Prove that every polynomial in $F[X]$ is solvable by radicals Im trying to do the following question:
Let $F$ be a field of characteristic zero, and $p$ a prime number. Suppose that $F$ has the property that every irreducible polynomial in $F$ have degree equals to a power of $p$. Prove that every polynomial in $F[X]$ is solvable by radicals over $F$.
My attempt at a solution was try to prove that, given $f$ a irreducible polynomial in $F[X]$, and $L$ a sppliting field of $f$ over $F$, we have $[L:F]=p^k$, so that $Gal(L/F)$ is a p-group, so it is solvable, but it seems that it will not work.
Any hints?

Your idea works with just a little modification.  Just note that by the primitive element theorem, the splitting field $L$ of any polynomial $f$ is generated by a single element $\alpha$ over $F$, and so $[L:F]$ is the degree of the minimal polynomial of $\alpha$.  Thus $[L:F]$ is a power of $p$ and so the Galois group is solvable.
