Let ${S}$ be a scheme. An elliptic curve over ${S}$ should be thought of as a continuously varying family of elliptic curves parametrized by ${S}$.

Definition 1 An elliptic curve over ${S}$ is a proper, flat morphism ${p: X \rightarrow S}$ whose geometric fibers are curves of genus one together with a section ${0: S \rightarrow X}$.

This is a reasonable notion of “family”: observe that a morphism ${T \rightarrow S}$ can be used to pull back elliptic curves over ${S}$. The flatness condition can be thought of as “continuity.” For an algebraically closed field, this reduces to the usual notion of an elliptic curve.

A basic property of elliptic curves over algebraically closed fields is that they imbed into ${\mathbb{P}^2}$ and are cut out by (nonsingular) Weierstrass equations of the form

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

This equation is unique up to an action of a certain four-dimensional group of transformations. The first goal is to show that, locally, the same is true for an elliptic curve over a base.

1. The zero section

Let ${p: X \rightarrow S}$ be an elliptic curve over the scheme ${S}$, with zero section ${0: S \rightarrow X}$. We are going to find an ample line bundle on ${X}$. To do so, observe that ${0}$ (as a section to a separated morphism) imbeds ${S}$ inside ${X}$ as a closed subscheme. Observe that ${X}$ is smooth over ${S}$, since it is flat and the fibers are smooth curves. The zero section ${0: S \rightarrow X}$ is thus cut out locally by one equation.

It follows now that ${0: S \rightarrow X}$ can be used to define a line bundle ${\mathcal{L}}$ on ${X}$, which is compatible with base-change on ${S}$. When restricted to each fiber (an elliptic curve), it is the line bundle associated to the divisor at the origin ${0}$. There is a natural morphism ${\mathcal{O}_X \rightarrow \mathcal{L}}$.

Now consider the line bundle ${\mathcal{L}^{3}}$. When restricted to each geometric fiber ${X_{\overline{s}}}$, this is a very ample line bundle on the elliptic curve ${X_{\overline{s}}}$ associated to three times the origin. By Riemann-Roch, we have

$\displaystyle h^0(\mathcal{L}^3|_{X_{\overline{s}}}) = 3 , \quad h^1(\mathcal{L}^3|_{X_{\overline{s}}}) = 0,$

for every geometric fiber.

By the theorem of cohomology and base change, we find that $$f_* \mathcal{L}^{3}$$ is a vector bundle of rank three on ${S}$, and that the maps

$\displaystyle (f_* \mathcal{L}^3)_s \otimes k(s) \rightarrow H^0(\mathcal{L}^3|_{X_s})$

are isomorphisms: that is, any section along ${X_s}$ extends to a section in some neighborhood. Moreover, we find that the formation of the pushforward ${f_*\mathcal{L}^3}$ commutes with base change ${S' \rightarrow S}$.

It follows now from very ampleness on the fibers that ${f^*f_* \mathcal{L}^3 \rightarrow \mathcal{L}^3}$ is a surjection of sheaves, so defines a map of ${S}$-schemes

$\displaystyle i: X \rightarrow \mathbb{P}(f_* \mathcal{L}^3).$

Since ${f_* \mathcal{L}^3}$ commutes with base change, we find that ${i}$ is a closed imbedding when restricted to any fiber. Hence, ${i}$ is an imbedding and ${\mathcal{L}^3}$ is very ample.

Thus, given an elliptic curve over ${S}$, we get a canonical imbedding ${X \hookrightarrow \mathbb{P}(f_*\mathcal{L}^3)}$ into the projectivization of a three-dimensional vector bundle on ${S}$.

