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# 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 $B$:

$m=\sum_{{i\in I}}r_{i}b_{i}$ |

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 $B_{1}\neq B_{2}$ for $M$, is

$|B_{1}|=|B_{2}|?$ |

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 $m,n<\infty$, $m=n$. 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 $(m,n)$.

## Mathematics Subject Classification

16P99*no label found*

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