 Solution

An $R$-module is free if it has a basis: a subset $B$ such that every element has a unique expression as a finite $R$-linear combination of elements of $B$.

Choose a maximal ideal $\mathfrak{m}$ of the nonzero ring $R$. If $R^n\cong R^m$, quotienting by $\mathfrak{m}$ givesThese are vector spaces over the field $R/\mathfrak{m}$, so equality of dimension gives $n=m$. This proves invariant basis number for a commutative ring.

A direct summand of a free module need not be free. Take $R=\mathbb{Z}/6\mathbb{Z}$. Its ideals $P=(2)$ and $Q=(3)$ satisfyso $P\oplus Q\cong R$. But $P$ has three elements, whereas a finite-rank free $R$-module has $6^r$ elements; hence $P$ is not free.