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 <fc3a501@GEO.math.uni-hamburg.de> 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