Let $x$ be transcendental over $k$. I wonder how to show:

The extension $k(x)$ over $k(x^p)$ has degree $p$.

I know $y^p-x^p$ is a polynomial over $k(x^p)$ that has a root $y=x$, but I don't know how to prove that $y^p-x^p$ is irreducible/minimal over $k(x^p)$.

Let $y^p-x^p=f(y)g(y)$, we need to show that one of them is trivial. But things get really messy from here and I don't see a slick proof

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    $\begingroup$ Eisenstein's criteria? $\endgroup$ – Don Thousand Apr 7 at 0:59
  • $\begingroup$ What about unique factorization in $k[x]$? $\endgroup$ – Ted Shifrin Apr 7 at 1:01
  • $\begingroup$ If the characteristic of $k$ is $p$, I think knowing the derivative is identically $0$ is helpful. It's been too long since I've really thought about these things, but you might find it helpful. $\endgroup$ – Clayton Apr 7 at 1:07
  • $\begingroup$ I agree with the commend about Eisenstein's Criterion. Can't we consider $k(x^p)$ as the rational functions in indeterminate $x^p.$ This is the field of fractions of $k[x^p],$ the polynomials in indeterminate $x^p$. Since $k[x^p]/\left<x^p\right>\cong k$ is an integral domain, $\left<x^p\right>$ is prime, therefore $y^p-x^p$ is irreducible by Eisenstein's Criterion with prime $x^p.$ $\endgroup$ – Melody Apr 7 at 1:50

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