resultant (alternative treatment)


Summary.

The resultant of two polynomialsMathworldPlanetmathPlanetmathPlanetmath is a number, calculated from the coefficients of those polynomials, that vanishes if and only if the two polynomials share a common root. Conversely, the resultant is non-zero if and only if the two polynomials are mutually prime.

Definition.

Let 𝕂 be a field and let

p⁢(x) =a0⁢xn+a1⁢xn-1+…+an,
q⁢(x) =b0⁢xm+b1⁢xm-1+…+bm

be two polynomials over 𝕂 of degree n and m, respectively. We define Res⁢[p,q]∈𝕂, the resultant of p⁢(x) and q⁢(x), to be the determinantMathworldPlanetmath of a n+m square matrixMathworldPlanetmath with columns 1 to m formed by shifted sequencesMathworldPlanetmath consisting of the coefficients of p⁢(x), and columns m+1 to n+m formed by shifted sequences consisting of coefficients of q⁢(x), i.e.

Res⁢[p,q]=|a00…0b00…0a1a0…0b1b0…0a2a1…0b2b1…0⋮⋮⋱⋮⋮⋮⋱⋮00…an-100…bm-100…an00…bm|
Proposition 1

The resultant of two polynomials is non-zero if and only if the polynomials are relatively prime.

Proof. Let p⁢(x),q⁢(x)∈𝕂⁢[x] be two arbitrary polynomials of degree n and m, respectively. The polynomials are relatively prime if and only if every polynomial — including the unit polynomial 1 — can be formed as a linear combinationMathworldPlanetmath of p⁢(x) and q⁢(x). Let

r⁢(x) =c0⁢xm-1+c1⁢xm-2+…+cm-1,
s⁢(x) =d0⁢xn-1+b1⁢xn-2+…+dn-1

be polynomials of degree m-1 and n-1, respectively. The coefficients of the linear combination r⁢(x)⁢p⁢(x)+s⁢(x)⁢q⁢(x) are given by the following matrix–vector multiplication:

[a00…0b00…0a1a0…0b1b0…0a2a1…0b2b1…0⋮⋮⋱⋮⋮⋮⋱⋮00…an-100…bm-100…an00…bm]⁢[c0c1c2⋮cm-1d0d1d2⋮dn-1]

In consequence of the preceding remarks, p⁢(x) and q⁢(x) are relatively prime if and only if the matrix above is non-singular, i.e. the resultant is non-vanishing. Q.E.D.

Alternative Characterization.

The following PropositionPlanetmathPlanetmath describes the resultant of two polynomials in terms of the polynomials’ roots. Indeed this property uniquely characterizes the resultant, as can be seen by carefully studying the appended proof.

Proposition 2

Let p⁢(x),q⁢(x) be as above and let x1,…,xn and y1,…,ym be their respective roots in the algebraic closureMathworldPlanetmath of K. Then,

Res⁢[p,q]=a0m⁢b0n⁢∏i=1n∏j=1m(xi-yj)

Proof. The multilinearity property of determinants implies that

Res⁢[p,q]=a0m⁢b0n⁢|10…010…0A11…0B11…0A2A1…0B2B1…0⋮⋮⋱⋮⋮⋮⋱⋮00…An-100…Bm-100…An00…Bm|

where

Ai =aia0,i=1,…⁢n,
Bj =bjb0,j=1,…⁢m.

It therefore suffices to prove the proposition for monic polynomialsMathworldPlanetmath. Without loss of generality we can also assume that the roots in question are algebraically independentMathworldPlanetmath.

Thus, let X1,…,Xn,Y1,…,Ym be indeterminates and set

F⁢(X1,…,Xn,Y1,…,Ym) =∏i=1n∏j=1m(Xi-Yj)
P⁢(x) =(x-X1)⁢…⁢(x-Xn),
Q⁢(x) =(x-Y1)⁢…⁢(x-Ym),
G⁢(X1,…,Xn,Y1,…,Ym) =Res⁢[P,Q]

Now by Proposition 1, G vanishes if we replace any of the Y1,…,Ym by any of X1,…,Xn and hence F divides G.

Next, consider the main diagonal of the matrix whose determinant gives Res⁢[P,Q]. The first m entries of the diagonal are equal to 1, and the next n entries are equal to (-1)m⁢Y1⁢…⁢Ym. It follows that the expansion of G contains a term of the form (-1)m⁢n⁢Y1n⁢…⁢Ymn. However, the expansion of F contains exactly the same term, and therefore F=G. Q.E.D.

Title resultant (alternative treatment)
Canonical name ResultantalternativeTreatment
Date of creation 2013-03-22 12:32:52
Last modified on 2013-03-22 12:32:52
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 6
Author Mathprof (13753)
Entry type Definition
Classification msc 12E05
Related topic DiscriminantMathworldPlanetmathPlanetmathPlanetmath