$\DeclareMathOperator\nn{\mathfrak{n}}\DeclareMathOperator\mm{\mathfrak{m}}\DeclareMathOperator\Supp{Supp}\newcommand\card[1]{\lvert#1\rvert}$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?


2- $f$ is a monomorphism?


1 Answer 1


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

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 $R_P/P$).


  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.

  • $\begingroup$ thank you David. lying over is by finiteness of $f$, am i right? $\endgroup$
    – 13571
    Nov 28, 2020 at 6:36
  • $\begingroup$ in the Note2 u said "A finite epimorphism of commutative rings is surjective". do u mean isomorphism? $\endgroup$
    – 13571
    Nov 28, 2020 at 6:38
  • $\begingroup$ @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) $\endgroup$ Nov 28, 2020 at 13:16
  • $\begingroup$ 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... $\endgroup$
    – 13571
    Dec 1, 2020 at 7:25
  • $\begingroup$ 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? $\endgroup$
    – 13571
    Dec 1, 2020 at 7:29

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