For every function field $F/K$ there exists $x\in F$ with $F/K(x)$ separable

Question

This is exercise $3.2$ from Stichtenoth's Algebraic Function Fields and Codes:

Let $F/K$ be function field in one variable such that $K$ is perfect and $\overline{K}\cap F=K$. Given $P_1,...,P_r$ places in $F$, prove that $\exists\,x\in F$ whose poles are exactely $P_1,...,P_r$.

I know that for $n\in\mathbb{N}$ big enough we may find $z_i\in F$ such that its pole divisor is $(z_i)_\infty=nP_i$ for all $i=1,...,r$. That way $z:=z_1+...+z_r$ has poles precisely at $P_1,...,P_r$.

If $\text{char}(K)=0$, we just take $z=x$ and the statement is obvious. So we may assume $\text{char}(K)=p>0$.

If $x=z$ doesn't work, I wonder if a different $K$-linear combination of $z_1,...,z_r$ could do the job, but that's just a wild guess.

I think $K$ being perfect should be useful here, but I don't know how to use it.

Besides, for a given $x$, how am I supposed to check whether or not $F/K(x)$ is separable?

Thank you in advance!

Answer 1

$F/K(t)$ is a finite extension, let $S = Hom_{K(t)}(F,\overline{F})$ be the set of $K(t)$-embeddings $F\to \overline{F}$ and $E=F^S$ the subfield fixed by every element of $S$.

$F/E$ is separable and $E/K(t)$ is purely inseparable, of degree $p^n$.

(the degree is a power of $p$ because $f\in L[y]$ is inseparable and irreducible implies $f'=0$ thus $f=g(y^p)$ with $g\in L[y]$ irreducible. Thus purely inseparable implies $f = y^{p^m}-a$ so that $f$ is the minimal polynomial of $a^{1/p^m}$ and hence $E\subset( K(t))^{1/p^n}$)

$K$ is perfect means $K^p = K$. Thus $(K(t^{1/p^n}))^{p^n}=K^{p^n}(t)=K(t)$ which means $K(t^{1/p^n})/K$ is the only one purely inseparable extension of degree $p^n$.

Whence $E=K(t^{1/p^n})$.