# Interesting algebras over non-commutative rings

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!

• "interesting" sounds vague. Can you be more precise? What do you find interesting in an $R$-algebra? Sep 26 '17 at 10:14
• 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?). Sep 26 '17 at 10:17
• 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. Sep 26 '17 at 12:02
• Thanks for the explanation. I will try to restrict my question. Sep 26 '17 at 14:50
• @user237522 Thanks: I think it's a big improvement. Sep 26 '17 at 18:57

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 $R$.
• 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). Sep 26 '17 at 20:01
• 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. Sep 26 '17 at 20:52