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  1. #1
    Behold, the power of SARDINES!
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    Galois Group

    I need to find the Galois Group of $\displaystyle (x^2-2)(x^2-3)(x^2-5)$

    So to start since the splitting field is a degree 6 extention $\displaystyle [K:F]=6$ so the Galois Group should have 6 isomporphisms.

    The splitting field looks like $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

    so the automorphisms look like

    $\displaystyle 1: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$ $\displaystyle \sigma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$ $\displaystyle \tau: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$

    $\displaystyle \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$ $\displaystyle \sigma \tau : \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$ $\displaystyle \sigma \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$

    $\displaystyle \tau \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    -\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$ $\displaystyle \sigma \tau \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$

    So I have 8, so I think I am misunderstanding something.

    Thanks
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  2. #2
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    $\displaystyle [K:F]$ is 8 not 6. ok, i really need to get some sleep now! haha
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by NonCommAlg View Post
    $\displaystyle [K:F]$ is 8 not 6. ok, i really need to get some sleep now! haha
    Wow I think I needed some too.

    The basis for field extention of $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$
    is

    $\displaystyle 1,\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{6},\sqrt{10},\s qrt{15},\sqrt{30}$

    and yes indeeded it does have 8 elements

    Thanks again
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    I need to find the Galois Group of $\displaystyle (x^2-2)(x^2-3)(x^2-5)$

    So to start since the splitting field is a degree 6 extention $\displaystyle [K:F]=6$ so the Galois Group should have 6 isomporphisms.

    The splitting field looks like $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

    so the automorphisms look like

    $\displaystyle 1: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$ $\displaystyle \sigma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$ $\displaystyle \tau: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$

    $\displaystyle \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$ $\displaystyle \sigma \tau : \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5}
    \end{cases}$ $\displaystyle \sigma \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$

    $\displaystyle \tau \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\
    -\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$ $\displaystyle \sigma \tau \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\
    \sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to -\sqrt{5}
    \end{cases}$

    So I have 8, so I think I am misunderstanding something.

    Thanks
    In general, let $\displaystyle p_1,...,p_n$ be distinct prime numbers. Then the Galois group of $\displaystyle (x^2 - p_1)...(x^2 - p_n)$ is $\displaystyle \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ and has order $\displaystyle 2^n$. It happens to be isomorphic to $\displaystyle \mathbb{Z}_2^n$ (since all non-trivial automorphisms have order $\displaystyle 2$).
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