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Given the galois extension $L/K$, where $L=\mathbb{Q}(\zeta, \sqrt[3]{2})$. I've already calculated that $[L:\mathbb{Q}]=6$. I'm trying to prove that the number of intermediate fields is also 6, including $L$ and $\mathbb{Q}$ as intermediate fields. I'm told that the fundamental theorem of galois theory can be useful here, but I can't find a use for it.

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Hint:

What is the Galois group $\operatorname{Gal}(L/\mathbb{Q})$?

Well, the minimal polynomial for which $L$ is a splitting field is $f(x) = x^3 - 2$, which has roots $\sqrt[3]{2}, \zeta \sqrt[3]{2}$, and $\zeta^2 \sqrt[3]{2}$. It is useful to remember that each element of the Galois group is uniquely determined by its action on the roots of $f$.

Now, complex conjugation is always a legitimate $\mathbb{Q}$-automorphism. That is, $\phi(\zeta) = \overline{\zeta}$. Further, I claim that another automorphism $\tau$ is defined by $\sqrt[3]{2} \mapsto \zeta\sqrt[3]{2}$. Can you prove this?

What happens when we take powers of $\phi$ and $\tau$ and compose them with each other? For me, it is especially illuminating to view the roots of $f$ as vertices of a triangle in the complex plane and see how these automorphisms are acting on it.


Alternative path:

It is a theorem that the Galois group will be a subgroup of the symmetric group acting on the roots of $f$. Therefore, we know $\operatorname{Gal}(L/\mathbb{Q}) \leq S_3$. However, we also know that $|\operatorname{Gal}(L/\mathbb{Q})| = [L:\mathbb{Q}] = 6$, so that certainly narrows down the possibilities for $\operatorname{Gal}(L/\mathbb{Q})$!


Once you've determined the Galois group, then what are its subgroups? Remember the fundamental theorem tells us that there is a one-to-one correspondence between subgroups of the Galois group and intermediate fields of the Galois extension.

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    $\begingroup$ Right. Count the subgroups, you’ve counted the intermediate fields, including the top and the bottom. $\endgroup$ – Lubin Nov 28 '14 at 1:58
  • $\begingroup$ I've calculated the galois group to be $S_3$, and according to the fundamental theorem of galois theory, I know that there's a bijection from subgroups of the galois group and the intermediate fields. so number of subgroups of galois group = number of intermediate fields. What I need to show now is that number of subgroups of galois group=6, but there's more than 6 subgroups of $S_3$, right? $\endgroup$ – Andrew Brick Nov 28 '14 at 2:43
  • $\begingroup$ Nope! There are only $6$. Two of them are simply $\{id\}$ and $S_3$ itself. What are the other $4$? Hint: Lagrange's theorem tells us they will be of order $2$ or of order $3$. $\endgroup$ – Kaj Hansen Nov 28 '14 at 2:44
  • $\begingroup$ ah I see, thanks $\endgroup$ – Andrew Brick Nov 28 '14 at 2:52
  • $\begingroup$ Glad I could help! $\endgroup$ – Kaj Hansen Nov 30 '14 at 5:19

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