Let ${M_{1, 1}}$ be the moduli stack of elliptic curves. Given a scheme ${S}$, maps ${S \rightarrow M_{1, 1}}$ are given by the groupoid of elliptic curves over ${S}$, together with isomorphisms between them. The goal of this post is to compute ${\mathrm{Pic}(M_{1, 1})}$ away from the primes ${2, 3}$. (This is done in Mumford’s paper “Picard groups of moduli problems.”)

In the previous post, we saw that ${M_{1, 1}}$ could be described as a quotient stack. Namely, consider the scheme ${B_1 = \mathrm{Spec} \mathbb{Z}[a_1, a_2, a_3, a_4, a_6]}$ and the Weierstrass equation

$\displaystyle Y^2 Z + a_1 XYZ + a_3 YZ^2 = X^3 + a_2 X^2 Z + a_4 XZ^2 + a_6 Z^3$

cutting out a subscheme ${E_1 \subset \mathbb{P}^2_{B_1}}$. This is a flat family of projective cubic curves over ${\mathbb{P}^2_{B_1}}$ with a section (the point at infinity given by ${[X: Y: Z] = [0 : 1 : 0]}$). There is an open subscheme ${B \subset B_1}$ over which the family ${E_1 \rightarrow B_1}$ is smooth, i.e., consists of elliptic curves. A little effort with cohomology and Riemann-Roch allows us to show that, Zariski locally, any elliptic curve ${X \rightarrow S}$ can be pulled back from one of these: that is, any elliptic curve locally admits a Weierstrass equation.

The Weierstrass equation was not unique, though; any change of parametrization (in affine coordinates here)

$\displaystyle x' = a^2 x + b, \quad y' = a^3 x + c + d, \ a \mathrm{\ invertible}$

preserves the form of the equation, and these are the only transformations preserving it. In other words, the map

$\displaystyle B \rightarrow M_{1, 1}$

exhibits ${B}$ as a torsor over ${M_{1,1}}$ for the group scheme ${\mathbb{G} = \mathrm{Spec} \mathbb{Z}[a^{\pm 1}, b, c, d]}$ with a multiplication law given by composing linear transformations. That is,

$\displaystyle M_{1, 1} \simeq B/\mathbb{G};$

that is, to give a map ${S \rightarrow M_{1, 1}}$, one has to choose an étale cover ${\left\{S_\alpha\right\}}$ of ${S}$ (Zariski is enough here), maps ${S_\alpha \rightarrow B}$ inducing elliptic curves over the ${S_\alpha}$, and isomorphisms (coming from maps to ${\mathbb{G}}$) over ${S_\alpha \times_S S_\beta}$.

1. The presentation away from 2 and 3

It turns out that ${M_{1, 1} \times_{\mathrm{Spec} \mathbb{Z}} \mathrm{Spec} \mathbb{Z}[1/2, 1/3]}$ admits a simpler and less “stacky” presentation: it is a quotient of of an open subset of affine space by ${\mathbb{G}_m}$.

Namely, consider any Weierstrass equation

$\displaystyle y^2+ a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$

over a ring ${R}$ where ${2, 3}$ are invertible. Then completing the square and the cube reduces the equation to the form

$\displaystyle y^2 = x^3 + Ax + B , \quad A, B \in R.$

This equation is not unique either, but there are far fewer automorphisms of an equation of this form. Namely, they are given by

$\displaystyle y' = u^3 y, \quad x' = u^2 x, \quad u \in R^*.$

This transformation acts by ${A \mapsto u^{-4} A, B \mapsto u^{-6} B}$.

There is a “universal” Weierstrass equation of the above form; namely, we consider the scheme

$\displaystyle C = \mathrm{Spec} \mathbb{Z}[A, B, \Delta^{-1}], \quad \Delta = 4A^3 + 27B^2,$

and the subscheme of ${\mathbb{P}^2_C}$ cut out by the equation ${Y^2 Z = X^3 + AX Z^2 + BZ^3}$. This is an elliptic curve over ${C}$ (the invertibility of the discriminant ${\Delta}$ precisely takes care of that). By the same analogy, we find that over a ${\mathrm{Spec} \mathbb{Z}[1/6]}$-scheme (i.e., one over which ${6}$ is invertible), any elliptic curve can Zariski locally be pulled back from this one in a manner unique up to ${\mathbb{G}_m}$-action.

Putting this together, we find:

Proposition 1 The stack ${M_{1,1} \times_{\mathrm{Spec} Z} \mathrm{Spec} \mathbb{Z}[1/6]}$ is isomorphic to the quotient stack ${\mathrm{Spec} \mathbb{Z}[1/6, A, B, \Delta^{-1}]/\mathbb{G}_m}$ where the multiplicative group acts as above.

This gives an alternative smooth cover of ${M_{1,1}}$ away from ${2}$ and ${3}$, which is a little easier to work with (one is quotienting by a smaller group).

