Resultant$(f,g)$ says when there exist $\phi,\psi$ such that $\psi f + \phi g = 0$. How do I actually find them?

Question

If $f$ and $g \in k[X]$ are two polynomials such that $\textrm{Res }(f,g)=0$ how do I find $\phi$ and $\psi$ with $\deg \phi < \deg f$ and $\deg \psi < \deg g$ such that $$\psi f +\phi g =0$$

Answer 1

Recall what the definition of the resultant of $f$ and $g$ is. That is, if $f = a_nx^n + \cdots a_1x +a_0$ and $g = b_mx^m + \cdots b_1x + b_0$ then $\operatorname{Res}(f,g)$ is the determinant of the Sylvester matrix $\operatorname{Syl}(f,g)$, where $\operatorname{Syl}(f,g)$ is the $(m+n)\times(m+n)$ matrix, $$\operatorname{Syl}(f,g) = \begin{bmatrix} a_0 & \ & \ & 0 & b_0 & \ & \ & 0 \\ a_1 & a_0 & \ & \ & b_1 & b_0 & \ & \ \\ a_2 & a_1 & \ddots & \ & b_2 & b_1 & \ddots \ \\ \vdots & \ & \ddots &a_0 & \vdots & \ & \ddots & b_0\\ \ & \vdots & \ & a_1 & \ & \vdots & \ & b_1 \\ a_n & \ & \ & \ & b_m & \ & \ & \ \\ \ & a_n & \ & \vdots & \ &b_m & \ & \vdots \\ \ & \ & \ddots & \ & \ & \ & \ddots & \ \\ 0 & \ & \ & a_n & 0 & \ & \ & b_m \end{bmatrix}$$ (where there are $m$ many "$a_i$" columns and $n$ many "$b_j$" columns)

Note that for any $m+n$ column vector $v:=(c_0, \ldots, c_{m-1}, d_0, \ldots, d_{n-1})^T$, $\operatorname{Syl}(f,g)v$ corresponds to the polynomial $(c_{m-1}x^{m-1} + \cdots + c_1x + c_0)f + (d_{n-1}x^{n-1} + \cdots + d_1x + d_0)g$. So, if $\operatorname{Res}(f,g)$ is zero, then by row reduction you can concretely compute $v$ such that $\operatorname{Syl}(f,g)v = 0$, hence giving you the $\phi$ and $\psi$ as desired.

Answer 2

Or shorter, since the resultant is zero, the polynomials $f$ and $g$ have a common factor $h$ which can be computed using the euclidean algorithm.

Then $f=h\phi$ and $g=h\psi$ and $f\psi-g\phi=0$.