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Thread: Discriminant

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Discriminant

    Consider $\displaystyle f(x)\in F[x] $. Define its discriminant to be $\displaystyle D=\prod_{i<j}(\alpha_i-\alpha_j)^2 $, where $\displaystyle \alpha_i $ are the roots of $\displaystyle f(x) $ in its splitting field $\displaystyle K $.

    I'm having trouble seeing why $\displaystyle D\in F $. I know it has to do with symmetric functions, but for some reason I can't quite see it. I think I'm missing something small here.

    Thanks!
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    Quote Originally Posted by chiph588@ View Post
    Consider $\displaystyle f(x)\in F[x] $. Define its discriminant to be $\displaystyle D=\prod_{i<j}(\alpha_i-\alpha_j)^2 $, where $\displaystyle \alpha_i $ are the roots of $\displaystyle f(x) $ in its splitting field $\displaystyle K $.

    I'm having trouble seeing why $\displaystyle D\in F $. I know it has to do with symmetric functions, but for some reason I can't quite see it. I think I'm missing something small here.

    Thanks!

    Hmmm...it's not hard but it's not easy, either: if you know that the resultant of f(x) is in F and that the discriminant of f(x) is the resultant times some power of the leading coefficient of f(x) times (-1) to some power then we're done, otherwise we can try:

    1) The discriminat is a symmetric function on the roots of f(x) (this is trivial)

    2) If s_1,...,s_n are the symmetric pol's on the roots of f(x), then any symmetric pol. on these roots is in F[s_1,...,s_n] = the ring of pol's on s_1,...,s_n.

    Of course, the problem is 2 and its proof isn't immediate (and I can't remember it right now, but some book must have it). I think the trick was about partial ordering the pol's in n variables in some way.

    Try to work on this, and I'll try to make some time to consult some books.

    Tonio
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    How about this:

    As you said, $\displaystyle D $ is symmetric in the roots of $\displaystyle f(x) $, so $\displaystyle D $ is fixed by all automorphisms of the Galois group of $\displaystyle f(x) $.

    This implies $\displaystyle D \in F $ since $\displaystyle F $ is the fixed field of $\displaystyle \text{Gal}(K/F) $.

    (We know $\displaystyle F $ is the fixed field of $\displaystyle \text{Gal}(K/F) $ since $\displaystyle [K:F]=|\text{Gal}(K/F)| $, so by the Fundamental Theorem of Galois Theory we get what we want.)
    Last edited by chiph588@; Mar 30th 2010 at 10:01 PM.
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    Quote Originally Posted by chiph588@ View Post
    How about this:

    As you said, $\displaystyle D $ is symmetric in the roots of $\displaystyle f(x) $, so $\displaystyle D $ is fixed by all automorphisms of the Galois group of $\displaystyle f(x) $.

    This implies $\displaystyle D \in F $ since $\displaystyle F $ is the fixed field of $\displaystyle \text{Gal}(K/F) $.

    (We know $\displaystyle F $ is the fixed field of $\displaystyle \text{Gal}(K/F) $ since $\displaystyle [K:F]=|\text{Gal}(K/F)| $, so by the Fundamental Theorem of Galois Theory we get what we want.)

    Excellent! I didn't even remember to mention Galois Theory because I tried to make it easier, but if you can deal with that then it's way
    easier and short...and elegant, imo.

    Tonio
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by tonio View Post
    Excellent! I didn't even remember to mention Galois Theory because I tried to make it easier, but if you can deal with that then it's way
    easier and short...and elegant, imo.

    Tonio
    Yeah, the 'simple' thing I was forgetting was that $\displaystyle D $ was symmetric with respect to $\displaystyle \alpha_i $.
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