TITLE: Galois Group of $(x^3-5)(x^2-3)$
QUESTION [12 upvotes]: I am having some trouble calculating the Galois group (over $\mathbb{Q}$) of $(x^3-5)(x^2-3)$. I can see the splitting field is $F:=\mathbb{Q}(\sqrt[3]{5},\omega,\sqrt{3})=\mathbb{Q}(\sqrt[3]{5},i,\sqrt{3})$, where $\omega$ is a primitive 3rd root of unity, and it has degree 12 over $\mathbb{Q}$. Since the extension is Galois (it is a splitting field extension for a separable polynomial), this makes $|\operatorname{Gal}(F/\mathbb{Q})|=12$ so it is either $S_3 \times S_2$ or $A_4$ as these are the only subgroups of of $S_5$ of order 12. It has a transposition (complex conjugation) so I conclude that it is $S_3 \times S_2$. I found all the subgroups of $S_3 \times S_2$, (all 16 of them), but I can't see to match them up with the intermediate fields. For example, there are three subgroups of order 6 in $S_3 \times S_2$ which should correspond to three field extensions of degree 2. One of them should be $\mathbb{Q}(i)$ since this is an intermediate field with a degree 2 minimal polynomial, but none of the subgroups of order 6 I found have every element fix $i$, so this is a problem. (A subgroup of order 6 has an element of order 3, which should be permuting the roots of $x^3-5$, no?)
Have I gone wrong somewhere? I would appreciate the help.

REPLY [8 votes]: It may help you to have a general remark on when to recognize that a Galois group splits as a direct product.
Suppose that $E/F$ is Galois with Galois group $G$, and that $G$ factors as a product $G = G_1 \times G_2$.  Then by the Galois correspondence we have Galois
subextensions $E_i/F$ contained in $E/F$, where $E_i = E^{G_i}$.  The fact that
$G_1$ and $G_2$ generate $G$ says that $E_1 \cap E_2 = E^G = F.$  The fact
that $G_1 \cap G_2 = \{1\}$ says that $E_1$ and $E_2$ together generate $E$
(in standard, if slightly flowery, terminology, we say that $E$ is the compositum of $E_1$ and $E_2$).
So to show that $Gal(E/F)$ is a product, we have to write $E$ as the compositum
of two subfields $E_1$ and $E_2$, each Galois over $F$, such  that $E_1\cap E_2 = F$.
In your case $E = \mathbb Q(5^{1/3}, \omega,\sqrt{3})$ can be written as the compositum of $E_1 = \mathbb Q(5^{1/3},\omega)$ and $E_2  = \mathbb Q(\sqrt{3})$.  This gives the product struture on $G$.