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Let $A$ be a unital, possibly noncommutative ring. Dischinger showed [1] that the following are equivalent:

  1. For every $a \in A$, there exists $n \in \mathbb N$ such that $a^n A = a^{n+1} A$;

  2. For every $a \in A$, there exists $n \in \mathbb N$ such that $A a^n = A a^{n+1}$;

  3. Every cyclic right module $M \in Mod_A$ is co-hopfian (i.e. every injective endomorphism of $M$ is an isomorphism);

  4. Every cyclic left module $M \in {}_A Mod$ is co-hopfian.

Such a ring $A$ is called strongly $\pi$-regular. Every strongly $\pi$-regular ring $A$ is $\pi$-regular in the sense that for every $a \in A$ there is some $n \geq 1$ such that $a^n$ is a von Neumann regular element of $A$ (i.e. $a^n = a^n b a^n$ for some $b \in A$). Every von Neumann regular ring is strongly $\pi$-regular, and if $A$ is commutative, then $A$ is strongly $\pi$-regular iff $A/nil(A)$ is von Neumann regular.

Dischinger also showed that the following are equivalent:

  1. Every finitely-generated right module $M \in Mod_A$ is co-hopfian;

  2. Every finitely-generated left module $M \in {}_A Mod$ is co-hopfian;

  3. Every finite-rank matrix ring over $A$ is strongly $\pi$-regular;

  4. Every ring right (or left) Morita equivalent to $A$ is strongly $\pi$-regular.

Let's call such a ring $A$ very strongly $\pi$-regular. Every von Neumann regular ring is very strongly $\pi$-regular.

Questions:

  1. Is there a direct ring-theoretic, rather than module-theoretic, characterization of very strongly $\pi$-regular rings which doesn't explicitly mention matrix rings?

  2. What is an example of a very strongly $\pi$-regular ring $A$ such that $A$ is not von Neumann regular but the center $Z(A)$ is von Neumann regular?

[1] Dischinger, Friedrich. “Sur Les Anneaux Fortement π-Réguliers.” CR Acad. Sci. Paris Sér. AB 283, no. 8 (1976): 571–573. link.

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    $\begingroup$ (2) Let $M$ be the monoid $\{1,x,y\}$ with $xy=xx=x$, $yx=yy=y$ and unit $1$. Let $K$ a field, and $A=KM$. Then the center of $A$ is reduced to $K$ (so is von Neumann regular). $A$ itself is not von Neumann regular because $x-y\notin (x-y)A(x-y)=0$. But since $A$ is a finite-dimensional $K$-algebra, clearly every finitely generated left $A$-module is cohopfian. $\endgroup$
    – YCor
    Commented May 8, 2020 at 16:29
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    $\begingroup$ The algebra suggested by @YCor is isomorphic to 2x2 upper triangular matrices over K which obviously has the desired properties since it has a non-trivial Jacobson radical the strictly upper triangular matrices $\endgroup$
    – Benjamin Steinberg
    Commented May 8, 2020 at 18:48
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    $\begingroup$ In general the path algebra of a finite connected acyclic quiver with at least one edge over K has center K and is not von Neumann regular but is strongly $\pi$-regular. $\endgroup$
    – Benjamin Steinberg
    Commented May 8, 2020 at 18:51
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    $\begingroup$ Just would like to +1 Benjamin's comment: von Neumann regular does not imply strongly $\pi$-regular. A strongly (von Neumann) regular ring is strongly $\pi$-regular, and a von Neumann regular ring is $\pi$-regular. Here are a few other examples of VNR rings that aren't strongly $\pi$-regular. $\endgroup$
    – rschwieb
    Commented May 8, 2020 at 19:33
  • $\begingroup$ Since "strong von Neumann regularity" isn't Morita invariant, it would not be possible to use it as a replacement as an example of "very strong $\pi$-regularity" $\endgroup$
    – rschwieb
    Commented May 8, 2020 at 19:39

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