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Question. Is there any standard name for a (commutative or non-commutative) unital ring $R$ with the property that, for every $a \in R$, the (descending) chain $R, aR, a^2 R, \ldots,$ is eventually constant?

Let me refer to this condition as the DCCPRP, that is, the "DCC on chains of principal right ideals generated by the powers of an element".

The DCCPRP is satisfied by left perfect rings, which, by a famous theorem of Bass (Theorem 28.4 in Anderson & Fuller's book), can be equivalently characterized as the rings satisfying the DCC on principal right ideals (DCCPR). However, the DCCPRP is much weaker than the DCCPR.

E.g., let $R$ be the commutative, boolean ring with identity obtained by endowing the power set $P(X)$ of a set $X$ with the operations of symmetric difference (as addition) and intersection (as multiplication). The DCCPRP is trivially verified in this case (as it would be in any boolean ring), while the DCCPR holds if and only if $X$ is finite: For, suppose $X$ is infinite. Accordingly, let $x_1, x_2, \ldots$ be an infinite sequence of pairwise distinct elements from $X$, and set $X_0 := X$ and $X_k := X \setminus \{x_1,\ldots, x_k\}$ for each $k \in \mathbf N^+$. Then $X_0 P(X), X_1 P(X), \ldots$ is a strictly decreasing sequence of principal right ideals of $R$, because $X_k P(X) = P(X_k)$ for every $k \in \mathbf N$ and, by construction, $X_{k+1} \subsetneq X_k$.

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Such rings are called strongly $\pi$-regular in the literature. The condition is left-right symmetric, as first proven by Dischinger.

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  • $\begingroup$ Ah, nice! Do you mind if I ask for a more precise reference to Dischinger's result? Maybe you could include a link in your answer, I guess I'm not the only one who would appreciate it. $\endgroup$ Jun 9, 2021 at 21:50
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    $\begingroup$ I would recommend T. Y. Lam's "Exercises in Classical Ring Theory". Dischinger's result appears as Exercise 23.5. There are also a few nice paragraphs (after Lam's proof of Dischinger's result) explaining the history of this class of rings, and some nice examples. In particular, it is mentioned that these rings are exactly the endomorphism rings of modules satisfying "Fitting's lemma". $\endgroup$ Jun 9, 2021 at 21:56
  • $\begingroup$ I couldn't have hoped for a better reference, thanks! $\endgroup$ Jun 9, 2021 at 22:01

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