Describe M/Tor(M) for a finitely generated module M over a PID

Let M be a finitely generated module over a principal ideal domain R. Describe the structure of M/\mathsf{Tor}(M).

We begin with a lemma.

Lemma: Let R be a domain, and let M and N be (left, unital) R-modules. Then \mathsf{Tor}(M \oplus N) = \mathsf{Tor}(M) \oplus \mathsf{Tor}(N). Proof: (\subseteq) Let (m,n) \in \mathsf{Tor}(M \oplus N). Then there exists a nonzero element r \in R such that r(m,n) = (0,0). Then (rm,rn) = (0,0), and so rm = 0 and rn = 0. So m \in \mathsf{Tor}(M) and n \in \mathsf{Tor}(N). (\supseteq) Let (m,n) \in \mathsf{Tor}(M) \oplus \mathsf{Tor}(N). Then there exist nonzero elements r,s \in R such that rm = 0 and sn = 0. Now rs is nonzero in R (since it is a domain), and we have rs(m,n) = (s(rm), r(sn)) = 0. So (m,n) \in \mathsf{Tor}(M \oplus N). \square

Now by the Fundamental Theorem of Finitely Generated Modules over a PID (FTFGMPID), if M is such a module, we have M \cong_R R^t \oplus \bigoplus R/(p_k^{e_k}) for some primes p_k \in R and natural numbers e_k. Now \mathsf{Tor}(M) = \mathsf{Tor}(R^t) \oplus \mathsf{Tor}(\bigoplus R/(p_k^{e_k})) = \bigoplus R/(p_k^{e_k}), since R^t is torsion-free (since R is a domain) while… the stuff inside the big sum is all torsion.

Using this previous exercise, we have that M/\mathsf{Tor}(M) \cong R^t.

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