Kollár's Example 11: the two hypersurfaces of maximal contact and the first blow-up #
[Kol07, Example 11]: S = (x² + y³ − z⁶ = 0) ⊂ 𝔸³, m = 2, H = (x), and a second hypersurface
of maximal contact H' = (x + y²). Computed exactly in ℚ[x, y, z] (MC(I) = D(I) = (x, y², z⁵)
and cosupp(I, 2) = {0} are derivative_example11 and cosupport_example11 of
HironakaExamples/Balanced/Example11.lean):
x,x + y² ∈ D(I); the restrictionsI|_H = (z⁶ − y³)(Kollár's trace of multiplicity3) andI|_{H'} = (z⁶ − y⁴ − y³)haveMC = (y², z⁵), and the origin is their only point of order2;- the blow-up of the origin: the
x- andy-charts have no point of order2. Thez-chart, which carries the whole locus of order2, and the second blow-up are inHironakaExamples/MaximalContact/Example11Stages.lean.
Tools: orders at a ℚ-point p of 𝔸² or 𝔸³ are memberships in powers of 𝔪_p, and from
g ∈ 𝔪_p^{k+1} the partial derivatives of g of order ≤ k vanish at p
(vanish_of_mem_m3_sq, vanish_of_mem_m3_cube and the 𝔸² versions); the resulting rational
equations are solved by cases.
Members of 𝔪_p = (X 0 − a, X 1 − b, X 2 − c) vanish at p.
Order ≥ 2 at p ∈ 𝔸³: the value and the first partials vanish at p.
Order ≥ 3 at p ∈ 𝔸³: the value, the first and the second partials vanish at p.
Members of 𝔪_p = (X 0 − b, X 1 − c) vanish at p ∈ 𝔸².
Order ≥ 2 at p ∈ 𝔸²: the value and the first partials vanish at p.
Order ≥ 3 at p ∈ 𝔸²: the value, the first and the second partials vanish at p.
Order ≥ 3 at p ∈ 𝔸² excluded by a nonzero second partial at p, the origin version:
g ∈ 𝔪₀³ kills every second partial at 0.
The same for 𝔸³.
g ∈ 𝔪₀⁴ in 𝔸² kills every third partial derivative of g at 0.
X 0 ∈ (X 0, X 1) in ℚ[y, z], the coordinate ring of a hypersurface H ⊂ 𝔸³.
The two hypersurfaces and the restrictions #
x ∈ D(I), from ∂f/∂x = 2x: H = (x) is a hypersurface of maximal contact.
x + y² ∈ D(I), from ∂f/∂y = 3y²: H' = (x + y²) is a hypersurface of maximal contact.
I|_H = (z⁶ − y³).
MC(I|_H) = D(z⁶ − y³) = (y², z⁵).
Kollár's "the trace S ∩ H = (y³ − z⁶ = 0) has multiplicity 3": order exactly 3 at the
origin (∂³/∂y³ = −6).
cosupp((I|_H, 2)) = {0}: z⁶ − y³ ∈ 𝔪₀², and a ℚ-point of order ≥ 2 has −3y² = 0,
6z⁵ = 0.
I|_{H'} = (z⁶ − y⁴ − y³) (x ↦ −y²).
MC(I|_{H'}) = D(z⁶ − y⁴ − y³) = (y², z⁵): the partials are −y²(4y + 3) and 6z⁵, and
y³ = (4y + 3)y³ − 4y⁴ ∈ ⟨y²(4y + 3), z⁶ − y⁴ − y³, …⟩ shows y² ∈ D.
cosupp((I|_{H'}, 2)) = {0}: z⁶ − y⁴ − y³ ∈ 𝔪₀²; at a ℚ-point of order ≥ 2 the partial
6z⁵ gives z = 0, the partial −y²(4y + 3) gives y ∈ {0, −3/4}, and the value −y³(y + 1)
excludes −3/4.
The blow-up of the origin: the x- and y-charts #
σ f = x₁²(1 + x₁y₁³ − x₁⁴z₁⁶) on the x-chart.
No point of order 2 on the x-chart: on x₁ = 0 the transform equals 1; off it,
∂/∂y₁ = 3x₁y₁² gives y₁ = 0, then ∂/∂x₁ = y₁³ − 4x₁³z₁⁶ gives z₁ = 0, and the value is 1
again.
σ f = y₁²(x₁² + y₁ − y₁⁴z₁⁶) on the y-chart.
No point of order 2 on the y-chart: ∂/∂y₁ = 1 − 4y₁³z₁⁶ and ∂/∂z₁ = −6y₁⁴z₁⁵ cannot
both vanish.