Stably free module

Remarks

A stably free module is finitely generated and projective, since it is a direct summand of $R^n$. Every free module $R^k$ is stably free, with $m = 0$.

Example: let $A = \R[x, y, z]/(x^2 + y^2 + z^2 - 1)$ and let $P$ be the kernel of the map $A^3 \to A$, $(u, v, w) \mapsto ux + vy + wz$. The map is onto and has the section $a \mapsto (ax, ay, az)$, because $x^2 + y^2 + z^2 = 1$ in $A$. So $P \oplus A \cong A^3$, and $P$ is stably free. It is not free: an isomorphism $P \cong A^2$ would give a nowhere-vanishing tangent vector field on the $2$-sphere, which the hairy ball theorem rules out.

Non-example: over $R = \Z/6\Z$, the projective module $P = \Z/2\Z$ is not stably free. The module $P \oplus R^m$ has $2 \cdot 6^m$ elements, which is never of the form $6^n$.