Relation between finite stable rank and IBN (invariant basis number)

Question

For R is any ring has an identity, we known if R has stable rank one, then R is "weakly finite" (or "stably finite," all matrix rings over R are Dedekind finite) and this implies R has IBN .

But when stable rank of R is two (or greater than, but finite) in general R is not Dedekind finite. Then R has IBN? I tried to prove it but I had many difficulties.

So do every rings has finite stable rank is IBN ? Thanks a lot.

Answer 1

The question you ask has a positive answer: If a ring has finite Bass stable rank, then it has IBN. The reasons are as follows, and the references below are in this paper of Rieffel. We define $$ Lg_m(R) = \{(a_1,a_2,\ldots, a_m)\in R^m : Ra_1+Ra_2+\ldots + Ra_m = R\} $$ Rieffel shows that :

1. [Discussion preceding Theorem 4.5] If $m\geq Bsr(R)+1$, then $GL_m(R)$ acts transitively on $Lg_m(R)$
2. [Proposition 10.5] If $GL_m(R)$ acts transitively on $Lg_m(R)$ and $W$ is a projective left $R$-module such that $W\oplus R \cong R^m$, then $W\cong R^{m-1}$.

Now the result follows from the familiar Eilenberg swindle: Suppose $Bsr(A)<\infty$ and $R$ does not have IBN, then choose the smallest $n\geq 1$ such that there is $m>n$ with $R^m \cong R^n$. From this it follows that $$ R^n \cong R^{m+k(m-n)} \quad\forall k\geq 0 $$ So choose $k$ such that $m+k(m-n) \geq Bsr(A)+1$, then by (1) and (2), one has $$ R^{n-1} \cong R^{m+k(m-n)-1} $$ This contradicts the minimality of $n$, thus completing the proof.