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?

$$|\operatorname{Supp}_R M| \lt \infty \iff |\operatorname{Supp}_S M| \lt \infty$$.

Combine the following:

1. 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.)

2. 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 $$Q \in \operatorname{Supp}_S S/I$$.

3. 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:

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

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

• thank you David. lying over is by finiteness of $f$, am i right? Nov 28, 2020 at 6:36
• in the Note2 u said "A finite epimorphism of commutative rings is surjective". do u mean isomorphism? Nov 28, 2020 at 6:38
• @13571 1. Right: lying over applies to finite injective ring homomorphism and $R/(I \cap R) \subset S/I$ is finite because $f$ is finite. (I write $I \cap R$ to mean $f^{-1}I$.) 2. A ring epimorphism may not be surjective e.g. localization; a finite ring epimporphism is surjective.(stacks.math.columbia.edu/tag/04VM) Nov 28, 2020 at 13:16
• i'm pretty confused. 1-by epimporphism i mean surjective ring homomorphism. 2-under what assumptions u proved this? did u assumed $f$ is injective/surjective? 3- to be continued... Dec 1, 2020 at 7:25
• u say: "$M_Q$ is a localization of $M_P$" but $P$ is not prime ideal of $S$. i also have problem with your reduction: $M$ is not f.g. can u please help in this cases? Dec 1, 2020 at 7:29