I asked a question a few days ago about where the function field $k(x,\sqrt{1-x^2})$ was purely transcendental over $k$, for $k$ algebraically closed. It turned out to be true, so I know this proves that the circle $x^2+y^2-1=0$ is actually a rational variety in $\mathbb{A}^2$.
Does this hold for all $n\geq 2$ also? I mean, is the $n$-spher given by $x_1^2+\cdots+x_n^2-1=0$ a rational variety in $\mathbb{A}^n$? I had a hard time generalizing the argument there to higher dimensions. I googled around, but no results turned up, but I suspect this is a standard question which is well known.
I tried: Let $S=V(x_1^2+\cdots+x_n^2-1)$. By stereographic projection, I get maps $\phi\colon S\to\mathbb{A}^{n-1}$ by $$ \phi(x_1,\dots,x_n)=\left(\frac{x_1}{1-x_n},\dots,\frac{x_{n-1}}{1-x_n}\right) $$ and $\psi\colon\mathbb{A}^{n-1}\to S$ defined by $$ \psi(u_1,\dots,u_{n-1})=\left(\frac{2u_1}{1+\sum_{i=1}^{n-1}u_i^2},\dots,\frac{2u_{n-1}}{1+\sum_{i=1}^{n-1}u_i^2},\frac{-1+\sum_{i=1}^{n-1} u_i^2}{1+\sum_{i=1}^{n-1}u_i^2}\right) $$ which are known inverses if I've written them down correctly. I have a birational map I think, does that suffices to show $S$ is a rational variety? Or maybe am I overlooking some subtleties?
Anyway, is $S$ a rational variety? And if so, what is the best to prove it? I'm also curious if there is an analogous approach with the function field like there is with the sphere. Cheers.
