Re: Algebraic dependency between polynomials
- From: jdolan@xxxxxxxxx (James Dolan)
- Date: Thu, 3 May 2007 05:27:42 +0000 (UTC)
in article <1178005970.077431.315430@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
keith ramsay <kramsay@xxxxxxx> wrote:
|On Apr 30, 11:00 pm, Denis Feldmann <denis.feldmann.asuppri...@club-
|internet.fr> wrote:
||If p and q are two polynomials (in one variable) of degree n (or
||less) with coefficients in a field K, it is easy to prove there
||exits a (non zero) polynomial in two variables, F, such that
||F(p,q)=0 (i.e) F(p(t),q(t)) is the null polynomial). Numerical
||solving (with Maple) shows that for n<=4, F is of total degreee <=n
||too (which is already quite surprising, as F(p(t),q(t)) has n^2+1=17
||coefficient to identify to 0, while F has only (n+1)(n+2)/2=15
||unknown coefficients). And F is (in general) unique. What is the
||explanation?
|
|Consider those pairs (x,y) for which p(t)-x=0 and q(t)-y=0 have a
|common root (in t).
|
|Polynomials' sharing of a root is characterized by the resultant,
|which is the determinant of a square matrix of side deg(p)+deg(q) =
|2n. It's a matrix representation of the transform (a,b)->ap+bq where
|a and b are polynomials, deg(a),deg(b)<n. If p and q share a root,
|then there are such polynomials such that ap+bq=0, and not otherwise,
|so they have a common root when the determinant is zero.
|
|But x and y only occur (linearly) in the last n columns, so the
|degree of the determinant in x and y is only n. For example, if
|p(t)=t^4+at^3+bt^2+ct+d and q(t)=t^4+et^3+ft^2+gt+h, the resultant of
|p(t)-x and q(t)-y is the determinant of:
|
|(1 a b c d-x 0 0 0)
|(0 1 a b c d-x 0 0)
|(0 0 1 a b c d-x 0)
|(0 0 0 1 a b c d-x)
|(1 e f g h-y 0 0 0)
|(0 1 e f g h-y 0 0)
|(0 0 1 e f g h-y 0)
|(0 0 0 1 e f g h-y)
|
|which has degree 4 in x and y (together).
it's probably also a good idea to think here about the idea that F is
the generator of the ideal obtained as the kernel of the homomorphism
sending x to p and y to q. i think that this is more closely related
to keith's answer than you might at first guess, though i didn't check
the details.
--
jdolan@xxxxxxxxxxxx
.
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- Algebraic dependency between polynomials
- From: Denis Feldmann
- Re: Algebraic dependency between polynomials
- From: Keith Ramsay
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