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.)

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