\magnification=\magstep1
\overfullrule=0pt
\def\Res{{\rm Res}}
\centerline{Mathematics 392 -- Seminar in Computational Commutative Algebra}
\centerline{Comment on Additional Problem, Problem Set 9}
\centerline{November 6, 2006}
\bigskip
\noindent
There was quite a bit of confusion about how the earlier
parts of the Additional Problem on this assignment would give
a way to compute resultants.  Here is the idea for part D and
an example that is somewhat more representative than part E.
\bigskip
\noindent
D)  Suppose we perform the following divisions in the 
course of computing $\gcd(f,g)$ by the Euclidean algorithm
(starting with $\deg(g) > \deg(f)$):
$$\eqalign{g &= q_1 f + r_1\cr
           f &= q_2 r_1 + r_2\cr
             &\,\vdots \cr
           r_{k-2} &= q_k r_{k-1} + r_k\cr}$$
where $r_k$ is the last nonzero remainder.  
Say the leading coefficients of $f,r_1,r_2$, etc.
are $a_0, a_0^{(1)}, a_0^{(2)}$, respectively.
Then applying parts B (division) and C (swap to put the 
polynomial of smaller degree first) repeatedly we have
$$\eqalign{
\Res(f,g,x) &= a_0^{\deg(g)-\deg(r_1)} \Res(f,r_1,x)\cr
&=a_0^{\deg(g)-\deg(r_1)} (-1)^{\deg(f)\deg(r_1)} \Res(r_1,f,x)\cr
&=a_0^{\deg(g)-\deg(r_1)} (-1)^{\deg(f)\deg(r_1)} \left(a_0^{(1)}\right)^{\deg(f)-\deg(r_2)}
\Res(r_1,r_2,x)\cr
&=a_0^{\deg(g)-\deg(r_1)} (-1)^{\deg(f)\deg(r_1)} \left(a_0^{(1)}\right)^{\deg(f)-\deg(r_2)}
(-1)^{\deg(r_1)\deg(r_2)}\Res(r_2,r_1,x)\cr
&= \cdots\cr}$$
The process terminates either when the remainder $r_k$ is a non-zero constant (and then we 
use part A), or else if we get $r_k$ of positive degree and the next remainder
is zero (in which case $\Res(f,g,x) = 0$ since $f,g$ have a common factor
of positive degree -- the polynomial $\gcd(f,g)$).  Here is an example showing 
how this works in practice.

Let $f = x^2 + 2x + 3$ and $g = x^3 + 7x^2 + x + 2$.  Then the steps of the Euclidean
algorithm look like 
$$\eqalign{g &= (x + 5)f + (-12 x - 13)\quad  \Rightarrow r_1 = -12x - 13\cr
           f &= (-1/12 x - 11/144) r_1 + 289/144 \quad \Rightarrow r_2 = 298/144\cr
           r_1 &= q_3 r_2 + 0\cr}$$
Then 
$$\eqalign{
\Res(f,g,x) &= (1)^{(3-1)} \Res(f,r_1,x) \quad {\rm (part\ B)}\cr
            &= (1)^{(3-1)} (-1)^{2\cdot 1} \Res(r_1,f,x) \quad {\rm (part\ C)}\cr
            &= (1)^{(3-1)} (-1)^{2\cdot 1} (-12)^{(2-0)} \Res(r_1,r_2,x) \quad {\rm (part\ B)}\cr
            &= (1)^{(3-1)} (-1)^{2\cdot 1} (-12)^{(2-0)} \cdot (289/144)^{(1-0)} \quad {\rm (part\  A)}\cr
            &= 289\cr}$$
It is easy to check that this is exactly the determinant of the Sylvester matrix
for $f,g$:
$$\det\pmatrix{1&0&0&1&0\cr
               2&1&0&7&1\cr
               3&2&1&1&7\cr
               0&3&2&2&1\cr
               0&0&3&0&2\cr} = 289$$
\bye
