I have a proof that shows that a stably finite ring (i.e. one where $AB=I\iff BA=I$ for any two square matrices $A,B$ with entries in the ring) has the Invariant Basis Number (IBN). Trouble is, it is used to prove that a commutative ring has IBN. So naturally the question is:
How do I prove that a commutative ring is stably finite?
And could you give me an example of a stably finite non-commutative ring?
For commutative rings, the ordinary determinant argument works.
From $\det(AB)=1$ you get that the determinant of $A$ is a unit, so you can form the adjugate and divide by the determinant with no problem to form the inverse of $A$. Thus any matrix ring over a commutative ring is Dedekind finite, hence the commutative ring is stably finite.
I guess you definitely want a ring that isn't commutative. How about the quaternions $\Bbb H$? Or any matrix ring over a field? Or any Noetherian ring of your choice?
Here's another example if those are too difficult to verify: try to show that finite rings are stably finite. Actually you can just show that finite rings are Dedekind finite, and then point out that matrix rings over finite rings are also finite, so they must be Dedekind finite by the same token. Thus a finite ring is stably finite.