Lemma 103.6.1. The morphism $W \xrightarrow {(E_ W, f_ W, 0_ W)} \mathcal{M}_{1, 1}$ is smooth and surjective.

## 103.6 A smooth cover

The last thing we have to do is find a smooth cover of $\mathcal{M}_{1, 1}$. In fact, in some sense the existence of a smooth cover *implies*^{1} the key fact! In the case of elliptic curves we use the Weierstrass equation to construct one.

Set

where $\Delta \in \mathbf{Z}[a_1, a_2, a_3, a_4, a_6]$ is a certain polynomial (see below). Set

Denote $f_ W : E_ W \to W$ the projection. Finally, denote $0_ W : W \to E_ W$ the section of $f_ W$ given by $(0 : 1 : 0)$. It turns out that there is a degree $12$ homogeneous polynomial $\Delta $ in $a_ i$ where $\deg (a_ i) = i$ such that $E_ W \to W$ is smooth. You can find it explicitly by computing partials of the Weierstrass equation – of course you can also look it up. You can also use pari/gp to compute it for you. Here it is

You may recognize the last two terms from the case $y^2 = x^3 + Ax + B$ having discriminant $-64A^3 - 432B^2 = -16(4A^3 + 27B^2)$.

**Proof.**
Surjectivity follows from the fact that every elliptic curve over a field has a Weierstrass equation. We give a rough sketch of one way to prove smoothness. Consider the sub group scheme

There is an action $H \times W \to W$ of $H$ on the Weierstrass scheme $W$. To find the equations for this action write out what a coordinate change given by a matrix in $H$ does to the general Weierstrass equation. Then it turns out the following statements hold

any elliptic curve $(E, f, 0)/S$ has Zariski locally on $S$ a Weierstrass equation,

any two Weierstrass equations for $(E, f, 0)$ differ (Zariski locally) by an element of $H$.

Considering the fibre product $S \times _{\mathcal{M}_{1, 1}} W = \mathit{Isom}_{S \times W}(E \times W, S \times E_ W)$ we conclude that this means that the morphism $W \to \mathcal{M}_{1, 1}$ is an $H$-torsor. Since $H \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is smooth, and since torsors over smooth group schemes are smooth we win. $\square$

Remark 103.6.2. The argument sketched above actually shows that $\mathcal{M}_{1, 1} = [W/H]$ is a global quotient stack. It is true about 50% of the time that an argument proving a moduli stack is algebraic will show that it is a global quotient stack.

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