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
\[ H = \left\{ \left( \begin{matrix} u^2
& s
& 0
\\ 0
& u^3
& 0
\\ r
& t
& 1
\end{matrix} \right) \middle | \begin{matrix} u\text{ unit}
\\ s, r, t\text{ arbitrary}
\end{matrix} \right\} \subset \text{GL}_{3, \mathbf{Z}} \]
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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