Lemma 29.45.7. Let $f : X \to S$ and $S' \to S$ be morphisms of schemes. Assume
$S' \to S$ is a closed immersion,
$S' \to S$ is bijective on points,
$X \times _ S S' \to S'$ is a closed immersion, and
$X \to S$ 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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)