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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?

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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.

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  • $\begingroup$ How do you know $(x)$ is a prime ideal? $\endgroup$ – abe May 27 '14 at 4:32
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    $\begingroup$ @abe: It's one of the easiest ideals to prove prime of all. Have you tried? Do you know what a prime ideal is? The only things not in $(x)$ are constants, and multiplying two constants gives another constant. Equivalently, $F[x]/(x)\cong F$ is a field whenever $F$ is. $\endgroup$ – blue May 27 '14 at 4:33
  • $\begingroup$ Oh Yeah I recall this, Thank You :D $\endgroup$ – abe May 27 '14 at 4:36
  • $\begingroup$ The factor ring $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$. $\endgroup$ – abe May 27 '14 at 4:38
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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.

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  • $\begingroup$ This is exactly what I was thinking, it just seems too easy! What bugs me is that it seems at first glance like it should show $x$ has no $r$-th roots for a lot of other $r$! $\endgroup$ – Robert Lewis May 27 '14 at 4:17
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    $\begingroup$ @RobertLewis That's correct, $x$ has no $r$th roots except when $r=\pm1$, by the same argument. The interesting thing about $r=p$ is that the characteristic can be used in an argument (as I mention in my answer). $\endgroup$ – blue May 27 '14 at 4:19

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