Example of a finitely generated faithful torsion module over a commutative ring 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.
 A: 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 $J$.
A: Here is a proof that no such examples exist over integral domains:
Let $M$ be finitely generated over an integral domain $R$.
Let $x_1,\dots,x_n$ be generators of $M$. Take $0 \neq a_i \in \operatorname{ann}(x_i)$ for 
every $i=1,\dots,n$. Define $a=a_1 \cdots a_n$. Since $R$ is an integral
domain, $a\neq 0$ and $a \in \operatorname{ann}(M)$. 
Let me now answer the question negatively in the case where
$R$ is Noetherian and  $0 \neq M= J$ is an ideal of $R$. So let us
assume that all elements of $J$ have zero annihilators. Then
all elements of $J$ are zero-divisors of $R$. Let $p_1,\dots,p_s$
be the minimal prime ideals of $R$. We know that the set of 
all zero-divisors of $R$ is precisely $p_1 \cup \cdots \cup p_s$.
Consequently, $ J \subset p_1 \cup \cdots \cup p_s$. Then by
prime-avoidance we must have that $J \subset p_1$ without loss of
generality. But the minimal prime ideals have the property that
$p_i = \operatorname{ann}(x_i)$ for some $0 \neq x_i \in R, \, i=1,\dots,s$.
This implies that $J \subset \operatorname{ann}(x_1)$, thus $x_1 \, J=0$ and so
$0 \neq x_1 \in \operatorname{ann}(J)$. 
A: What if you take $C=\{i^r | r\in \mathbb Q\}$, where $i\in\mathbb C$.  $C$ is closed under multiplication.  Let $\mathbb Q$ act on it by $r\cdot z = z^r$.  Then it's finitely generated as a $\mathbb Q$ module (generated by $i$).  Then $C$ is a $\mathbb Q$ module (I think) and every element has non-zero annihilator (I think), but no non-zero rational can annihilate everything (I think).