2. Weierstrass equations

Suppose now given a trivialization ${ \mathcal{O}_S^3 \simeq f_* \mathcal{L}^3}$, given by three sections ${a, x, y}$ generating ${f_* \mathcal{L}^3}$. Assume that ${a = 1}$, in fact (recall that there is a map ${\mathcal{O}_X \rightarrow \mathcal{L}^3}$). Locally, we can always choose such a trivialization, but it is not unique. Then we have an imbedding of ${S}$-schemes

$\displaystyle X \hookrightarrow \mathbb{P}^2_S$

given by the sections ${1, x, y \in \Gamma(X, \mathcal{L}^3)}$ generating ${\mathcal{L}^3}$. Suppose ${x}$ is, further, in the image of ${f_* \mathcal{L}^2}$ (which is locally free of rank two).

The claim now is that ${X}$ is cut out as a subscheme of ${\mathbb{P}^2_S}$ by a Weierstrass equation. Here the reasoning is analogous as in the ordinary case when ${S = \mathrm{Spec} k}$. In fact, we note that

$\displaystyle h^0(\mathcal{L}^6|_{X_{\overline{s}}}) = 6,$

for each ${s}$, by a Riemann-Roch calculation (and the ${h^1}$‘s are zero). It follows that ${f_* \mathcal{L}^6}$ is locally free of rank six. The elements ${\{ y^2, xy, y, x^3, x^2, x, 1 \}}$ are seven sections and consequently must be linearly dependent. They generate the vector bundle, too: this can be checked stalkwise and then one notes that the orders at zero are anywhere between ${0}$ and ${6}$.

Locally on ${S}$, we can find a relation of linear dependence (unique up to scaling) among all these involving functions of ${S}$. That is, we can find a relation

$\displaystyle a_0 y^2 + a_1 xy + a_3 y = a_0'x^3 + a_2 x^2 + a_4 x + a_6.$

Now, the claim is that ${a_0, a_0'}$ are necessarily invertible. In fact, we can check this after restricting to the geometric fibers: if when restricted to one geometric fiber exactly one of ${a_0, a_0'}$ vanished, then one side would have order six at the origin and one side would have order ${\leq 5}$. vanished, the left side and the right side could not have the same order either.

Anyway, rescaling we can get the appropriate equation where ${a_0, a_0' = 1}$.

Proposition 2 Let ${f: X \rightarrow S}$ be an elliptic curve over ${S}$. Then, for each ${s \in S}$, there is an open neighborhood ${S'}$ containing ${s}$ such that the elliptic curve ${X \times_S S'}$ has a Weierstrass equation as above.

In other words, locally on ${S}$, ${X \rightarrow S}$ looks like the zero locus of a Weierstrass equation when the ${a_i}$ are taken as functions on (an open subset of) ${S}$.

There is thus a “Zariski locally universal” elliptic curve. Consider the scheme ${B_1 = \mathrm{Spec} \mathbb{Z}[a_1, a_2, a_3, a_4, a_6]}$ and the subscheme ${E_1}$ of ${\mathbb{P}^2_B}$ cut out by the Weierstrass equation ${ Y^2 Z + a_1 XYZ + a_3 YZ^2 = X^3 + a_2 X^2 Z + a_4 XZ^2 + a_6 Z^3 }$. We have then a proper, flat morphism

$\displaystyle E_1 \rightarrow B_1.$

Properness is clear, and flatness follows from the fact that the fibers all have the same Hilbert polynomial (they are cubic curves in ${\mathbb{P}^2}$). There is an open locus ${B \subset B_1}$ over which the fibers are smooth (i.e., where suitable discriminant polynomials are nonzero), and the restriction ${E \rightarrow B}$ of ${E_1 \rightarrow B_1}$ is then an elliptic curve. The zero section ${B \rightarrow E}$ is given by the point at infinity.

Here the line bundle ${\mathcal{L}}$ satisfies ${\mathcal{L}^3 \simeq \mathcal{O}(1)}$ restricted to ${E}$, and we take the sections ${X, Y, Z \in \Gamma(E, \mathcal{O}(1))}$ to imbed ${E}$ in ${\mathbb{P}^2_B}$.

