Let \( A \) and \( B \) be two square matrices. It is well-known that the equation \( AB = I \) is equivalent to \( BA = I \). This equivalence holds even for matrices whose entries lie in a commutative ring. However, I am curious if there is a counterexample to this claim in a non-commutative ring, whether straightforward or complex.
Thank you!
There is a nice proof here that there is no counterexample if a and b are not zero divisors: https://math.stackexchange.com/questions/352992/does-ab-1-imply-ba-1-in-a-ring
However, if you do allow zero divisors, then there are counterexamples. For example, an element may have a left inverse, but no right inverse at all:
https://math.stackexchange.com/questions/70777/a-ring-element-with-a-left-inverse-but-no-right-inverse
Mohammad,
A ring R is said to be n-finite if, in the matrix ring M_n(R), AB=I implies BA=I.
In his response, James Tuite provided you with a simple example of a ring that is not 1-finite, and therefore also not n-finite for all n>1. However, it is important to realize that there are also rings that are 1-finite but fail to be n-finite for some n>1. An example of a domain enjoying this property for n=2 is described in the paper "Inverses and zero divisors in matrix rings" by J. C. Shepherdson (Proc. London Math. Soc. 1 (1951), 71-85).
In his paper "Some remarks on the invariant basis property" (Topology 5 (1966), 215-228), Paul Cohn generalizes Shepherdson's construction to produce, for each integer n>1, an example of a domain that is not n-finite but is k-finite for k
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