On the annihilator of a module

Question

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 $\mathrm{Ann}(a)=\mathrm{Ann}(I)$?

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.