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Let $R$ be a (commutative or non-commutative, associative) ring with unity, and let $a$ be an element of $R$ such that $r(a) = r(a^2)$, where $r(\cdot)$ denotes a right annihilator. It follows that $r(a) \cap aR = \{0_R\}$ and $r(a) = r(a^k)$ for every $k \in \mathbf N^+$, and I've the following question:

Q1. If $r(a)$ is a direct summand of $R_R$ and $R$ is , does there exist an idempotent $e \in R$ such that $r(a) = eR$ and $aR \subseteq (1_R-e)R$? In case it helps, feel free to assume that (i) $R$ is directly irreducible (i.e., $R$ is a non-zero ring and not the direct product of two non-zero rings) and (ii) $r(a)$ is indecomposable as a right $R$-module.

This is closely related to a question in the same spirit that I asked some weeks ago with $r(1-a)$ in place of $aR$ (here): That question was answered affirmatively by Pace Nielsen (here), and it turned out that the conclusion is even independent from the condition that $r(a) = r(a^2)$.

As per the present question, the answer is yes when $R$ is right artinian or, more generally, right strongly $\pi$-regular (meaning that $R$ satisfies the ACC on descending chains of the form $bR, b^2 R, \ldots$ with $b \in R$); and it is still yes when $R$ is von Neumann regular, or a domain, or a local ring (and this is true independently from the condition that $r(a) = r(a^2)$), with the last two cases being sort of trivial. Yet, I expect the answer to Q1 to be negative in general. And this leads to:

Q2. Are there any other "interesting classes" of rings (apart from those mentioned in the previous paragraph) for which the answer to Q1 is yes?

Edit (8 July 2021): The answer to Q1 is still yes (and in a strong form) provided $R$ is an abelian ring, meaning that every idempotent of $R$ is central (i.e., lies in the center of $R$). In fact, let $e \in R$ be any idempotent such that $r(a) = eR$ (recall that $r(a)$ is assumed to be a direct summand of $R_R$). If $R$ is abelian, then $a = ea + (1-e)a = ae + (1-e)a = (1-e)a \in (1-e)R$, because $ea = ae = 0_R$.

By Exercise 22.4A in Lam's Exercises in Classical Ring Theory, left (resp., right) duo rings (that is, rings where every left (resp., right) ideal is actually a two-sided ideal) are abelian rings, so this is ultimately a generalization of an observation made by Luc Guyot in the comments under the OP.

This is all fine, except for the fact that, in the applications I've in mind, $R$ is directly irreducible, and a ring is directly irreducible if and only if it has no non-trivial central idempotents. Therefore, I edited Question Q1 so as to mention that, in case it helps, one should feel free to assume that $R$ is directly irreducible.

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    $\begingroup$ Trivial remark: if the right ideals of $R$ are two-sided ($R$ is termed a "duo ring" in this case), then you can always find an idempotent $e$ with the desired property, and for this you don't need the hypothesis $r(a) = r(a^2)$. Indeed, any idempotent $e$ such that $r(a) = eR$ will do, and such an element always exists (see e.g., math.stackexchange.com/questions/578801/…). $\endgroup$
    – Luc Guyot
    Commented Jul 8, 2021 at 14:14
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    $\begingroup$ @LucGuyot Do you mean right duo rings? The property is not left-right symmetric (see, e.g., Exercise 22.4A in Lam's Exercises in Classical Ring Theory). In any case, +1 for the comment: I didn't know this kind of rings had a name. $\endgroup$
    – Salvo Tringali
    Commented Jul 8, 2021 at 15:16
  • $\begingroup$ @LucGuyot How do you show that $R$ being a right duo ring yields $aR \subseteq (1-e)R$ for every idempotent $e$ s.t. $r(a) = eR$? If $r(a) \cap aR = \{0_R\}$ (which, in the OP, is guaranteed by the hp that $r(a) = r(a^2)$), then $eaR \subseteq RaR \subseteq aR$ (by the hp that $R$ is right duo) and hence $ea \in r(a) \cap aR$, implying $ea = 0_R$ and hence $a=ea+(1-e)a\in (1-e)R$. But you mentioned in your comment that assuming $r(a) = r(a^2)$ is unnecessary. $\endgroup$
    – Salvo Tringali
    Commented Jul 8, 2021 at 15:48
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    $\begingroup$ $R = eR \oplus (1 - e)R$ hence $aR = a(1 - e)R \subseteq (1 - e)R$ when $(1 - e)R$ is two-sided. (Did I miss anything?) $\endgroup$
    – Luc Guyot
    Commented Jul 8, 2021 at 15:50
  • $\begingroup$ Nope, it's me who had missed something! On the positive side, this exchange made me realize that right duo rings are only a special case of a wider class for which the answer to Q1 is still true in the "strong form" (I edited the OP to include the details). $\endgroup$
    – Salvo Tringali
    Commented Jul 8, 2021 at 16:29

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