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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.

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  • $\begingroup$ Can you do the analogue for $M_2(\mathbb Q)$? $\endgroup$
    – Kimball
    Dec 3, 2022 at 12:01
  • $\begingroup$ @Kimball Yes, it is available in Voight's textbook (example 10.1.1) and Clark's notes (example 1). $\endgroup$
    – rationalbeing
    Dec 3, 2022 at 22:50
  • $\begingroup$ 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. $\endgroup$
    – Kimball
    Dec 4, 2022 at 11:04
  • $\begingroup$ 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! $\endgroup$
    – Jyrki Lahtonen
    Dec 19, 2022 at 20:57

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – rationalbeing
    Dec 20, 2022 at 2:00
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    $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Dec 20, 2022 at 3:59
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    $\begingroup$ 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). $\endgroup$
    – rationalbeing
    Dec 20, 2022 at 5:03

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