# Prove that $M$ is a free module if and only if $M$ is a projective module over $PID$.

Let $R$ be a principal ideal domain and $M$ a finitely generated $R$ module. Prove that $M$ is a free $R$-module if and only if $M$ is a projective $R$-module.

I am quite confused and totally not clear about projective modules defined by the universal lift property in commutative diagram.

• One equivalent condition for $M$ being a projective is that it is a direct summand of a free module. In your finitely generated case, this means that there exists another module $N$ such that $M\oplus N = R^k$ for some natural number $k$. Do you know the classification of finitely generated modules over a principal ideal domain? – user2055 Dec 26 '13 at 7:54

## 1 Answer

In general, over any ring, for any module, the following implications hold:

free $\implies$ projective $\implies$ flat $\implies$ torsionfree

For finitely generated modules over a PID, the structure theorem says such a module is a direct sum of a free module and a torsion module. Thus, for such modules, torsionfree implies free, so all 4 properties are equivalent.