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I'm having a hard time finding information about the quotient rings of the Lipschitz quaternions and the Hurwitz quaternions.

The Lipschitz quaternions are defined as the quaternions with integral coordinates. The Hurwitz quaternions are those with all coordinates in $\mathbb Z$ or in $\mathbb Z + \frac{1}{2}$. It is quite well-known that the former lacks unique factorization in irreducible elements, while the latter admits a Euclidean division with respect to the norm $N(a+ib+jc+kd) = a^2+b^2+c^2+d^2$.

I'm interested in the quotient rings of these rings. Is it, in principle, possible to determine the quotient of the (left)-ideal generated by (say) $a+ib+jc+kd$? My guess would be that this ideal has index $a^2+b^2+c^2+d^2$ in the corresponding ring, based on the analogous result in the Gaussian integers. And what about two-sided ideals?

My motivation is to realize some given $\mathbb Z/n\mathbb Z$ as such a quotient, but I'm afraid I can only get noncommutative quotients.

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    $\begingroup$ As you can see from the example of the ideal (2) -- which is even 2-sided -- the index can't be given by a^2 + ... + d^2. Rather it is the square of that norm. One way is to compute the volume of the parallelepiped (which in this case is a hypercube) formed by the multiples of your generator by 1,i,j,k. $\endgroup$
    – Noam D. Elkies
    Commented Apr 3 at 1:31
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    $\begingroup$ In a commutative quotient, you have $i=jk=kj=-i$, so $2i=0$ and hence $2=0$ because $i$ is a unit. So if you want $\mathbb{Z}/n\mathbb{Z}$ as a quotient then $n=2$. $\endgroup$
    – Dave Benson
    Commented Apr 3 at 7:07
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    $\begingroup$ @DaveBenson or $n=1$! $\endgroup$
    – Kimball
    Commented Apr 3 at 12:38
  • $\begingroup$ Indeed, thank you all for making this clearer. $\endgroup$
    – A. Bailleul
    Commented Apr 4 at 14:49

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It's overkill, but this generalizes: Main Theorem 16.1.3 in my book gives $\#(O/I) = [O : I]=\mathrm{nrd}(I)^2$ (cardinality and indices as abelian groups) as the square of the reduced norm of the (left or right) ideal $I$ when $I$ is locally principal--something which is automatic when $O$ is maximal by Proposition 16.1.2. Even more generally, see Main Theorem 16.7.7.

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