In this post, we shall see how the resultant, a determinant containing coefficients of two polynomials is useful in determining the existence of their common roots.

Suppose we are working in a ring {A} and {f(y), g(y)} are two polynomials in {A[y]}:

\displaystyle f(y) = a_0 y^m + a_1 y^{m-1} + \cdots + a_m, \qquad a_i \in A \text{ and } a_0 \neq 0.

\displaystyle g(y) = b_0 y^n + b_1 y^{n-1} + \cdots + b_n, \qquad b_i \in A \text{ and } b_0 \neq 0.

Consider the system of equations obtained by multiplying the first equation by {y^{n-1}, y^{n-2}, \cdots, y, 1} and the second by {y^{m-1}, y^{m-2}, \cdots, y, 1}. Make the following change of variables:

\displaystyle z_0 = 1,

\displaystyle z_1 = x,

\displaystyle z_2 = x^2,

\displaystyle \vdots

\displaystyle z_{m+n-1} = x^{m+n-1}.

The {(m+n)} equations: {f=0, \;yf=0, \cdots, \;y^{n-1}f=0, \; g=0, \;yg=0, \cdots, \;y^{m-1}g=0} can be denoted in matrix form as:

\displaystyle \begin{bmatrix} a_0 & a_1 & \cdots & \cdots & a_{m-1} & a_m & 0 & 0 & \cdots & 0\\ 0 & a_0 & a_1 & \cdots & \cdots & a_{m-1} & a_m & 0 & \cdots & 0\\ 0 & 0 & a_0 & a_1 & \cdots & \cdots & a_{m-1} & a_m & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & a_0 & a_1 & \cdots & \cdots & \cdots & a_m\\ b_0 & b_1 & \cdots & \cdots & b_{n-1} & b_n & 0 & 0 & \cdots & 0\\ 0 & b_0 & b_1 & \cdots & \cdots & b_{n-1} & b_n & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & b_0 & b_1 & \cdots & \cdots & \cdots & b_n\\ \end{bmatrix} \begin{bmatrix} z_0 \\ z_1 \\ z_2\\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ z_{m+n} \end{bmatrix} = \begin{bmatrix} \;\;0\;\; \\ 0 \\ 0\\ \vdots \\ \vdots \\ \vdots \\ \vdots \\ 0 \end{bmatrix}.

Call the above matrix as {R}. If there was a common solution {x = \alpha} so that {f(\alpha) = g(\alpha) = 0}, then above system of equations would have a non-zero solution (non-zero since {z_0 = 1}), so the determinant of {R} would have to be zero. If the ring {A} is an integral domain, then the vanishing of the determinant of {R} is equivalent to {f} and {g} having a common root.

This determinant is known as the resultant of {f} and {g}. It has some interesting consequences that I state below:

Theorem 1 For two non-constant irreducible polynomials {f(x), g(x) \in k[x]} to have a common root, it is necessary and sufficient that their resultant vanishes.

Assume that {f(x,y), g(x,y) \in A[x,y]} are irreducible and consider {f(y) , g(y) \in B[y]} where {B = A[x]}. Thus the coefficients of {f} and {g} are polynomials in {x}. The matrix similar to above has coefficients polynomials in {x}, so here the resultant is a polynomial in {x}, say {R(x)}. If {(\alpha, \beta)} is a common root of {f} and {g}, then {R(\alpha) = 0}. Hence the {x}-coordinate of any common root must be a root of {R(x)}. In particular, there can be at most finitely many {\alpha}‘s so that {(\alpha, \beta)} is a common root.

Further, for every value of {\alpha} so that {R(\alpha) = 0}, there are only finitely many values of {\beta} so that {\beta} is a root of {f(\alpha, y) = 0}. Hence {f} and {g} can intersect in finitely many points! We record this as one version of Bezout’s theorem:

Theorem 2 (Bezout) If {f} and {g} are irreducible polynomials of degrees {m, n>1} in {k[x,y]}, then they can intersect in at most {mn} points.

Here, we have proved only finiteness of common points. For the refinement {\leq mn}, see Abhyankar‘s book Lectures in Algebra.

Theorem 3 Any curve {f(x,y) \in k[x,y]} can have at most finitely many singular points.

Here, we define a point {(\alpha, \beta)\in k^2} to be singular if it satisfies {f(x,y) = f_x(x,y) = f_y(x,y) = 0}. The proof follows from Bezout’s theorem when we set {g=f_x} or {f_y}.

Resultants can also be used to compute the discriminant of a polynomial. The discriminant of a polynomial

\displaystyle f(x) = a_0 x^n + a_1 x^{n-1} + \cdots + a_n = a_0 \displaystyle\prod_{i=1}^n (x-\alpha_i)
is given by

\displaystyle \text{Disc}(f) = \displaystyle \prod_{i<j} (\alpha_i - \alpha_j).

Upto a sign factor {(-1)^{\frac{n(n-1)}{2}}}, we have

\displaystyle \text{Disc}(f) = \text{Resultant of } f \text{ and } f'.

The Wikipedia article on Resultants takes the Theorem 1 to be the definition and states different results.

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