Proof.
Assumptions (2) and (3) imply that X_{red} = X \times _ Y Y_{red} = Y_{red}. We will use this without further mention.
If Y' \to Y is an étale morphism of formal algebraic spaces over S, then the base change f' : X \times _ Y Y' \to Y' satisfies conditions (1) – (4). Hence by Lemma 87.27.4 we may assume Y is an affine formal algebraic space.
Say Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda as in Definition 87.9.1. Then X_\lambda = X \times _ Y Y_\lambda is an algebraic space endowed with a monomorphism f_\lambda : X_\lambda \to Y_\lambda which induces an isomorphism X_{\lambda , red} \to Y_{\lambda , red}. Thus X_\lambda is an affine scheme by Limits of Spaces, Proposition 70.15.2 (as X_{\lambda , red} \to X_\lambda is surjective and integral). To finish the proof it suffices to show that X_\lambda \to Y_\lambda is a closed immersion which we will do in the next paragraph.
Let X \to Y be a monomorphism of affine schemes such that X_{red} = X \times _ Y Y_{red} = Y_{red}. In general, this does not imply that X \to Y is a closed immersion, see Examples, Section 110.36. However, under our assumption (4) we know that in the previous parapgrah either X_\lambda \to Y_\lambda is of finite type or Y_\lambda is Noetherian. This means that X \to Y corresponds to a ring map R \to A such that R/I \to A/IA is an isomorphism where I \subset R is the nil radical (ie., the maximal locally nilpotent ideal of R) and either R \to A is of finite type or R is Noetherian. In the first case R \to A is surjective by Algebra, Lemma 10.126.9 and in the second case I is finitely generated, hence nilpotent, hence R \to A is surjective by Nakayama's lemma, see Algebra, Lemma 10.20.1 part (11).
\square
Comments (0)