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?


1 Answer 1


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.


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