# Calculating the class number of a maximal order from a Quaternion Algebra

I am currently reading Quaternion Algebras from John Voight and saw how he calculated the class number of a maximal order at example 17.6.3. His quaternion algebra is $$B' = \left( \frac{-1,-23}{\mathbb{Q}}\right)$$ and he has maximal order $$O = \mathbb{Z} + \mathbb{Z} i + \mathbb{Z} \frac{1+j}{2} + \mathbb{Z} i \frac{1+j}{2}$$

He defines $$\alpha := i, \beta := \frac{1+j}{2}$$ and then uses the Embedding $$O \rightarrow M_2(\mathbb{Z}_2)$$ $$\alpha, \beta \mapsto \begin{pmatrix}0 & -1 \\\ 1 & 0 \end{pmatrix}, \begin{pmatrix}1 & 0 \\\ 0 & b_0\end{pmatrix}$$ where $$b_0$$ satisfies $$b_0^2 - b_0 + 6 = 0$$ and $$b_0 \equiv 0 \pmod{2}$$. After that he uses

Let $$e \in \mathbb{Z}_{\geq 0}$$. Then every principal right $$M_2(\mathbb{Z}_p)$$-ideal $$I$$ with $$\operatorname{nrd}(I) = p^e$$ is of the form $$I = \alpha M_2(\mathbb{Z}_p)$$ where $$\alpha \in \left\{ \begin{pmatrix}p^u & 0 \\\ c & p^v\end{pmatrix} : u, v \in \mathbb{Z}_{\geq 0}, u + v = e, \text{ and } c \in \mathbb{Z}/p^v\mathbb{Z} \right\}.$$

to calculate Isomorphism.

I wanted to try it too so I chose $$B = \left( \frac{-1,-19}{\mathbb{Q}}\right)$$ which has the same maximal order as $$B'$$. I tried using the same $$\alpha$$ and $$\beta$$ as he did but for my example nrd($$\beta) = 5$$. This means that $$\beta$$ has minimal polynomial $$m_\beta (x) = x^2-x+5$$ which has no solution (mod 2). I don't really know what to do now without an embedding and would appreciate if someone could help me. If this is just a hard example I would also appreciate other suggestions of quaternion algebras with low class number that are easier to calculate.

• I don't really know what to do now without an embedding - You can either do calculations in terms of your basis directly or embed B in a matrix algebra. For the latter, e.g., you use the right regular representation (but nicer to do over quadratic field). Commented Jan 7 at 15:52

Here is some Magma code to get you started:

> B := QuaternionAlgebra(Rationals(),-1,-19);
> O := MaximalOrder(B);
> B<i,j,ij> := QuaternionAlgebra(Rationals(),-1,-19);
> O := MaximalOrder(B);
> Basis(O);
[ 1, i, 1/2*i + 1/2*ij, 1/2 + 1/2*j ]
> _, iotap := pMatrixRing(O,2);
> iotap(i);
[O(2^16) -1 + O(2^16)]
[1 + O(2^16) O(2^16)]
> iotap((j+1)/2);
[5*2 + O(2^16) -31919 + O(2^16)]
[-31919 + O(2^16) -9 + O(2^16)]


Hopefully that's intelligible, but the point is that you can send $$(j+1)/2$$ to the matrix $$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$ modulo $$2$$.

In case you're curious about how the rest of it goes:

> Is := RightIdealClasses(O);
> #Is;
2
> Is[1];
Ideal with basis Pseudo-matrix over Integer Ring
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
> Is[2];
Right Ideal with basis Pseudo-matrix over Integer Ring
[2 0 0 0]
[0 2 0 0]
[1 0 1 0]
[1 1 0 1]


I hope that helps!

P.S. There's a typo in the example, fixed in the online version: the map for discriminant $$19$$ should be $$\beta \mapsto \begin{pmatrix} 1-b_0 & 0 \\ 0 & b_0 \end{pmatrix}$$, which gives the same thing modulo $$2$$.