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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).

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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
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1 Answer 1

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).

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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
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