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

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

• Eisenstein's criteria? – Don Thousand Apr 7 at 0:59
• What about unique factorization in $k[x]$? – Ted Shifrin Apr 7 at 1:01
• 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. – Clayton Apr 7 at 1:07
• 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 $x^p.$ – Melody Apr 7 at 1:50