Lemma 29.45.7. Let $f : X \to S$ and $S' \to S$ be morphisms of schemes. Assume

1. $S' \to S$ is a closed immersion,

2. $S' \to S$ is bijective on points,

3. $X \times _ S S' \to S'$ is a closed immersion, and

4. $X \to S$ is of finite type or $S' \to S$ is of finite presentation.

Then $f : X \to S$ is a closed immersion.

Proof. Assumptions (1) and (2) imply that $S' \to S$ is a universal homeomorphism (for example because $S_{red} = S'_{red}$ and using Lemma 29.45.6). Hence (3) implies that $X \to S$ is homeomorphism onto a closed subset of $S$. Then $X \to S$ is affine by Lemma 29.45.4. Let $U \subset S$ be an affine open, say $U = \mathop{\mathrm{Spec}}(A)$. Then $S' = \mathop{\mathrm{Spec}}(A/I)$ by (1) for a locally nilpotent ideal $I$ by (2). As $f$ is affine we see that $f^{-1}(U) = \mathop{\mathrm{Spec}}(B)$. Assumption (4) tells us $B$ is a finite type $A$-algebra (Lemma 29.15.2) or that $I$ is finitely generated (Lemma 29.21.7). Assumption (3) is that $A/I \to B/IB$ is surjective. From Algebra, Lemma 10.126.9 if $A \to B$ is of finite type or Algebra, Lemma 10.20.1 if $I$ is finitely generated and hence nilpotent we deduce that $A \to B$ is surjective. This means that $f$ is a closed immersion, see Lemma 29.2.1. $\square$

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