Consider the subring $\mathcal{O}:=\mathbb{Z}+\mathbb{Z}i+\mathbb{Z}j+\mathbb{Z}k$ of the ring $\mathbb{H}=\mathbb{R}+\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ of real Hamiltonians. Let $I$ be a right ideal of $\mathcal{O}$. I claim that $I$ must be of the form $I=m\mathbb{Z}+m\mathbb{Z}i+m\mathbb{Z}j+m\mathbb{Z}k$ for some $m\in \mathbb{Z}$. Is this true ?!. I noticed that, by the closeness of $I$ under multiplication by elements of $\mathcal{O}$ on the right, $I$ must contain non-zero integers, multiplies of $i$, multiplies of $j$, and multiplies of $k$. I defined the four nonempty subsets of $\mathbb{Z}$:
\begin{align*}
\mathfrak{A}_1 &= \bigl\lbrace a\in \mathbb{Z}: a+bi+cj+dk\in I \text{ for some } b,c,d\in \mathbb{Z} \bigr\rbrace, \\
\mathfrak{A}_2 &= \bigl\lbrace b \in \mathbb{Z}: a+bi+cj+dk\in I \text{ for some } a,c,d\in \mathbb{Z} \bigr\rbrace, \\
\mathfrak{A}_3 &= \bigl\lbrace c \in \mathbb{Z}: a+bi+cj+dk\in I \text{ for some } a,b,d \in \mathbb{Z} \bigr\rbrace, \\
\mathfrak{A}_4 &= \bigl\lbrace d \in \mathbb{Z}: a+bi+cj+dk\in I \text{ for some } a,b,c \in \mathbb{Z} \bigr\rbrace.
\end{align*}
Indeed, $\mathfrak{A}_1, \mathfrak{A}_2, \mathfrak{A}_3$, and $\mathfrak{A}_4$ are all right ideals of $\mathbb{Z}$. Can we show that $I=\mathfrak{A}_1+\mathfrak{A}_2i+\mathfrak{A}_3j+\mathfrak{A}_4k$.
If the claim is wrong, can anyone please explain why ?!. Thanks in advance.
If $I=m\mathbb{Z}+m\mathbb{Z}i+m\mathbb{Z}j+m\mathbb{Z}k$ that says $I=m\mathcal O$. That would mean the ring is a principal right ideal ring, but the ring of Lipschitz quaternions is not a principal right ideal ring...
From this thesis you can learn, however, that all right ideals are at most 2-generated.