## Every finitely generated torsion module over an integral domain has a nonzero annihilator

Let $R$ be an integral domain and let $M$ be a finitely generated torsion left $R$-module. Prove that $\mathsf{Ann}_R(M) \neq 0$.

Since $M$ is finitely generated, we have $M = (A)$ for some finite set $A = \{a_i\}_{i \in I}$. Since $M$ is torsion, for each $a_i$ there exists $z_i \in R$ nonzero such that $z_i \cdot a_i = 0$. Since $R$ is an integral domain, $z = \prod_{i \in I} z_i$ is nonzero. Now if $m = \sum r_i \cdot a_i \in M$, $z \cdot m = \sum zr_i a_i$ $= 0$. Thus $z \in \mathsf{Ann}_R(M)$, and in particular $\mathsf{Ann}_R(M) \neq 0$.

Now consider the abelian group (hence left $\mathbb{Z}$-module) $\mathbb{Q}/\mathbb{Z}$. Suppose there exists a nonzero element $z \in \mathsf{Ann}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z})$. Choose $w \in \mathbb{Z}$ such that $w$ does not divide $z$. Now $z \cdot \overline{1/w} = \overline{z/w} = \overline{0}$, so that $z/w$ is an integer. But then $z/w = k$, and thus $z = kw$, so that $w$ divides $z$, a contradiction. So $\mathsf{Ann}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z}) = 0$. (So we can conclude that $\mathbb{Q}/\mathbb{Z}$ is not finitely generated as an abelian group, a result we proved previously with a bit more effort.)