The Stacks project

Lemma 80.5.3. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Assume that

  1. the presheaf $F$ is a sheaf,

  2. there exists an algebraic space $X$ and a map $X \to F$ which is representable by algebraic spaces, surjective, flat and locally of finite presentation.

Then $\Delta _ F$ is representable (by schemes).

Proof. Let $U \to X$ be a surjective étale morphism from a scheme towards $X$. Then $U \to X$ is representable, surjective, flat and locally of finite presentation by Morphisms of Spaces, Lemmas 67.39.7 and 67.39.8. By Lemma 80.4.3 the composition $U \to F$ is representable by algebraic spaces, surjective, flat and locally of finite presentation also. Thus we see that $R = U \times _ F U$ is an algebraic space, see Lemma 80.3.7. The morphism of algebraic spaces $R \to U \times _ S U$ is a monomorphism, hence separated (as the diagonal of a monomorphism is an isomorphism, see Morphisms of Spaces, Lemma 67.10.2). Since $U \to F$ is locally of finite presentation, both morphisms $R \to U$ are locally of finite presentation, see Lemma 80.4.2. Hence $R \to U \times _ S U$ is locally of finite type (use Morphisms of Spaces, Lemmas 67.28.5 and 67.23.6). Altogether this means that $R \to U \times _ S U$ is a monomorphism which is locally of finite type, hence a separated and locally quasi-finite morphism, see Morphisms of Spaces, Lemma 67.27.10.

Now we are ready to prove that $\Delta _ F$ is representable. Let $T$ be a scheme, and let $(a, b) : T \to F \times F$ be a morphism. Set

\[ T' = (U \times _ S U) \times _{F \times F} T. \]

Note that $U \times _ S U \to F \times F$ is representable by algebraic spaces, surjective, flat and locally of finite presentation by Lemma 80.4.4. Hence $T'$ is an algebraic space, and the projection morphism $T' \to T$ is surjective, flat, and locally of finite presentation. Consider $Z = T \times _{F \times F} F$ (this is a sheaf) and

\[ Z' = T' \times _{U \times _ S U} R = T' \times _ T Z. \]

We see that $Z'$ is an algebraic space, and $Z' \to T'$ is separated and locally quasi-finite by the discussion in the first paragraph of the proof which showed that $R$ is an algebraic space and that the morphism $R \to U \times _ S U$ has those properties. Hence we may apply Lemma 80.5.2 to the diagram

\[ \xymatrix{ Z' \ar[r] \ar[d] & T' \ar[d] \\ Z \ar[r] & T } \]

and we conclude. $\square$


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