Intermediate fields of degree 6 of $\mathbb{Q}(i,\sqrt{3}, 3^{1/3}):\mathbb{Q}$ Consider the field extension of $\mathbb{Q}(i,\sqrt{3}, 3^{1/3}):\mathbb{Q}$ which is the splitting field of $f(x) = (x^3-3)(x^2+1)$ over $\mathbb{Q}$. 
I am trying to find the intermediate fields of degree $6$ over $\mathbb{Q}$.
I can see that $\mathbb{Q}(\sqrt{3}, 3^{1/3} ), \mathbb{Q}(\sqrt{3}i, 3^{1/3} ),\mathbb{Q}(\sqrt{3}, i3^{1/3} ), \mathbb{Q}(i, 3^{1/3} ),\mathbb{Q}( 3^{1/6} ),\mathbb{Q}(i3^{1/6} ), \mathbb{Q}(\sqrt{3}i \cdot 3^{1/6} )$ 
are all intermediate fields with degree 6 over $\mathbb{Q}$. 
Am I missing any?
 A: First prove that the splitting field is $\mathbb{Q}(3^{1/6}, i)$, with degree $12$. Then this will give a convenient representation of $Gal(\mathbb{Q}(3^{1/6}, i)/\mathbb{Q})$ as 
$$\{\sigma_{j,\pm}|\sigma_{j,\pm}:3^{1/6}\mapsto3^{1/6}\zeta_6^j, i\mapsto \pm i\}$$
Then, it's clear that it is the dihedral group of order 12, which has 7 subgroups of order 2, so there should be 7 intermediate fields of degree 6 over $\mathbb{Q}$. They are $$<\sigma_{1+}>, <\sigma_{3+}>, <\sigma_{5+}>, <\sigma_{0-}>, <\sigma_{2-}>, <\sigma_{3-}>, <\sigma_{4-}>$$
Then we can simply calculate their fixed fields as a linear algebra problem, using $\{1, 3^{1/6}, 3^{2/6}, 3^{3/6}, 3^{4/6}, 3^{5/6}, i, 3^{1/6}i, 3^{2/6}i, 3^{3/6}i, 3^{4/6}i, 3^{5/6}i\}$ as a basis for $\mathbb{Q}(3^{1/6}, i)$, or we can try to guess some order 6 subfields, then verify that each one is fixed by one of these subgroups.
By inspection, we can see that $<\sigma_{0-}>, <\sigma_{2-}>, <\sigma_{4-}>$ are conjugate subgroups, and $<\sigma_{1+}>, <\sigma_{3+}>, <\sigma_{5+}>$ are also conjugate subgroups, and $<\sigma_{3-}>$ is a normal subgroup. These will come in handy later. Now we guess some order 6 subfields:
Easiest: $\mathbb{Q}(3^{1/6})$. This is clearly order 6, and fixed by $<\sigma_{0-}>$.
Then, $\mathbb{Q}(3^{1/3}, i)$ is also clearly order 6, and different from the previous one since it's not real. It's fixed by $<\sigma_{3+}>$.
Now, since conjugate subgroups of Galois group corresponds to conjugate subfields, we calculate conjugate subfields of $\mathbb{Q}(3^{1/6})$, $\mathbb{Q}(3^{1/3}, i)$, to get these:
$\mathbb{Q}(3^{1/6})$ is conjugate to $\mathbb{Q}(3^{1/6}\zeta_6)$, $\mathbb{Q}(3^{1/6}\zeta_6^2)$. Then verify that $\mathbb{Q}(3^{1/6}\zeta_6)$ is fixed by $<\sigma_{2-}>$, $\mathbb{Q}(3^{1/6}\zeta_6^2)$ is fixed by $<\sigma_{4-}>$,
$\mathbb{Q}(3^{1/3}, i)$ is conjugate to $\mathbb{Q}(3^{1/3}\zeta_6, i)$, $\mathbb{Q}(3^{1/3}\zeta_6^5, i)$. Then verify that $\mathbb{Q}(3^{1/3}\zeta_6, i)$ is fixed by $<\sigma_{1+}>$, $\mathbb{Q}(3^{1/3}\zeta_6^5, i)$ is fixed by $<\sigma_{5+}>$.
Finally, there's $<\sigma_{3-}>$ left. By some luck(?), we find $\mathbb{Q}(3^{1/6}i)$ which is fixed by it.

In conclusion, we found 7 subfields: $\mathbb{Q}(3^{1/6}i)$ is a normal subfield. $\mathbb{Q}(3^{1/3}, i)$, $\mathbb{Q}(3^{1/3}\zeta_6, i)$, $\mathbb{Q}(3^{1/3}\zeta_6^5, i)$ are conjugate subfields. $\mathbb{Q}(3^{1/6})$, $\mathbb{Q}(3^{1/6}\zeta_6)$, $\mathbb{Q}(3^{1/6}\zeta_6^2)$are conjugate subfields.

