Here are nine examples of projective modules that are not free, some of which are finitely generated.

## Direct Products

Consider the ring $R= \Z/2\times\Z/2$ and the submodule $\Z/2\times \{0\}$. It is by construction a direct summand of $R$ but certainly not free. And it's finitely generated! Another example is the submodule $\Z/2\subset \Z/6$, though this is the same kind of thing because $\Z/6\cong\Z/2\times\Z/3$. This was the first example I ever saw of a nonfree projective module.

## Infinite Direct Products

One can modify the above construction for infinite direct products of rings, too. For instance, $R = \prod_{i=1}^\infty \Z$ contains $\Z$ as a direct summand. Hence $\oplus_{i=1}^\infty\Z$ is a projective $R$ module, yet cannot be free since nonzero free modules are uncountable.

## Ideals in Dedekind Domains

In a Dedekind domain $R$, take an ideal representing a nontrivial element in the class group. It will then be projective. As an example, the class number of $\Z[\sqrt{5}]$ is two, and the ideal $(2,1+\sqrt{5})$ represents the nontrivial element in the class group. It is not free since it is not principal, and it is finitely generated projective since it is invertible.

More generally, for any ring extension of commutative rings $R\subseteq S$, one may define invertible $R$-submodules of $S$ as it is done for Dedekind domains. Then any invertible $R$-submodule of $S$ will be finitely-generated and projective. For more details and a further example, see Lam's 'Lectures on Modules and Rings', Sections 2B-2C.