A ringR has invariant basis number if for all positive integersm and n, Rmisomorphic to Rn implies that. Equivalently, this means there do not exist distinct positive integers m and n such that Rm is isomorphic to Rn. Rephrasing the definition of invariant basis number in terms of matrices, it says that, whenever A is an m-by-n matrix over R and B is an n-by-m matrix over R such that and, then. This form reveals that the definition is left–right symmetric, so it makes no difference whether we define IBN in terms of left or right modules; the two definitions are equivalent. Note that the isomorphisms in the definitions are not ring isomorphisms, they are module isomorphisms.
Discussion
The main purpose of the invariant basis number condition is that free modules over an IBN ring satisfy an analogue of the dimension theorem for vector spaces: any two bases for a free module over an IBN ring have the same cardinality. Assuming the ultrafilter lemma, this result is actually equivalent to the definition given here, and can be taken as an alternative definition. The rank of a free module Rn over an IBN ring R is defined to be the cardinality of the exponentm of any R-module Rm isomorphic to Rn. Thus the IBN property asserts that every isomorphism class of free R-modules has a unique rank. The rank is not defined for rings not satisfying IBN. For vector spaces, the rank is also called the dimension. Thus the result above is in short: the rank is uniquely defined for all free R-modules iff it is uniquely defined for finitely generated free R-modules.
Examples
Any field satisfies IBN, and this amounts to the fact that finite-dimensional vector spaces have a well defined dimension. Moreover, any commutative 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. Let the canonical basis of An, which means is all zeros except a one in the i-th position. By Krull's theorem, let I a maximalproper ideal of A and. An A-module morphism means because I is an ideal. So f induces an A/I-module morphism, 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 ring that does not satisfy IBN is the ring of column finite matrices, the matrices with coefficients in a ring R, with entries indexed by 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 is given by: This infinite matrix ring turns out to be isomorphic to the endomorphisms of a right free module over R of countable rank, which is found on page 190 of. From this isomorphism, it is possible to show that for any positive integern, 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 as seen in.
