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Thread: 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 $\displaystyle f(x)=x^4 - 7$ over $\displaystyle F= \mathbb{Q}$and find all the intermediate fields of the extension.

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

    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 $\displaystyle f(x)=x^4 - 7$ over $\displaystyle F= \mathbb{Q}$and find all the intermediate fields of the extension.

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

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

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

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

    Proceeding in almost the same way I get:

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

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

    Proceeding in almost the same way I get:

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

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