Invariant basis number

Remarks

Example: every nonzero commutative ring $R$ has IBN. To check it, take a maximal ideal $\mathfrak{m}$ of $R$; an isomorphism $R^m \cong R^n$ gives an isomorphism $(R/\mathfrak{m})^m \cong (R/\mathfrak{m})^n$ of vector spaces over the field $R/\mathfrak{m}$, so $m = n$ by dimension.

Non-example: let $V$ be a vector space of countably infinite dimension over a field $k$ and let $R = \End_k(V)$. Since $V \cong V \oplus V$, one gets $R \cong R^2$ as left $R$-modules, so $R$ does not have IBN.

The zero ring does not have IBN, since $0 = R^m$ for all $m$. The condition can be stated with matrices: $R$ has IBN if and only if there are no matrices $A \in R^{m \times n}$ and $B \in R^{n \times m}$ with $m \ne n$, $AB = I_m$ and $BA = I_n$. This form is symmetric in left and right, so the definition does not depend on the side.