If $R$ is a ring, it is known that $\operatorname {MaxSpec}R$ can be equipped with a topology: the one induced by $\operatorname{Spec}R$. Furthermore, if $A\to B$ is a morphism of finitely generated algebras over an algebraically closed field $k$, the map $f:\operatorname {Spec}B\to \operatorname {Spec}A$ induces a map $\operatorname {MaxSpec}B\to \operatorname {MaxSpec}A$, by restriction. As usual denote with $V(I)\subseteq \operatorname {Spec}R$ the vanishing set of the ideal $I\subseteq R$, and write $Z(I)$ for $V(I)\cap \operatorname {MaxSpec}R$. It makes sense to think if, for ideals $I,J$ in respectively $A,B$, the following equalities hold: $$fZ(J)=fV(J)\cap \operatorname{MaxSpec} B;$$ $$f^{-1}Z(I)\cap \operatorname{MaxSpec} A=f^{-1}V(I)\cap \operatorname{MaxSpec} A.$$
Observe also that the radical of an ideal in a $k$-algebra is the intersection of the maximal ideal over it, so in this setting the max spectrum is a dense set in the spectrum.
Therefore the problem can be reduced as it follows. Let $X'\subseteq X,Y'\subseteq Y$ be dense subsets of two topological space, and let $g:X\to Y$ be a map such that $gX'\subseteq Y'$. So $(g^{-1}(S^r))^r=(g^{-1}S)^r$, where $S\subseteq Y$ and $-^r$ denotes the restriction to $X'$; this is clear even without the hypothesis that $X',Y'$ are dense, and shows the second equality. Instead the first equality becomes $g(S^r)=(gS)^r$, where $S\subseteq X$ and $-^r$ denotes the restriction to $Y'$; this is true if $S$ is closed, because if a prime $\mathfrak p$ contracts to a maximal ideal $\mathfrak m$, also the contraction of any maximal containing $\mathfrak p$ is $\mathfrak m$. I don't know precisely what topological result I used, I know very little of that branch, I'd say something similar to: if every point of $V$ is closed and $U$ is a space with a dense subset $U'$, whose points are all closed, for any closed $S\subset U$ and $S\cap U'$ has the same image.
What do you think about this argument, does it hold? The motivation for my answer is that I often don't know how the results that I see for the spectrum can be used when working with the max spectrum, for the $k$-algebras. Thank you in advance
Best Answer
You argument looks mostly fine to me.
Here is one way to see this: If $f: X \to Y$ is a continuous map of topological spaces, and $x, x' \in X$ are two points with $x \in \overline{\{x'\}}$, then $f(x) \in \overline{\{f(x')\}}$, where $\bar{ }$ denotes the closure of a subset. This is true because $f^{-1}(\overline{\{f(x')\}})$ is a closed set containing $x'$, hence it has to contain $\overline{\{x'\}}$ and $x$.
Consequently, if $\mathfrak p$ is any prime mapping to a maximal ideal $\mathfrak m$, then any $\mathfrak n \supset \mathfrak p$ also maps to $\mathfrak m$.