A division quaternion algebra in which the integral elements don't form a ring

I wish to find a division quaternion algebra $$B$$ over $$\mathbb{Q}$$ and elements $$\alpha, \beta\in B$$ such that $$\alpha, \beta$$ are integral over $$\mathbb{Z}$$ but both $$\alpha + \beta$$ and $$\alpha \beta$$ are not.

I know the following facts:

• Let $$B=\{\alpha=t+xi+yj+zij \mid t,x,y,z\in \mathbb{Q}, i^2=a, j^2=b, ji=-ij\}$$ where $$a,b\in \mathbb{Q}^\times$$. Then $$\alpha$$ is integral if and only if $$\mathrm{trd}(\alpha) = \alpha+\overline{\alpha},\mathrm{nrd}(\alpha)=\alpha\overline{\alpha}\in \mathbb{Z}$$, where $$\overline{\alpha} = t-(xi+yj+zij)$$.
• Since $$B$$ is a division quaternion algebra, it is NOT isomorphic to $$M_2(\mathbb{Q})$$. That is, we can't have $$a=1$$ or $$b=1$$.
• If $$a=b=-1$$ then we get a restriction of Hamiltonians which have ring of integers given by Hurwitz order.
• Since $$B$$ is noncommutative, the set of elements in $$B$$ integral over $$\mathbb{Z}$$ is no longer necessarily itself a ring because although $$\mathbb{Z}[\alpha]$$ and $$\mathbb{Z}[\beta]$$ may be finitely generated as $$\mathbb{Z}$$-modules for $$\alpha,\beta\in B$$, this need not be the case for the $$\mathbb{Z}$$-algebra generated by $$\alpha$$ and $$\beta$$.

Thank you for reading. Any help you could provide would be much appreciated.

Source: Exercise 10.10 of Voight's textbook.

• Can you do the analogue for $M_2(\mathbb Q)$? Dec 3, 2022 at 12:01
• @Kimball Yes, it is available in Voight's textbook (example 10.1.1) and Clark's notes (example 1). Dec 3, 2022 at 22:50
• Well, you can use a similar idea, but it might help if you say what you tried. Or you can just find some random integral elements of your favorite $B$ and try them. Dec 4, 2022 at 11:04
• I have always thought the Hurwitz order refers to the ring described here. But the construction described in your WP link sounds really cool. Thanks for sharing! Dec 19, 2022 at 20:57

Posting the following to take this out of the unanswered pile.

Let $$B=\Bbb{H}_{\Bbb{Q}}$$ be the usual division algebra of Hamilton's quaternions but limited to rational coefficients of $$1,i,j,k$$.

• Show that $$u=\dfrac35i+\dfrac45j$$ is a zero of $$x^2+1$$ and hence integral.
• Show that $$iu$$ is not integral even though both $$i$$ and $$u$$ are.

The maximal orders are not unique in non-commutative division algebras over $$\Bbb{Q}$$, in sharp contrast to the case of number fields.

Still, every integral element belongs to some maximal order. In the above case we see that the $$\Bbb{Z}$$-span of $$1,u,k,uk$$ forms an order isomorphic to the ring $$L$$ of Lipschitz integer quaternions. That is not a maximal order, but there is the obvious analogue of the Hurwitz quaternions (loc. cit.), $$\Bbb{Z}$$-spanned by $$u,k,uk$$ and $$(1+u+k+uk)/2$$ that is a maximal order.

• A follow-up question: For $B = M_2(\mathbb{Q})$ and $B=\left(\dfrac{-1,-3}{\mathbb{Q}}\right)$, every maximal order is conjugate (and thus isomorphic) in $B$ to $O=M_2(\mathbb{Z})$ and $O = \mathbb{Z}\left<i,(i+j)/2\right>$, respectively (Voight corollary 10.5.5 and exercise 11.11). Is this what you mean by "every integral element belongs to some maximal order"? However, in case of Hamilton's quaternions we don't get a partition via conjugacy classes? Dec 20, 2022 at 2:00
• IIRC maximal orders always share the same discriminant. I vaguely recall that there is a way of linking any maximal orders together by some process that involved taking left/right orders related to a given lattice. First by finding common suborders of a finite index and such. I learned a few such things from Reiner's book Maximal orders, but I was no longer in my prime when learning, so the details have been forgotten. IIRC the scene is simpler over a completion of a local ring (like the $p$-adic integers. Dec 20, 2022 at 3:59
• Thank you very much for sharing your thoughts. Yes, you are correct. Given a complete DVR $R$ and its field of fractions $K$, if $D$ is a central (simple) division algebra over $K$ with $\mathrm{dim}_{F} D= n^2$, then $D$ contains a unique maximal $R$-order (Reiner, Theorem 12.8). Dec 20, 2022 at 5:03