I think that for any field $k$ and any $k$-schemes $X$, $Y$ locally of finite type, any $k$-morphism $X\rightarrow Y$ sends closed points to closed points (https://math.stackexchange.com/a/116636/691259). I want to understand what happens if we do not assume that both $X$ and $Y$ are locally of finite type over $k$.
Assume $k$ is algebraically closed.
- Does there exist a $k$-scheme $X$ and a $k$-scheme locally of finite type $Y$ and a $k$-morphism sending a closed point to a non-closed point? I think the answer is yes, consider the map $\mathrm{Spec}\:Frac(\mathbb{C}[x])\rightarrow \mathrm{Spec}\:\mathbb{C}[x]$.
- Does there exist a $k$-scheme locally of finite type $X$ and a $k$-scheme $Y$ and a $k$-morphism sending a closed point to a non-closed point?
Best Answer
I wrote this quickly--read it skeptically to make sure I didn't make any silly mistakes.
Let $X$ be a finite type $k$-scheme and let $Y$ be any $k$-scheme. Let $x$ be a closed point in $X$. Then, $k(x)/k$ is finite. Let $f:X\to Y$ be a $k$-morphism and set $y:=f(x)$.
To show that $y$ is closed it suffices to show that for any affine open $U$ containing $x$ we have that $x$ is closed in $U$. Indeed, note then that
$$X-\{x\}=\bigcup_{\stackrel{U\text{ affine open}}{x\in U}}(U-\{x\})\cup \bigcup_{\stackrel{U\text{ affine open}}{x\notin U}}U$$
By assumption each element in the first union is open in $U$ and thus open in $X$, and every element of the second union is obviously open thus $X-\{x\}$ is open as desired.
So, we may assume that $Y$ is affine, say $Y=\mathrm{Spec}(A)$. Let us then write $y=\mathfrak{p}$. Note then that we have natural embeddings of $k$-algebras
$$A/\mathfrak{p}\hookrightarrow \mathrm{Frac}(A/\mathfrak{p})=k(y)\hookrightarrow k(x)$$
This implies that $A/\mathfrak{p}$ is finite dimensional over $k$. This then implies that $A/\mathfrak{p}$ is a field. Indeed, this is classical: let $B$ be a domain finite dimensional over a field $k$, then for any $0\ne b\in B$ the multiplication by $b$ map $B\to B$ is injective, so surjective, so $b$ has a multiplicative inverse. Thus, we see that $\mathfrak{p}$ is maximal so that $y$ is closed.