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My experience with non commutative rings is limited to 2 by 2 matrices and the quaternions. The first of which is not a domain, and the latter is a division ring. I'm looking for an example of a domain that is not a division ring.

Invertible matrices do not produce an example, as they must be division rings. Is there an example within some ring of matrices? Similarly, is there an example within the quaternions?

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    $\begingroup$ What about $\mathbb{H}[X]$? $\endgroup$ Dec 12 '13 at 16:35
  • $\begingroup$ Actually I too had the same doubt, thanks ! $\endgroup$ Aug 17 '19 at 15:41
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Wow, three answers and a comment with the same example, and not even the easiest example, IMO! (It's still a good example, though!)

Similarly, is there an example within the quaternions?

Absolutely! There are lots of nonzero subrings of $\Bbb H$ that would already work. Consider for example $\{a+bi+cj+dk\mid a,b,c,d\in \Bbb Z\}\subseteq\Bbb H$.

It's easy to check that this is a subring of $\Bbb H$, which automatically makes it a domain. Clearly it isn't commutative since $ij\neq ji$. Finally, it doesn't contain $2^{-1}$, so it can't be a division ring.

Is there an example within some ring of matrices?

Absolutely! Actually this will be cheating since it will be the same as my first solution, but you might be interested in what happens anyway.

It turns out that the quaternions can be represented by complex matrices in this way:

$a+bi+cj+dk\mapsto\begin{bmatrix}a+bi&c+di\\-c+di&a-bi\end{bmatrix}\in M_2(\Bbb C)$. So by just taking the image of the ring I described in the first part, you have a subring of $M_2(\Bbb C)$ which is a noncommutative domain, not a division ring.

You could take it even further to real matrices:

$a+bi+cj+dk\mapsto\begin{bmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\\\end{bmatrix}\in M_4(\Bbb R)$ to get another (the same) example. In fact, you can look upon the final example as using a matrix representation of elements of $\Bbb C$: $a+bi\mapsto\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ and plugging that representation into the second example. All three are interlinked.

These are both matrix representations of the original example I suggested.

For a final exotic example that you might be interested in, check out the Hurwitz quaternions. They are an interesting subdomain of $\Bbb H$.

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    $\begingroup$ I agree this is the simplest answer. $\endgroup$
    – clem
    Dec 12 '13 at 17:45
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nik's example of $\mathbb H[X]$, where $\mathbb H$ is the quaternions is a good one. The set of invertible matrices is not a ring at all (it is not closed under addition). Another example is the Weyl algebra $$ W_1 = k\langle x,y\rangle / (xy - y x=1) $$ Of course, in general if $D$ is an arbitrary division ring, then $D[X]$ is a domain but not a division ring.

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  • $\begingroup$ I believe it is more standard to write something like $W_1$ since the Weyl algebra is the ring of differential operators on a polynomial ring of 1 variable. $\endgroup$
    – J. Gaddis
    Dec 12 '13 at 16:45
  • $\begingroup$ You're quite right. $\endgroup$ Dec 12 '13 at 18:49
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The easiest example is of course the quaternion polynomial ring, $\mathbb{H}[x]$. However, there are more examples but they are not easy to create. For example, any noncommutative domains which satisfy the right Ore condition give you a way of building such rings. Actually, it turns out to be very similar to the construction of fraction fields for commutative domains.

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I would say the simplest example, in the sense of most elementary from some perspective, is the unital free ring in two noncommuting variables, $\mathbb{Z}\langle X,Y\rangle$. You can think about it as the ring of "polynomials" in two variables which do not commute, with integer coefficients. It is clearly noncommutative, it is not a division ring ($p(X,Y)X=1$ would add a nontrivial relation to the ring) and is a domain (again, $p(X,Y)q(X,Y)=0$ would add a nontrivial relation to the ring).

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