## 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.