We define a functorial scheme as in "Two functorial definitions of schemes". I want to define
a closed map but we only have the definition of schemes as functors in category theory, so we don't know underlying space. My question is : Can we define a closed map in this case? Thanks in advance.
The definition of a closed map on a functorial scheme in category theory
algebraic-geometrycategory-theory
Best Answer
Here's one attempt, though perhaps a disappointing one since it's not much different from the usual definition in terms of maps of ringed spaces. $\newcommand{\spec}{\operatorname{spec}}\newcommand\mf\mathfrak$
Say $f:X\to Y$ is a map of schemes (i.e. sheaves on $\mathbf{Ring}^{\mathrm{op}}$ covered by affine schemes). Choose an affine cover $\spec A\to Y$ of $Y$ (exists since $Y$ is a scheme), and let $U:=X\times_Y(\spec A)$ be its pullback to $X$. Let $\spec B\to U$ be an affine cover of $U$. We say that $f$ is a closed map if the ring map $\phi:A\to B$ induced by $\spec B\to U\to\spec A$ satisfies the following property: (*) for any ideal $J\subset B$ and any prime $\mf p\subset A$ containing $\ker(A\xrightarrow\phi B\to B/J)$, there exists a prime $\mf q\subset B$ containing $J$ so that $\phi^{-1}(\mf q)=\mf p$.
Let's unpack a little. Property (*) above is just a translation of what is means for a map $\spec B\to\spec A$ of affine schemes to be closed (assuming I didn't make a mistake). For the rest, keep in mind the following diagram $\require{AMScd}$ \begin{CD} \spec B\\ @VVV\\ U @>>> \spec A\\ @VVV @VVV\\ X @>f>> Y \end{CD} $\spec B,\spec A$ are affine (open) covers of $X,Y$. Since, topologically speaking, a subspace is closed iff it is in (every member of an) open cover, we're safe to pass to the cover $\spec B\to\spec A$ in order to define closedness of the map $X\to Y$ (e.g. this definition should be independent of the choice of covers and should exactly recover the usual definition when $X,Y$ are themselves affine).
You may worry that all my affine covers are by a single object instead of being of the form $\{\spec A_i\to Y\}_{i\in I}$. This was just to simplify notation. Really, you'd make the above definition with covers like this and then say for every $\spec B_{ij}$ in a cover of $U_i:=X\times_Y(\spec A_i)$ the resulting ring maps $A_i\to B_{ij}$ satisfies (*).