The Stacks project

Lemma 65.38.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

  1. If $f$ is unramified, then the diagonal morphism $\Delta _{X/Y} : X \to X \times _ Y X$ is an open immersion.

  2. If $f$ is locally of finite type and $\Delta _{X/Y}$ is an open immersion, then $f$ is unramified.

Proof. We know in any case that $\Delta _{X/Y}$ is a representable monomorphism, see Lemma 65.4.1. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose a scheme $U$ and a surjective étale morphism $U \to X \times _ Y V$. Consider the commutative diagram

\[ \xymatrix{ U \ar[d] \ar[rr]_-{\Delta _{U/V}} & & U \times _ V U \ar[d] \ar[r] & V \ar[d]^{\Delta _{V/Y}} \\ X \ar[rr]^-{\Delta _{X/Y}} & & X \times _ Y X \ar[r] & V \times _ Y V } \]

with cartesian right square. The left vertical arrow is surjective étale. The right vertical arrow is étale as a morphism between schemes étale over $Y$, see Properties of Spaces, Lemma 64.16.6. Hence the middle vertical arrow is étale too (but it need not be surjective).

Assume $f$ is unramified. Then $U \to V$ is unramified, hence $\Delta _{U/V}$ is an open immersion by Morphisms, Lemma 29.35.13. Looking at the left square of the diagram above we conclude that $\Delta _{X/Y}$ is an étale morphism, see Properties of Spaces, Lemma 64.16.3. Hence $\Delta _{X/Y}$ is a representable étale monomorphism, which implies that it is an open immersion by Étale Morphisms, Theorem 41.14.1. (See also Spaces, Lemma 63.5.8 for the translation from schemes language into the language of functors.)

Assume that $f$ is locally of finite type and that $\Delta _{X/Y}$ is an open immersion. This implies that $U \to V$ is locally of finite type too (by definition of a morphism of algebraic spaces which is locally of finite type). Looking at the displayed diagram above we conclude that $\Delta _{U/V}$ is étale as a morphism between schemes étale over $X \times _ Y X$, see Properties of Spaces, Lemma 64.16.6. But since $\Delta _{U/V}$ is the diagonal of a morphism between schemes we see that it is in any case an immersion, see Schemes, Lemma 26.21.2. Hence it is an open immersion, and we conclude that $U \to V$ is unramified by Morphisms, Lemma 29.35.13. This in turn means that $f$ is unramified by definition. $\square$


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