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

Leave a Reply

Your email address will not be published. Required fields are marked *