2. ${\mathbb{G}_m}$-actions and gradings

From here it will be straightforward to obtain a description of ${\mathrm{Pic}(M_{1, 1} \times_{\mathrm{Spec} \mathbb{Z}} \mathrm{Spec} \mathbb{Z}[1/6])}$. Namely, we saw in the previous section that ${M_{1, 1} \times_{\mathrm{Spec} \mathbb{Z}} \mathrm{Spec} \mathbb{Z}[1/6]}$ was the quotient of ${ \mathrm{Spec} \mathbb{Z}[1/6, A, B, \Delta^{-1}]}$ under the action of the multiplicative group ${\mathbb{G}_m}$, and thus gave a faithfully flat cover of the stack.

Now, by descent theory, we can identify line bundles on ${M_{1, 1} \times_{\mathrm{Spec} \mathbb{Z}} \mathrm{Spec} \mathbb{Z}[1/6])}$ with ${\mathbb{G}_m}$-equivariant line bundles on ${\mathrm{Spec} \mathbb{Z}[1/6, A, B, \Delta^{-1}]}$: in other words, one should have a projective module of rank one over ${\mathbb{Z}[1/6, A, B, \Delta^{-1}]}$ together with a coaction of the Hopf algebra ${\mathbb{Z}[t, t^{-1}]}$ associated to the multiplicative group.

What can such an equivariant line bundle be? ${\mathbb{Z}[1/6, A, B, \Delta^{-1}]}$ is a unique factorization domain, so any line bundle is trivial. The only question is that of the ${\mathbb{G}_m}$-action and how that can vary. Here a fruitful point of view is that a ${\mathbb{G}_m}$-action on an abelian group ${A}$ (that is, the structure of a comodule over ${\mathbb{Z}[t, t^{-1}]}$ on ${A}$) is the same thing as a ${\mathbb{Z}}$grading of ${A}$. If we imagine the situation analytically, the intuition is that a ${S^1}$-action on a topological vector space gives a grading of the vector space by decomposition into character eigenspaces. That is, given an ${S^1}$-action on the topological vector space ${V}$, we have

$\displaystyle V \simeq \bigoplus V[k],$

where a vector ${v \in V}$ belongs to ${V[k]}$ if and only if ${v}$ is an eigenvector for ${S^1}$ with eigenvalue given by the character ${S^1 \rightarrow \mathbb{C}^*, z \mapsto z^k}$.

In the algebraic situation, one has a structure map

$\displaystyle \phi: A \rightarrow A \otimes \mathbb{Z}[t, t^{-1}]$

if ${A}$ is given a structure of comodule over ${\mathbb{Z}[t, t^{-1}]}$. Given an element ${a \in A}$, we have

$\displaystyle \phi(a) = \sum_{i \in \mathbb{Z}} a_i \otimes t^i .$

Coassociativity and the fact that the comultiplication sends ${t \mapsto t \otimes t}$ implies that

$\displaystyle \sum_{i \in \mathbb{Z}} \phi( a_i )\otimes t^i \simeq \sum_{i \in \mathbb{Z}} a_i \otimes t^i \otimes t^i \in A \otimes \mathbb{Z}[t, t^{-1}] \otimes \mathbb{Z}[t, t^{-1}].$

It follows that ${\phi(a_i ) = a_i \otimes t^i}$; this is the analog of being an eigenvector in the earlier sense. If we define ${A[i]}$ to consist of all vectors ${a}$ satisfying ${\phi(a) = a \otimes t^i}$, then we get a decomposition

$\displaystyle A = \bigoplus_{i \in \mathbb{Z}} A[i],$

which is to say a grading of ${A}$.

3. The Picard group

We saw in the previous section that ${\mathrm{Pic}(M_{1,1}[1/6])}$ (where I will abbreviate ${M_{1, 1}[1/6] = M_{1,1} \times_{\mathrm{Spec} \mathbb{Z}} \mathrm{Spec} \mathbb{Z}[1/6]}$) can be identified with the group of isomorphism classes of ${\mathbb{G}_m}$-equivariant line bundles on ${\mathrm{Spec} \mathbb{Z}[1/6, A, B, \Delta^{-1}]}$. We also saw two things that simplify the problem considerably:

1. Every (nonequivariant) line bundle on ${\mathrm{Spec} \mathbb{Z}[1/6, A, B, \Delta^{-1}]}$ is trivial.
2. A ${\mathbb{G}_m}$-action is the same as a grading.

So, in other words, ${\mathrm{Pic}(M_{1,1}[1/6])}$ is just the group of isomorphism classes of free ${\mathbb{Z}[1/6, A, B, \Delta^{-1}]}$-modules of rank 1. Any such is a shift of ${\mathbb{Z}[1/6, A, B, \Delta^{-1}]}$, so the Picard group is a quotient of ${\mathbb{Z}}$: we just need to see when two shifts can be isomorphic.

