TITLE: Domains with isomorphic field of fractions
QUESTION [5 upvotes]: Given the domain $\mathbb{Q}[r,s,t]/(s^2 - (r-1)(r-2)(r-3), t^2 - (r+1)(r+2)(r+3))$, find a domain $\mathbb{Q}[x,y]/(f)$ with isomorphic field of fractions.

REPLY [3 votes]: Unless I'm mistaken the field of fractions of the first domain is
$$
K=\mathbf{Q}(r)[s,t]=\mathbf{Q}(r)[\sqrt{(r-1)(r-2)(r-3)},\sqrt{(r+1)(r+2)(r+3)}]
$$
a biquadratic extension of $\mathbf{Q}(r)$. 
Hint: A possibility that comes to mind is to try and express $K$ as a simple field extension $K=\mathbf{Q}(r)(y)$ for some element $y\in K$. Obviously $y$ must be algebraic of degree 4 over
$\mathbf{Q}(r)$. By a well-known theorem a tower of (separable) simple algebraic extensions is simple, so this should be possible! Have you seen examples of elements $\alpha$ such that
$\mathbf{Q}[\sqrt2,\sqrt3]=\mathbf{Q}[\alpha]$ or some such? If you have, then try to mimic that!