# Finitely generated torsion module homomorphism in PID local ring

Problem Statement: Let R be a commutative local ring, PID, but not a field. Given two nonzero finitely generated torsion $$R$$-modules, show that there exists a nonzero $$R$$-module homomorphism $$M \to N$$.

Call the generator of maximal ideal $$p$$, it is the only prime in $$R$$. Using structural theorem, each torsion module must be of the form $$\bigoplus_{i \in I} R/(p^i)$$ for some $$I \subseteq \mathbb{N}$$. But how do I go from here?

By the structure theorem and the biadditivity of the Hom-functor, it suffices to show this for cyclic torsion modules, which are all of the form $$R/(p^n)$$. To construct a non-zero homomorphism $$R/(p^n) \to R/(p^m)$$ we can do the following: first note that $$(R/(p^n))/((p)/(p^n)) \cong R/(p)$$ by the third isomorphism theorem, so we get a surjective homomorphism $$R/(p^n) \to R/(p)$$. To construct a non-zero homomorphism $$R/(p) \to R/(p^m)$$, just take $$r+(p) \mapsto rp^{m-1}+(p^m)$$ (you need to check this is well-defined and non-zero, I'll leave that to you.) Now if we compose those two maps, we get a non-zero homomorphism $$R/(p^n) \to R/(p^m)$$
• Since both $M$ and $N$ are direct sum of these singletons. When number of singletons of $M$ is $\geq$ that of $N$, we just map the rest to vector 0. And if $\leq$, we just keep those coodinate in $N$ zero? Commented Sep 6, 2021 at 0:13