Splitting field of $f(x)=x^7-3$ over $\mathbb{Q}$ and its degree

Question

The roots of $f$ has the form $\sqrt[7]{3}\,\gamma^{n}$, where $\gamma=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$ and $n\in\{0,1,2,3,4,5,6\}$.

I saw that, if $K$ is the root field of $f$ over $\mathbb{Q}$, so $K=\mathbb{Q}(\sqrt[7]{3},\sqrt[7]{3}\,\gamma)$, because
$$\sqrt[7]{3}\gamma^{2}=\frac{\sqrt[7]{3}\gamma\cdot\sqrt[7]{3}\gamma}{\sqrt[7]{3}}\quad\textrm{and}\quad \sqrt[7]{3}\gamma^{n}=\frac{\sqrt[7]{3}\gamma^{n-1}\cdot\sqrt[7]{3}\gamma}{\sqrt[7]{3}}. $$ I'm trying to work with the extensions $\mathbb{Q}\subset\mathbb{Q}(\sqrt[7]{3})\subset K$.

We have $[\mathbb{Q}(\sqrt[7]{3}):\mathbb{Q}]=\deg (x^7-3)=7$ because of Eisenstein's criterion.

But I'm having some problems with calculating $[K:\mathbb{Q}(\sqrt[7]{3})].$

Obviously, $f\in\mathbb{Q}(\sqrt[7]{3})[x]$ satisfies $f(\sqrt[7]{3}\gamma)=0$, but I don't know if this polynomial is irreducible over $\mathbb{Q}(\sqrt[7]{3})[x]$. What can I do?

Answer 1

You have already found that $$K=\Bbb{Q}(\sqrt[7]{3},\sqrt[7]{3}\gamma)=\Bbb{Q}(\sqrt[7]{3},\gamma),$$ and that $[\Bbb{Q}(\sqrt[7]{3}):\Bbb{Q}]=7$. Next note that $[\Bbb{Q}(\gamma):\Bbb{Q}]=6$ and that $K$ is the compositum of $\Bbb{Q}(\sqrt[7]{3})$ and $\Bbb{Q}(\gamma)$, so $[K:\Bbb{Q}]=7\times6=42$.