Find $[\mathbf{Q}(\sqrt{2},\sqrt[3]{7},\zeta_7):\mathbf{Q}(\sqrt{2},\sqrt[3]{7})]$ As part of a bigger problem that I am solving, I need to find $[\mathbf{Q}(\sqrt{2},\sqrt[3]{7},\zeta_7):\mathbf{Q}(\sqrt{2},\sqrt[3]{7})]$. I think this must be equal to $6$, but I don't know how to prove this, for example that the cyclotomic polynomial $\Phi_7$ stays irreducible.
Could someone provide any help?
 A: You can prove that $[\mathbb{Q}(\sqrt{2},\zeta_7):\mathbb{Q}] = 12$ and $[\mathbb{Q}(\sqrt[3]{7},\zeta_7):\mathbb{Q}] = 18$. Then from here we conclude that $36 =\text{lcm}(12,18) \mid [\mathbb{Q}(\sqrt{2},\sqrt[3]{7},\zeta_7):\mathbb{Q}]$. But $[\mathbb{Q}(\sqrt{2},\sqrt[3]{7},\zeta_7):\mathbb{Q}] \le 36$, so we must have that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{7},\zeta_7):\mathbb{Q}] = 36$. Finally using the fact that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{7}):\mathbb{Q}] = 6$ we get what we want. 
For the first claim we consider $\mathbb{Q}(\zeta_7)$, which is a Galois extension with Galois group $\mathbb{Z}/6\mathbb{Z}$. Now if $x^2 - 2$ isn't irreducible in $\mathbb{Q}(\zeta_7)[x]$ we have that it splits into linear factors, hence $\sqrt{2}$ is an element of $\mathbb{Q}(\zeta_7)$. However $\mathbb{Q}(\zeta_7)$ has a unique quadratic subfield and it is $\mathbb{Q}(i\sqrt{7})$. So as $\mathbb{Q}(i\sqrt{7}) \not = \mathbb{Q}(\sqrt{2})$ we conclude  $\sqrt{2} \not \in \mathbb{Q}(\zeta_7)$ and so $[\mathbb{Q}(\sqrt{2},\zeta_7):\mathbb{Q}] = 12$
For the second claim we again consider $\mathbb{Q}(\zeta_7)$. Now if $x^3-7$ isn't irreducible it has a linear factor in $\mathbb{Q}(\zeta_7)[x]$. This means that $\sqrt[3]{7}$, $\zeta_3\sqrt[3]{7}$ or $\zeta_3^2\sqrt[3]{7}$ is an element of $\mathbb{Q}(\zeta_7)$. Hence one of $\mathbb{Q}(\sqrt[3]{7})$,$\mathbb{Q}(\zeta_3\sqrt[3]{7})$,$\mathbb{Q}(\zeta_3^2\sqrt[3]{7})$ is a subfield of $\mathbb{Q}(\zeta_7)$. But this is impossible as its Galois group is abelian, hence all subfields of $\mathbb{Q}(\zeta_7)$ are Galois, but neither of the above fields is. Thus we have $[\mathbb{Q}(\sqrt[3]{7},\zeta_7):\mathbb{Q}] = 18$.
Hence the proof.

REMARK: To deduce that $\mathbb{Q}(i\sqrt{7})$ is the only quadratic subfield of $\mathbb{Q}(\zeta_7)$ you can compute the discriminant of $\Phi_7$, which is $-7^5$, as described here. Then use the fact that its square root is an element of $\mathbb{Q}(\zeta_7)$. Indeed, a more general statement is true, i.e that $\mathbb{Q}(i^{\frac{p-1}{2}}\sqrt{p})$ is the unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ for a prime $p \ge 3$
A: On one side, the cyclotomic field $\mathbf Q(\zeta_7)$ is cyclic of degree $6$ over $\mathbf Q$, hence has 2 unique subextensions of respective degrees 2 and 3, which are $\mathbf Q(\sqrt {-7})$ (this results from a classical calculation of discriminant) and $\mathbf Q(\zeta_7 + {\zeta_7}^{-1})$. On the other side, $\mathbf Q(\sqrt 2)$  and $\mathbf Q(\sqrt [3] 7)$ have degrees 2 and 3 over $\mathbf Q$ by Eisenstein criterion, and none of them is contained in $\mathbf Q(\zeta_7)$. So their compositum has degree 6 and is linearly disjoint from $\mathbf Q(\zeta_7)$. It follows that $[\mathbf Q(\zeta_7,\sqrt 2, \sqrt [3] 7):\mathbf Q]=36 $ and $[\mathbf Q(\zeta_7,\sqrt 2, \sqrt [3] 7):\mathbf Q (\sqrt 2, \sqrt [3] 7)]=6 $.
