Mathematics 392 – Seminar in Computational Commutative Algebra Comment on Additional Problem, Problem Set 9 November 6, 2006

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.

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)):

g = q_{1}f + r_{1}
f = q_{2}r_{1} + r_{2}
⋮
r_{k−2} = q_{k}r_{k−1} + r_{k}

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

Res(f, g, x) = a_{0}^{deg(g)−deg(r_{1})}Res(f, r_{1}, x)
= a_{0}^{deg(g)−deg(r_{1})}(−1)^{deg(f) deg(r_{1})}Res(r_{1}, f, x)
= a_{0}^{deg(g)−deg(r_{1})}(−1)^{deg(f) deg(r_{1})} (a_{0}^{(1)})^{deg(f)−deg(r_{2})} Res(r_{1}, r_{2}, x)

= a_{0}^{deg(g)−deg(r_{1})}(−1)^{deg(f) deg(r_{1})} (a_{0}^{(1)})^{deg(f)−deg(r_{2})} (−1)^{deg(r_{1}) deg(r_{2})}Res(r_{2}, r_{1}, x)

= ⋯

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

\begin{aligned}g & = (x + 5)f + (−12x − 13) ⇒ r_{1} = −12x − 13 \\ f & = (−1/12x − 11/144)r_{1} + 289/144 ⇒ r_{2} = 298/144 \\ r_{1} & = q_{3}r_{2} + 0 \end{aligned}

Then

Res(f, g, x) = (1)^{(3−1)}Res(f, r_{1}, x) (part B)

= (1)^{(3−1)}(−1)^{2·1}Res(r_{1}, f, x) (part C)

= (1)^{(3−1)}(−1)^{2·1}(−12)^{(2−0)}Res(r_{1}, r_{2}, x) (part B)

= (1)^{(3−1)}(−1)^{2·1}(−12)^{(2−0)} · (289/144)^{(1−0)} (part A) = 289 It is easy to check that this is exactly the determinant of the Sylvester matrix for f, g:

det ( \begin{matrix}1 & 0 & 0 & 1 & 0 \\ 2 & 1 & 0 & 7 & 1 \\ 3 & 2 & 1 & 1 & 7 \\ 0 & 3 & 2 & 2 & 1 \\ 0 & 0 & 3 & 0 & 2 \end{matrix} ) = 289