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Math Help - resolvent cubic

  1. #1
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    resolvent cubic

    Let f(x) =x^4+ax^2+bx+c, and r₁, r₂, r₃, r₄ be the roots of f(x). Consider the elements
    Ѳ₁= (r₁+r₂) (r₃+r₄)
    Ѳ₂= (r₁+r₃) (r₂+r₄)
    Ѳ₃= (r₁+r₄) (r₂+r₃)
    Show that Ѳ₁, Ѳ₂, Ѳ₃ are the roots of R(x) =x^3-2ax^2+ (a^2-4c) x+b^2, called the resolvent cubic of f(x).
    I try to show that Ѳ₁+ Ѳ₂+ Ѳ₃=2a, Ѳ₁ Ѳ₂+ Ѳ₁Ѳ₃+Ѳ₂Ѳ₃=a^2-4c; Ѳ₁Ѳ₂ Ѳ₃=-b^2, after the computation I only find Ѳ₁+ Ѳ₂+ Ѳ₃=2a, and didnít get the other two equal.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by apple2009 View Post
    Let f(x) =x^4+ax^2+bx+c, and r₁, r₂, r₃, r₄ be the roots of f(x). Consider the elements
    Ѳ₁= (r₁+r₂) (r₃+r₄)
    Ѳ₂= (r₁+r₃) (r₂+r₄)
    Ѳ₃= (r₁+r₄) (r₂+r₃)
    Show that Ѳ₁, Ѳ₂, Ѳ₃ are the roots of R(x) =x^3-2ax^2+ (a^2-4c) x+b^2, called the resolvent cubic of f(x).
    I try to show that Ѳ₁+ Ѳ₂+ Ѳ₃=2a, Ѳ₁ Ѳ₂+ Ѳ₁Ѳ₃+Ѳ₂Ѳ₃=a^2-4c; Ѳ₁Ѳ₂ Ѳ₃=-b^2, after the computation I only find Ѳ₁+ Ѳ₂+ Ѳ₃=2a, and didnít get the other two equal.
    Brute force computation my friend! This is the only way I know how to verify this...

    I see your solving the quartic!
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  3. #3
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    Quote Originally Posted by apple2009 View Post
    Let f(x) =x^4+ax^2+bx+c, and r₁, r₂, r₃, r₄ be the roots of f(x). Consider the elements
    Ѳ₁= (r₁+r₂) (r₃+r₄)
    Ѳ₂= (r₁+r₃) (r₂+r₄)
    Ѳ₃= (r₁+r₄) (r₂+r₃)
    Show that Ѳ₁, Ѳ₂, Ѳ₃ are the roots of R(x) =x^3-2ax^2+ (a^2-4c) x+b^2, called the resolvent cubic of f(x).
    I try to show that Ѳ₁+ Ѳ₂+ Ѳ₃=2a, Ѳ₁ Ѳ₂+ Ѳ₁Ѳ₃+Ѳ₂Ѳ₃=a^2-4c; Ѳ₁Ѳ₂ Ѳ₃=-b^2, after the computation I only find Ѳ₁+ Ѳ₂+ Ѳ₃=2a, and didnít get the other two equal.
    You just need to plug Ѳ, Ѳ₂and Ѳ₃into the forms of elementary symmetric polynomials and verify that it indeed generates R(x).

    See here and verify that e_1 = 2a, e_2= a^2-4c, e_3=-b^2

    It requires a lengthy computation.
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  4. #4
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    Quote Originally Posted by TheArtofSymmetry View Post
    You just need to plug Ѳ, Ѳ₂and Ѳ₃into the forms of elementary symmetric polynomials and verify that it indeed generates R(x).

    See here and verify that e_1 = 2a, e_2= a^2-4c, e_3=-b^2

    It requires a lengthy computation.
    This computation is indeed an art of symmetry !!
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