Any field satisfies IBN, and this amounts to the fact that finite-dimensional vector spaces have a well defined dimension. Moreover, any
commutative ring (except the
zero ring) satisfies IBN, as does any
left-Noetherian ring and any
semilocal ring. Let
A be a commutative ring and assume there exists an
A-module isomorphism f \colon A^n \to A^p. Let (e_1,\dots,e_n) the canonical basis of
An, which means e_i\in A^n is all zeros except a one in the
i-th position. By
Krull's theorem, let
I a
maximal proper
ideal of
A and (i_1,\dots,i_n)\in I^n. An
A-module morphism means f(i_1,\dots,i_n) = \sum_{k=1}^n i_k f(e_k) \in I^p because
I is an ideal. So
f induces an
A/
I-module morphism f' \colon \left(\frac{A}{I}\right)^n\to \left(\frac{A}{I}\right)^p, that can easily be proven to be an isomorphism. Since
A/
I is a field,
f' is an isomorphism between finite dimensional vector spaces, so . An example of a nonzero ring that does not satisfy IBN is the ring of
column finite matrices \mathbb{CFM}_\mathbb{N}(R), the matrices with coefficients in a ring
R, with entries indexed by \mathbb{N}\times\mathbb{N} and with each column having only finitely many non-zero entries. That last requirement allows us to define the product of infinite matrices
MN, giving the ring structure. A left module isomorphism \mathbb{CFM}_\mathbb{N}(R)\cong\mathbb{CFM}_\mathbb{N}(R)^2 is given by: \begin{array}{rcl} \psi : \mathbb{CFM}_\mathbb{N}(R) &\to & \mathbb{CFM}_\mathbb{N}(R)^2 \\ M &\mapsto & (\text{odd columns of } M, \text{ even columns of } M) \end{array} This infinite matrix ring turns out to be isomorphic to the
endomorphisms of a right
free module over
R of
countable rank. From this isomorphism, it is possible to show (abbreviating \mathbb{CFM}_\mathbb{N}(R)=S) that for any positive integer
n, and hence for any two positive integers
m and
n. There are other examples of non-IBN rings without this property, among them
Leavitt algebras. ==Other results==