Is $\{0\}$ a free module (over any ring $R$) ?
A free module is isomorphic to $R^n$, but is $n=0$ allowed?
Alternatively, a free module is defined to have a set of linearly independent generating sets, in this case it would be $0$ but I recall by definition in the case of vector spaces that $0$, by itself, is dependent.
However $\{0\}$ is torsion free as the only torsion element is $0$. (I know torsion free does not imply free, but this fact seemed relevant to me).
The thing is, there is a result I need to use which has free modules among the hypothesis and I am too lazy now to check if the proof of that result is ok for $\{0\}$.
Best Answer
I would say yes, it has the empty set as a basis.