If \(R\) has an identity, then a nonzero unitary \(R\dash\)module is simple if its only submodules are \(0\) and \(A\).
Show that every simple \(R-\)module is cyclic.
If \(A\) is simple, every \(R-\)module endomorphism is either the zero map or an isomorphism.
Solution
1. Every simple \(R\)-module \(A\) is cyclic; indeed \(A=Ra\) for every nonzero \(a\in A\).
Since \(A\ne0\), there is \(a\in A\) with \(a\ne0\). The cyclic submodule \(Ra=\{ra : r\in R\}\) contains \(a=1_R\cdot a\) because \(A\) is unitary, so \(Ra\ne0\). As \(A\) is simple, its only submodules are \(0\) and \(A\), hence \(Ra=A\).
2. (Schur’s lemma) If \(A\) is simple, every \(f\in\operatorname{End}_R(A)\) is either \(0\) or an isomorphism.
2.1. Either \(f=0\) or \(f\) is injective.
The kernel \(\ker f\) is a submodule of \(A\), so \(\ker f=A\) or \(\ker f=0\). In the first case \(f=0\); in the second \(f\) is injective.
2.2. If \(f\) is injective, then \(f\) is surjective.
The image \(\operatorname{im} f\) is a submodule of \(A\). Since \(A\ne0\) and \(f\) is injective, \(\operatorname{im} f\ne0\), so simplicity gives \(\operatorname{im} f=A\).
2.3. Q.E.D.
By steps 1 and 2, a nonzero \(f\) is a bijective \(R\)-module homomorphism, hence an isomorphism.