We warn the reader right away that the result of this section will be superseded by the stronger Theorem 80.10.1. On the other hand, the theorem in this section is quite a bit easier to prove and still provides quite a bit of insight into how things work, especially for those readers mainly interested in Deligne-Mumford stacks.
In Spaces, Section 65.6 we defined an algebraic space as a sheaf in the fppf topology whose diagonal is representable, and such that there exist a surjective étale morphism from a scheme towards it. In this section we show that a sheaf in the fppf topology whose diagonal is representable by algebraic spaces and which has an étale surjective covering by an algebraic space is also an algebraic space. In other words, the category of algebraic spaces is an enlargement of the category of schemes by those fppf sheaves $F$ which have a representable diagonal and an étale covering by a scheme. The result of this section says that doing the same process again starting with the category of algebraic spaces, does not lead to yet another category.
Another motivation for the material in this section is that it will guarantee later that a Deligne-Mumford stack whose inertia stack is trivial is equivalent to an algebraic space, see Algebraic Stacks, Lemma 94.13.2.
Here is the main result of this section (as we mentioned above this will be superseded by the stronger Theorem 80.10.1).
Theorem 80.6.1. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Assume that
the presheaf $F$ is a sheaf,
the diagonal morphism $F \to F \times F$ is representable by algebraic spaces, and
there exists an algebraic space $X$ and a map $X \to F$ which is surjective, and étale.
or assume that
the presheaf $F$ is a sheaf, and
there exists an algebraic space $X$ and a map $X \to F$ which is representable by algebraic paces, surjective, and étale.
Then $F$ is an algebraic space.
Proof. We will use the remarks directly below Definition 80.4.1 without further mention.
Assume (1), (2), and (3) and let $X \to F$ be as in (3). By Lemma 80.5.1 the morphism $X \to F$ is representable by algebraic spaces. Thus we see that (a) and (b) hold.
Assume (a) and (b) and let $X \to F$ be as in (b). Let $U \to X$ be a surjective étale morphism from a scheme towards $X$. By Lemma 80.3.8 the transformation $U \to F$ is representable by algebraic spaces, surjective, and étale. Hence to prove that $F$ is an algebraic space boils down to proving that $\Delta _ F$ is representable (Spaces, Definition 65.6.1). This follows immediately from Lemma 80.5.3. On the other hand we can circumvent this lemma and show directly $F$ is an algebraic space as in the next paragraph.
Namely, let $U$ be a scheme and let $U \to F$ be representable by algebraic spaces, surjective, and étale. Consider the fibre product $R = U \times _ F U$. Both projections $R \to U$ are representable by algebraic spaces, surjective, and étale (Lemma 80.4.2). In particular $R$ is an algebraic space by Lemma 80.3.6. 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). Since $R \to U$ is étale, we see that $R \to U$ is locally quasi-finite, see Morphisms of Spaces, Lemma 67.39.5. We conclude that also $R \to U \times _ S U$ is locally quasi-finite by Morphisms of Spaces, Lemma 67.27.8. Hence Morphisms of Spaces, Proposition 67.50.2 applies and $R$ is a scheme. By Lemma 80.4.6 the map $U \to F$ is a surjection of sheaves. Thus $F = U/R$. We conclude that $F$ is an algebraic space by Spaces, Theorem 65.10.5. $\square$
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Comment #8847 by Simon Vortkamp on