Prove $f=x^p-a$ either irreducible or has a root. (arbitrary characteristic) (without using the field norm) [duplicate]

Let $K$ be an arbitrary field, $p$ a prime and $a\in K$. Show $f=x^p-a$ is either irreducible in $K[x]$ or has a root in $K$.

My strategy was to split this up into a case for each characteristic.

The characteristic $p$ case is easy. Assume $f$ is reducible $=gh$ both smaller degree $g$ irreducible, with $\alpha$ a root of $g$. Then in $K(\alpha)[x]$, $f=(x-\alpha)^p$. So $g=(x-\alpha)^k$, and all the other factors of $f$ are of the same form. Hence they are all equal to the $minpoly(\alpha)$ and the degree of $minpoly(\alpha)|p$ so it is $1$ or $p$ and we are done.

The characteristic $0$ case and characteristic $q\ne p$ I get stuck on. If $K$ contains a primitive $p$th root of unity $\zeta$, then over $K(\alpha)$, $f$ splits into linear factors. So the degree of the field extension for any root of $f$ is the same, so the degree of such a root is either $1$ or $p$. If $K$ does not have a primitive $p$th root of unity, then simply adjoin one. We know $[K(\zeta):K]|p-1$ and by the previous step either $f$ is irreducible over $K(\zeta)$ or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form $x-\zeta^ia^{1/p}$, but I am unsure what to do from here. The characteristic $q$ case seems similar to the characteristic $0$ case but again I am stuck trying to finish the proof.

Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did.

Solution 1:

Starting where you have left off, if $f=gh$, then $g$ must be a product of terms of the form $x-\zeta^i\beta$, where my $\beta$ is your $\alpha^{1/p}$ (which I think is supposed to be $a^{1/p}$). Given $i$ and $j$ where $x-\zeta^i\beta$ is a factor of $g$ and $x-\zeta^j\beta$ isn't, there's an automorphism taking $\zeta^i$ to $\zeta^j$. The automorphism fixes $g$, since the coefficients of $g$ are in $K$, but it doesn't fix the factors of $g$, contradiction, since we have unique factorization of polynomials over a field.