Kollár's Example 11: the z-chart and the second blow-up #
[Kol07, Example 11] after the blow-up of the origin, on the z-chart: f₁ = x₁² + y₁³z₁ − z₁⁴
(Kollár writes x₁² + (y₁³ − z₁³)z₁), H₁ = (x₁), H₁' = (x₁ + y₁²z₁), E₁ = (z₁). Computed
exactly: I₁ = (f₁) has order exactly 2 at the origin, which is its only ℚ-point of order
≥ 2; MC(I₁) = D(I₁) = (x₁, y₁²z₁, y₁³ − 4z₁³, z₁⁴), so x₁, x₁ + y₁²z₁ ∈ MC(I₁); the
restrictions I₁|_{H₁} = (y₁³z₁ − z₁⁴) and I₁|_{H₁'} = (y₁⁴z₁² + y₁³z₁ − z₁⁴) have the origin as
their only point of order 2. The second point blow-up on the z-chart gives
f₂ = x₂² + (y₂³ − 1)z₂², Kollár's display, whose locus of order 2 is the line
x₂ = z₂ = 0 = E₂ ∩ H₂; the trace I₂|_{H₂} = ((y₂³ − 1)z₂²) has order 3 at the three points
y₂³ = 1 and order 2 at the origin.
The z-chart #
The z-chart of the blow-up of the origin: σ f = z₁² f₁ with f₁ Kollár's
x₁² + (y₁³ − z₁³)z₁, σ x = z₁x₁, σ(x + y²) = z₁(x₁ + y₁²z₁).
I₁ = (f₁) has order exactly 2 at the origin (∂²f₁/∂x₁² = 2).
The origin is the only ℚ-point of order ≥ 2: ∂f₁/∂x₁ = 2x₁, ∂f₁/∂y₁ = 3y₁²z₁,
∂f₁/∂z₁ = y₁³ − 4z₁³ vanish together only at 0.
MC(I₁) = D(f₁) = (x₁, y₁²z₁, y₁³ − 4z₁³, z₁⁴): the partials are 2x₁, 3y₁²z₁, y₁³ − 4z₁³,
and 3z₁⁴ = f₁ − x₁·x₁ − z₁(y₁³ − 4z₁³).
x₁ ∈ MC(I₁): the strict transform H₁ is of maximal contact.
x₁ + y₁²z₁ ∈ MC(I₁): the strict transform H₁' is of maximal contact.
I₁|_{H₁} = (y₁³z₁ − z₁⁴).
cosupp((I₁|_{H₁}, 2)) = {0}: y₁³z₁ − z₁⁴ ∈ 𝔪₀², and the partials 3y₁²z₁, y₁³ − 4z₁³
vanish together only at 0.
I₁|_{H₁'} = (y₁⁴z₁² + y₁³z₁ − z₁⁴) (x₁ ↦ −y₁²z₁).
cosupp((I₁|_{H₁'}, 2)) = {0}: at a ℚ-point of order ≥ 2, ∂/∂y₁ = y₁²z₁(4y₁ + 3) and
∂/∂z₁ = 2y₁⁴z₁ + y₁³ − 4z₁³; with z₁ = 0 this gives y₁ = 0; with z₁ ≠ 0 and y₁ = 0 it
gives z₁ = 0; with y₁z₁ = −3/4 the value becomes −(3/16)y₁² − z₁⁴ < 0.
The second blow-up on the z-chart #
The second blow-up: σ f₁ = z₂² f₂ with f₂ Kollár's x₂² + (y₂³ − 1)z₂², σ x₁ = z₂x₂,
σ(x₁ + y₁²z₁) = z₂(x₂ + y₂²z₂²).
The locus of order 2 contains the line x₂ = z₂ = 0: f₂ ∈ 𝔪_{(0,b,0)}² for every b.
The locus of order 2 lies in the line x₂ = z₂ = 0: ∂f₂/∂x₂ = 2x₂, ∂f₂/∂y₂ = 3y₂²z₂²,
∂f₂/∂z₂ = 2(y₂³ − 1)z₂ force x₂ = z₂ = 0 (a point with z₂ ≠ 0 would need y₂ = 0 and
y₂³ = 1).
I₂|_{H₂} = ((y₂³ − 1)z₂²).
At the ℚ-rational one of Kollár's three points y₂³ = 1: (y₂³ − 1)z₂² has order exactly
3 at (1, 0), since it is (y − 1)(y² + y + 1)z² and ∂³/∂y∂z² = 6y² = 6 ≠ 0.
Kollár's three points y₂³ = 1, over any field: for a field of characteristic zero and any
ω with ω³ = 1, (y³ − 1)z² has order exactly 3 at (ω, 0).
Away from the three points the trace has order 2: (y₂³ − 1)z₂² has order exactly 2 at
0.