# 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 $$F$$.

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

Definition. A lattice in a finite-dimensional $$\mathbb{Q}$$-algebra $$V$$ is a finitely generated $$\mathbb{Z}$$-submodule $$\mathcal{L} \subset V$$ such that $$\mathcal{L}\mathbb{Q}=V$$.

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 $$1\in B$$.

Let's take an example. Consider the quaternion algebra $$\mathbb{H}(\mathbb{Q}):=\mathbb{Q}+\mathbb{Q}i+\mathbb{Q}j+\mathbb{Q}k$$ subject to $$i^2=j^2=-1,k=ij,ij=-ji$$. By the theorem mentioned above, an example of an order in $$\mathbb{H}(\mathbb{Q})$$ is $$\mathcal{O}=\mathbb{Z}+\mathbb{Z}i+\mathbb{Z}j+\mathbb{Z}k$$.

I ask if we can always take $$1$$ as a generator of $$\mathcal{O}$$. In other words, If $$\mathcal{O} \subset \mathbb{H}(\mathbb{Q})$$ is any order, can we write $$\mathcal{O}=\mathbb{Z}+\mathbb{Z}u+\mathbb{Z}v+\mathbb{Z}w$$ for some $$u,v,w\in \mathcal{O}$$ ?!

I really appreciate any help. Thanks in advance.

• Yes, $O\cap \Bbb{Q}$ cannot be bigger than $\Bbb{Z}$ to be finitely generated, so $O\cap \Bbb{Q}=\Bbb{Z}$. This is enough to claim that there is a basis starting with $1$ ( $\ell=O/\Bbb{Z}$ is a lattice in $V=\Bbb{H}/\Bbb{Q}$ a $3$-dimensional vector space, so you can repeat taking some element such that $\Bbb{Q}a \cap \ell = a \Bbb{Z}$, there is a basis of $\ell$ starting with $a$, $\ell/a\Bbb{Z}$ is a lattice in $V/a\Bbb{Q}$, ...) Mar 14, 2023 at 22:26
• I do not see why $\mathcal{O} \cap \mathbb{Q}=\mathbb{Z}$. Indeed, $\supset$ holds. How does the inclusion $\subset$ hold?. I will be grateful if you can clarify what you typed above. Thanks. Mar 14, 2023 at 23:24
• $O\cap \Bbb{Q}$ is a subring of $O$. If it contains any non-integer element then it is not finitely generated as group, contradicting that we have a surjective linear map $\Bbb{H}\to \Bbb{Q}$ which is the identity on $\Bbb{Q}$. This restricts to a map $O\to \Bbb{Q}$ which gives that $O\cap \Bbb{Q}$ is a finitely generated group. Mar 14, 2023 at 23:28
• Excuse me. It is not obvious to me why $O\cap \mathbb{Q}=\mathbb{Z}$ implies that there exists a basis of $O$ starting with 1. Mar 15, 2023 at 22:21
• I tried to explain it above, this is due to how we show that finitely generated subgroups $G$ of a $\Bbb{Q}$ vector spaces $V$ are free, by constructing a basis $G=\bigoplus b_j \Bbb{Z}$, starting with any element $b_1$ such that $b_1\Bbb{Q}\cap G= b_1\Bbb{Z}$. This is often called en.wikipedia.org/wiki/… Mar 15, 2023 at 22:39

Let me provide a proof in a bit more generality: Let $$R$$ be a PID, $$\Bbb F={\rm Frac}(R)$$, $$B$$ a finite-dimensional $$\Bbb F$$-algebra with $$\dim_{\Bbb F}V=n$$ and let $${\cal O}\subseteq B$$ be an $$R$$-order.
Claim: $${\cal O}\cap\Bbb F=R$$.
Proof. Certainly, since $$1\in{\cal O}$$, we have that $$R\subseteq {\cal O}\cap\Bbb F$$. Let then $$\alpha\in{\cal O}\cap\Bbb F$$ and suppose that $$\alpha\not\in\Bbb F$$. Observe then that we have a chain of injections of $$R$$-modules $$R[\alpha]\hookrightarrow {\cal O}\cap{\Bbb F}\hookrightarrow{\cal O}$$ Then, since $$R$$ is Noetherian (as a PID), $$R[\alpha]$$ is finitely generated as an $$R$$-submodule of a finitely generated $$R$$-module. Hence, $$\alpha\in{\Bbb F}$$ is integral over $$R$$, but since $$R$$ is integrally closed (again as $$R$$ is a PID), it follows that $$\alpha\in R$$. Hence, the claim.
Now, as above, since $$1\in{\cal O}$$ we have that $$R\subseteq{\cal O}$$ and so we can consider the short exact sequence of $$R$$-modules $$0\to R\to{\cal O}\to {\cal O}/R\to 0$$ We prove that $${\cal O}/R$$ is torsion-free. Indeed, if otherwise, there exist $$r\in R\setminus\{0\}$$ and $$[\alpha]\in{\cal O}/R$$ not zero such that $$r\alpha\in R$$. But then, $$\alpha\in r^{-1}R\subseteq{\Bbb F}$$ and hence by the Claim it follows that $$\alpha\in R$$, which gives that $$[\alpha]=0$$,a contradiction. Therefore, by the structure theorem for modules over a PID, we get that $${\cal O}/R$$ is free. Hence, the sequence above splits and thus we can extend the basis of $$R$$, i.e. the unit element $$1\in R$$ into a basis of $${\cal O}$$.
• Why is $R[\alpha]$ is a Noetherian as an $R$-module if $R$ is Noetherian?!. It's known that (for me) if $R$ is Noetherian then so is the ring $R[\alpha]$. May 13, 2023 at 17:11
• I don't claim that $R[\alpha]$ is Noetherian (even though it is). I claim that it is finitely generated. It is known that if $R$ is a Noetherian ring then every finitely generated $R$-module is a Noetherian module (not a Noetherian ring). Thus, ${\cal O}$ is a Noetherian $R$-module which implies that every $R$-submodule of it, namely $R[\alpha]$, is finitely generated. See here: en.wikipedia.org/wiki/Noetherian_module May 13, 2023 at 20:27