Do Lipschitz/Hurwitz quaternions satisfy the Ore condition? The Lipschitz quaternions $L$ are the quaternions with integer coefficients and the Hurwitz quaternions $H$ are the quaternions with coefficients from $\Bbb Z\cup(\Bbb Z+1/2).$ A ring satisfies the right Ore condition when for $x,y\in R$ the right principal ideals generated by them have a non-empty intersection.
I can infer from what I've seen in various places that $H$ satisfies the condition, but I would like to see a proof that actually gives $a,b$ such that $xa=yb$ for each $x,y\in H.$ I don't know the answer for $L.$ 
Also, for either of them, what is the ring of fractions if it exists? Is it the quaternions with rational coefficients?
Please give as elementary answers as possible -- my knowledge of ring theory is limited.
 A: Hint: If $q$ is a non-zero element of either $H$ or $L$, then so is the conjugate quaternion $\overline{q}$. Furthermore, $q\overline{q}$ is a (rational) integer in the right ideal generated by $q$. Thus every right ideal contains a non-zero integer. Can you use what you know about intersections of non-trivial ideals of the ring of integers?

The trick was that multiplying by a conjugate took us to the center.
A: 
I would like to see a proof that actually gives $a,b$ such that $xa=yb$ for each $x,y\in H$.

Notice that $a\overline{a}$ is a scalar in the ring (integer or half integer) and so $a\overline{a}$ commutes with everything in $H$. Then if $x,y$ are nonzero elements, $x(y\overline{y})=y(\overline{y}x)$. The same argument works for $L$.

Also, for either of them, what is the ring of fractions if it exists? Is it the quaternions with rational coefficients?

Yes, you can check that the Lipschitz quaternions are an order in the rational quaternions (temporarily denoted by $S$) by noting that:


*

*$S$ is of course a domain, and for a quaternion $a\in S$, $\overline{a}$ is also in $S$, and $a\overline{a}\in \Bbb Q$. Therefore when $a\neq 0$, $a^{-1}=\frac{\overline{a}}{a\overline{a}}\in S $, so $S$ is a division ring.

*For any $a\in S$, it's easy to see that there exists an integer $z$ such that $zs=sz\in L$. (Just use the least common multiple of the denominators of the coefficients.)


So $S$ is a division ring which $L$ is dense (on both sides.) Then $H$, being sandwiched between $L$ and $S$, is also dense in $S$. So, $H$ is an order in $S$ as well.

Apologies if this takes you too far afield. An excellent introduction to this sort of thing appears in chapter 4 of Lam's book Lectures on rings and modules. Be sure to check it out :)
Among the things you can learn there is that a domain $R$ is right Ore iff there exists a division ring $D\supseteq R$ with the following "density" property:


*

*For any $d\in D$, there exists $r\in R$ such that $dr\in R$.


This is the essential property that makes $R\subseteq D$ more special than merely being a subring: $R$ is "large" inside $D$ so that you can always translate something in $D$ into $R$ with an element of $R$. You can see that this is a necessary condition if you are doing something like the fraction field for the integers. (Multiplying by the denominator would shift a fraction into the things with denominator $1$.)
