Lemma 105.4.1 (Key fact). The functor $\mathit{Sch}^{opp} \to \textit{Sets}$, $T \mapsto \{ (a, a', \alpha )\text{ as above}\} $ is representable by a scheme $S \times _{\mathcal{M}_{1, 1}} S'$.

## 105.4 Fibre products

The question we pose here is what should be the fibre product

The answer: A morphism from a scheme $T$ into $?$ should be a triple $(a, a', \alpha )$ where $a : T \to S$, $a' : T \to S'$ are morphisms of schemes and where $\alpha : E \times _{S, a} T \to E' \times _{S', a'} T$ is an isomorphism of elliptic curves over $T$. This makes sense because of our definition of composition and commutative diagrams earlier in the discussion.

**Proof.**
Idea of proof. Relate this functor to

and use Grothendieck's theory of Hilbert schemes. $\square$

Remark 105.4.2. We have the formula $S \times _{\mathcal{M}_{1, 1}} S' = (S \times S') \times _{\mathcal{M}_{1, 1} \times \mathcal{M}_{1, 1}} \mathcal{M}_{1, 1}$. Hence the key fact is a property of the diagonal $\Delta _{\mathcal{M}_{1, 1}}$ of $\mathcal{M}_{1, 1}$.

In any case the key fact allows us to make the following definition.

Definition 105.4.3. We say a morphism $S \to \mathcal{M}_{1, 1}$ is *smooth* if for every morphism $S' \to \mathcal{M}_{1, 1}$ the projection morphism

is smooth.

Note that this is compatible with the notion of a smooth morphism of schemes as the base change of a smooth morphism is smooth. Moreover, it is clear how to extend this definition to other properties of morphisms into $\mathcal{M}_{1, 1}$ (or your own favorite moduli stack). In particular we will use it below for *surjective* morphisms.

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