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Math Help - Application of the Fund. Theorem of Galois Theory

  1. #1
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    Application of the Fund. Theorem of Galois Theory

    I'm having trouble with this question: Find the splitting field of f(x)=x^4 - 7 over F= \mathbb{Q}and find all the intermediate fields of the extension.

    The splitting field is \mathbb{Q} (\sqrt[4]{7} , \zeta) = K where \zeta = e^{\frac{2i \pi}{4}}=i . Now this extension has degree 8 since 2, 4\vert [K:F] and [K:f] \leq 8. By the isomorphis extension theorem there are \sigma , \tau: K\longrightarrow K automorphisms such that \sigma (\sqrt[4]{7}) = i\sqrt[4]{7} and \sigma (i) =i and \tau (\sqrt[4]{7}) = \sqrt[4]{7} \tau (i)=-i
    from this we know that G=Gal(\frac{K}{F}) \cong D_8 . Now the fixed field of < \sigma> is F(i) and the fixed field of < \tau>= F(\sqrt[4]{7}), but I'm having trouble with the other subgroups <\sigma ^2> , <\sigma \tau>, <\sigma ^2 \tau>, < \sigma ^3 \tau>. I know they must be degree 4 extensions of F, but using the trace I get either trivial fixed elements or ones that I don't know how to check their degree over F.

    Tnaks in advance
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  2. #2
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    Quote Originally Posted by Jose27 View Post
    I'm having trouble with this question: Find the splitting field of f(x)=x^4 - 7 over F= \mathbb{Q}and find all the intermediate fields of the extension.

    The splitting field is \mathbb{Q} (\sqrt[4]{7} , \zeta) = K where \zeta = e^{\frac{2i \pi}{4}}=i . Now this extension has degree 8 since 2, 4\vert [K:F] and [K:f] \leq 8. By the isomorphis extension theorem there are \sigma , \tau: K\longrightarrow K automorphisms such that \sigma (\sqrt[4]{7}) = i\sqrt[4]{7} and \sigma (i) =i and \tau (\sqrt[4]{7}) = \sqrt[4]{7} \tau (i)=-i
    from this we know that G=Gal(\frac{K}{F}) \cong D_8 . Now the fixed field of < \sigma> is F(i) and the fixed field of < \tau>= F(\sqrt[4]{7}), but I'm having trouble with the other subgroups <\sigma ^2> , <\sigma \tau>, <\sigma ^2 \tau>, < \sigma ^3 \tau>. I know they must be degree 4 extensions of F, but using the trace I get either trivial fixed elements or ones that I don't know how to check their degree over F.

    Tnaks in advance
    the fixed field of <\sigma^2> is \mathbb{Q}(\sqrt{7},i). this is quite easy to see: let \sqrt[4]{7}=a. then an element of K is in the form z=r_1+r_2a+r_3a^2+r_4a^3+r_5i + r_6ia+r_7ia^2+r_8ia^3, where r_j \in \mathbb{Q}.

    then since \sigma^2(a)=-a, we see immediately that \sigma^2(z)=z if and only if r_2=r_4=r_6=r_8=0. therefore \sigma^2(z)=z if and only if z=r_1+r_3a^2+r_5i + r_7ia^2 \in \mathbb{Q}(a^2,i).
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  3. #3
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    Okay, I think I got it:

    The fixed field of < \sigma \tau >= F((1+i) \sqrt[4]{7}). This is easily seen to be fixed by \sigma \tau and since it cannot be a root of a degree 2 polynomial over F, it's minimal polynomial must have degree 4 (since otherwise it would generate the whole extension which cannot be).

    Proceeding in almost the same way I get:

    Fixed field of < \sigma ^3 \tau> = F((1-i) \sqrt[4]{7}) and < \sigma ^2 \tau>= F(i \sqrt[4]{7}).

    Is this right?
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  4. #4
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    Quote Originally Posted by Jose27 View Post
    Okay, I think I got it:

    The fixed field of < \sigma \tau >= F((1+i) \sqrt[4]{7}). This is easily seen to be fixed by \sigma \tau and since it cannot be a root of a degree 2 polynomial over F, it's minimal polynomial must have degree 4 (since otherwise it would generate the whole extension which cannot be).

    Proceeding in almost the same way I get:

    Fixed field of < \sigma ^3 \tau> = F((1-i) \sqrt[4]{7}) and < \sigma ^2 \tau>= F(i \sqrt[4]{7}).

    Is this right?
    you're right about the fixed field of <\sigma \tau>. i didn't check the other ones.
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