The Stacks project

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 35.23.4, and universally injective by Lemma 35.23.8. 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

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 35.22.4 applies and we win. $\square$