# Can an $R$-module have bases of different cardinality? [duplicate]

This question already has an answer here:

I am not so very familiar with the mathematics of modules over rings. I heard of them as a generalization of vector spaces with a ring instead of a field. I also heard of them as providing many pathological examples when thinking in the familiar term of vector spaces (there was an MO thread discussing examples, but I cannot find it). E.g. I learned that an $R$-module (over some ring $R$) does not necessarily have a basis, even though it has independent and generating sets. My immediate next question then was:

Question: Can an $R$-module have bases of different cardinality?

I am especially interested in different finite cardinalities. Can you give me an example of such a module? Or is it that the dimension of an $R$-module (if at all) can be still defined uniquely?

## marked as duplicate by rschwieb, Dando18, B. Goddard, user354271, EnnarSep 19 '17 at 19:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• There are several questions linked to the duplicate that explain examples where $R\cong R^2$ as $R$ modules. For such a ring, $R^n\cong R^m$ for every positive integers $n,m$. Here is a second link. You should just search for "invariant basis number". – rschwieb Sep 19 '17 at 16:25
• @rschwieb Thank you very much ;) – M. Winter Sep 19 '17 at 16:25

## 1 Answer

Yes, it is possible with a non-commutative ring $R$ that the left $R$-modules $R^m$ and $R^n$ to be non-isomorphic for some finite $m\ne n$.

An example of such a ring is the endomorphism ring of an infinite-dimensional vector space over a field.

A ring such that $R^m\cong R^n$ entails $m=n$ is said to have Invariant Basis Number.