Compute $\dim_{\mathbb{Q}}(\mathbb{Q}[s^{\pm 1},t^{\pm 1}]/(t^2-t+1,s^2t-st+s-1))$ In the paper the homological algebra of Artin groups by Craig C. Squire it is stated that the following ring:
$$R=\mathbb{Z}[s^{\pm 1},t^{\pm 1}]/(t^2-t+1,(st+1)(s-1))$$
seen as an abelian group has rank $4$. It is obvious that, as an abelian group with the sum, it is generated by $1,t,s,st$. Indeed, computing the ranks of quotients of $\mathbb{Z}[s^{\pm 1},t^{\pm 1}]$ is fairly simple. Now, I was trying to pass from $\mathbb{Z}$ to $\mathbb{Q}$ and to compute the dimension over $\mathbb{Q}$ of the following rings
$$R_1=\mathbb{Q}[s^{\pm 1},t^{\pm 1}]/(t^2-t+1,(st+1)(s-1))$$
$$R_2=\mathbb{Q}[s^{\pm 1},t^{\pm 1}]/(s+1,s^{n-1}t+1)$$
Here I'm not sure how to proceed, I don't know if $\dim_\mathbb{Q}R_i$ coincides with the Krull dimension and if there is some easy way to compute these dimensions.
Any help will be highly appreciated.
 A: The intuition is clear: $t^2-t+1=0$ means $t$ is a primitive third root $\omega$ of unity, which is invertible hence the presence of $t^{-1}$ is not a problem. And $st+1=0\Leftrightarrow t=-s^{-1}$ or $s-1=0$ are exclusive and in either case $s^{\pm 1}$ are already present in $\mathbb Q[\omega]$, hence by the Chinese remainder theorem, $R_1\simeq \mathbb Q[\omega]\times \mathbb Q[\omega]$. But to justify this is not completely trivial. Sometimes I wonder to what extent the above is not good enough.
First show that $I=(t^2-t+1, st+1)$ and $J=(t^2-t+1, s-1)$ are coprime, and conclude $$IJ=I\cap J=(t^2-t+1, (st+1)(s-1))$$
$IJ=I\cap J$ is part of the CRT, and it's clear that both of $t^2-t+1, (st+1)(s-1)\in I\cap J$. On the other hand, $IJ$ is generated by $(t^2-t+1)^2, (t^2-t+1)(s-1), (t^2-t+1)(st+1), (st+1)(s-1)$, which are clearly all in $(t^2-t+1, (st+1)(s-1))$.
Therefore by CRT, $R_1\simeq \mathbb Q[s^{\pm 1}, t^{\pm 1}]/I \times \mathbb Q[s^{\pm 1}, t^{\pm 1}]/J$.
We define the homomorphism $\rho$ from $\mathbb Q[s^{\pm 1}, t^{\pm 1}]$ to $\mathbb  Q[\omega]$ by sending $s$ to $\omega$ and $t$ to $-\omega^{-1}$. Note that this is possible, because $\mathbb Q[s^{\pm1}, t^{\pm 1}]$ is precisely the ring that satisfies the following universal property: For any $\mathbb Q$-algebra $R$ and invertible elements $a,b\in R$, there is a unique $\mathbb Q$-algebra homomorphism from $\mathbb Q[s^{\pm 1}, t^{\pm 1}]$ to $R$ that takes $s$ to $a$ and $t$ to $b$.
Now it's easy to check that $\rho(t^2-t+1)=\rho(st+1)=0$. Therefore $I\subset\ker \rho$, and we have the induced homomorphism $\tilde{\rho}:\mathbb Q[s^{\pm 1}, t^{\pm1}]/I\rightarrow\mathbb Q[\omega]$. On the other hand, consider the subring $\mathbb Q[t]$ which is mapped to $\mathbb Q[t]/(I\cap\mathbb Q[t])$ in $\mathbb Q[s^{\pm 1}, t^{\pm1}]/I$. Note that $(t^2-t+1)\subset I\cap\mathbb Q[t]$ and $(t^2-t+1)$ is maximal, hence $I\cap\mathbb Q[t]=(t^2-t+1)$. And it's easy to check that $\mathbb Q[\omega]\simeq \mathbb Q[t]/(t^2-t+1)\rightarrow \mathbb Q[s^{\pm 1}, t^{\pm1}]/I$ by sending $\omega$ to $t+I$ is the inverse of $\tilde{\rho}$.
Therefore $\mathbb Q[s^{\pm 1}, t^{\pm 1}]/I\simeq\mathbb Q[\omega]$ and similarly we have $\mathbb Q[s^{\pm 1}, t^{\pm 1}]/J\simeq\mathbb Q[\omega]$.
$R_2$ can be figured out in a similar but easier way, as CRT is not needed.
