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Discriminants of cubics and quartics and general polynomials Watch

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    Apparently, I'm supposed to be able to determine that the discriminant of the polynomial X^4 + p X + q is -3^3 p^4 + 4^4 q^3 by exploiting properties of discriminants and choosing p and q carefully. It's easy enough to determine that the discriminant is some multiple of the given expression, using the fact that the discriminant is 0 iff the polynomial has any repeated roots, but I don't see how to fix the multiple without going into calculations for general roots. Unless, by clever choice of p and q, the lecturer means there's a choice which yields an easy factorisation and no repeated roots...?

    Similarly, I can see that the discriminant of the polynomial X^5 + p X + q is some multiple of 4^4 p^5 + 5^5 q^4, but I don't see how to make it exactly that expression.
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    (Original post by Zhen Lin)
    Apparently, I'm supposed to be able to determine that the discriminant of the polynomial X^4 + p X + q is -3^3 p^4 + 4^4 q^3 by exploiting properties of discriminants and choosing p and q carefully. It's easy enough to determine that the discriminant is some multiple of the given expression, using the fact that the discriminant is 0 iff the polynomial has any repeated roots, but I don't see how to fix the multiple without going into calculations for general roots. Unless, by clever choice of p and q, the lecturer means there's a choice which yields an easy factorisation and no repeated roots...?
    By multiple do you mean constant? In which case is the discriminant hard to find when p=0 and q = -1?
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    (Original post by RichE)
    By multiple do you mean constant? In which case is the discriminant hard to find when p=0 and q = -1?
    Oops. I was confining myself to integers only. I guess Gaussian integers are nice enough - which handles the quartic and quintic case with p = -1, q = 0. Thanks!
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    Stuck again - this time with a general case. Let P be an irreducible separable quartic polynomial, and let Q be its resolvent cubic. (I had to look up what this was - I'm pretty sure it wasn't lectured, but it's in the problems without definition. :-/) I need to show that P and Q have the same discriminant. I have a heuristic argument, but it's probably full of holes...

    Again, it's clear that the discriminant of P is zero whenever Q is, and they are both homogeneous polynomials of the roots of P of the degree 12. I haven't convinced myself that the discriminants are polynomials of the same degree in the same variable (fixing the other three), but I think this should be reasonably easy. Then by appealing to the fundamental theorem of algebra, I think I can conclude one discriminant is a non-zero multiple of the other. A simple calculation for P(X) = X^4 - 1 shows that they are in fact exactly equal.

    One problem I can immediately see is that I haven't used the hypothesis that P is irreducible anywhere. I also suspect that it isn't quite so easy to prove that the discriminants are linearly dependent, even if I fix three of the roots of P and regard the discriminants as polynomials in the remaining root. So is there a better (preferably with fewer calculations?) way of doing this, or can this be made to work somehow?
 
 
 
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