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Math Help - 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 x^4-5 over \mathbb{Q}, \mathbb{Q}[\sqrt{5}], \mathbb{Q}[\sqrt{-5}].

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

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

    Is this irreducible or not?


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

    Thank you!
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  2. #2
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    Quote Originally Posted by marianne View Post
    So G is isomorphic to D_4 or \mathbb{Z}_4.
    To determine which, we check is f irreducible over \mathbb{Q}[\sqrt{-5}]
    <br />
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 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 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 x^4-5 over \mathbb{Q}, \mathbb{Q}[\sqrt{5}], \mathbb{Q}[\sqrt{-5}].

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

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

    Is this irreducible or not?


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