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resultant (alternative treatment)
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(Definition)
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The resultant of two polynomials 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.
Let $\kfield$ be a field and let
be two polynomials over $\kfield$ of degree $n$ and $m$ , respectively. We define $\Res[p,q]\in\kfield$ , 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 of coefficients of $q(x)$ , i.e.
Proposition 1 The resultant of two polynomials is non-zero if and only if the polynomials are relatively prime.
Proof. Let $p(x), q(x)\in \kfield[x]$ 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 combination of $p(x)$ and $q(x)$ . Let
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:
$$\begin{bmatrix} a_0 & 0 & \ldots & 0 & b_0 & 0 & \ldots & 0 \\ a_1 & a_0 & \ldots & 0 & b_1 & b_0 & \ldots & 0 \\ a_2 & a_1 & \ldots & 0 & b_2 & b_1 & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots &\vdots &\vdots &\ddots & \vdots \\ 0 & 0 & \ldots & a_{n-1} & 0 & 0 &\ldots & b_{m-1} \\ 0 & 0 & \ldots & a_n & 0 & 0 &\ldots & b_m \\ \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ \vdots \\ c_{m-1} \\ d_0 \\ d_1 \\ d_2 \\ \vdots \\ d_{n-1} \end{bmatrix} $$ 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.
The following Proposition 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 $x_1,\ldots,x_n$ and $y_1,\ldots,y_m$ be their respective roots in the algebraic closure of $\kfield$ . Then, $$\Res[p,q] = a_0^m\, b_0^n \prod_{i=1}^n \prod_{j=1}^m (x_i-y_j)$$
Proof. The multilinearity property of determinants implies that
where
It therefore suffices to prove the proposition for monic polynomials. Without loss of generality we can also assume that the roots in question are algebraically independent.
Thus, let $X_1,\ldots,X_n, Y_1,\ldots,Y_m$ be indeterminates and set
Now by Proposition 1, $G$ vanishes if we replace any of the $Y_1,\ldots,Y_m$ by any of $X_1,\ldots,X_n$ 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 $(-1)^m Y_1\ldots Y_m$ . It follows that the expansion of $G$ contains a term of the form $(-1)^{mn} Y_1^n\ldots Y_m^n$ . However, the expansion of $F$ contains exactly the same term, and therefore $F=G$ . Q.E.D.
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Cross-references: contains, diagonal, divides, indeterminates, algebraically independent, without loss of generality, monic polynomials, implies, multilinearity, algebraic closure, property, terms, proposition, non-singular, matrix, consequence, multiplication, combination, linear combination, unit, proof, relatively prime, sequences, columns, square matrix, determinant, degree, field, prime, conversely, root, vanishes, coefficients, number, polynomials, resultant
This is version 3 of resultant (alternative treatment), born on 2002-03-14, modified 2006-10-24.
Object id is 2792, canonical name is Resolvent.
Accessed 4977 times total.
Classification:
| AMS MSC: | 12E05 (Field theory and polynomials :: General field theory :: Polynomials ) |
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Pending Errata and Addenda
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![\begin{displaymath}\mathrm{Res}[p,q] = \left \vert \begin{array}{cccccccc} a_0 &... ... \ldots & a_n & 0 & 0 &\ldots & b_m \ \end{array}\right\vert \end{displaymath}](http://images.planetmath.org:8080/cache/objects/2792/js/img5.png "\begin{displaymath}\mathrm{Res}[p,q] = \left \vert \begin{array}{cccccccc} a_0 &... ... \ldots & a_n & 0 & 0 &\ldots & b_m \ \end{array}\right\vert \end{displaymath}")
![\begin{displaymath}\mathrm{Res}[p,q] = a_0^m\, b_0^n \left \vert \begin{array}{c... ...& \ldots & A_n & 0 & 0 &\ldots & B_m \ \end{array}\right\vert\end{displaymath}](http://images.planetmath.org:8080/cache/objects/2792/js/img10.png "\begin{displaymath}\mathrm{Res}[p,q] = a_0^m\, b_0^n \left \vert \begin{array}{c... ...& \ldots & A_n & 0 & 0 &\ldots & B_m \ \end{array}\right\vert\end{displaymath}")
![$\displaystyle = \mathrm{Res}[P,Q]$](http://images.planetmath.org:8080/cache/objects/2792/js/img22.png "$\displaystyle = \mathrm{Res}[P,Q]$"){kind=small}