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?
Best Answer
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.