Proposition 96.10.4. Let $S$ be a scheme. Let $Z \to B$ and $X \to B$ be morphisms of algebraic spaces over $S$. If $Z \to B$ is finite locally free then $\mathit{Mor}_ B(Z, X)$ is an algebraic space.

**Proof.**
Choose a scheme $B' = \coprod B'_ i$ which is a disjoint union of affine schemes $B'_ i$ and an étale surjective morphism $B' \to B$. We may also assume that $B'_ i \times _ B Z$ is the spectrum of a ring which is finite free as a $\Gamma (B'_ i, \mathcal{O}_{B'_ i})$-module. By Lemma 96.10.2 and Spaces, Lemma 64.5.5 the morphism $\mathit{Mor}_{B'}(Z', X') \to \mathit{Mor}_ B(Z, X)$ is surjective étale. Hence by Bootstrap, Theorem 79.10.1 it suffices to prove the proposition when $B = B'$ is a disjoint union of affine schemes $B'_ i$ so that each $B'_ i \times _ B Z$ is finite free over $B'_ i$. Then it actually suffices to prove the result for the restriction to each $B'_ i$. Thus we may assume that $B$ is affine and that $\Gamma (Z, \mathcal{O}_ Z)$ is a finite free $\Gamma (B, \mathcal{O}_ B)$-module.

Choose a scheme $X'$ which is a disjoint union of affine schemes and a surjective étale morphism $X' \to X$. By Lemma 96.10.3 the morphism $\mathit{Mor}_ B(Z, X') \to \mathit{Mor}_ B(Z, X)$ is representable by algebraic spaces, étale, and surjective. Hence by Bootstrap, Theorem 79.10.1 it suffices to prove the proposition when $X$ is a disjoint union of affine schemes. This reduces us to the case discussed in the next paragraph.

Assume $X = \coprod _{i \in I} X_ i$ is a disjoint union of affine schemes, $B$ is affine, and that $\Gamma (Z, \mathcal{O}_ Z)$ is a finite free $\Gamma (B, \mathcal{O}_ B)$-module. For any finite subset $E \subset I$ set

By More on Morphisms, Lemma 37.66.1 we see that $F_ E$ is an algebraic space. Consider the morphism

Each of the morphisms $F_ E \to \mathit{Mor}_ B(Z, X)$ is an open immersion, because it is simply the locus parametrizing pairs $(a, b)$ where $b$ maps into the open subscheme $\coprod \nolimits _{i \in E} X_ i$ of $X$. Moreover, if $T$ is quasi-compact, then for any pair $(a, b)$ the image of $b$ is contained in $\coprod \nolimits _{i \in E} X_ i$ for some $E \subset I$ finite. Hence the displayed arrow is in fact an open covering and we win^{1} by Spaces, Lemma 64.8.5.
$\square$

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