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 Incoming links: Degree drop in a specialized resultant

Eliminating a variable with a resultant 2026-10-18

Consider the original pair of polynomial equations

F (x, y) = y - x^{2} = 0, G (x, y) = y + x^{2} - 2 = 0.

Treat them as polynomials in

y

with coefficients in

R [x]

. Their Sylvester matrix and resultant are

S_{y} (F, G) = (11 - x^{2} x^{2} - 2), R (x) = Res_{y} (F, G) = 2 (x^{2} - 1) .

Every common solution must therefore have

x = - 1

or

x = 1

. Back-substitution into

F = 0

gives

y = x^{2} = 1

, and both pairs also satisfy

G = 0

. The complete real solution set is

{(- 1, 1), (1, 1)} .

Here elimination and direct substitution agree: subtracting

F

from

G

gives

2 x^{2} - 2

, exactly the same condition.

Figure 1.
Polynomial elimination keeps the possible first coordinates of intersections
. Original parabola example. Left: both equations vanish at the two marked intersections; dashed lines retain their $x$ -coordinates on an auxiliary horizontal reference line at $y = - 1/2$ . Right: the resultant has zeros at those coordinates. Substitution recovers $y = 1$ and verifies each full solution pair. Details.

In general, first solve the one-variable resultant equation, then recover and check the other coordinate in both original polynomial equations. A real coordinate candidate can correspond only to nonreal values of the eliminated variable. Also, vanishing leading coefficients require the separate check in degree drop in a specialized resultant. The general elimination-and-back-substitution method is described in the lecture's §3.1.^[ref]

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