# Are the quaternions a domain?

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.

• 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. – Matt Samuel Jan 12 '16 at 3:05
• Subrings of domains are domains, hence $R$ is a domain since $\mathbb{H}$ is a domain (being a division ring). – Hayden Jan 12 '16 at 3:05
• by a domain I mean that if $ab = 0$ then $a=0$ or $b=0$. – Jack Jan 12 '16 at 3:07
• @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 $a=0$. – Hayden Jan 12 '16 at 3:12
• @Tim an integral domain requires commutativity but in $R$, $ij\ne ji$. – 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.