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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?

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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 $R/(p^n) \to R/(p^m)$

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  • $\begingroup$ 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? $\endgroup$
    – SmoothKen
    Commented Sep 6, 2021 at 0:13
  • $\begingroup$ @SmoothKen I think the easiest way to do it is if you take the zero map on all but one cyclic summands $\endgroup$ Commented Sep 6, 2021 at 8:11
  • $\begingroup$ Good idea........ $\endgroup$
    – SmoothKen
    Commented Sep 6, 2021 at 17:38

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