Let \(R\) be a Principal Ideal Domain (PID). (1) If \(R\) is not a field, prove that the irreducible (simple) \(R\)-modules are precisely \(R/(p)\) where \(p\in R\) is irreducible. If \(R\) is a field, show that the only simple \(R\)-module up to isomorphism is \(R\). (2) Prove that the finitely generated indecomposable \(R\)-modules are precisely \(R\) and \(R/(p^n)\) where \(p\in R\) is irreducible and \(n\ge1\).
Solution
Let \(M\) be a nonzero simple \(R\)-module and choose \(0\ne m\in M\). Then \(Rm=M\), so \[ M\cong R/\operatorname{Ann}(m). \] Simplicity says that \(\operatorname{Ann}(m)\) is maximal. If \(R\) is not a field, every maximal ideal of the PID \(R\) is nonzero and therefore has the form \((p)\) with \(p\) irreducible. Thus \(M\cong R/(p)\). Conversely, \((p)\) is maximal for irreducible \(p\), so \(R/(p)\) is simple. If \(R\) is a field, its only maximal ideal is \((0)\), and the only simple module up to isomorphism is \(R\) itself.
For part (2), the structure theorem gives every finitely generated \(R\)-module in the form \[ R^r\oplus\bigoplus_i R/(p_i^{n_i}). \] If such a module is indecomposable, at most one nonzero summand can occur. Hence an indecomposable module must be either \(R\) or \(R/(p^n)\).
The module \(R\) is indecomposable: a decomposition \(R=I\oplus J\) would give \(1=e+(1-e)\) with \(e^2=e\); a domain has only the idempotents \(0,1\), so one summand is zero.
Finally, the submodules of \(R/(p^n)\) are exactly \[ (p^j)/(p^n),\qquad 0\le j\le n, \] which form a chain. Hence two nonzero submodules cannot have zero intersection, so \(R/(p^n)\) cannot decompose as a direct sum of two nonzero submodules. Therefore the finitely generated indecomposable modules are exactly \(R\) and the \(R/(p^n)\).