For finitely generated modules over a PID, projective and free are equivalent

Prove that a finitely generated module over a PID is projective if and only if it is free.


We begin with a lemma. (I probably already proved this somewhere, but I can’t find it now.)

Lemma: Let M and N be modules over a domain R. Then \mathsf{Tor}(M \oplus N) = \mathsf{Tor}(M) \oplus \mathsf{Tor}(N). Proof: (\subseteq) If (m,n) \in \mathsf{Tor}(M \oplus N), then we have r(m,n) = (rm,rn) = 0 for some nonzero r \in R. So rm = 0 and rn = 0, and we have (m,n) \in \mathsf{Tor}(M) \oplus \mathsf{Tor}(N). (\supseteq) If (m,n) \in \mathsf{Tor}(M) \oplus \mathsf{Tor}(N), then for some nonzero r,s \in R, we have rm = sn = 0. Since r is a domain, rs is nonzero, and now rs(m,n) = 0. So (m,n) \in \mathsf{Tor}(M \oplus N). \square

Let M be a finitely generated module over a principal ideal domain R.

If M is projective, then M \oplus N is free over R for some module N, by Proposition 30 on page 389 of D&F. (This is one of the equivalent definitions of projectivity for modules.) If M \oplus N is free, then it is torsion free, and we have \mathsf{Tor}(M \oplus N) = \mathsf{Tor}(M) \oplus \mathsf{Tor}(N) = 0 by the lemma. So \mathsf{Tor}(M) = 0. By Theorem 5 in D&F (FTFGMPID), we have that M \cong R^t for some t \in \mathbb{N}, so that M is free. By Corollary 31 on page 390 of D&F, (every free module is projective), M is projective.

So for finitely generated modules over a PID, free and projective are equivalent.

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