# Generalized Nakayama's lemma over a non-commutative ring

Let $$R$$ be a ring, $$J(R)$$ its Jacobson radical, $$M$$ is a finite $$R$$-module. The following statement is usually called Nakayama's lemma: if $$IM=M$$ then $$M=0$$. This is true over any ring (commutative or not).

Over a commutative ring the generalized version of Nakayama's lemma holds: if $$IM=M$$ for some ideal $$I$$ and finite $$M$$ then there is $$x \in I$$ such that $$(1-x)M=0$$.

The only proof I know is based on determinant trick, which doesn't generalize to non-commutative situation. I doubt that this is true over non-commutative ring. What are the simplest examples of a non-commutative ring and finite module over it that generalized Nakayama's lemma is false?

I don't know about the simplest example, but here's a low-dimensional one. Take $$R$$ to be the path algebra of the quiver with vertices 1,2 and arrows $$a\colon 1\to 2$$ and $$b\colon 2\to 2$$, modulo the relation $$b^2$$. Thus $$R$$ has basis $$e_1,e_2,a,b,ba$$. Take $$M$$ to be the module $$Re_1/Rba$$, so having basis $$e_1,a$$. Take $$I$$ to be the two-sided ideal generated by $$e_1$$, so having basis $$e_1,a,ba$$. Now $$IM=M$$, since $$e_1^2=e_1$$ and $$ae_1=a$$, but $$(1-x)M\neq0$$ for all $$x\in I$$, since $$(1-x)a=a$$.
EDIT: Actually one can remove the $$b$$ to get a smaller example: Take $$A$$ to be the path algebra of the quiver $$1\xrightarrow{a}2$$, so having basis $$e_1,e_2,a$$ such that $$e_1,e_2$$ are orthogonal idempotents, and $$a=e_2ae_1$$. Take $$I=M=Ae_1$$, a projective left ideal. Then $$I$$ has basis $$e_1,a$$ and $$I=IA$$ is actually a two-sided ideal. Now $$I^2=I$$, but $$(1-x)a=a$$ for all $$x\in I$$, so $$(1-x)I$$ is nonzero.