In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element?

Question

In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element?

Answer 1

Yes, it does.

If $ba=1$ but $ab\neq 1$, then you can show $Rb\supseteq Rb^2\supseteq Rb^3\supseteq\ldots$ is an infinite descending sequence of left summands, and the complements of these summands are an infinite ascending chain.

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Added later

Looking back, I'm a little appalled at not being clear about the complements.

Firstly, here is a hint about why they are summands: $b^na^nb^n=b^n$. Secondly, let me point out that $aR\supseteq a^2R\supseteq a^3R\supseteq \ldots$ is a descending chain of right summands for similar reasons, because $a^nb^na^n=a^n$.

Thirdly, I had a specific complement in mind for each summand $Re$, and that was $R(1-e)$. Perhaps at the time it seemed clear to me that $Rf\subseteq Re$ implied $R(1-f)\supseteq R(1-e)$, but now it seems not so clear to me.

Now, it does seem that given a descending chain of summands, it is indeed possible to find an ascending chain of complements for them, but it takes a little explaining.

Instead, I would prefer to suggest an easier route:

1. As hinted above, the chain of $a^nR$ is a strictly descending chain of right summands.

2. For an idempotent $e$, the left annihilator of $eR$ is $R(1-e)$.

3. The chain $\ell.ann(a^nR)$ is a strictly ascending chain of left ideals.