It would be nice to have several examples of an interesting $R$-algebra $A$, where $R$ is a non-commutative ring (plausible definitions can be found here).

One example is a polynomial ring over any non-commutative ring, see this question.

Also, I wonder if some properties of commutative ring extensions are still valid for two non-commutative rings, for example, the definition of a separable ring extension $R \subseteq A$, where $R$ is a non-commutative ring (hence $A$ is also non-commutative). The usual definition where $R$ is a commutative ring, can be found in this post (in his notations, is it possible to just replace $k$ by a non-commutative ring? Is there a problem with tensoring over a non-commutative ring?).

Edit: After receiving two comments requesting me to be more precise/pick a definition, my new question is:

Let $R \subseteq A$, $R$ is a non-commutative ring ($R$ and $A$ are associative with $1$). It would be nice to have examples of such rings with $A$ being a flat $R$-module. My example: $R$ is a division ring.

Thank you very much!

  • $\begingroup$ "interesting" sounds vague. Can you be more precise? What do you find interesting in an $R$-algebra? $\endgroup$
    – Crostul
    Sep 26, 2017 at 10:14
  • $\begingroup$ ok, I will try to be more precise. By an interesting algebra $A$, I mean an algebra that was studied and there exist some theorems concerning it. For example, the first Weyl algebras $A_1(k)$ ($k$ is a field; what if we replace $k$ by a non-commutative ring?). $\endgroup$
    – user237522
    Sep 26, 2017 at 10:17
  • 1
    $\begingroup$ I think at the very least you're going to have to pick the definition you want to use. Asking responders to take their pick from a basket of random definitions, then provide "interesting" picks, whatever that means, is just too broad. Basically posters shouldn't have to make so many decisions about what constitutes an answer. $\endgroup$
    – rschwieb
    Sep 26, 2017 at 12:02
  • $\begingroup$ Thanks for the explanation. I will try to restrict my question. $\endgroup$
    – user237522
    Sep 26, 2017 at 14:50
  • $\begingroup$ @user237522 Thanks: I think it's a big improvement. $\endgroup$
    – rschwieb
    Sep 26, 2017 at 18:57

1 Answer 1


This is a quick and dirty way to get a lot of examples:

A ring is von Neumann regular iff all of its modules are flat. So, you can pick any ring $A$ containing a von Neumann regular ring $R$, and you have an example.

A few other obvious constructions that would work for any ring $R$ include regarding $\prod_{i=1}^n R$ and $M_n(R)$ and $R[x]$ as $R$ algebras, since all are free hence flat modules over $R$.

  • $\begingroup$ I like your idea! (However, if $A$ is a domain, then $R$ is a domain, and von Neumann regularity of $R$ would imply that $R$ is a division ring, if I am not wrong). $\endgroup$
    – user237522
    Sep 26, 2017 at 20:01
  • $\begingroup$ @user237522 You are correct, a VNR domain is a division ring. I'm not very good with flatness, so i don't know what can be worked in the situation is, say for flat extensions of noncommutative domains. $\endgroup$
    – rschwieb
    Sep 26, 2017 at 20:40
  • $\begingroup$ In my question there was no requirement that $A$ is a domain, so your answer is really ok (and nice). Nevertheless, I am still curious to see examples when $R$ is a domain which is not a division ring. $\endgroup$
    – user237522
    Sep 26, 2017 at 20:52
  • $\begingroup$ @user237522 Mentioning also several obvious constructions which are based on free modules, and they'll work for any ring. $\endgroup$
    – rschwieb
    Sep 26, 2017 at 21:37

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