mathbb{F}_p$. $\endgroup$ Mar 4, 2015 at 21:36
  • $\begingroup$ @Tyler and Question Mark: Thanks for the suggestion and for the correction. I have incorporated both into the answer. $\endgroup$ Mar 6, 2015 at 17:57

Here are some modern references: [1, Theorem 13.1.15] and [2, Lemma 2.1]

[1] Olsson, "Algebraic Spaces and Stacks", Colloquium Publications 62, AMS (2016)

[2] Fulton, Olsson, "The Picard group of $\mathscr{M}_{1,1}$", Algebra & Number Theory, vol. 4, no. 1 (2010) link


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