3. Uniqueness

The choice of a Weierstrass equation is not unique. Given an elliptic curve ${f: X \rightarrow S}$, certain things were canonical: the line bundle ${\mathcal{L} \in \mathrm{Pic}(X)}$, the three-dimensional vector bundle ${f_* \mathcal{L}^3}$ on ${S}$, and the imbedding

$\displaystyle X \hookrightarrow \mathbb{P}(f_* \mathcal{L}^3).$

The element ${1 \in \Gamma(f_* \mathcal{L}^3)}$ is canonical, but the choice of ${x, y}$ is not (over a sufficiently small base that ${f_* \mathcal{L}^3}$ is free). Given another choice ${x', y'}$ which satisfy a Weierstrass equation, we have

$\displaystyle x' = u_1 x + u_2, \quad y' = u_3 y + u_4 x + u_5$

where ${u_1, u_3}$ are units. This follows because ${x, x'}$ generate the same two-dimensional subbundle ${f_* \mathcal{L}^2}$. Since ${x', y'}$ are required to satisfy a Weierstrass equation, we must have

$\displaystyle u_1^3 = u_2^2;$

this means that we can parametrize the pair ${(u_1, u_2)}$ via ${(t^2, t^3)}$. It follows that the general change of variables is

$\displaystyle x' = t^2 x + r, \quad y' = t^3 y + s t^2 + q.$

These transformations are parametrized by a group scheme, smooth and of finite type over ${\mathrm{Spec} \mathbb{Z}}$.

4. The stack of elliptic curves

We define the following (pseudo)-functor

$\displaystyle \mathrm{Ell}: \mathrm{Aff}^{op} \rightarrow \mathrm{Gpd}.$

To a ring ${R}$, we set ${\mathrm{Ell}(R)}$ to be the collection of all elliptic curves over ${R}$, together with all ${R}$-isomorphisms between them; the pull-back is the pull-back of schemes. The claim is that ${\mathrm{Ell}(R)}$ is a Deligne-Mumford stack.

First, we should see that ${\mathrm{Ell}}$ is even a stack: given an étale morphism ${R \rightarrow R'}$, we should be able to “descend” elliptic curves over ${R'}$ to elliptic curves over ${R}$.

Proposition 3 ${\mathrm{Ell}}$ is a stack for the flat topology.

Proof: Suppose given a faithfully flat morphism ${\mathrm{Spec} R' \rightarrow \mathrm{Spec} R}$, and suppose given an elliptic curve ${X' \rightarrow \mathrm{Spec} R'}$ together with an isomorphism over ${\mathrm{Spec} R' \otimes_R R'}$ satisfying the cocycle condition. Let ${\mathcal{E}'}$ be the three-dimensional vector bundle over ${\mathrm{Spec} R'}$ associated functorially to ${X'}$; this comes equipped with the descent data to become canonically the base-change of an ${R}$-module ${\mathcal{E}}$. It follows that we really have a descent problem for subschemes of ${\mathbb{P}(\mathcal{E}) \rightarrow \mathrm{Spec} R}$, which we can solve using standard flat descent. $\Box$

In order to see this, we will need to produce a cover of ${\mathrm{Ell}}$. We will just produce a smooth cover here. Consider the map

$\displaystyle B \rightarrow \mathrm{Ell}$

classifying the elliptic curve ${E \rightarrow B}$ which was “universal” for Weierstrass elliptic curves. The claim is that this is smooth and surjective. It is surjective because any elliptic curve is locally Weierstrass: stated stackily, any map ${\mathrm{Spec} R \rightarrow \mathrm{Ell}}$ factors locally through ${B}$.

To see that it is smooth, we will argue more strongly, that ${\mathrm{Ell} }$ is the quotient stack ${B/G}$ for ${G}$ a suitable (smooth) group scheme over ${\mathbb{Z}}$. In fact, this follows from our earlier analysis: étale (or Zariski) locally, any elliptic curve comes from a Weierstrass equation, and this Weierstrass equation is unique up to the action of ${G}$, where $G$ is the group of transformations of Weierstrass equations as above.

(Actually, so far we have only proved that it is an Artin stack.)