# Are there non commutative rings with no zero divisors?

If there are, Are there unity (but not division) rings of this kind? Are there non-unity rings of this kind?

Sorry, I forgot writting the non division condition.

• why not? Take the "polynomials" with integer coefficients in two non-commuting variables $x$ and $y$. Nov 5, 2013 at 3:54
• @anon:Ah, yes, I was only thinking of multiplication. My mistake. Let me delete that; it was a stupid comment. Nov 5, 2013 at 4:01

Yes, in fact much more can be said: There are rings with $1$ such that every non-zero element has a multiplicative inverse. These rings are called division rings or skew-fields.

The real quaternions are an example of a non-commutative division ring.

Take the "polynomials" with integer coefficients in two non-commuting variables $x$ and $y$. If you don't want a unit, use even integers only.

A related example replaces integer coefficients by coefficients in $\mathbb{Z}_2$.

• ... or the ideal of all nonconstant polynomials.
– user14972
Nov 5, 2013 at 4:00
• Do you know anything about the first (corrected) question? Nov 5, 2013 at 4:29
• My example happens not to be a division ring, so it happens to answer the revised question. The quaternion answers can be modified to be a non-division ring, by using the integer quaternions $a+bi+cj+dk$ where $a,b,c,d$ are integers. Nov 5, 2013 at 4:32

Yes. In fact not only can we force every nonzero element to not be a zero divisor, we can force every nonzero element to be invertible. You get a division ring (also called skew field).

Perhaps one of the most famous historical examples: the quaternions. They are defined by

$$\Bbb H=\{a+bi+bj+ck:i^2=j^2=k^2=ijk=-1,a,b,c,d\in\Bbb R\}.$$

As for nonunital rings: simply take a unital ring with no zero divisors at hand (say $\Bbb H$) and look at a non-unital subring. For example the elements $2\Bbb Z+2i\Bbb Z+2j\Bbb Z+2k\Bbb Z\subseteq\Bbb H$.

Another kind of example: non-commutative deformations of commutative integral domains, e.g. $\mathbb C[x,y]$ with the commutation relation $[x,y] = 1$.