# Counterexample request: a surjective endomorphism of a finite module which is not injective

It's a classical and useful result that over a commutative, unital ring $A$, a surjective endomorphism of a finite module $M$ is an isomorphism. The standard proof seems to require commutativity in that one needs determinants and the adjugate matrix. So I imagine there are simple counterexamples over noncommutative rings, even over connected commutative graded algebras. What are they?

• By "a finite module" you mean, I suppose, f.g. = finitely generated...right? Feb 9 '18 at 23:59
• Any ring A for which $A\cong A\oplus A$ works. There are several such examples in this site. Search for "IBN'. Feb 10 '18 at 0:01
• Right! I'd forgotten about these rings. I'll keep searching the site in case I'm wrong, but the examples I know of are not what I'd call "simple": they tend to involve endomorphism algebras of infinite-dimensional free modules or the like. So, with respect, these aren't really what I'm looking for. I'm hoping for something along the lines of a finitely-generated CGA.
– jdc
Feb 10 '18 at 0:14
• An endomorphism algebra is about the simple sort of ring in existence. Feb 10 '18 at 0:37
• There are also famous examples known as Leavitt algebras. Feb 10 '18 at 0:38

I think your intuition is entirely wrong here: usually, graded-commutative rings behave basically the same as commutative rings (as long as you restrict to graded modules, graded homomorphisms, etc). So you should expect that the result does still hold for graded-commutative rings (again, assuming your modules and homomorphisms are graded), and in fact it does. For instance, the proof in Martin Brandenburg's answer in the post you linked to still works--the only place it uses commutativity is in the cyclic case (to say that $A/I$ is a ring and the statement is true when $M=A$), and these arguments still work for graded-commutative rings (any graded left ideal is two-sided and any homogeneous element with a one-sided inverse is a unit).
As for a natural source of examples, just consider any ring $A$ which has an element $a\in A$ which has a left inverse but not a right inverse. Then right multiplication by $a$ is a homomorphism of left $A$-modules $A\to A$ which is surjective but not an isomorphism.