What is the Krull dimension of the ring $A:=\mathbb{Q}[\pi,\sqrt{11},x,y]/I$ where $I=\langle x^2-y^3+xy+3\rangle$?
My idea is to use the Krull's Principal Ideal Theorem. Since $a=x^2-y^3+xy+3 \in \mathbb{Q}[\pi,\sqrt{11},x,y]$ is neither a zero-divisor nor a unit, then every minimal prime ideal $\mathfrak{p}$ of $\langle a\rangle=I$ has $\mathrm{ht}(\mathfrak{p})=1$. But I don't get evolve the idea.
What stands out to me here is what does a transcendence basis of $A/\mathbf{Q}$ look like? Clearly $\pi$ should be in there and if we add $x$ then we are done because now the ring $R = \mathbf{Q}[\pi, x]$ is a 2-dimensional polynomial ring and both $\sqrt{11}$ and $y$ are integral over $R$. Hence $A/R$ is finite which means that $A$ has dimension $2$.