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Is it possible to find two integral domains $R,S$ such that $R[X] \cong S[X]$ but $\mathrm{Frac}(R)[X] \not \cong \mathrm{Frac}(S)[X]$ ? $\renewcommand{\Frac}{\mathrm{Frac}}$

Here $\Frac(R)$ denotes the fraction field of $R$. Obviously, $R[X] \cong S[X] \implies \Frac(R[X]) = \Frac(R)(X) \cong \Frac(S)(X)$, but it doesn't follow that $\Frac(R) \cong \Frac(S)$ (see (1) or (2)). So we can't conclude $\mathrm{Frac}(R)[X] \cong \mathrm{Frac}(S)[X]$.

However, the restriction of an isomorphism $f : \mathrm{Frac}(R)[X] \to \mathrm{Frac}(S)[X]$ to $(\mathrm{Frac}(R))^*$ has range lying in $(\Frac(S))^*$, and since $f(0)=0$ it follows that $f\vert_{\Frac(R)} : \Frac(R) \to \Frac(S)$ is an isomorphism. It doesn't follow that $R \cong S$.

That's why I think that a counter-example may come from such examples. See also here. Let $$R=\frac{\mathbb{C}[x,y,z]}{(xy - (1 - z^2))}, \quad S=\frac{\mathbb{C}[x,y,z]}{(x^2y - (1 - z^2))}$$ Apparently, $R \not \cong S$ (see the previous link). It looks like $R,S$ are domains. But do we have $\Frac(R) \not \cong \Frac(S)$ ? It would solve my problem. Or is there an easier way to solve it?

Notice that the "reverse" question is easy to solve $\Bbb Z \not \cong \Bbb Q$, but $\Frac(\Bbb Z)[X] \cong \Bbb Q[X]$.

Thank you for your help.

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    $\begingroup$ Unfortunately, in your example, $\operatorname{Frac}(R)\cong\operatorname{Frac}(S)$: both can just be identified with $\mathbb{C}(x,z)$, with $y$ being $\frac{1-z^2}{x}$ and $\frac{1-z^2}{x^2}$, respectively. $\endgroup$ – Eric Wofsey Dec 5 '16 at 8:57
  • $\begingroup$ Another example is given by Hochster for $R[x] \cong S[x]$ but $R \not \cong S$, where $$ R = \left( \dfrac{\Bbb R[X,Y,Z]}{(X^2+Y^2+Z^2-1)} \right)[P,Q]$$ and $$ S = \left( \dfrac{\Bbb R[X,Y,Z]}{(X^2+Y^2+Z^2-1)} \right)[U,V,W] \;/\; (xU+yV+zW) $$ But do we have $$\Frac(R) = \Bbb R(x,y,z ; p,q) \cong \Frac(S) = \Bbb R(x,y,z ; u,v,w)$$? $\endgroup$ – Watson Feb 11 '17 at 13:16

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