resultant (alternative treatment)

Summary.

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

Definition.

Let $𝕂$ be a field and let

$p(x)$$=a_0{x}^n+a_1{x}^{n-1}+\mathrm{\dots }+a_n,$

$q(x)$$=b_0{x}^m+b_1{x}^{m-1}+\mathrm{\dots }+b_m$

be two polynomials over $𝕂$ of degree $n$ and $m$, respectively.We define $\mathrm{Res}[p,q]\in 𝕂$, the resultant of $p(x)$ and $q(x)$,to be the determinant of a $n+m$ square matrix with columns 1 to $m$formed by shifted sequences consisting of the coefficients of $p(x)$,and columns $m+1$ to $n+m$ formed by shifted sequences consisting ofcoefficients of $q(x)$, i.e.

Proposition 1

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

Proof. Let $p(x),q(x)\in 𝕂[x]$ be two arbitrarypolynomials of degree $n$ and $m$, respectively. The polynomials arerelatively prime if and only if every polynomial — including theunit polynomial 1 — can be formed as a linear combination of $p(x)$and $q(x)$. Let

$r(x)$$=c_0{x}^{m-1}+c_1{x}^{m-2}+\mathrm{\dots }+c_{m-1},$

$s(x)$$=d_0{x}^{n-1}+b_1{x}^{n-2}+\mathrm{\dots }+d_{n-1}$

be polynomials of degree$m-1$ and $n-1$, respectively. The coefficients of the linearcombination$r(x)p(x)+s(x)q(x)$ are given by the following matrix–vectormultiplication:

In consequence of the preceding remarks, $p(x)$ and $q(x)$ arerelatively 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 Proposition describes the resultant of two polynomialsin terms of the polynomials’ roots. Indeed this property uniquelycharacterizes the resultant, as can be seen by carefully studying theappended proof.

Proposition 2

Let $p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{,}q\mathit{}\mathrm{(}x\mathrm{)}$ be as above and let $x_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n$ and$y_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{y}_m$ be their respective roots in the algebraic closureof $\mathrm{K}$. Then,

$$\mathrm{Res}[p,q]=a_0^m{b}_0^n\prod _{i=1}^n\prod _{j=1}^m(x_i-y_j)$$

Proof. The multilinearity property of determinants implies that

where

$A_i$$={\displaystyle \frac{a_i}{a_0}},i=1,\mathrm{\dots }n,$

$B_j$$={\displaystyle \frac{b_j}{b_0}},j=1,\mathrm{\dots }m.$

It therefore suffices to prove the proposition for monic polynomials.Without loss of generality we can also assume that the roots inquestion are algebraically independent.

Thus, let $X_1,\mathrm{\dots },X_n,Y_1,\mathrm{\dots },Y_m$ be indeterminates and set

$F(X_1,\mathrm{\dots },X_n,Y_1,\mathrm{\dots },Y_m)$$={\displaystyle \prod _{i=1}^n}{\displaystyle \prod _{j=1}^m}(X_i-Y_j)$

$P(x)$$=(x-X_1)\mathrm{\dots }(x-X_n),$

$Q(x)$$=(x-Y_1)\mathrm{\dots }(x-Y_m),$

$G(X_1,\mathrm{\dots },X_n,Y_1,\mathrm{\dots },Y_m)$$=\mathrm{Res}[P,Q]$

Now by Proposition 1, $G$ vanishes if we replace any of the$Y_1,\mathrm{\dots },Y_m$ by any of $X_1,\mathrm{\dots },X_n$ and hence $F$ divides $G$.

Next, consider the main diagonal of the matrix whose determinant gives$\mathrm{Res}[P,Q]$. The first $m$ entries of the diagonal are equal to $1$,and the next $n$ entries are equal to ${(-1)}^m{Y}_1\mathrm{\dots }{Y}_m$. Itfollows that the expansion of $G$ contains a term of the form${(-1)}^{mn}{Y}_1^n\mathrm{\dots }{Y}_m^n$. However, the expansion of $F$ containsexactly the same term, and therefore $F=G$. Q.E.D.