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