Question. Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module. Does there always exist an element $s\in M$ such that $\mathrm{Ann}(s)=\mathrm{Ann}(M)$?

Remark. The annihilator of a module is a lower bound of the annihilators of elements in the module. The question asks whether this lower bound can be reached at some element. It appears that (using primary decomposition,) one is only able to show the existence of $s$ satisfying $\sqrt{\mathrm{Ann}(s)}=\sqrt{\mathrm{Ann}(M)}$.

Update. Thanks for the the counter-example provided by Dao. He also provides a criterion when $A$ is Artinian and $M$ are confined to finitely generated $A$-modules with $\mathrm{Ann}(M)=(0)$.

Special case: if $I$ is an ideal of a Noetherian ring $A$, does there always exist $a\in I$ such that $\mathrm{Ann}(a)=\mathrm{Ann}(I)$?

  • 1
    $\begingroup$ In the case $M$ is an ideal, the answer is positive: this is proved in the answer of Manos to this MSE question (note that quotienting by $\operatorname{Ann}(M) $ reduces to the case $M$ is faithful). I do not know the answer for the general case. $\endgroup$
    – abx
    Feb 25, 2021 at 10:53
  • $\begingroup$ @abx: be careful, an ideal of $A$ might no longer be an ideal of $A$ mod the annihilator. $\endgroup$ Feb 25, 2021 at 15:07
  • $\begingroup$ @Hailong Dao: You are right, what I meant is that for the general problem (about modules) it is enough to consider faithful modules. But I agree that you cannot do that directly with ideals. $\endgroup$
    – abx
    Feb 25, 2021 at 15:51

1 Answer 1


Here is a big class of negative examples to your question. Let $A$ be a non-Gorenstein Artinian ring and $M=w_A$ the canonical module of $A$.

Then $ann(M)=0$, as can be seen because $Hom(M,M)\cong A$. Suppose $M$ contains $s$ with annihilator $=(0)$. Then $As\cong A$ sits inside $M$. But since the length of $M$ is equal to the length of $A$, $M=As\cong A$, contradicting our choice of a non-Gorenstein $A$.

Simplest concrete example is $A=k[[x,y]]/(x,y)^2$.

In fact, in the Artinian case one has

Proposition: For an Artinian ring $A$, the following are equivalent:

  1. $A$ is Gorenstein.
  2. Any finitely generated faithful module $M$ over $A$ contains an element with $(0)$ annihilator.

Proof: (2) implies (1) is above. Assume (1) and let $M$ be a faithful module. Let $s_1,...,s_n$ be a set of generators of $M$. Then we have an injection $A\to M^n$ taking $1$ to $(s_1,...,s_n)$. Since $A$ is Gorenstein, this injection splits. As Krull-Schmidt holds over $A$, $M$ contains $A$ as a free summand, which implies what we need.

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    $\begingroup$ In your example, can $k[[x,y]]$ be replaced by $k[x,y]?$ It seems in either case you end up killing off all monomials of degree at least $2,$ ending up with linear polynomials in $x,y.$ $\endgroup$ Feb 25, 2021 at 16:06
  • $\begingroup$ @ChrisLeary: yes, the quotient will be the same. $\endgroup$ Feb 25, 2021 at 16:28
  • $\begingroup$ By the way, we don't need local, any non-Gorenstein Artinian ring would work. $\endgroup$ Feb 25, 2021 at 16:53
  • $\begingroup$ Thank you for your input, Hailong. The Gorenstein condition is a beautiful criterion of cohomological nature in the case you mentioned. $\endgroup$
    – Chris
    Feb 28, 2021 at 4:44

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