TITLE: Function field in one variable over a finite field.
QUESTION [11 upvotes]: Let $K$ be a finite extension over $\mathbb F_p(t)$. How to prove that $K$ is isomorphic to a finite separable extension of $\mathbb F_p(u)$ for some $u\in K$? If I take $K=\mathbb F_p(t)$, then I know that $char(K)=p$ and $K$ is not a perfect field. So there exists an element $u\in K$ such that $u\notin K^p$. Thus $\exists u\in K$ such that $K/\mathbb F_p(u)$ is a finite separable extension, because $\mathbb F_p$ is a perfect field with characteristic $p$. Now, I am not being able to generalise this idea to prove the result for any finite extension $K$ of $\mathbb F_p(t)$.

REPLY [2 votes]: This proof is inspired by the proof of Proposition 3.2.15 from Liu's Algebraic Geometry and Arithmetic Curves (in fact, this statement follows directly from this proposition and Corollary 3.2.14, since a reduced variety over a perfect field is geometrically reduced). Let me call a field of the form $\newcommand{\F}{\mathbb F}\F_p(u)$ for some $u\in K$ a purely transcendental field. We wish to show $K$ is a separable extension of a purely transcendental field.
Let $L$ be a subfield of $K$ which is separable over some purely transcendental field $\F_p(u)$ for which $[K:L]$ is minimal (this value is finite e.g. for $L=\F_p(t)$ so this makes sense). We wish to show $L=K$. If this were not the case then $K/L$ is purely inseparable, hence there is some element $f\in K\setminus L$ such that $f^p\in L$. We'll show $L(f)$ is separable over a purely transcendental field, which will contradict minimality.
Let $T^n+a_{n-1}T^{n-1}+\dots+a_1T+a_0\in \F_p(u)[T]$ be the minimal polynomial of $f^p$ (hence it is separable). Multiplying $f$ by an element of $\F_p[u]$ we can assume all $a_i\in\F_p[u]$. We claim not all of them lie in $\F_p[u^p]$, for otherwise $f$ would be a root of $T^n+a_{n-1}^{1/p}T^{n-1}+\dots+a_1^{1/p}T+a_0^{1/p}\in\F_p(u)[T]$ which is still separable. Therefore in one of the $a_i$ we have a power of $u$ not divisible by $p$. Viewing the relation $f^{pn}+a_{n-1}f^{p(n-1)}+\dots+a_1f^p+a_0=0$ as a polynomial in $u$ with coefficients in $\F_p(f)$, we get that $u$ is separable over $\F_p(f)$, and hence $L(f)$ is separable over $\F_p(f)$, as we wanted.