Étale and universally injective

This is the second in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains the actual result, and the next post will contain a fun application.

Remark. Recall that a morphism $f \colon X \ra Y$ of schemes is étale if $f$ is flat and locally of finite presentation, and $\Omega_{X/Y} = 0$ .

Lemma. Suppose $f \colon X \ra S$ is étale and universally injective. Then $f$ is an open immersion.

Proof. Since étale morphisms are open and surjectivity is stable under base change, we may assume that $f$ is universally bijective; then we have to prove that $f$ is an isomorphism. Since étale morphisms are open, $f$ is in fact a universal homeomorphism. By Tag 04DE, $f$ is affine.

The question is local on $S$ , so we may assume $S$ is affine, and hence so is $X$ . Say $f$ is induced by $g \colon A \ra B$ . Now $f$ is proper and affine, hence finite. Moreover, since $B$ is finitely presented and finite as $A$ -algebra, and $B$ is a finitely presented $B$ -module, it is also a finitely presented $A$ -module (Tag 0564).

Now $B$ is flat of finite presentation over $A$ , hence locally free (actually, we need the slightly stronger result that I mention in the first remark; see Tag 00NX for statement and proof). Since the question is local, we may assume $B$ is free of rank $r$ .

Now let $\bar{s} \ra S$ be a geometric point; that is, let $A \ra \bar{K}$ be a map to an algebraically closed field. Then the tensor product $B \tens_A \bar{K}$ is étale of dimension $r$ over $\bar{K}$ . Hence, $X_{\bar s} = X \times_{S} \bar{s}$ is a union of $r$ points. Since $f$ is universally bijective, we have $r = 1$ . Then the result follows from my previous post. $\qedsymbol$

Lovely little lemmas

Or maybe ‘amusing observations’

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