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.

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