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Thread: Algebra

  1. #1
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    Algebra

    Let
    $\displaystyle f(x) = x^4+ax^3+bx^2+cx+d$
    be a polynomial in $\displaystyle \mathbb Q[x] $ which has no rational root.

    Show that $\displaystyle f(x) $ is irreducible in $\displaystyle \mathbb Q [x] $

    if the cubic resolvent
    $\displaystyle g(y)=y^3-by^2+(ac-4d)y-a^2d+4bd-c^2 $ has no rational root.
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  2. #2
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    Quote Originally Posted by GaloisGroup View Post
    Let
    $\displaystyle f(x) = x^4+ax^3+bx^2+cx+d$
    be a polynomial in $\displaystyle \mathbb Q[x] $ which has no rational root.

    Show that $\displaystyle f(x) $ is irreducible in $\displaystyle \mathbb Q [x] $

    if the cubic resolvent
    $\displaystyle g(y)=y^3-by^2+(ac-4d)y-a^2d+4bd-c^2 $ has no rational root.
    If the cubic resolvent has no rational roots then it is irreducible. Therefore, the Galois group of $\displaystyle f(x)$ must contain $\displaystyle A_4$ and so it splitting field is at least a $\displaystyle 12$-th degree extension. Now if the original polynomial was reducible then its will factor as a product of two quadradics which would mean its Galois group would be at most a 4 degree extension. Which is a contradiction.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Therefore, the Galois group of $\displaystyle f(x)$ must contain $\displaystyle A_4$
    How can you conclude that the galois group contain $\displaystyle A_4? $ what is the reason for it?
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  4. #4
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    Quote Originally Posted by GaloisGroup View Post
    How can you conclude that the galois group contain $\displaystyle A_4? $ what is the reason for it?
    I think that is a theorem about resolvents. If you never seen it then I do not know of anything else to say.
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  5. #5
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    do you remember where you have seen this theorem in which book.
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