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.

  • $\begingroup$ 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 $V/a\Bbb{Q}$, ...) $\endgroup$
    – reuns
    Mar 14, 2023 at 22:26
  • $\begingroup$ 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. $\endgroup$ Mar 14, 2023 at 23:24
  • $\begingroup$ $O\cap \Bbb{Q}$ 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. $\endgroup$
    – reuns
    Mar 14, 2023 at 23:28
  • $\begingroup$ 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. $\endgroup$ Mar 15, 2023 at 22:21
  • $\begingroup$ 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/… $\endgroup$
    – reuns
    Mar 15, 2023 at 22:39

1 Answer 1


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 ${\cal O}$.

  • $\begingroup$ 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 $R[\alpha]$. $\endgroup$ May 13, 2023 at 17:11
  • $\begingroup$ 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 $\endgroup$ May 13, 2023 at 20:27

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