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

If the associated primes of $M$ are all isolated primes, the existence of such $s$ is easy to show. However, when $M$ has embedded primes, it seems that a similar argument can only show the existence of $s$ satisfying $\sqrt{\mathrm{Ann}(s)}=\sqrt{\mathrm{Ann}(M)}$. (Do I miss anything, or does there exist a counter-example?)

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

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

If the associated primes of $M$ are all isolated primes, the existence of such $s$ is easy to show. However, when $M$ has embedded primes, it seems that a similar argument can only show the existence of $s$ satisfying $\sqrt{\mathrm{Ann}(s)}=\sqrt{\mathrm{Ann}(M)}$.

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

If the associated primes of $M$ are all isolated primes, the existence of such $s$ is easy to show. However, when $M$ has embedded primes, it seems that a similar argument can only show the existence of $s$ satisfying $\sqrt{\mathrm{Ann}(s)}=\sqrt{\mathrm{Ann}(M)}$. (Do I miss anything, or does there exist a counter-example?)

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

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

If the associated primes of $M$ are all isolated primes, the existence of such $s$ is easy to show. However, when $M$ has embedded primes, it seems that a similar argument can only show the existence of $s$ satisfying $\sqrt{\mathrm{Ann}(s)}=\sqrt{\mathrm{Ann}(M)}$.

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

If the associated primes of $M$ are all isolated primes, the existence of such $s$ is easy to show. However, when $M$ has embedded primes, it seems that a similar argument can only show the existence of $s$ satisfying $\sqrt{\mathrm{Ann}(s)}=\sqrt{\mathrm{Ann}(M)}$.