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
\xymatrix{ & ? \ar@{..>}[rd] \ar@{..>}[ld] \\ S \ar[rd]_{(E, f, 0)} & & S' \ar[ld]^{(E', f', 0')} \\ & \mathcal{M}_{1, 1} }
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
\mathit{Isom}_{S \times S'}(E \times S', S \times E')
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
S \times _{\mathcal{M}_{1, 1}} S' \longrightarrow S'
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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