I proved that $K=\mathbb{Q}(\sqrt[3]{3},\sqrt{2},\sqrt{3}i)$ is a splitting field of the polynomial $(x^3 -3)(x^2 -2)$ whose galois group is isomorph to $S_3 \times \mathbb{Z}_2 $ and this group have a one element of order 1, 7 elements of order 2, 1 element of order 3, 3 elements of order 4, 3 elements of order 6 and 1 element of order 12. Hence i have the subfields of $K$ by the by galois correspondence are
order 1 is $K$
order 3 is $\mathbb{Q}(\sqrt[3]{3})$
order 6 are $\mathbb{Q}(\sqrt[3]{3},\sqrt2)$, $\mathbb{Q}(\sqrt[3]{3},\sqrt{3}i)$, $\mathbb{Q}(\sqrt[3]{3},\sqrt{6}i)$.
order 4 are $\mathbb{Q}(\sqrt{2},\sqrt{3}i)$, $\mathbb{Q}(\sqrt{2},\sqrt{6}i)$,$\mathbb{Q}(\sqrt{3}i,\sqrt{6}i)$.
order 12 is $\mathbb{Q}$
and I do not know which are the 7 fields of degree 4. am I doing well? thanks.
Let us check the situation by asking the engine...
... and we can also ask for the subfields...
So the splitting field of the given polynomial $f=(x^3-3)(x^2-2)$ is a field $K$ of degree $12$ over $\Bbb Q$. It is Galois over $\Bbb Q$. The Galois field has order $12$, and its structure is
D6, the dihedral group with $12$ elements.The subfields above correspond to the subgroups of the $D_6$ (sometimes denoted with sub index $12$, reflecting the order but...)
(Although the most information comes here in code and from code, i hope that the structure is clear enough to put explicitly the hands on the objects... Of course, sage can give more, if we really need more...)