Example of modules that are projective but not free; torsion-free but not free

Let $R=\mathbb{Z}/6\mathbb{Z}$. Obviously, $R$ is a free module over itself. Because $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}\cong R$, we have that $\mathbb{Z}/2\mathbb{Z}$, considered as an $R$-module, is projective, but it cannot be free - any non-trivial direct sum of $R$'s would have at least 6 elements.

The Baer-Specker group is an example of a torsion-free $\mathbb{Z}$-module which is not free.

The issue of when projective modules are free is discussed in $\S 3.5.4$ of my commutative algebra notes. In particular one gets very easy (but not very satisfying) examples by looking at disconnected rings: e.g. $\mathbb{C} \times \{0\}$ is quite clearly projective but not free over $\mathbb{C} \times \mathbb{C}$.

It is more interesting to ask for examples over domains. One huge class of examples comes from invertible fractional ideals which are not principal. In particular a Dedekind domain possesses such ideals iff it is not a PID. This is worked out much later in my notes in the section on Dedekind domains (currently $\S 22$, but this is subject to change). However, in $\S 3.5.4$ I take some time to exhibit from scratch a specific projective, nonfree module over $R=\mathbb{Z}[\sqrt{-5}]$.

Note that free implies torsionfree holds for modules over a domain $R$. (In fact the usual definition of "torsionfree module" is only useful over domains, so far as I know. I seem to recall that Lam's book Lectures on modules and rings gives a more sophisticated definition for modules over a general ring...) To be more precise, over any commutative ring $R$, free $\implies$ projective $\implies$ flat, and over a domain flat $\implies$ torsionfree. So any example of a projective nonfree module over a domain is also an example of a torsionfree nonfree module.

Over a general domain there is a vast gap between torsion free modules and flat modules. For instance, a prime ideal $\mathfrak{p}$ in a Noetherian domain is always a torsionfree module, but it is flat only if it has height at most one. However, a torsionfree module over a PID is flat -- see e.g. $\S 3.11$ of my notes -- but need not be free. For instance, let $R$ be any PID which is not a field and let $K$ be its field of fractions. Then $K$ is a nontrivial torsionfree $R$-module which is divisible, hence not free.

If $k$ is a field and $A = \operatorname{Mat}_n(k) = \operatorname{End}_k(V)$ is the ring of $n\times n$ matrices with coefficients in $k$, then the (simple, left) $A$-module $V = k^n$ of "column vectors" is a projective $A$-module, but it is not free as an $A$-module as soon as $n>1$ (this non-freeness is easy to see by comparing the dimension of $V$ and of $A$ as $k$-vector spaces).

To see the fact that $V$ is projective $A$-module, fix a basis $b_1,\dots,b_n$ for $V$ as a vector space, and for each $j=1,\dots,n$, let $e_j \in A$ be the "idempotent" given by projection on the 1 dimensional linear subspace of $V$ spanned by $b_j$. Form the left ideal $I_j = Ae_j$. The mapping $$X \mapsto Xb_j$$ defines an $A$-module isomorphism $I_j \to V$, and it is straightforward to check that $$A = \bigoplus_{j=1}^n I_j,$$ so indeed $V \simeq I_1$ is a direct summand of a free $A$-module, hence projective.