matrices - Looking for a proof that the resultant is the product of the differences of roots

I'm trying to find a general proof to an exercise given in Garrity et al's book, Algebraic Geometry: A problem-solving approach.

The problem is this: Given two polynomials f and g, show that for each pair of roots, f(r) = 0, g(s) = 0, that (r - s) divides the resultant.

There is a book of selected answers, but somewhat disappointingly, the solution is given as a brutal appeal to algebra. Moreover, the result is only given for quadratic polynomials.

It seems cited in a few places that the resultant, defined as the determinant of the Sylvester matrix of two polynomials $f = \lambda_1\prod (x - r_i)$ and $g = \lambda_2 \prod (x - s_i)$, is equal to the product $\prod r_i - s_i$. But so far, I have been unable to find a general proof of this fact.

Would anyone either mind sketching the proof, or else pointing me to a resource which does?

Answer

The following is really only a sketch. Feel free to ask for more details.

The coefficients of a polynomial $f$ are equal to elementary symmetric polynomials in the roots of $f$. Since the resultant is a polynomial function in the coefficients of two polynomials $f$ and $g$, it is a symmetric polynomial function in the roots of $f$ and $g$. The definition of the resultant is made in such a way that $\operatorname{res}(f,g)=0$ if (and only if) $f$ and $g$ share a common root. Now, we use the following:

Lemma. Let $R$ be an integral domain and denote by $K$ the algebraic closure of its quotient field. Let $p\in R[X,Y]$ a polynomial such that $p(a,a)=0$ for all $a\in K$. Then, $(Y-X)$ divides $p$.

Proof. Write $p\in R[X][Y]$ as a polynomial in $Y$, i.e. $p=\sum_{i=0}^n p_i Y^i$ with certain $p_i\in R[X]$. In this integral domain, perform division with remainder of $p$ by $Y-X$ to obtain $p=q(Y-X)+r$ for $q\in R[Y,X]$ and $r\in R[X]\subseteq R[X,Y]$. Since
$r(a)=r(a,a)=p(a,a)=0$ for all $a\in K$, we must have $r=0$. Indeed, $K$ is an infinite field because it is algebraically closed and any nonzero polynomial has only finitely many roots. Consequently, $p=(X-Y)\cdot q$ as required.

Applying this lemma to the resultant as a polynomial in the zeros of $f$ and $g$, you get the desired statement.