## Exhibit a nonzero torsion module over an integral domain whose annihilator is trivial

Exhibit an integral domain $R$ and a nonzero torsion module $M$ over $R$ such that $\mathsf{Ann}(M) = 0$. Prove that if $N$ is a finitely generated torsion $R$-module ($R$ a domain), then $\mathsf{Ann}(N) \neq 0$.

First consider the direct sum $M = \bigoplus_{\mathbb{N}^+} \mathbb{Z}/(2^k)$ as a $\mathbb{Z}$-module in the usual way. Every element $\alpha$ of $M$ is annihilated by $2^t$, where $t$ is the largest nonzero component of $\alpha$. However, for all $k \in \mathbb{Z}$, there exist elements not annihilated by $k$. For example, if $2^t \leq k < 2^{t+1}$, then the element with $1$ in the $t+1$ component an 0 elsewhere is not annihilated by $k$.

Suppose now that $N$ is a finitely generated torsion $R$-module; say $N = (A)_R$, with $A = \{a_i\}_{i=1}^n$. For each $a_i$, there exists a nonzero element $r_i \in R$ such that $r_ia_i = 0$, since $N$ is torsion. Let $r = \prod r_i$. Since $R$ is an integral domain, $r \neq 0$, and certainly $rN = 0$. Thus $\mathsf{Ann}(N) \neq 0$.