An abelian group $A$ is a left $E = {\rm End}(A)$-module via $f*a = f(a)$. If $B$ is a direct summand of $A$ as an abelian group, then ${\rm Hom}(B,A)$ is also a left $E$-module and is in fact a direct summand of $E$ as an $E$-module, so it is $E$-projective. In particular, if $B = \Z$, then ${\rm Hom}(B,A)\cong A$ as $E$-modules. *Thus $A$ is a projective $E$-module whenever $A$ has $\Z$ as a direct-summand.*

These observations allow us to construct projective modules that often aren't free over interesting rings. Take the abelian group $A = \Z\oplus \Z$ for instance. Its endomorphism ring $E$ is the ring $M_2(\Z)$ of $2\times 2$ matrices with coefficients in $\Z$. As we have remarked, $\Z\oplus \Z$ must be projective as an $M_2(\Z)$-module.

Is $\Z\oplus\Z$ free as an $M_2(\Z)$-module? On the surface, it seems not to be, but of course we need proof. And here it is: for each element of $\Z\oplus \Z$, there exists an element of $M_2(\Z)$ annihilating it. Such a thing can't happen for free modules.

One might wonder, is every $M_2(\Z)$-module projective? Or in other words, is $M_2(\Z)$ semisimple? Let's hope not! But $M_2(\Z)$ is thankfully not semisimple: $\Z/2\oplus\Z/2$ is a $M_2(\Z)$-module that is not projective: any nonzero element of $M_2(\Z)$ spans a submodule of infinite order, and therefore so must any nonzero element of a nonzero projective.