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It is well known that the finite fields $\mathbb{F}_q$ are perfect i.e. every finite extension of $\mathbb{F}_q$ is separable. I am wondering if a pure transcendental extension $\mathbb{F}_q(x) \supseteq \mathbb{F}_q$ remains perfect? Can someone tell me if this is true and where can I find a proof? Thanks in advance!

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A purely transcendental extension $F(x)$ is never perfect when $F$ is a field of characteristic $p \neq 0$; the polynomial $f(t) = t^p - x$ is an irreducible over $F(x)$, but has repeated roots.

One way to see this is that if we let $F(y)$ be another purely transcendental extension, then $F(x) \cong F(y^p)$. Since $F(y)$ is generated over $F(x)$ by a root of $f(t)$ and $\deg(f) = [F(y) : F(x)]$ (which, for example, can be seen by looking at formal Laurent series), we see that $f$ is irreducible over $F(x)$. And over $F(y)$, we have the factorization

$$ f(t) = t^p - x = (t - y)^p $$

so $f$ has repeated roots.

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$\Bbb F_p(\sqrt[p]{x}) / \Bbb F_p(x)$ is not separable.

A field of characteristic $p$ is perfect iff the Frobenius endomorphism $x \mapsto x^p$ is surjective. In the case of $\Bbb F_p(x)$, $x$ cannot be contained in the image of the Frobenius endomorphism for degree reasons.

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