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Thread: just one more question on Galois group

  1. #1
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    just one more question on Galois group

    So, I have to determine the Galois group of $\displaystyle x^4-5$ over $\displaystyle \mathbb{Q}$, $\displaystyle \mathbb{Q}[\sqrt{5}]$, $\displaystyle \mathbb{Q}[\sqrt{-5}]$.

    $\displaystyle f(x)=x^4-5$
    That is irreducible over $\displaystyle \mathbb{Q}$ by Esienstein's criterion.
    Resolvent is:$\displaystyle g(x)=x^3+20x=x(x^2+20)=x(x-\sqrt{-20})(x+\sqrt{-20})$
    $\displaystyle \alpha=0, \beta=\sqrt{-20}, \gamma=-\sqrt{-20}$
    $\displaystyle
    \mathbb{Q}(\alpha, \beta, \gamma)=\mathbb{Q}(0, \sqrt{-20}, -\sqrt{-20})=\mathbb{Q}(\sqrt{-20})=\mathbb{Q}[\sqrt{-5}]$
    $\displaystyle
    [\mathbb{Q}(\alpha, \beta, \gamma):\mathbb{Q}]=2$

    So $\displaystyle G$ is isomorphic to $\displaystyle D_4$ or $\displaystyle \mathbb{Z}_4$.
    To determine which, we check is $\displaystyle f$ irreducible over $\displaystyle \mathbb{Q}[\sqrt{-5}]$
    $\displaystyle
    f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$

    Is this irreducible or not?


    And what to do with
    $\displaystyle \mathbb{Q}[\sqrt{5}]$, $\displaystyle \mathbb{Q}[\sqrt{-5}]$?

    Thank you!
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  2. #2
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    Quote Originally Posted by marianne View Post
    So $\displaystyle G$ is isomorphic to $\displaystyle D_4$ or $\displaystyle \mathbb{Z}_4$.
    To determine which, we check is $\displaystyle f$ irreducible over $\displaystyle \mathbb{Q}[\sqrt{-5}]$
    $\displaystyle
    f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$

    [COLOR=Red][COLOR=Black]Is this irreducible or not?
    Argue that the splitting field has degree 8 over Q and so the Galois group has degree 8 as well.
    This means the Galois group is $\displaystyle D_4$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Argue that the splitting field has degree 8 over Q and so the Galois group has degree 8 as well.
    This means the Galois group is $\displaystyle D_4$.
    Oh, I missed that, thank you!!!

    Could you possibly help me out with , ?
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  4. #4
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    Quote Originally Posted by marianne View Post
    So, I have to determine the Galois group of $\displaystyle x^4-5$ over $\displaystyle \mathbb{Q}$, $\displaystyle \mathbb{Q}[\sqrt{5}]$, $\displaystyle \mathbb{Q}[\sqrt{-5}]$.

    $\displaystyle f(x)=x^4-5$
    That is irreducible over $\displaystyle \mathbb{Q}$ by Esienstein's criterion.
    Resolvent is:$\displaystyle g(x)=x^3+20x=x(x^2+20)=x(x-\sqrt{-20})(x+\sqrt{-20})$
    $\displaystyle \alpha=0, \beta=\sqrt{-20}, \gamma=-\sqrt{-20}$
    $\displaystyle
    \mathbb{Q}(\alpha, \beta, \gamma)=\mathbb{Q}(0, \sqrt{-20}, -\sqrt{-20})=\mathbb{Q}(\sqrt{-20})=\mathbb{Q}[\sqrt{-5}]$
    $\displaystyle
    [\mathbb{Q}(\alpha, \beta, \gamma):\mathbb{Q}]=2$

    So $\displaystyle G$ is isomorphic to $\displaystyle D_4$ or $\displaystyle \mathbb{Z}_4$.
    To determine which, we check is $\displaystyle f$ irreducible over $\displaystyle \mathbb{Q}[\sqrt{-5}]$
    $\displaystyle
    f(x)=(x^2-\sqrt{5})(x^2+\sqrt{5})$

    Is this irreducible or not?