For this, let’s determine the grading on the ring ${\mathbb{Z}[1/6, A, B, \Delta^{-1}]}$ (which we need even to make sense of this problem). The grading came from the ${\mathbb{G}_m}$-action on ${\mathbb{Z}[1/6, A, B, \Delta^{-1}]}$: here ${A}$ was multiplied by ${u^{-4}}$ and ${B}$ by ${u^{-6}}$. It follows that ${A}$ is in degree ${-4}$ and ${B}$ in degree ${-6}$.

Now, the only way a grading shift by ${i}$ of the module ${T = \mathbb{Z}[1/6, A, B, \Delta^{-1}]}$ (considered as graded from the graded ring) could be isomorphic to ${T}$ is if there is an invertible element, homogeneous of degree ${i}$. This happens precisely if ${i}$ is a multiple of ${12}$: the invertible elements in the ring are just ${\pm \Delta^{\mathbb{Z}}}$, and ${\Delta}$ lives in degree twelve.

We find:

Theorem 2 We have an isomorphism ${\mathrm{Pic}(M_{1,1}[1/6]) \simeq \mathbb{Z}/12}$.

We don’t yet have a very geometric interpretation of the generator of this Picard group (other than saying that it corresponds to a suitable module with a ${\mathbb{G}_m}$-action). It can be constructed as follows (even when ${6}$ is not invertible). In order to define a line bundle ${\omega}$ over ${M_{1, 1}}$, we need to specify, for each ring ${R}$ and elliptic curve ${X \rightarrow \mathrm{Spec} R}$, an ${R}$-module projective of rank one together with certain naturality isomorphisms. Given ${X \rightarrow \mathrm{Spec} R}$, we define the ${R}$-module

$\displaystyle \omega(R) = H^0(X, \Omega_{X/R}),$

where ${\Omega_{X/R}}$ is the sheaf of Kähler differentials on ${X}$.

Proposition 3 ${\omega}$ defines a line bundle over ${M_{1, 1}}$, which when restricted to the locus where ${6}$ is invertible is a generator for the Picard group.

To see this, we will show more generally that if ${f: X \rightarrow S}$ is any elliptic curve, then formation of the pushforward ${f_* \Omega_{X/S}}$ commutes with base change in ${S}$ and is always a locally free sheaf of rank one. This follows from the general machine of cohomology and base change, and the fact that it is true for an algebraically closed field (by Riemann Roch). (I think: there are a few details to check which I’m omitting here when the base is nonreduced.)

Now, given an elliptic curve defined by a Weierstrass equation ${y^2 = x^3 + Ax + B}$, there is a global nonzero holomorphic differential

$\displaystyle \frac{dx}{y}$

on the curve. (One has to check that it is actually holomorphic at ${\infty}$.) For instance, this would define a differential on the elliptic curve over ${\mathrm{Spec} \mathbb{Z}[1/6, A, B, \Delta^{-1}]}$. Under action of the multiplicative group, ${x}$ is multiplied by ${u^2}$ and ${y}$ by ${u^3}$, so ${\frac{dx}{y}}$ lives in grading ${-1}$. Since ${\frac{dx}{y}}$ is a generator, it follows that the line bundle ${\omega}$ corresponds to ${-1}$ in the previous isomorphism ${\mathrm{Pic} (M_{1,1}[1/6]) \simeq \mathbb{Z}/12}$.

4. Projective case

The idea that ${\mathbb{G}_m}$-actions are the same as gradings can be used in a simpler example, where one knows the answer by other means: projective space. Let ${R}$ be a ring. Consider the open subscheme ${U \subset \mathrm{Spec} R[x_0, x_1, \dots, x_n]}$ given by the locus where any one of the coordinates is invertible: to give a map of ${R}$-schemes

$\displaystyle \mathrm{Spec} R' \rightarrow U$

(for ${R'}$ an ${R}$-algebra) is the same as choosing ${n+1}$ elements of ${R'}$, one of which is invertible. There is an action of ${\mathbb{G}_m}$ on ${\mathrm{Spec} R[x_0, \dots, x_n]}$, which places each of the polynomial generators in degree 1; the subscheme ${U}$ is stable and inherits a ${\mathbb{G}_m}$-action.

We have:

$\displaystyle U / \mathbb{G}_m \simeq \mathbb{P}^n_R.$

The left-hand-side here means a stacky quotient, although in fact no automorphisms are introduced by it. From this point of view, it is straightforward to see that ${\mathrm{Pic}(\mathbb{P}^n_R) \simeq \mathbb{Z}}$ if ${R}$ is, say, a field. For in this case any line bundle over ${U}$ is trivial (${U}$ is, modulo codimension ${\geq 2}$, affine space). So the only data necessary to specify a line bundle on ${\mathbb{P}^n_R}$ is a grading; this corresponds to the classical isomorphism ${\mathrm{Pic}(\mathbb{P}^n_R) \simeq \mathbb{Z}}$ (valid for a field ${R}$). In addition, this observation makes more apparent the connection between quasi-coherent sheaves on projective space and graded ${R[x_0, \dots, x_n]}$-modules.