IBN
Bases of a Module
Like a vector space^{} over a field, one can define a basis of a module $M$ over a general ring $R$ with 1. To simplify matter, suppose $R$ is commutative^{} with $1$ and $M$ is unital. A basis of $M$ is a subset $B=\{{b}_{i}\mid i\in I\}$ of $M$, where $I$ is some ordered index set^{}, such that every element $m\in M$ can be uniquely written as a linear combination^{} of elements from $B$:
$$m=\sum _{i\in I}{r}_{i}{b}_{i}$$ 
such that all but a finite number of ${r}_{i}=0$.
As the above example shows, the commutativity of $R$ is not required, and $M$ can be assumed either as a left or right module of $R$ (in the example above, we could take $M$ to be the left $R$module).
However, unlike a vector space, a module may not have a basis. If it does, it is a called a free module^{}. Vector spaces are examples of free modules over fields or division rings. If a free module $M$ (over $R$) has a finite basis with cardinality $n$, we often write ${R}^{n}$ as an isomorphic^{} copy of $M$.
Suppose that we are given a free module $M$ over $R$, and two bases ${B}_{1}\ne {B}_{2}$ for $M$, is
$${B}_{1}={B}_{2}\mathrm{?}$$ 
We know that this is true if $R$ is a field or even a division ring. But in general, the equality fails. Nevertheless, it is a fact that if ${B}_{1}$ is finite, so is ${B}_{2}$. So the finiteness of basis in a free module $M$ over $R$ is preserved when we go from one basis to another. When $M$ has a finite basis, we say that $M$ has finite rank (without saying what rank is!).
Now, even if $M$ has finite rank, the cardinality of one basis may still be different from the cardinality of another. In other words, ${R}^{m}$ may be isomorphic to ${R}^{n}$ without $m$ and $n$ being equal.
Invariant Basis Number
A ring $R$ is said to have IBN, or invariant basis number if whenever ${R}^{m}\cong {R}^{n}$ where $$, $m=n$. The positive integer $n$ in this case is called the rank of module $M$. To rephrase, when $F$ is a free $R$module of finite rank, then $R$ has IBN iff $F$ has unique finite rank. Also, $R$ has IBN iff all finite dimensional invertible matrices over $R$ are square matrices^{}.
Examples

1.
If $R$ is commutative, then $R$ has IBN.

2.
If $R$ is a division ring, then $R$ has IBN.

3.
An example of a ring $R$ not having IBN can be found as follows: let $V$ be a countably infinite^{} dimensional vector space over a field. Let $R$ be the endomorphism ring^{} over $V$. Then $R=R\oplus R$ and thus ${R}^{m}={R}^{n}$ for any pairs of $(m,n)$.
Title  IBN 
Canonical name  IBN 
Date of creation  20130322 14:51:45 
Last modified on  20130322 14:51:45 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  12 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 16P99 
Synonym  invariant basis number 
Synonym  invariant dimension property 
Related topic  ExampleOfFreeModuleWithBasesOfDiffrentCardinality 
Defines  basis of a module 
Defines  finite rank 
Defines  rank of a module 