Corollary 4

— Let A be a principal ideal domain which is not a field, and let M be an A-module. Then M is semi-simple if and only if M is a torsion module and M, = M(π) for every irreducible element π of A.

First suppose that M is semi-simple; let x ∈ M and let π be an irreducible element of $\mathscr{A}$. If N is a complement of $Annx$ in M, then we can write $x = \alpha \pi x + y$, with $\alpha \in \mathscr{A}$ and $y \in N$; but that implies $y = (1 - \alpha n)x$, so

It follows first of all that M is a torsion module; if moreover $x \in M_\pi$, then $\pi(1 - \alpha \pi)x = 0$, thus $\pi x = \alpha \pi^2 x = \alpha^2 \pi^3 x = \cdots = \alpha^n \pi^{n+1} x$ is zero and $M_\pi = M(\pi)$.

Conversely, by Cor. 2 it is enough to prove that an A-module M annihilated by an irreducible element π is semi-simple; but that is clear, since M then has a natural structure of a vector space over the field $\mathscr{A}/(\pi)$, and the submodules of M are precisely the vector subspaces under this structure.