Can a finitely generated module $M$ over a commutative ring have $\operatorname{Ann}(x) \neq 0$ for all $x \in M$ while $\operatorname{Ann}(M) = 0$?
It's not difficult to show that there is no such module if the ring is a integral domain. For general, I guess the answer is yes. But I failed to find a desired example.
Best Answer
Let $R$ be a UFD which is not a PID, e.g. $R=\mathbb Z[X]$, and $M=\bigoplus_{p\text{ prime}} R/(p)$. Note that every non-invertible element of $R$ is a zero-divisor on $M$. Let $I=(p_1,p_2)$ with $p_1,p_2$ primes such that $I\ne R$. Since $I$ does not contain invertible elements, every element of $I$ is a zero-divisor on $M$. Moreover $(0:_MI)=0$.
Now consider the idealization $A=R(+)M$ of the $R$-module $M$. Let $J=IA$. We have that $J$ is finitely generated ideal and consists of zero-divisors, but no non-zero element of $A$ annihilates $J$.