    And what to do with
    $\displaystyle \mathbb{Q}[\sqrt{5}]$, $\displaystyle \mathbb{Q}[\sqrt{-5}]$?
    Let $\displaystyle K$ be the splitting field for the resolvent, $\displaystyle x(x^2+20)$. As you have shown $\displaystyle K = \mathbb{Q}(\sqrt{-5})$ and so $\displaystyle [K:\mathbb{Q}]=2$. Therefore, the Galois group is either $\displaystyle \mathbb{Z}_4$ or $\displaystyle D_4$. It all depends whether or not $\displaystyle f$ is irreducible over $\displaystyle K$. First $\displaystyle f$ has NO roots in $\displaystyle K$ because the roots of $\displaystyle f$ are: $\displaystyle \pm\sqrt[4]{5}, \pm i\sqrt[4]{5}$. These numbers have degree four over $\displaystyle \mathbb{Q}$ while $\displaystyle K$ is only a degree two extension. Thus, you need to show that $\displaystyle (x^2 + ax + b)(x^2 + cx + d) = x^4 - 5$ is impossible for $\displaystyle a,b,c,d\in K$. To simplify things we will appeal to a result from number theory, that the ring of integers in $\displaystyle \mathbb{Q}(\sqrt{d})$ is $\displaystyle \mathbb{Z}[\sqrt{d}]$ if $\displaystyle d\equiv 2,3(\bmod 4)$. Thus, the ring of integers in $\displaystyle \mathbb{Q}(\sqrt{-5})$ is $\displaystyle \mathbb{Z}[\sqrt{-5}]$ i.e. a field of fractions of $\displaystyle \mathbb{Z}[\sqrt{-5}]$ is $\displaystyle \mathbb{Q}(\sqrt{-5})$. By Gauss' Lemma (the general one) to prove irreduciblity of $\displaystyle x^4 - 5$ over $\displaystyle K$ it sufficies to prove irreducibility of $\displaystyle x^4 - 5$ over $\displaystyle \mathbb{Z}[\sqrt{-5}]$. Since we have $\displaystyle (x^2 + ax + b)(x^2 + bx + d) = x^4 - 5$ where $\displaystyle a,b,c,d,\in \mathbb{Z}[\sqrt{-5}]$ if we expand we get $\displaystyle x^4+(a+c)x^3+(b+ac + d)x^2 +(ad+bc)x+bd = x^4 - 5$. I think (I did not check this!) that it is only possible to get $\displaystyle bd = -5$ if $\displaystyle b=d=\pm \sqrt{-5}$, now show that both those cases lead to impossibilities in the coefficiensts $\displaystyle a,b,c,d$. Can you be the mathematician of the gaps and fill them in?
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    To simplify things we will appeal to a result from number theory, that the ring of integers in $\displaystyle \mathbb{Q}(\sqrt{d})$ is $\displaystyle \mathbb{Z}[\sqrt{d}]$ if $\displaystyle d\equiv 2,3(\bmod 4)$. Thus, the ring of integers in $\displaystyle \mathbb{Q}(\sqrt{-5})$ is $\displaystyle \mathbb{Z}[\sqrt{-5}]$ i.e. a field of fractions of $\displaystyle \mathbb{Z}[\sqrt{-5}]$ is $\displaystyle \mathbb{Q}(\sqrt{-5})$.
    I have something similar, but it's$\displaystyle x^4-7$ over$\displaystyle \mathbb{Q}[\sqrt{-7}]$. How do I check is $\displaystyle f$ irreducible? I can't use the number theory result you provided.
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  6. #6
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    Quote Originally Posted by georgel View Post
    I have something similar, but it's$\displaystyle x^4-7$ over$\displaystyle \mathbb{Q}[\sqrt{-7}]$. How do I check is $\displaystyle f$ irreducible? I can't use the number theory result you provided.
    If $\displaystyle d\equiv 1(\bmod 4)$ the ring of integers of $\displaystyle \mathbb{Q}(\sqrt{d})$ is given by $\displaystyle \{ \tfrac{a+b\sqrt{d}}{2} | a\equiv b(\bmod 2) \}$.
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