Let $(R,m)$ be a Noetherian local ring and $M$ an $R$-module. If $M$ is a finite $R$-module, then we know that $Ext^i_R(R/I,M)$ is a finite $R$-module for all $i\geq 0$; Now suppose $supp(M)\subseteq V(I)$ and $Ass_R(M)$ is a finite set, then $Ass_R(Hom(R/I,M))=Ass_R(M)$ is finite. For $i>0$, is $Ass_R(Ext_R^i(R/I,M))$ still a finite set?

  • $\begingroup$ In your questions $M$ is not a finite module? $\endgroup$ Nov 10, 2015 at 22:57

2 Answers 2


Take $R = \mathbb{Z}[X]$, so $\mathbb{Z} = R/(X)$. Let $I = (X)$. For each prime number $p$ we consider the module $M_p := (p,X)/(X^2)$. We have $\mathrm{Supp}M_p = V(I)$ and $\mathrm{Ass}_R(M_p) = (X)$. Consider the short exact sequence $$0 \to M_p \to R/(X^2) \to R/M_p \cong \mathbb{Z}/p\mathbb{Z} \to 0.$$ Applying the $\mathrm{Ext}^i_R(R/(X),-)$ to the above sequence with note that $$\mathrm{Hom}_R(R/(X), M_p) \cong (X)/(X^2) \cong \mathrm{Hom}_R(R/(X), R/(X^2)$$ we have $$0 \to \mathrm{Hom}_R(R/(X), \mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z} \to \mathrm{Ext}^1_R(R/(X),M_p) \to \cdots$$ Thus $(p,X) \in \mathrm{Ass}_R(\mathrm{Ext}^1_R(R/(X),M_p))$. Now set $$M = \bigoplus_{p: prime} M_p$$ we have $\mathrm{Supp}(M) = V(I)$, $\mathrm{Ass}_R(M) = (X)$ but $$\mathrm{Ass}_R(\mathrm{Ext}^1_R(R/(X),M)) = \{(p,X): p \text{ is prime}\}$$ is an infinite set.

  • $\begingroup$ Your $R$ isn't local. $\endgroup$
    – user91132
    Dec 11, 2015 at 10:20
  • $\begingroup$ the local condition is not important in this question. you can change $\mathbb{Z}[X]$ by $\mathbb{Q}[X,Y,Z]_{(X,Y,Z)}$ and the prime ideals $\{(p,X): p \text{ is prime}\}$ by $\{(X, Y+nZ) : n = 0, 1, ...\}$. $\endgroup$ Dec 11, 2015 at 16:05

One can show that if $R$ is a commutative ring, $S$ is a multiplicative system, and $M$ and $N$ are $R$-modules such that $M$ is finitely generated, then the natural map $S^{-1}Hom_R(M,N)\rightarrow Hom_{S^{-1}R}(S^{-1}M,S^{-1}N)$ is an isomorphism. We need to use the finite generation of $M$ twice, both for the injectivity and the surjectivity. Now, Write a projective resolution $P$ for $R/I$ over $R$. Since $R$ is Noetherian, all the terms in the resolution and all their submodules will be finitely generated. Let $p$ be a prime of $R$ and let $S = R-p$. Then $S^{-1}P$ is a projective resolution for $S^{-1}R/I$ over $S^{-1}R$ (we use here the fact that localization is exact). We can then show, using the above isomorphism, that $S^{-1}Ext^i_R(R/I,M)\cong Ext^i_{S^{-1}R}(S^{-1}R/I,S^{-1}M)$. Since $Ass(M)$ is finite, the last group will be zero for almost all primes $p$. But this means that the number of associated primes of $Ext^i_R(R/I,M)$ is also finite.

  • 2
    $\begingroup$ By the finitenes of $Ass(M)$, can we deduce $Ext^i_{S^{-1}R}(S^{-1}R/I,S^{-1}M)$ is zero for alomost all prime $p$? Seldomly $Supp(M)$ is a finite set. $\endgroup$
    – yaxuan
    Sep 12, 2015 at 3:16
  • $\begingroup$ Given Pham Hung Quy's example, your answer is seems to be wrong. Could you please explain where your argument fails. $\endgroup$ Dec 12, 2015 at 18:08
  • $\begingroup$ The argument breaks at the end: we have that $S^{-1}R/q\neq 0$ if and only if $q\subseteq p$. So if we vary over prime ideals containing $q$, and we can get infinitely many ideals such that the localization is not zero. $\endgroup$
    – Ehud Meir
    Dec 13, 2015 at 23:12
  • $\begingroup$ Thanks for the explanation. There's one more question: Does the stated isomorphism for $S^ {-1}Hom_R(M,N)$ need $R$ to be noetherian ? In the question $R$ is supposed to be noetherian anyway, but it's not clear if it is required in your statement. This is of interest to me because I only knew this isomorphism for finitely presented $M$ (being equivalent to finitely generated in the noetherian case, of course). $\endgroup$ Dec 14, 2015 at 14:06

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