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.
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 ${\cal O}$.