Suppose $X\stackrel f\to Y$ be a morphism of finite type $k$-schemes, where $k$ is a field; for the time being let me say that $k$ is algebraically closed.

Then one knows that $f$ takes $k$-valued points to $k$-valued points. Now suppose the scheme morphism is a closed map i.e. takes closed subsets to closed subsets. Take the restriction of $f$ to closed points i.e. $k$-valued points. We get a map of topological spaces $\tilde f:X_0\to Y_0$ where $X_0$, $Y_0$ are the subsets of closed points. We have topology on $X$ and $Y$ since they are schemes, so we get induced topology on $X_0,Y_0$ also. The question is, whether the restriction map $\tilde f$ is still a closed map.

I think it is easy for $X$ and $Y$ affine. But especially when $X$ is not affine, I have no idea.

Best regards, Saurav


I think you can just reduce to the case of of $X,Y$ affine by covering $Y$ by open affines.

Alternatively, the sets $X_{cl}, Y_{cl}$ of closed (i.e., $k$-valued) points in $X,Y$ map quasi-homeomorphically onto $X,Y$ (cf. EGA 0.3). Thus a constructible set $E \subset X$ is closed iff $E \cap X_{cl}$ is closed, and similarly for $Y$. Thus by Chevalley's theorem and this fact, it follows that $\tilde{f}$ is closed. In detail, given a closed $E_c \subset X_{cl}$, there is a unique constructible $E \subset X$ with $E \cap X_{cl} = E_c$ (by quasi-homeomorphism). Then $f(E)$ is closed in $Y$, so $f(E_c)$ must be closed in $Y_{cl}$.

  • $\begingroup$ But $f(E_c)[=f(E\cap X_c)]$ may not be equal to $f(E)_c[=f(E)\cap Y_c]$. To say that $f(E_c)$ is closed in $Y_c$, we probably need this equality. $\endgroup$ – saurav Mar 19 '10 at 13:18
  • $\begingroup$ This argument applies verbatim to any finite-type map between Jacobson schemes (EGA IV$_3$, section 10); e.g. finite type schemes over Dedekind domain with infinitely many primes. But since formation of images and intersections don't generally commute in set theory, it seems appropriate to note (at least for this generalization!) that "pass to underlying set of closed points" commutes with images for constructible sets (because non-empty constructible sets in Jacobson schemes have closed points; clearly false for noetherian schemes in general). $\endgroup$ – BCnrd Mar 19 '10 at 13:22
  • $\begingroup$ @saurav: your objection overlooks that $E$ is not a random subset, but a constructible one. (Feel free to delete this comment once it makes sense to you.) $\endgroup$ – BCnrd Mar 19 '10 at 13:24

Dear saurav, yes the restriction is closed and you don't have at all to assume $k$ algebraically closed.

Reminder: For a scheme $S$ of finite type over $k$, the set $T_0$ of closed points of $T$ is very dense in $T$, i.e. dense in every closed subset of $T$ .

Here then is the statement you need :

Let $f:X\to Y$ be a closed morphism between schemes of finite type over a field $k$. Then the restriction $f_0:X_0\to Y_0$ to the subspaces of respective closed points is closed.


1) We may assume $X$ and $Y$ reduced. We have to show that, for $F$ closed in $X$, the subset $f(F\cap X_0)$ is closed in $Y_0$. By endowing $F$ and $f(F)$ with their reduced scheme structure we can assume $F=X, f(F)=Y$ and we have reduced (!) the problem to showing that $Y_0=f(X_0)$ : call this "closed surjectivity".

2) Here is why "closed surjectivity" is true. Take any closed point $y_0\in Y_0$. It has a residual field $\kappa (y_0)$ which is a finite extension of $k$.The fibre $Z=f^{-1}(y_0)$ is a closed non-empty subscheme of $X$ (recall that after our reduction $f$ is assumed surjective!). But $Z$ is also a $\kappa (y_0)$-scheme of finite type. Hence it has a closed point $x_0$, since those closed points are even very dense in $Z$. Since $Z$ is closed in $X$, $x_0$ is closed also in $X$ i.e. $x_0\in X_0$. So we have shown "closed surjectivity" and everything is proved.

(Everything I used is contained in EGA I)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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