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Lennart Meier
Lennart Meier
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I have not actually checked itEdit: Using the suggestion of Tyler and the correction by Question Mark, but this should be rather doable in a number of waysread like this:

Say, we already know this result for $R=\mathbb{Z}$. We onlyIn general, we have to check it for $R = \mathbb{Z}$ andshow that the $R= \mathbb{F}_p$ as everything else is flat over these(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 get itcan actually show this for freeevery abelian group $R$. It is clear if $p \geq 5$ because then$R$ is a free abelian group. As $$\Gamma(\mathcal{M}_{1,\mathbb{F}_p},\mathcal{O}) \cong \Gamma(\mathcal{M}_{1,\mathbb{Z}},\mathcal{O})/p$$ as$H^1(\mathcal{M},\mathcal{O}) = 0$ and $H^1$ vanishes. Say you want to find out the case$\mathcal{O}$ is flat over $\mathbb{F}_3$$\mathbb{Z}$, then you just have to compute the fixed points $R[x_2,y_2, \Delta^{-1}]^{GL_2(\mathbb{Z}/2)}_0$, wheresequence $0$ stands$0 \to \mathbb{Z}^? \to \mathbb{Z}^? \to R\to 0$ induces short exact sequences of source and target. This implies the result for degreeevery $0$$R$.

The result $H^1(\mathcal{M},\mathcal{O}) = 0$ is only formal after inverting $2$ and $3$ -- the group action is completely explicit. Or youintegrally it can also use directly the Weierstrass equationsessentially be found in Sections 5 and 7 of http://arxiv.org/pdf/math/0311328.pdf (only that all transformations between them are knownBauer 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 feel that I have omitted the most difficult part, but I hope it is of some help anyhow.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...

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...

I feel that I have omitted the most difficult part, but I hope it is of some help anyhow.

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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Lennart Meier
Lennart Meier
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  • 107

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

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...

I feel that I have omitted the most difficult part, but I hope it is of some help anyhow.