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Discriminant is a symmetric polynomial

  1. Nov 19, 2005 #1
    I've got to proof the following:

    Let f be a monic polynomial in Q[X] with deg(f) = n different complex zeroes. Show that the sign of the discriminant of f is equal to (-1)^s, with 2s the number of non real zeroes of f.

    I know the statement makes sense, because the discriminant is a symmetric polynomial over Q, so it can be written as a polynomial in elementary symmetric polynomials.

    The question seems to suggest that the complex zeroes always come in pairs of a zero and its conjugate. But even if this is true, i still don't know how to proceed :(

    Can anyone give me a hint?
  2. jcsd
  3. Nov 19, 2005 #2


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    Prove that if z is a root of the polynomial, then so is [itex]\bar{z}[/itex]. Use the fact that rational numbers are self-conjugate, and that

    [tex]\overline{ab + c} = \overline{a}\overline{b} + \overline{c}[/tex]

    Next, go to the definition of the descriminant as a product of squares of differences of roots. You know that if ri is a non-real root and if rj is a real root, then both [itex](r_i - r_j)^2[/itex] and [itex](\overline{r_i} - r_j)^2[/itex] will occur in the product, so consider what the product of these two factors will be (well, just consider the sign). Consider the case when both roots under consideration are real, when both are complex and not conjugate to one another, and when they are complex and conjugate to one another.
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