Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is the converse of this question. Let $X$(resp. $Y$) be a scheme of finite type over a field $k$. Let $f\colon X \rightarrow Y$ be a morphism. Let $X_0$(resp. $Y_0$) be the set of closed points of $X$(resp. $Y$). Then $f$ induces a map $f_0\colon X_0 \rightarrow Y_0$. We consider $X_0$(resp. $Y_0$) as a subspace of $X$(resp. $Y$). Suppose $f_0$ is a closed map. Is $f$ closed(*)?

(*) A morphism is closed if the image of any closed subset is closed(Hartshorne, p.100).

share|cite|improve this question
Dear Makoto, please define "closed morphism". If you mean by closed map (only in the topological sense), then the answer is yes. – user18119 Nov 27 '12 at 9:50
@QiL I edited the question. – Makoto Kato Nov 27 '12 at 19:58
up vote 3 down vote accepted

First, $f$ maps indeed $X_0$ to $Y_0$: if $x\in X_0$, then the residue field $k(x_0)$ is a finite extension of $k$ by Hilbert Nullstellensatz. Hence $k(y_0)$, where $y_0=f(x_0)$, which is a sub-extension of $k(x_0)$, is also finite over $k$. So $\dim\overline{\{ y_0\}}=\mathrm{trdeg}_k k(y_0)=0$ (transcendental degree) and $y_0$ is closed.

Now suppose $f$ is a closed map. Let's show $f_0$ is a closed map. Let $F_0$ be a closed subset of $X_0$. It is the trace of some closed subset $F$ of $X$: $F_0=F\cap X_0$. Then $f_0(F_0)\subset f(F)\cap Y_0$. Let's show the inverse inclusion. This will imply that $f_0(F_0)$ is closed in $Y_0$. Let $y_0\in f(F)\cap Y_0$. Consider $F\cap f^{-1}(y_0)$. This is a non-empty closed subset of $X$, so it contains a closed point $x_0$ (because $X$ is noetherian). By definition, we have $x_0\in F_0$ and $f_0(x_0)=y_0$.

Conversely, to answer the actual question, suppose $f_0$ is closed. We have to show $f$ is closed. A bit more technology is needed in my proof. Let $F$ be closed subset of $X$. Then $f(F_0)=f_0(F_0)$ is closed in $Y_0$, so is equal to $E\cap Y_0$ for some closed subset $E$ of $Y$. We have $F_0\subseteq f^{-1}(E)$, so $F=\overline{ F_0} \subseteq f^{-1}(E)$. In other words, $f(F)\subseteq E$.

Now we need Chevalley's theorem which states that $f(F)$ is constructible in $E$ (EGA IV, § 1.8). So the complement $E\setminus f(F)$ is constructible. As a constructible subset is a finite union of locally closed subsets, and any non-empty locally closed subset of $E$ contains a closed point of $E$, we have $E\setminus f(F)=\emptyset$ as $f(F)\supseteq E_0=f(F_0)$. Therefore $f(F)=E$ is closed in $Y$.

Unfortunaltely I don't see a more direct proof (without Chevalley).

share|cite|improve this answer
Dear QiL, aren't you answering Makoto's previous, converse question? In the present question here on this page he assumes that $f_0$ is closed. – Georges Elencwajg Nov 27 '12 at 20:06
Ah you are right. I wrote the answer in a hurry. – user18119 Nov 27 '12 at 20:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.