Given a field $K$ of characteristic 0, which contains a primitive root of order 3, I would like to show that the extension $K(\sqrt2+3^{\frac{1}{3}})/K$ is cyclic.
My attempt was to look at the "bigger" extension $K(\sqrt2, 3^{\frac{1}{3}})/K$. The problem is that I don't know it's size, because $\sqrt2$ might be in $K$ for example. But I do know that it's embedded in $\mathbb{Z}/2\mathbb{Z} \ X \ \mathbb{Z}/3\mathbb{Z}$ by the diamond theorem.
Anyway, we still have to show that the initial extension is Galois, and we know it's enough to show its normality, but I can't figure it out.
Any help would be appreciated, Thanks.
Both $K(\sqrt2)/K$ and $K(\sqrt[3]3)$ are Kummer extensions: the latter as $K$ has a cube root of unity. The first has Galois group trivial or $C_2$ the second has Galois group trivial or $C_3$. Those Galois groups have coprime orders, so their compositum $K(\sqrt2,\sqrt[3]3)$ is Galois over $K$ with group $G=\text{Gal}(K(\sqrt2)/K)\times\text{Gal}(K(\sqrt[3]3)/K)$. In all possible cases $G$ is cyclic.
The subfield $K(\sqrt2+\sqrt[3]3)$ is Galois with group a quotient of $G$ (by the Galois correspondence, since $G$ is Abelian). A quotient of a cyclic group is cyclic.