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A ring $R$ is called strongly $\pi$-regular if for every element $r \in R$ there exists an element $x \in R$ such that $r^{n+1}x = r^n$ for some positive integer $n$. Meanwhile, a ring $R$ is said to have artinian prime factors if $R/P$ is artinian for every prime ideal $P$ of $R$. It is known that every ring with artinian prime factors is strongly $\pi$-regular. Does every strongly $\pi$-regular ring have artinian prime factors?

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No.

Note that a full matrix ring $M_n(k)$ over a field is simple, so its only prime ideal is $\{0\}$, and it is artinian. Hence it has artinian prime factors; hence is strongly $\pi$-regular. The latter property is preserved under passing to direct limits, hence the union $R$ of a chain of such matrix rings, say $M_2(k)\subset M_4(k)\subset\dots\subset M_{2^n}(k)\subset\dots$ (where $M_{2^n}(k)$ is embedded in $M_{2^{n+1}}(k)$ by sending each matrix to the diagonal sum of two copies of itself) is strongly $\pi$-regular. This ring $R$ is again simple, but no longer artinian, so it does not have artinian prime factors.

(A variant example is the $k$-algebra $R$ of $\mathbb{N}\times\mathbb{N}$ matrices over $k$ spanned by the identity matrix and the matrices $e_{ij}$. This has two prime ideals, $\{0\}$ and the span of the $e_{ij}$. It can be shown to be strongly $\pi$-regular by writing every element $r$ as the sum of an $n\times n$ matrix for some $n$, and a scalar multiple of "the identity $(\mathbb{N}-n)\times(\mathbb{N}-n)$ matrix", and using the fact that $M_n(k)$ is artinian. But again, for the prime ideal $P=\{0\}$, the ring $R/P=R$ is non-artinian.)