morphism of schemes that is closed at topological space level - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:59:27Z http://mathoverflow.net/feeds/question/18720 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18720/morphism-of-schemes-that-is-closed-at-topological-space-level morphism of schemes that is closed at topological space level saurav 2010-03-19T08:53:24Z 2010-03-19T13:28:28Z <p>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.</p> <p>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.</p> <p>I think it is easy for $X$ and $Y$ affine. But especially when $X$ is not affine, I have no idea.</p> <p>Best regards, Saurav</p> http://mathoverflow.net/questions/18720/morphism-of-schemes-that-is-closed-at-topological-space-level/18725#18725 Answer by Akhil Mathew for morphism of schemes that is closed at topological space level Akhil Mathew 2010-03-19T11:03:07Z 2010-03-19T11:03:07Z <p>I think you can just reduce to the case of of $X,Y$ affine by covering $Y$ by open affines.</p> <p>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 $Y_{cl}$.</p> http://mathoverflow.net/questions/18720/morphism-of-schemes-that-is-closed-at-topological-space-level/18741#18741 Answer by Georges Elencwajg for morphism of schemes that is closed at topological space level Georges Elencwajg 2010-03-19T13:28:28Z 2010-03-19T13:28:28Z <p>Dear saurav, yes the restriction is closed and you don't have at all to assume $k$ algebraically closed. </p> <p><em>Reminder:</em> 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 $T$ .</p> <p>Here then is the statement you need :</p> <p>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.</p> <p><strong>Proof</strong>:</p> <p>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". </p> <p>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.</p> <p>(Everything I used is contained in EGA I)</p>