From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math.research Subject: Re: On resultants Date: 4 Nov 1996 18:44:20 GMT In article <55ksgt\$o3j@rzsun02.rrz.uni-hamburg.de>, Hauke Reddmann wrote: >Suppose you have something like > >p=y^2+1 q=2*y^2+3*y > >where y stands short for x^2-1. > >You eliminate x to get f(p,q)=0. Am I right that >the resultant factors into two equal factors? The resultant of two monic polynomials f(p,q) may be expressed as the product Prod( p(x_i) ), the product taken over all roots of q. If the roots of q fall into pairs for which p(x_i) is equal, then clearly the resultant is a square. The setting you have may be generalized: if there is a polynomial r for which p = p' o r and q = q' o r for some other polynomials p' and q', then resultant(p,q) = resultant(p',q')^d where d = degree of r. >And the other way, suppose I only known that >this happens, how can I compute the polynome in x >on which p and q depend on? I take it you mean you wish to compute what I called r. I don't see in general why there needs to be an r of the appropriate degree. The resultant of p=x and q=x-4 is 4, a square, but there is no possibility of composites here. Arguably, the source of the problem is that your calculations are being done over too small a ring. If your p and q had symbolic coefficients and a resultant which was a square in the corresponding polynomial ring, that would perhaps tell you something. dave