|
|
|
|
|
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 $B=\lbrace b_i\mid i\in I\rbrace$ 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 $B$ : $$m=\sum_{i\in I}r_ib_i$$ 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 $B_1\neq B_2$ 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.
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
- If $R$ is commutative, then $R$ has IBN.
- If $R$ is a division ring, then $R$ has IBN.
- 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 $(m,n)$ .
|
"IBN" is owned by CWoo.
|
|
(view preamble | get metadata)
Cross-references: endomorphism ring, countably infinite, square matrices, matrices, invertible, finite dimensional, iff, integer, positive, rank, equality, even, bases, isomorphic, cardinality, division rings, examples of free module, free module, module, right module, number, finite, linear combination, index set, subset, basis, unital, commutative, ring, field, vector space
There are 9 references to this entry.
This is version 9 of IBN, born on 2004-11-29, modified 2008-06-05.
Object id is 6537, canonical name is IBN.
Accessed 5873 times total.
Classification:
| AMS MSC: | 16P99 (Associative rings and algebras :: Chain conditions, growth conditions, and other forms of finiteness :: Miscellaneous) |
|
|
|
|
|
|
Pending Errata and Addenda
|
|
|
|
|
|
|
|
|
|
|
