Characterization of irreducible modules

Let R be a ring with 1. A (left, unital) R-module M is called irreducible if M \neq 0 and 0 and M are the only R-submodules of M. Prove that M is irreducible if and only if M \neq 0 and if x \in M is nonzero, then M = (x). Describe the irreducible \mathbb{Z}-modules.

(\Rightarrow) Suppose M is irreducible. Now M \neq 0 by definition. Now let x \in M be nonzero. Note that Rx \subseteq M is a nonzero submodule since 1 \cdot x = x \in Rx; since M is irreducible, Rx = M. Thus any nonzero element of M generates M.

(\Leftarrow) Now suppose M \neq 0 and that if x \neq 0, then Rx = M. Let N \subseteq M be a nonzero submodule. Now there exists some nonzero x \in N, and N \supseteq Rx = M. Thus N = M. Since the only submodules of M are 0 and M, M is irreducible.

Recall that \mathbb{Z}-modules are precisely the abelian groups, and that \mathbb{Z}-submodules are precisely the subgroups of abelian groups. Now let M be an irreducible \mathbb{Z}-module. In particular, M is cyclic as an abelian group. So M must be finite of prime order, as otherwise it has nontrivial proper subgroups. Conversely, every cyclic group of prime order is irreducible as a \mathbb{Z}-module.

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