# Galois Group

• Apr 5th 2010, 10:53 AM
mathman88
Galois Group
Find the Galois group of $\displaystyle f(x)=(x^3-2)(x^3-3)$.
• Apr 5th 2010, 01:30 PM
chiph588@
Quote:

Originally Posted by mathman88
Find the Galois group of $\displaystyle f(x)=(x^3-2)(x^3-3)$.

I could have this wrong, but here goes:

The splitting field for $\displaystyle f(x)$ is $\displaystyle K=\mathbb{Q}(\sqrt[3]2,\sqrt[3]3,\rho)$, where $\displaystyle \rho$ is a third root of unity.

We then have $\displaystyle [K:\mathbb{Q}]=18$

Since we can determine all automorphisms based on the roots of $\displaystyle f(x)$ and $\displaystyle \alpha\mapsto\alpha'$, where $\displaystyle \alpha'$ is another root of $\displaystyle m_{\alpha,\mathbb{Q}}(x)$, let's see where that takes us.
Define the automorphisms as follows:

$\displaystyle \sigma:\begin{cases} \sqrt[3]2\mapsto\rho\sqrt[3]2\\ \rho\mapsto\rho\\ \sqrt[3]3\mapsto\sqrt[3]3 \end{cases} \quad\quad \tau:\begin{cases} \sqrt[3]2\mapsto\sqrt[3]2\\ \rho\mapsto\rho^2\\ \sqrt[3]3\mapsto\sqrt[3]3 \end{cases} \quad\quad \gamma:\begin{cases} \sqrt[3]2\mapsto\sqrt[3]2\\ \rho\mapsto\rho\\ \sqrt[3]3\mapsto\rho\sqrt[3]3 \end{cases}$

Now note $\displaystyle |\sigma|=3,\; |\tau|=2,\; |\gamma|=3$, so $\displaystyle |<\sigma,\tau,\gamma>|=18= [K:\mathbb{Q}] = |\text{Gal}(K/\mathbb{Q})|$.

I'll leave it to you to see why $\displaystyle <\sigma,\tau> \cong S_3$.

So I'm thinking $\displaystyle \text{Gal}(K/\mathbb{Q}) = <S_3,\gamma>$, where $\displaystyle \gamma$ is a three-cycle.