# Is a surjective endomorphism of a finitely generated module over a *non-commutative* ring with unity necessarily an isomorphism?

As is well known, any surjective endomorphism of a finitely generated module $$M$$ over a commutative ring with unity $$R$$ must be an isomorphism.

What about the non-commutative case? In other words, is a surjective endomorphism of a finitely generated module $$M$$ over a non-commutative ring with unity $$R$$ necessarily an isomorphism?

Further, is there a finitely generated projective module $$P$$ such that there exists a surjective endomorphism of $$P$$ which is not an isomorphism?

• I think this wikipedia page may help you. Under IBN, the argument here should apply (at least for f.g. projective modules). Notice that when IBN is not satisfied, the result you ask is definitely false even for f.g. free modules. Commented May 3 at 9:36
• @Antonio Lorenzin Thanks for sharing the information with me. Commented May 3 at 11:01

One says an $$R$$ module $$M$$ is Hopfian if every surjective endomorphism is an automorphism.
One also says that a ring is Dedekind finite (or directly finite) if $$ab=1$$ implies $$ba=1$$.
Putting these together, you can see that a module with your property is such that $$End(M_R)$$ is a Dedekind finite ring (although it is known the converse does not hold.)
A simple counterexample would be to use a ring $$R$$ which is not Dedekind finite, so that $$R\cong End(R_R)$$, and therefore $$R_R$$ is a free cyclic module that is not Hopfian (i.e. it doesn't have the condition you seek.)
The most famous example is probably the endomorphism ring of an infinite dimensional vector space, where the "shift operators" provide an example of $$ab=1$$ but $$ba\neq 1$$.