From: Max Lieblich <lieblich@cheese.fas.harvard.edu>
Newsgroups: sci.math
Subject: Re: Nonabelian rings
Date: 2 Nov 1998 20:25:55 GMT

: 1) Are there any nonabelian rings?

If you are talking about rings with units, nope (assuming that both left
and right distributive laws are in your axioms).  Indeed, by the left
distributive law (and the associative law, and the fact that 1 is a
multiplicative unit), (1+1)(a+b)=(a+b)+(a+b)=(a+(b+a))+b. On the other
hand, by the right distributive law (etc.), (1+1)(a+b)=(a+(a+b))+b.  Using
additive inverses, etc., we see that a+b=b+a for all a,b in the ring.

If the "ring" doesn't have a unit, there a billion examples: take any
nonabelian group R, define + to be the group operation, and define ab=0
for all a,b in R (where 0 is the group identity element).

Max