# A note on orders in quaternion algebras

Definition. An algebra $$B$$ over a field $$F$$ is a quaternion algebra if there exists $$i,j\in B$$ such that $$1,i,j,ij$$ is a basis for $$B$$ as a vector space over $$F$$, where $$i^2=a,j^2=b;a,b\in F^\times$$.

Throughout, we fix $$F=\mathbb{Q}$$.

Definition. A lattice in a finite-dimensional $$\mathbb{Q}$$-algebra $$V$$ is a finitely generated $$\mathbb{Z}$$-submodule $$\mathcal{L} \subset V$$ that contains a $$\mathbb{Q}$$-basis of $$V$$ (as a vector space over $$\mathbb{Q}$$).

I proved that

Let $$V_\mathbb{Q}$$ be a finite-dimensional vector space. $$\mathcal{L} \subset V$$ is a lattice if and only if $$\mathcal{L}=x_1\mathbb{Z} \oplus \ldots \oplus x_n \mathbb{Z}$$ where $$x_1,\ldots,x_n$$ is a basis for $$V_\mathbb{Q}$$.

Definition. An order $$\mathcal{O} \subset B$$ is a lattice that is also a subring having $$1\in B$$.

I have two questions:

1. Are any two orders isomorphic as rings?!.
2. What are the orders in the quaternion algebra $$\mathbb{M}_2(\mathbb{Q})$$ (the matrix ring over $$\mathbb{Q}$$)?!. I found that the following subrings are all orders: \begin{align} &\begin{pmatrix} \mathbb{Z} &n\mathbb{Z} \\ m\mathbb{Z} & \mathbb{Z} \end{pmatrix}, \quad m,n\in \mathbb{Z}^* \\ &\begin{pmatrix} \mathbb{Z} & \frac{a}{d}\mathbb{Z} \\ n\mathbb{Z} & \mathbb{Z} \end{pmatrix}, \quad a,d,n\in \mathbb{Z}^* \text{ and } d \mid n, \\ &\begin{pmatrix} \mathbb{Z} & \frac{a}{b}\mathbb{Z} \\ \frac{c}{d}\mathbb{Z} & \mathbb{Z} \end{pmatrix}, \quad a,b,c,d\in \mathbb{Z}^* \text{ and } bd\mid ac. \end{align} Are there more orders?!. Observe that these three types of orders are all isomorphic as rings.

Note: The second question is my main question.

I appreciate any help. Thanks in advance.

• You might send an email to John Voight at Dartmouth, who is an expert in these matters and very approachable. May 17 at 21:43
• @PaceNielsen - That was basically the opposite of my first thought, which was this question would be better suited for Math Stack Exchange. May 17 at 23:24
• Reading Voight's book is probably the best way to find the answers to these May 18 at 4:58
• This is the link of the question on Math Stack Exchange. It's unsolvable till now. math.stackexchange.com/questions/4699250/… @Kimball May 18 at 16:02
• There is no mention to such a question or such an idea in the book of John Voight. I have the book and I use it frequently in my work. @WatsonLadd May 20 at 19:28

Two orders need not be isomorphic.

First of all, in number fields $$K$$ other than $$\mathbf Q$$ not all orders are isomorphic rings (even if they are isomorphic abelian groups): the full ring of integers $$\mathcal O_K$$ is integrally closed, but no other order in $$K$$ is integrally closed.

In the quaternion algebra $$\mathbf H(\mathbf Q)$$ we have the two orders $$L = \mathbf Z + \mathbf Zi + \mathbf Zj + \mathbf Zk$$ (called the Lipschtz quaternions) and the larger order $$H = \mathbf Z + \mathbf Zi + \mathbf Zj + \mathbf Z\frac{1+i+j+k}{2}$$ (called the Hurwitz quaternions). In $$H$$ there is a left (and right) division algorithm, so all left (and right) ideals in $$H$$ are principal. But in $$L$$ there are nonprincipal left ideals and nonprincipal right ideals (see here). So $$H$$ and $$L$$ are not isomorphic rings.

• This is great. Can we determine all orders in $\mathbb{M}_2(\mathbb{Q})$?. In fact, this is my main problem. May 18 at 16:40
• In ${\rm M}_2(\mathbf Q)$ each maximal order is conjugate to ${\rm M}_2(\mathbf Z)$ (Corollary 10.5.5 in Voight's book), so you can focus on orders in ${\rm M}_2(\mathbf Z)$. I don't know if they are all classified. Orders in quadratic fields $K$ are known: for each $n \geq 1$ there is a unique order with index $n$ in $\mathcal O_K$ and it is $\mathbf Z+n\mathcal O_K$. In $\mathbf Q^2$ (the "split case"), the integral closure of $\mathbf Z$ is $\mathbf Z^2$, with a unique order of each index $n$: $\mathbf Z(1,1) + n\mathbf Z^2$. May 18 at 17:37
• In your comment, you mentioned that there exist non-isomorphic (as rings) orders in $\mathbb{H}(\mathbb{Q})$. But what about the quaternion algebra $\mathbb{M}_2(\mathbb{Q})$. May it possible to show that all orders are isomorphic (as rings)? Or can we find two non-isomorphic orders (as rings) ?!. May 18 at 20:03
• "not every order $\subset \mathbb{M}_2(\mathbb{Z})$ is contained in a maximal order $\subset \mathbb{M}_2(\mathbb{Z})$" - what do you mean with that? $\mathbb{M}_2(\mathbb{Z})$ itself is a maximal order. May 18 at 20:10
• @HusseinEid It is true in this case. See here. May 18 at 21:05

The classification of orders is the topic of chapters 22-24 in my book. Yes, there are plenty more orders than the ones you listed.

One cheeky answer is that isomorphism classes of orders in $$\mathrm{M}_2(\mathrm{Z})$$ are in bijection (inverse to the Clifford map) with similarity classes of nondegenerate, isotropic, ternary quadratic forms over $$\mathbb{Z}$$. Examples are $$xy-Nz^2$$ for every nonzero $$N \in \mathbb{Z}$$, but I bet you can easily dream up many more. So the classification you seek is as hard as classifying ternary quadratic forms. (But at least there is no multiplication law to check!)

Not sure what your goal is, but I'll try my best to give a quick outline. The first step in the classification is to use the local-global dictionary for lattices, which reduces the question to classifying local orders up to isomorphism; the failure of the relevant local-global principle (reconstructing a global order from its local orders) is handled by strong approximation.

Local orders are first classified by their Eichler invariant. For the orders you wrote, at a prime $$p \mid mn$$, the local order at $$p$$ is residually split = Eichler; these admit a nice classification, as do the residually inert orders. The residually ramified orders are more complicated; but at least we can classify the Bass orders quite explicitly, and in general an order either has a nontrivial Gorenstein saturation, or it is Gorenstein and then it has a unique minimal superorder obtained from the radical idealizer chain, giving a recursive way to classify them.