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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