Take the tour ×  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|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
add comment  1 Answer  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