How to factorize $x^p-a$ in $K[x]$ over $K(\xi,\sqrt[p]a)$ where $a\in K$ and char$K=p$, $\xi$ is a primitive $p$-th root of unity and $p$ is a prime?
What I know is $ x^p - a = (x-\sqrt[p]a)^p$ over $K(\sqrt[p]a)$ by the Freshman's Dream while it is also equal to $\prod\limits_{k=0}^{p-1} (x-\xi^k\sqrt[p]a)$ over $K(\xi, \sqrt[p]a)$ but since $K(\sqrt[p]a) < K(\xi, \sqrt[p]a)$ and $x^p -a$ is splits in the both fields but into different linear factors, I'm now confused it with the fact, $F[x]$ is a UFD if $F$ is a field.
Can anyone address this confusion?
If $K$ is of characteristic $p$, $x^p = 1$ will split over some finite fiels extension of $\mathbb{F}_p$. Hence we need an element of order $p$ in this finite field but every finite fields's multiplicative group is of order $p^m - 1$ ad hence cannot have $p$ as a factor. Hence the only solution to $x^p - 1 = 0$ is $x = 1$. So your $\zeta$ is actually equal to $1$ i.e., $\zeta = 1$. So one question is what if we go to $K[x]/\langle x^p-1 \rangle$. This will not yield anything since $x^p-1 = (x-1)^p$.
So answer is $\zeta = 1$ in $K$. There does not exist a primitive $p^{th}$ root of unity in characeritic $p$ fields. If you want non-trivial $\zeta$, you should consider other than characteristic $p$ fields in which case you cannot factorize as $(x^p-a) = (x-a^{1/p})^p$.