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Let $V$ be an $n$-dimensional variety over $k$. The function field $k(V)$ doesn't have to be separable over $k$ but I'd like to know which conditions imply that we can find a finite bijective morphism from $V$ to a variety $W$ with separable function field. In particular I want to prove this for proper irreducible varieties.

So far I've proved it for affine irreducible varieties: If $V=\text{Spec}(A)$ is affine irreducible then by Noether normalisation there exists a finite surjective morphism $k[X_1,\ldots,X_n]\hookrightarrow A$. The function field Frac$(A)$ is a finite extension of $k(X_1,\ldots,X_n)$ which splits into a separable extension $F/k(X_1,\ldots,X_n)$ followed by a purely inseparable extension Frac$(A)/F$. Then the inclusion $A\cap F\subseteq A$ corresponds finite bijective morphism $\text{Spec}(A)\rightarrow\text{Spec}(A\cap F)=W$ and $W$ has separable function field $F$ over $k$.

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up vote 1 down vote accepted

Suppose $k$ has characteristic $p>0$ (otherwise everything is separable).

Consider the relative Frobenius $V\to V^{(p)}$, and iterate it $n$ times to get $V\to V^{(p^n)}$. This is a finite homeomorphism.

Let us prove $k(V^{(p^n)})$ is separable over $k$. As the function field can be computed with a dense affine open subset, we can suppose $V=\mathrm{Spec}(A)$ is affine. You already showed the existence of a subextension $F\subseteq k(V)$ such that $F$ is separable and $k(V)/F$ is finite and purely inseparable. So there exists $n\ge 1$ such that $(k(V))^{p^n}\subseteq F$, therefore $k(k(V))^{p^n}$ (the sub-$k$-extension generated by $(k(V))^{p^n}$) is separable over $k$. But this is nothing but the function field of $k(V^{(p^n)})$, so we are done.

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Thanks, that's elegant! – Michalis Mar 8 '13 at 11:49

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