We want to show that every simple \(R\dash\)module \(M\) is cyclic, i.e. if the only ideals of \(M\) are \((0)\) and \(M\) itself, that \(M = \generators{m}\) for some element \(m\in M\).
Towards a contradiction, let \(M\) be a simple \(R\dash\)module and suppose \(M\) is not cyclic, so \(M\neq \generators{m}\) for any \(m\in M\). But then let \(a\in M\) be an arbitrary nontrivial element; then \((a)\) is a non-empty ideal (since it contains \(a\)), so \((a) \neq 0\). Since \(M\) is simple, we must have \((a) = M\), a contradiction.
Solution
Let \(R\) be a ring with identity. Recall that an \(R\)-module \(M\) is simple if \(M\neq0\) and its only submodules are \(0\) and \(M\).
1. For every \(0\neq m\in M\), \(M=Rm\).
\(Rm\) is a submodule of \(M\), and \(m=1\cdot m\in Rm\), so \(Rm\neq0\). By simplicity, \(Rm=M\).
2. For every \(0\neq m\in M\), \(M\cong R/\operatorname{ann}_R(m)\), and \(\operatorname{ann}_R(m)\) is a maximal left ideal.
The map \(R\to M\), \(r\mapsto rm\), is \(R\)-linear, surjective by step 1, and has kernel \(\operatorname{ann}_R(m)\). Submodules of \(R/\operatorname{ann}_R(m)\) correspond to left ideals of \(R\) containing \(\operatorname{ann}_R(m)\), and \(M\) has exactly two submodules, so \(\operatorname{ann}_R(m)\) is maximal.
3. Q.E.D.
A simple module is nonzero, so it has some \(m\neq0\), and \(M=Rm\) is cyclic by step 1.