Function field in one variable over a finite field.

Question

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 $\mathbb F_p(t)$.

Answer 1

We know that every extension of global function fields splits into a separable part and an inseparable one (see Stichtenoth "Algebraic Function Fields and Codes", Appendix A). Let $F$ be the "purely inseparable subfield" i.e. the field such that $K/F$ is separable and $F/\mathbb F_p(t)$ is purely inseparable i.e. $F^{p^e}\subseteq \mathbb F_p(t)$ for some $e$. We can assume that $F\neq \mathbb F_p(t)$ as otherwise we are done directly.

It is well known that there always exists a primitive element of a finite extension of $\mathbb F_p(t)$ see for example the answer to Is a field perfect iff the primitive element theorem holds for all extensions, and what about function fields .

Let $u\in K$ be such that $F=\mathbb F_p(u,t)$. We claim that $K/\mathbb F_p (u)$ is a separable extension. Since $K/F$ is separable, to show the separability of $K/\mathbb F_p (u)$ it is enough to show that $F/\mathbb F_p(u)$ is separable (see for example Theorem 3.13 of https://kconrad.math.uconn.edu/blurbs/galoistheory/separable1.pdf ).

By standard field theory this reduces to show that $t$ is separable over $\mathbb F_p (u)$. Let $e$ be the minimal integer such that $u^{p^e}=f(t)/g(t)\in \mathbb F_p(t)$ (with $f,g$ coprime), which exists because $F/\mathbb F_p(t)$ is purely inseparable. It follows that $t$ is a root of $H(T)=f(T)-g(T)u^{p^e}\in \mathbb F_p (u)[T]$. Since at least one between $f(T)$ and $g(T)$ is not a $p$-power by the choice of $e$, then $H'(T)=f'(T)-g'(T)u^{p^e}\neq 0$ (the derivative is taken with respect to $T$).

If I show you now that $t$ is a simple root for $H(T)$, then its minimal polynomial will be separable as it has to divide $H(T)$.

Suppose by contradiction that $H'(t)=f'(t)-g'(t)u^{p^e}=0$. If $g(t)$ is separable, then $g'(t)\neq 0$ and therefore $f'(t)/g'(t)=u^{p^e}=f(t)/g(t)$ which implies that $f(t)/g(t)$ is a $p$-power (as one can deduce that its $t$ derivative is zero), which is a contradiction by the minimality of $e$. If $g(t)$ is not separable we have a contradiction as $f'(t)=0$ and $t$ is trascendental over $\mathbb F_p$.

Answer 2

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.