Given a field $F$ containing all the roots of unity I'm trying to show that $f(x) = x^p - \alpha^p$ is irreducible over $F$ (where $\alpha$ is not in $F$).

It's clear that $f$ splits in $F(\alpha)$ and the other roots are $\alpha \omega^i$ where $\omega$ is a primitive pth root of unity. Hence $F(\alpha)$ is the splitting field of $f$ - now if we could show this was also a separable extension then the Galois group would be transitive and $f$ would be irreducible. However I'm not sure if this is possible - it doesn't work when $charF = p$ for example because the extension won't be separable.

I'm guessing I probably need to use the fact that $p$ is a prime somehow but I'm struggling to see how!

Thanks for any tips


If $p$ is the characteristic, then $X^p-a^p=(X-a)^p$, and if there were any nontrivial $F$-factorization, one of the factors would be $g=(X-a)^m$ for $1\le m<p$. But the $m-1$-degree term of $g$ is $-maX^{m-1}$, implying that $ma\in F$, and thus that $a\in F$.

If $p$ is prime to the characteristic, then the $p$-th roots of unity are $p$ in number, and the roots of $f$ are distinct, so that $f$ is separable, and its roots generate a separable extension.


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