If $\mathbb{Q}[t_1,\ldots, t_n] = A[x_1,\ldots, x_{n-1}]$ has it been proven that $A\cong\mathbb{Q}[t]$?
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$\begingroup$ This is an easy exercise. $\endgroup$– nafCommented Sep 15, 2020 at 10:48
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$\begingroup$ I can only prove in case of algebraically closed field. Any hints to not closed field case? $\endgroup$– A.SkutinCommented Sep 15, 2020 at 11:10
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$\begingroup$ @ulrich : what easy argument do you have in mind? $\endgroup$– Steven LandsburgCommented Sep 15, 2020 at 20:56
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1$\begingroup$ $\mathrm{Spec}(A)$ is obviously a curve which is unirational over $\mathbb{Q}$. It is also clear that the space of complex points of $\mathrm{Spec}(A)$ is contractible (in the analytic topology). There is only one curve over $\mathbb{Q}$ which has these two properties. $\endgroup$– nafCommented Sep 16, 2020 at 14:02
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$\begingroup$ Is this argument valid for any other field which isn't algebraically closed but with zero characteristic? $\endgroup$– A.SkutinCommented Sep 16, 2020 at 15:25
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