Prove this polynomial to be either irreducible or competely splitting in K[x]. The following question was asked in my quiz of field theory(over now) and I was unable to solve it and so I am posting here. So, I tried it again today but couldn't make any progress.

Let p be a prime and assume either (i) char $K = p$ or (ii) char $K \neq  p$ and $K$
contains a primitive $p$th root of unity. Then $x^p - a \in K[x]$ is either irreducible or
splits in $K[x]$.

So, 1st I let that char $K= p$ and that $x^p-a$  is  reducible in $K[x]$ so , let a be it's root Now I have to prove that $x^p -a$ actually splits in $K[x]$, $x^p -a = f(x) (x-a)$ and degree of $f(x)$ is $n-1$ but I am unable to think using what result I should proceed now.
And alas similar problem is occuring when I tried to solve (ii) of the question when I assumed that polynomial is not irreducible then how I should do it.
I think I will not be able to do this without any help and so do you mind giving your time on this.
 A: Fix a root $\omega$ of $f = X^p-a$ in some algebraic closure.
First suppose $\mathbf{char}(K) = p$. Then $f = X^p-\omega^p = (X-\omega)^p$, as the comments indicate. Hence splitting and having a root in $K$ are equivalent conditions.
Likewise, if $\mathbf{char}(K) \neq p$ and $\xi_p \in K$ is a $p$-th primitive root of unity, then $f = X^p-a$ is separable its roots are $\xi_p^j \omega$ for all $0\leq j \leq p-1$. Thus if some root $\xi_p^j\omega$ lies in $K$, dividing by $\xi_p^j \in K$ we get $\omega \in K$ and so $p$ has all its roots in $K$, hence it splits.
What we have to do now, in both cases, is showing that splitting and reducibility are equivalent. By our previous discussion, this is the same as showing that if $p$ is reducible then $\omega \in K$ in each case.
The case $\mathbf{char}(X) = p$ is  the easiest: if $f = gh$ in $K[X]$ for $g,h \in K[X]$ of positive degree, then - in some algebraic closure -  we have $g = (X-\omega)^s$ with $s < p$. In particular, we know that $\omega^s \in K$. But then $s$ and $p$ are coprime so choosing $k,l$ such that $ks+pl = 1$ we obtain
$$
\omega = \omega^1 = \omega^{sk}+\omega{pl} = (\omega^s)^k + a^l \in K.
$$
On the other hand, in the characteristic $\neq p$ setting having $f = gh$ for positive degree $g$ and $h$ shows that a proper product of $\{\xi_p^j\omega\}_j$ lies in $K$, but since we are assuming $\xi_p \in K$, dividing by its powers we still have $\omega^s \in K$ with $s <p$.
A: We write $b=\sqrt[p]{a}$, $\omega$ a $p$-th root of unity and $h(x)=x^p-a$.
If char $K=p\ (\neq 0)$, then $p$ is prime as $K$ is a field. So $$h(x)=(x-b)^p$$ (by freshman's dream). Thus, if there is a $b\in K$ our polynomial splits. If there is no $b\in K$, then $h(x)$ must be irreducible as it can only be factorised into powers of $(x-b)$, contradicting $b\not\in K$.
If char $K\neq p$, then we don't have the freshman's dream factorisation $h(x)=(x-b)^p$, so the approach must be different. We're given $\omega\in K$.

*

*If there is a $b\in K$, then $h(x)$ splits: $$h(x)=(x-b)(x-\omega b)(x-\omega^2 b)\cdots (x-\omega^{p-1}b).$$


*If there is no $b\in K$, we need to show $h(x)$ is irreducible. We have $[K(b):K]=p$ because we have a basis $\{1,b,b^2,\cdots,b^{p-1}\}$ (using $\omega\in K$). Suppose $h(x)$ does factorise, say, $h(x)=g(x)k(x)$. Then take any root $\alpha$ of $g(x)$, by above we know that $\alpha=\omega^ib$, for some $0\leq i<p$, then we have $$p=[K(b):K]=[K(b):K(\alpha)][K(\alpha):K]$$ by the tower law. Since $p$ is prime, $[K(\alpha):K]$ can only be $1$ or $p$. It cannot be $1$ as that would imply $K(\alpha)=K$, contradicting the assumption that $b\not\in K$. If it is $p$, then in fact $K(b)=K(\alpha)$, so all roots of $h(x)$ are in $K(\alpha)$, so $h(x)$ splits. So to summarise, we've shown that if $h(x)$ factorises over $K$, then it must split.
