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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.

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Is this homework? What have you tried? –  lhf Oct 5 '11 at 12:07
    
Well, I don't have a clear idea as to how to go about solving this, or problems of a similar nature. I tried to find some polynomial in ($s^2 -(r-1)(r-2)(r-3), t^2 -(r+1)(r+2)(r+3)$) that is a polynomial in two variables. One such polynomial exists. It is in the variables $s$ and $t$. But, then I don't know what to do with that, or if that polynomial actually qualifies as the polynomial $f$. I feel like I am missing some key insight as to how to think about this problem. –  Rankeya Oct 5 '11 at 12:16

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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!

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Thanks. I don't know how I missed this. –  Rankeya Oct 5 '11 at 13:05

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