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This paper roughly claims that given a projective variety $X$ over a finite field, there is a finite map $f:X\rightarrow \mathbb{P}^n$ such that if $H$ is a hyperplane in $\mathbb{P}^n$ and $U=\mathbb{A}^n$ is the complement of $H$, then the induced map $f:f^{-1}(U)\rightarrow U$ is étale.

So let us consider that $X$ is a rational smooth projective variety. Since $f:f^{-1}(U)\rightarrow U$ is étale and $X$ and $\mathbb{P}^n$ have isomorphic function fields then it implies that $f:f^{-1}(U)\rightarrow U$ is an isomorphism. So this implies that there is actually an affine $f^{-1}(U)$ in $X$ which is $\mathbb{A}^n$.

I am not sure whether I'm making any mistakes or not, but this seems a little bit odd.

Question. Is there any way to understand the complement of $f^{-1}(U)$? Like whether it is a smooth variety or not?

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    $\begingroup$ You are indeed making a mistake. That $k(X)$ and $k(\mathbb{P}^n)$ are isomorphic doesn't say that $f^*:k(\mathbb{P}^n)\rightarrow k(X)$ is an isomorphism. $\endgroup$
    – abx
    Commented Feb 19, 2021 at 7:07

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