If $p$ is a prime number and $a$ is an element of the field $K$, what can be said about the Galois group of $K(\sqrt[p]a)$ over $K$? (That is, $\sqrt[p]a$ is an element whose $p$-th power is $a$)
I was hoping to show it's a cyclic group of order $p$, and then use that to show that the Galois group of a polynomial solvable in radicals is solvable. I thought I could do this by showing that $[K(\sqrt[p]a):K]=p$, but I think that might actually be false. I think it's true for primitive roots of unity, but it's not clear to me what general conditions need to be imposed on $\sqrt[p]a$ in order to make this work.
Can anything like what I want be proven?
When the field $K$ contains the p-th roots of unity, then $K(\sqrt[p]{a})$ is a splitting field of the irreductible polynomial $P=X^p-a$ (of course we suppose that a is not a p-th power), so is galoisien of degree p. The roots of $P$ are $\mu ^j \sqrt[p]{a}$ where $\mu$ is a generator of the group of p-th roots of unity. In this case the galois group is cyclic of order p (sent $\sqrt[p]{a}$ to $\mu ^k \sqrt[p]{a}$).