One-sided version of the Nakayama lemma?

Question

The Nakayama lemma is often used to show that finitely generated idempotent ideals are generated by an idempotent. What remains true if we go to non-commutative rings? In other words, given a unital non-commutative ring $R$ and a left ideal $L$ of $R$ which is finitely generated as a left ideal. Suppose that $L\cdot L=L$. Must $L$ be generated (as a left ideal) by an idempotent?

Answer 1

No, I don't believe that this version of Nakayama's lemma is valid for noncommutative rings in general.

There is a notion of a right weakly regular ring (or right fully idempotent ring) where all right ideals are idempotent. Now, all von Neumann regular ring are right and left weakly regular, but there must be examples of right weakly regular rings that aren't von Neumann regular. (I just don't have one in mind at the moment.)

Now, one of the characterizations of von Neumann regular rings is that all of their finitely generated right ideals are generated by an idempotent. So:

if you have an example of a right weakly regular ring that isn't von Neumann regular, it must have a finitely generated right ideal that isn't generated by an idempotent.

This invalidates that particular commutative version of Nakayama's lemma in the general case.

However, it's simple to prove that among commutative rings, weakly regular rings are exactly the von Neumann regular rings: I think you are probably perfectly capable of proving it already, so it's an exercise!

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Update: I finally found an example available online: Right fully idempotent rings need not be left fully idempotent by Andruszkiewicz and Puczylowski.

Be careful: the first example given is a simple ring without identity, and they note that every simple ring with identity is necessarily left and right fully idempotent. They go on to describe the modifications that make an example with identity. To paraphrase:

Let $A$ be a simple, non-Artinian von Neumann regular ring that's an algebra over a finite prime field $F$. Let $L$ be a maximal left ideal of $A$. Then the ring $R=\begin{bmatrix}L&A\\L&A\end{bmatrix}$ can be embedded as an ideal into a ring $R^\ast$ with identity such that $R^\ast/R\cong F$, and $R^\ast$ is right but not left fully idempotent.

Answer 2

This is not true in general. In a recent joint work with N. Laustsen (Remark 1.10), we construct a Banach space $X$ and a maximal two-sided ideal $\mathscr{M}$ of the algebra of all bounded operators on $X$ which is generated as a left ideal by two elements but not by one. This ideal also has a bounded left approximate identity, so by Cohen's factorisation theorem we have $\mathscr{M}^2 = \mathscr{M}$.

I am aware that this counter-example is very complicated but it since it comes from analysis one may consider it quite natural in certain respect.