Example of a finitely generated faithful torsion module over a commutative ring

Solution 1:

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 $J$.

Solution 2:

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 $0 \neq x_1 \in \operatorname{ann}(J)$.