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Math Help - galois group and intermediate fields

  1. #1
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    galois group and intermediate fields

    Here is the problem I have:
    Let E be a splitting field of f = x^4 - 4 in Q[x]

    1. determine the galois group of E/Q (up to isomorphism)

    2. Determine all intermediate fields between E and Q. Which of them are Galois over Q?

    so far I have this:
    letting a = 2^(1/2) then the roots are +/-a and +/- ai

    I know Q[a] < E and |Q[a]:Q| = 2 (minimal polynomal is x^2 - 2) and ai is not in Q[a].

    Letting E = Q[a, i] we have all roots of f in E and |Q[a, i]: Q[a]| = 2 since minimal polynomial is x^2 + 1.

    thus |E| = |Q[a, i]: Q[a]||Q[a]:Q| = 4

    from here I start drawing a big blank. Any help is appreciated.
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  2. #2
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    Quote Originally Posted by sporkmonkey View Post
    Here is the problem I have:
    Let E be a splitting field of f = x^4 - 4 in Q[x]

    1. determine the galois group of E/Q (up to isomorphism)

    2. Determine all intermediate fields between E and Q. Which of them are Galois over Q?

    so far I have this:
    letting a = 2^(1/2) then the roots are +/-a and +/- ai

    I know Q[a] < E and |Q[a]:Q| = 2 (minimal polynomal is x^2 - 2) and ai is not in Q[a].

    Letting E = Q[a, i] we have all roots of f in E and |Q[a, i]: Q[a]| = 2 since minimal polynomial is x^2 + 1.

    thus |E| = |Q[a, i]: Q[a]||Q[a]:Q| = 4

    from here I start drawing a big blank. Any help is appreciated.
    well, if \sigma \in \text{Gal}(E/\mathbb{Q}), then \sigma(\sqrt{2}) has to be a root of x^4 - 4. clearly we cannot have \sigma(\sqrt{2})=\pm i \sqrt{2} because then 2=\sigma(2)=(\sigma(\sqrt{2}))^2=(\pm i\sqrt{2})^2=-2, which is not possible.

    similarly we cannot have \sigma(i \sqrt{2})=\pm \sqrt{2}. so every non-identity element of \text{Gal}(E/\mathbb{Q}) has order two and thus \text{Gal}(E/\mathbb{Q}) \cong C_2 \times C_2. (here C_2 is the cyclic group of order two)
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