Lemma 105.6.1. The morphism $W \xrightarrow {(E_ W, f_ W, 0_ W)} \mathcal{M}_{1, 1}$ is smooth and surjective.

**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. $\square$

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