I have to give an example of a non-commutative domain that is not a division ring. My first thought was $R = \big\{ a + bi + cj + dk \mid a,b,c,d \in \mathbb{Z} \big\}$ since $R$ is clearly non-commutative and not a division ring (take the number 2 for example). I have seen conflicting accounts for if it is a domain though. One source said it is since it is a subring of the quaternions. However my book does not specify that quaternions are a domain, and I read a paper online that said that in fact they weren't a domain. Then I read an answer on stack exchange from someone with a 62k rep that said the quaternions are a domain. I tried proving it myself but I got stuck in a very long cycle.

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    $\begingroup$ It depends what you mean by "domain." Some require commutativity in the term. If all it means is that there are no zero divisors, then of course the quaternions form a domain, because they form a division ring. $\endgroup$ – Matt Samuel Jan 12 '16 at 3:05
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    $\begingroup$ Subrings of domains are domains, hence $R$ is a domain since $\mathbb{H}$ is a domain (being a division ring). $\endgroup$ – Hayden Jan 12 '16 at 3:05
  • $\begingroup$ by a domain I mean that if $ab = 0$ then $a=0$ or $b=0$. $\endgroup$ – Jack Jan 12 '16 at 3:07
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    $\begingroup$ @Tim In a division ring, For each $a\neq 0$ there exists $a^{-1}$ such that $aa^{-1}=1$. If $ab=0$, then either $b\neq 0$, or there is $b^{-1}$ with $0=0b^{-1}=abb^{-1}=a1=a$, showing that $a=0$. $\endgroup$ – Hayden Jan 12 '16 at 3:12
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    $\begingroup$ @Tim an integral domain requires commutativity but in $R$, $ij\ne ji$. $\endgroup$ – Vim Jan 12 '16 at 3:48

I think your confusion is coming from the fact that different authors use the word "domain" with different meanings. In the context of your problem, "domain" clearly just means a (nonzero) ring in which $xy=0$ implies $x=0$ or $y=0$. With that definition, any division ring is a domain (if $x\neq 0$ you can multiply $xy=0$ on the left by $x^{-1}$ to get that $y=0$), and since a subring of a domain is clearly a domain, it follows that $R$ is a domain.

On the other hand, some authors (especially those that think mainly about commutative rings) take "domain" to mean that in addition to the condition above, the ring is commutative. With this definition, of course the quaternions are not a domain, because they are not commutative.

  • $\begingroup$ Ahhh yes I think that is why I was confused. Thank you, I see it clearly now! $\endgroup$ – Jack Jan 12 '16 at 3:12

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