Determine the Galois group of the polynomial $(x^3-2)(x^3-3)(x^2-2)$ over $\mathbb{Q}(\sqrt {-3})$

Solution 1:

$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\R}{\mathbb{R}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Note that if $a \in \Q$, then the roots of $x^{3} - a$ are $\alpha, \alpha \omega, \alpha \omega^{2}$, where $\alpha = \sqrt[3]{a} \in \R$, and $\omega$ is a primitive $3$rd root of unity, that is, a root of $x^{2} + x + 1$. Thus $$ \omega = \frac{-1 + \sqrt{-3}}{2} \in \Q[\sqrt{-3}]. $$ This tells you that the first two factors of $G$ are not $S_{3}$ but only $A_{3}$, that is, cyclic.

There is a little extra work to do to show that each $x^{3} - a$ is irreducible over $\Q[\sqrt{-3}]$, for $a = 2, 3$. Clearly $$ \Size{\Q[\sqrt[3]{a}] : \Q} = 3. $$ Now note that $\Q[\sqrt[3]{a}] \subseteq \R$, so that $$ \Size{(\Q[\sqrt[3]{a}])[\sqrt{-3}] : \Q[\sqrt[3]{a}]} = 2, $$ which yields $$ \Size{(\Q[\sqrt[3]{a}])[\sqrt{-3}] : \Q} = \Size{(\Q[\sqrt[3]{a}])[\sqrt{-3}] : \Q[\sqrt[3]{a}]} \cdot \Size{\Q[\sqrt[3]{a}] : \Q} = 6. $$ But then \begin{align} 6 &= \Size{(\Q[\sqrt[3]{a}])[\sqrt{-3}] : \Q} \\&= \Size{(\Q[\sqrt{-3}])[\sqrt[3]{a}] : \Q}\\&= \Size{(\Q[\sqrt{-3}])[\sqrt[3]{a}] : \Q[\sqrt{-3}]} \cdot \Size{\Q[\sqrt{-3}] : \Q} \\&= \Size{(\Q[\sqrt{-3}])[\sqrt[3]{a}] : \Q[\sqrt{-3}]} \cdot 2, \end{align} whence $$ \Size{(\Q[\sqrt{-3}])[\sqrt[3]{a}] : \Q[\sqrt{-3}]} = 3, $$ which means $x^{3} - a$ is irreducible over $\Q[\sqrt{-3}]$.