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Math Help - Galois Group

  1. #1
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    Galois Group

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

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

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

    so the automorphisms look like

    1: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases} \sigma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases} \tau: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
\sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases}

    \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases} \sigma \tau : \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases} \sigma \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases}

    \tau \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
-\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases} \sigma \tau \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases}

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

    Thanks
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  2. #2
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    [K:F] is 8 not 6. ok, i really need to get some sleep now! haha
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    [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 \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})
    is

    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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    Quote Originally Posted by TheEmptySet View Post
    I need to find the Galois Group of (x^2-2)(x^2-3)(x^2-5)

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

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

    so the automorphisms look like

    1: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases} \sigma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases} \tau: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
\sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases}

    \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases} \sigma \tau : \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to \sqrt{5} <br />
\end{cases} \sigma \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases}

    \tau \gamma: \begin{cases} \sqrt{2 } \to \sqrt {2} \\<br />
-\sqrt{3} \to \sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases} \sigma \tau \gamma: \begin{cases} \sqrt{2 } \to -\sqrt {2} \\<br />
\sqrt{3} \to -\sqrt{3 } \\ \sqrt{5} \to -\sqrt{5} <br />
\end{cases}

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

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