Finiteness of support of a module as $S$- and $R$- module $\DeclareMathOperator\nn{\mathfrak{n}}\DeclareMathOperator\mm{\mathfrak{m}}\DeclareMathOperator\Supp{Supp}\newcommand\card[1]{\lvert#1\rvert}$Let $f:R\to S$ be a homomorphism  of commutative Noetherian  rings with identity that makes
$S$ a finite $R$-module.
Let $M$ be a (not necessarily finite) $S$-module. So $M$ is also an $R$-module.

If $\card{\Supp_S M}\lt \infty$, is $\card{\Supp_R M}\lt \infty$?
If $\card{\Supp_R M}\lt \infty$, is $\card{\Supp_S\ M}\lt \infty$?


What if:
1- $f$ is epimorphism?
or
2- $f$ is a monomorphism?
 A: $|\operatorname{Supp}_R M| \lt \infty \iff |\operatorname{Supp}_S M| \lt \infty$.
Combine the following:

*

*If $Q \in \operatorname{Supp}_S M$ then $P=Q \cap R \in \operatorname{Supp}_R M$, because $M_Q$ is a localization of $M_P$ (as $S$-modules). (This statement doesn't require $f$ finite.)


*If $P \in \operatorname{Supp}_R M$ then there is at least one $Q \in \operatorname{Supp}_S M$ which lies over $P$ (i.e. $Q \cap R = P$):  The support of a sum of modules is the union of their supports, so the statement reduces to $S$-modules generated by one element i.e. $M=S/I$. Then $P \in \operatorname{Supp}_R S/I \implies P \supset I \cap R \implies \exists Q \supset I$ such that $Q \cap R = P$ (by "lying over" for the finite ring injection $R/(I \cap R) \subset S/I)$, and such $Q \in \operatorname{Supp}_S S/I$.


*For each $P \subset R$ the set of $Q \subset S$ lying over $P$ is finite, because the fiber $S_P/PS$ is an Artinian ring (finite over the field $R_P/P$).
Notes:

*

*It is not required that rings $R, S$ are Noetherian.


*A finite epimorphism of commutative rings is surjective, so $\operatorname{Supp}_R M = \operatorname{Supp}_S M$ in this case.
