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Thread: Galois Group

  1. #1
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    Galois Group

    Find the Galois group of $\displaystyle f(x)=(x^3-2)(x^3-3) $.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by mathman88 View Post
    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.
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