Let $R$ be a finitely generated subring of the ring of real quaternions $\mathbb{H}$, that is, $R$ is the subring generated by a finite subset of $\mathbb{H}$. I want to show (or find a counterexample) that $R$ is Noetherian.
My attempt is the same as the attempt in this question. However, this attempt is not correct for general rings as pointed out in the answers. I wonder if there is some special property of $\mathbb{H}$ that would make this attempt work.
Any help is much appreciated.
Let $a,b,c,d\in\mathbb{R}$ be algebraically independent over $\mathbb{Q}$, let $x=a+bi$ and let $y=c+dj$. Then the subring $R\subset\mathbb{H}$ generated by $x$ and $y$ is not Noetherian.
To prove this, consider the left ideal $I\subseteq R$ generated by the elements $xy^n$ for each $n\in\mathbb{N}$. I claim that for each $n$, $xy^n$ is not in the left ideal generated by the $xy^m$ for $m<n$, and so $I$ cannot be finitely generated.
Note first that $R$ is a subring of the quaternion algebra $S$ over the polynomial ring $\mathbb{Z}[a,b,c,d]$. We may consider $\mathbb{Z}[a,b,c,d]$ as a bigraded ring where the first degree is respect to $a$ and $b$ and the second degree is with respect to $c$ and $d$. This then makes $S$ a bigraded ring, with $x$ being homogeneous of bidegree $(1,0)$ and $y$ being homogeneous of bidegree $(0,1)$. Since $x$ and $y$ are homogeneous, the ring $R$ is bigraded as well.
Now suppose we could write $xy^n$ as an $R$-linear combination of the $xy^m$ for $m<n$. The only way to take a multiple of $xy^m$ that has the same bidegree as $xy^n$ is to take an integer multiple of $y^{n-m}xy^m$, so this in fact means that we can write $$xy^n=\sum_{m<n}k_my^{n-m}xy^m$$ where each $k_m$ is an integer. Now consider the homomorphism $S\to\mathbb{H}$ given by mapping $a$ to $0$, $b$ to $1$, $c$ to $1$, and $d$ to $2$ (and $i,j,k$ to themselves). This turns the equation above into $$i(1+2j)^n=\sum_{m<n}k_m(1+2j)^{n-m}i(1+2j)^m.$$ Multiplying both sides by $-i$ and noting that $-i(u+vj)i=u-vj$, we get $$(1+2j)^n=\sum_{m<n}k_m(1-2j)^{n-m}(1+2j)^m.$$ But this now violates unique factorization in the Gaussian integer ring $\mathbb{Z}[j]$, since the right-hand side is divisible by $1-2j$ and the left-hand side is not.