I wonder, if the ring without unity $2\mathbb{Z}$, considered as a modul over itself, is a free modul.
For a ring with unity, which is not the nullring the answer is clearly yes, because one can write every element r as r $\cdot$ 1, so {1} is a linearly independed spanning set. Oh, I think I don't have to exclude the nullring.
For a ring $R$ without a unit, we can make it into a ring with unit by forming the ring $\mathbb{Z}\ltimes R$ with underlying abelian group $\mathbb{Z}\oplus R$ and multiplication $(n,r)(n',r') = (nn',nr'+n'r' + rr')$.
Note that a module over a ring without a unit satisfies one less axiom than a module defined over a ring with a unit (i.e $1x = x$ does not make sense for modules over a ring without a unit). There is an isomorphism of categories between $R$-modules and $(\mathbb{Z}\ltimes R)$-modules. An $R$-module $M$ can be made into a $(\mathbb{Z}\ltimes R)$-module by defining $(n,r)m = nm+rm$, and a $(\mathbb{Z}\ltimes R)$-module can be made into an $R$-module by defining $rm = (0,r)m$. Since we know that for rings with unit the free module with a single generator is the ring we conclude the free $R$-module (where $R$ is a ring without unit) with a single generator is $\mathbb{Z}\ltimes R$. In particular $2\mathbb{Z}$ is not a free $2\mathbb{Z}$-module, while $\mathbb{Z} \ltimes 2\mathbb{Z}$ is.