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# Discriminants of cubics and quartics and general polynomials Watch

1. Apparently, I'm supposed to be able to determine that the discriminant of the polynomial is 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 is some multiple of , but I don't see how to make it exactly that expression.
2. (Original post by Zhen Lin)
Apparently, I'm supposed to be able to determine that the discriminant of the polynomial is 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?
3. (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!
4. 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 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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Updated: December 20, 2010
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