RefacteTwo quadratic equations can hide a cubic obstruction. The useful move is to let the first equation choose a coordinate, then eliminate that coordinate from the second. The order at which a motion fails becomes a coefficient we can calculate.
1. Start with the equations, not their degrees
Work over a field k, such as the rationals, and study the origin in
y − x² = 0, xy = 0.
The first equation says y = x². The second says that their product must vanish. Both displayed equations are quadratic. That observation does not tell us which order of motion will first fail.
To test a motion, substitute polynomials with no constant term into the equations and keep only a chosen number of powers of a symbol ε. In k[ε]/(εN+1), equality means equality of coefficients through degree N. This is a finite calculation; no convergence of a series is involved.
2. Let the solvable equation choose the coordinate
The first equation determines y once x is known. Keep x, and look for a new second coordinate that is zero precisely when this equation holds. A translation z = y + h(x) is a useful form to try: it has the polynomial inverse y = z − h(x), whatever polynomial h we choose. On the graph y = x², the requirement z = 0 reads x² + h(x) = 0. It determines the translation:
z = y − x², y = z + x².
The choice of a translation was ours; its correction h(x) = −x² was fixed by the job we gave it. The displayed inverse proves that no information was lost. The first equation is now z = 0. Substitution turns the other one into xz + x³ = 0. To remove its dependence on the already constrained coordinate, subtract x times the first equation from the second:
(z, xz + x³) ↔ (z, x³).
This operation is reversible: add x times the first equation to recover the former second equation. Thus the original solution problem is exactly z = 0, x³ = 0, over every test algebra. The cubic was produced by carrying out the elimination.
3. Watch the first two tests pass
Start with x = a₁ε, y = b₁ε and impose ε² = 0. The first equation forces b₁ = 0; the product equation imposes nothing further. Every value of a₁ passes this first-order test.
Keep one more order and allow every possible correction:
x = a₁ε + a₂ε², y = b₂ε² modulo ε³.
The first equation forces b₂ = a₁². The second still vanishes, since xy starts at degree three. The coefficient a₂ is free. In particular, the direction a₁ = 1 continues through second order as x = ε, y = ε².
4. Keep the term that can no longer be repaired
Now retain degree three and permit arbitrary new coefficients:
x = a₁ε + a₂ε² + a₃ε³, y = a₁²ε² + 2a₁a₂ε³ modulo ε⁴.
The formula for y is forced by y = x². Multiplication then gives
Neither a₂ nor a₃ changes this coefficient. There is no additional free y-coefficient to use: the first equation has already determined it. Since k is a field, a nonzero a₁ has nonzero cube. Every nonzero first-order direction therefore passes the second-order test and fails the third-order test.
For the earlier example, x = ε, y = ε² gives xy = ε³. The failure has a location and a reason: a product of three first-order copies survives, while later corrections arrive too late to affect it.
5. Derive a procedure for any truncation
Write x = ∑aᵢεⁱ and y = ∑bᵢεⁱ, with indices starting at one and ending at the desired order. Comparing coefficients in the first equation gives
bₙ = ∑ aᵢaⱼ over i + j = n, rₙ = ∑ aᵢaⱼaₗ over i + j + l = n.
Choose the aᵢ, compute each bₙ by the first sum, and test the second sums through the retained order. These finite convolutions are precisely the coefficients of y = x² and xy = x³. They produce all solutions over the chosen truncated algebra, including motions whose first coefficient is zero.
Over the field k, we can read all these tests at once. If the first nonzero term of x is a_dεd, before truncation the cube has leading term a_d³ε3d. This term survives exactly when 3d ≤ N. Hence a nonzero truncated motion satisfies the equations through degree N exactly when 3d > N, with y = x². Equivalently, the first ⌊N/3⌋ coefficients of x must vanish; every later coefficient is free. The zero motion is included by taking all coefficients zero.
Further context
The earlier extended treatment also polarizes the cubic and expresses it using higher operations.