Lemma 35.20.16. The property $\mathcal{P}(f) =$“$f$ is an open immersion” is fpqc local on the base.

Proof. The property of being an open immersion is stable under base change, see Schemes, Lemma 26.18.2. The property of being an open immersion is Zariski local on the base (this is obvious).

Let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is an open immersion. We claim that $f$ is an open immersion. Then $f'$ is universally open, and universally injective. Hence we conclude that $f$ is universally open by Lemma 35.20.4, and universally injective by Lemma 35.20.8. In particular $f(X) \subset S$ is open. If for every affine open $U \subset f(X)$ we can prove that $f^{-1}(U) \to U$ is an isomorphism, then $f$ is an open immersion and we're done. If $U' \subset S'$ denotes the inverse image of $U$, then $U' \to U$ is a faithfully flat morphism of affines and $(f')^{-1}(U') \to U'$ is an isomorphism (as $f'(X')$ contains $U'$ by our choice of $U$). Thus we reduce to the case discussed in the next paragraph.

Let $S' \to S$ be a flat surjective morphism of affine schemes, let $f : X \to S$ be a morphism, and assume that the base change $f' : X' \to S'$ is an isomorphism. We have to show that $f$ is an isomorphism also. It is clear that $f$ is surjective, universally injective, and universally open (see arguments above for the last two). Hence $f$ is bijective, i.e., $f$ is a homeomorphism. Thus $f$ is affine by Morphisms, Lemma 29.45.4. Since

$\mathcal{O}(S') \to \mathcal{O}(X') = \mathcal{O}(S') \otimes _{\mathcal{O}(S)} \mathcal{O}(X)$

is an isomorphism and since $\mathcal{O}(S) \to \mathcal{O}(S')$ is faithfully flat this implies that $\mathcal{O}(S) \to \mathcal{O}(X)$ is an isomorphism. Thus $f$ is an isomorphism. This finishes the proof of the claim above. Therefore Lemma 35.19.4 applies and we win. $\square$

Comment #2831 by Ko Aoki on

Typo in the proof: "we may replace $S$ by $f(S)$" should be replaced by "we may replace $S$ by $f(X)$".

Comment #4212 by Sean Cotner on

Small point: after replacing S by f(X), you still use the assumption that S is affine. I think this can be fixed by instead base changing to an arbitrary affine open contained in f(X), after which the rest of the proof goes through.

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