Example of a ring such that $R^2\simeq R^3$, but $R\not\simeq R^2$ (as $R$-modules)

The usual example of unitary ring without the IBN property is the ring of column finite matrices, and in this case we have $R\simeq R^2$ as (left) $R$-modules. (See also here.) In particular, we have $R^2\simeq R^3$.

I wonder if there is an example of unitary ring (without the IBN property) such that $R^2\simeq R^3$, but $R\not\simeq R^2$ (as $R$-modules).

Solution 1:

The only other examples of non IBN rings that I am aware of are built using Leavitt path algebras, and they do exactly what you want. It is possible to specify positive integers $n<m$ such that $R^i\ncong R^j$ for $(i,j)$ less than $(n,m)$ in the lexicographic order, but $R^n\cong R^m$ as modules. (The $(m,n)$ pair is called the 'module type' in the Leavitt article.)

Unfortunately, it is a bit unwieldy to reproduce the entire construction here. Here is one paper discussing the construction.

I think the other Abrams reference in the wiki article on IBN must be useful as well (I was the one who put the citation up, but I was a long time ago.)

Here is another article that addresses it much more directly. It cites Leavitt's original result which is what I'm describing (theorem 1.3.3)

For everyone's enjoyment, here's the citation for Leavitt's article: