1.1 A brief introduction to computational algebraic geometry  (Page 17/20)

Let $V_k$ be the vector space of polynomials of degree $k$ or less. Then $V_k$ has dimension $k+1$ , since $1,t,t^2,...,t^k$ is basis. Multiplication by $f$ defines a linear transformation from $V_{m-1}$ to $V_{n+m-1}$ . With respect to these standard bases for $V_{m-1}$ and $V_{n+m-1}$ , the $(n+m)\times m$ matrix

represents multiplication by $f$ . Similarly, multiplication by $g$ defines a linear transformation from $V_{n-1}$ to $V_{n+m-1}$ , represented by the $(n+m)\times n$ matrix

with respect to the standard bases for $V_{n-1}$ and $V_{n+m-1}$ . Now, let $V_{m-1}\oplus V_{n-1}$ be the vector space of pairs of polynomials $(u,v)$ where $degu\le m-1$ and $degv\le n-1$ . The vector space $V_{m-1}\oplus V_{n-1}$ has dimension $m+n$ , since

is a basis; this is essentially the standard basis for $V_{m-1}$ followed by the standard basis for $V_{n-1}$ . The function $T(u,v)=uf+vg$ defines a linear transformation from $V_{m-1}\oplus V_{n-1}$ to $V_{n+m-1}$ whose matrix with respect to the above basis on $V_{m-1}\oplus V_{n-1}$ and the standard basis on $V_{n+m-1}$ is the $(n+m)\times (n+m)$ matrix

which we call the Sylvester matrix . The determinant of the Sylvester matrix we call the resultant :

of $f$ and $g$ with respect to the variable $t$ . The resultant is a polynomial in the coefficients of $f$ and $g$ with integer coefficients.

The determinant of a matrix is non-zero precisely when the corresponding linear transformation is one-to-one (and equivalently, if and only if the linear transformation is onto). Thus, since we know that solutions to $uf+vg=h$ with $u\in V_{m-1}$ and $v\in V_{n-1}$ exist (and are unique) for all $h\in V_{n+m-1}$ precisely when $f$ and $g$ have no common factor, we have the following:

Theorem Suppose that $f,g\in k\left[t\right]$ are polynomials over a field $k$ of degrees $n>0$ and $m>0$ respectively. Then $\text{Res}(f,g,t)=0$ if and only if $f$ and $g$ have a common factor in $k\left[t\right]$ .

One thing we need to be careful of is that this theorem only applies when the degrees of $f$ and $g$ are actually $n$ and $m$ . If $a_n=b_m=0$ , applying the resultant as if $f$ and $g$ had degree $n$ and $m$ will always yield zero (Why?) even though $f$ and $g$ may not have a common factor.

Remark There's another way to define the resultant. If we assume that $f,g\in \mathbf{C}\left[t\right]$ are monic polynomials, then by the fundamental theorem of algebra, we can factor them over the complex numbers as

Then if we form the product

then it will certainly have the property that $R(f,g,t)=0$ if and only if $f$ and $g$ have a root (or equivalently, a factor) in common. Also, it's easy to see that permuting the ${\alpha }_j$ or permuting the ${\beta }_k$ has no effect on this product. It turns out that this means it's possible to rewrite $R(f,g,t)$ as a polynomial in the coefficients of $f$ and $g$ , The general statements is that a symmetric polynomial in the roots ${\alpha }_1,...,{\alpha }_n$ of a monic polynomial $f\left(t\right)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$ can be written as a polynomial in the elementary symmetric polynomials