Field $K (x)$ of rational functions over $K$, the element $x$ has no $p$th root.

Question

Let $p$ be a prime number, and let $K = \mathbb{F}_p$. Show that in the field $K (x)$ of rational functions over $K$, the element $x$ has no $p$th root.

I am having trouble understanding what $x$ is. I know that it is just an indeterminate
and that $x$ is transcendental over $K$ but how do I show $x$ has no $p$th root?

I did something similar to this but without the field of fractions. What if I let $f(t) = t^p-x$ where $t$ is just another indeterminate? If $f$ is irreducible then $x$ has no $p$-th root. Am I allowed to use Eisenstein's criterion on $f$? Sadly, I only know the proof of Eisenstein's Criterion over the field of rational numbers. But I've read about a more generalized criterion. Is $x$ a prime element?

Answer 1

It seems you don't have any trouble understanding what $x$ is: you know it's an indeterminate that is transcendental over $\Bbb F_p$. You're correct that Eisenstein's criterion works: since $(x)$ is a prime ideal of the ring $\Bbb F_p[x]$, the polynomial $t^p-x\in \Bbb F_p(x)[t]$ has no roots in $\Bbb F_p(x)$ by the criterion.

Alternately, if $f(x)\in\Bbb F_p(x)$ then $f(x)^p=f(x^p)$ is rational in $x^p$ by little Fermat, whereas clearly $x$ is not rational in $x^p$ so the equation $f(x)^p=x$ is impossible.

Answer 2

I think this argument should do it, but maybe I am missing something:

For a contradiction, suppose there are two non-zero polynomials $f,g \in k[x]$ such that $(\frac{f}{g})^p=x$. Then $f^p=g^px$. The degree of the polynomial on the left is $\deg(f)*p$ and the degree of the polynomial on the right is $\deg(g)*p+1$. So these are equal. I.e. $$\deg(f)*p=\deg(g)*p+1$$ But this is outrageous and impossible by divisibility considerations.