Left- and right-sided principal ideals of quaternions have same index? One fact about the Lipschitz integers (quaternions of the form $a + bi + cj + dk$ where $a, b, c, d$ are integers) is that the left-sided ideal generated by any element $Q$ has the same index in the additive group as does the right-sided ideal generated by $Q$.
I know this is a fact, but I don't know a simple abstract reason for it. But surely there must be some embarrassingly simple reason I'm overlooking! Thus, I ask at exactly what level of abstraction this sort of thing cleanly holds:


*

*Is this the case in any ring with an involution which preserves ring structure except for reversing the order of multiplication (like the conjugation operation on quaternions)?

*Failing that, is this the case in any ring of the form $R[i, j, k \; | \; i^2 = j^2 = k^2 = ijk = -1]$ where $R$ itself is a commutative ring?

*Failing all that... what is the right level of abstraction for this result? What is the short and sweet reason it is true?

 A: Here's something reasonably quick: if $R$ is a domain and $\mathcal{O}$ is an $R$-order in an $F$-algebra $B$ with $F=\mathrm{Frac}(R)$, then the $R$-index $[\mathcal{O}:\mathcal{O}\alpha]_R$ is (either by definition, or after some argument) equal to the $R$-ideal generated by $\det_{B/F}(\cdot \alpha)$, the determinant of right multiplication by $\alpha$ on $B$.  (This is also written $\det_{B/F}(\cdot \alpha)=\mathrm{Nm}_{B/F}(\alpha)$ as a norm, but to be precise one should say `right norm', and when it isn't usually it is because the property you are asking about is known so it doesn't matter.)  So the thing you want to know in your algebra is that the determinant of left multiplication is the same as the determinant of right multiplication.  Equality between these doesn't hold in general, but a large class of algebras where it does hold are the separable (i.e., geometrically semisimple) algebras over $F$, including your quaternion algebra: in this case, you can compute the determinant over the algebraic closure $F=\overline{F}$, in which case $B$ is isomorphic as an $F$-algebra to the product of matrix rings, and where on each matrix factor we have $\det_{\mathrm{M}_n(F)/F}(\cdot \alpha)=\det_{\mathrm{M}_n(F)/F}(\alpha\cdot)=\det(\alpha)^n$, where the latter $\det$ is taken in the usual sense.  
