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.)

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