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HironakaExamples.Balanced.Example11

Kollár's Example 11: the tuned ideal of I = (x² + y³ − z⁶) #

[Kol07, Example 11] resolves S := (x² + y³ − z⁶ = 0) ⊂ 𝔸³ through the hypersurface H := (x = 0) and the trace S ∩ H = (y³ − z⁶ = 0) ⊂ 𝔸², which has multiplicity 3 although "we came from a multiplicity 2 situation". The tuning step of the order reduction on this example, computed exactly in ℚ[x, y, z]:

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Kollár's I = (x² + y³ − z⁶).

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    D(I) = (x, y², z⁵).

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      The tuned ideal W(I) = (x², xy², xz⁵, y⁴, y²z⁵, y³z⁴, z⁶ − y³).

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        The maximal ideal of the origin, 𝔪₀ = (x, y, z).

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          D(x² + y³ − z⁶) = (x, y², z⁵).

          z¹⁰ = z⁴(z⁶ − y³) + y³z⁴ ∈ W(I).

          (x, y², z⁵)² ⊆ W(I).

          W(I) = I + D(I)² = (x², xy², xz⁵, y⁴, y²z⁵, y³z⁴, z⁶ − y³).

          x² ∉ 𝔪₀³: differentiate twice in x and evaluate at the origin.

          I = (x² + y³ − z⁶) has order exactly 2 at the origin.

          W(I) ⊆ 𝔪₀²: every generator has degree ≥ 2.

          The tuned ideal W(I) has order exactly 2 at the origin.

          theorem Hironaka.Examples.Example11.cosupport_example11 (a b c : ℚ) (h : ∀ g ∈ Ideal.derivative ℚ I11, (MvPolynomial.eval ![a, b, c]) g = 0) :
          a = 0 ∧ b = 0 ∧ c = 0

          cosupp(I, 2) = V(D(I)) has the origin as its only ℚ-point.

          x² ∉ I: at (1, 0, 1), a zero of x² + y³ − z⁶, x² evaluates to 1.

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          The common restriction (y⁴, y²z⁵, y³z⁴, z⁶ − y³) in ℚ[y, z] (y = X 0, z = X 1).

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            W(I)|_H for H = (x = 0) is (y⁴, y²z⁵, y³z⁴, z⁶ − y³).

            W(I)|_{H'} for H' = (x + y² = 0) is the same ideal (y⁴, y²z⁵, y³z⁴, z⁶ − y³): x ↦ −y² sends x² ↦ y⁴, xy² ↦ −y⁴, xz⁵ ↦ −y²z⁵.

            The order-2 locus of (y⁴, y²z⁵, y³z⁴, z⁶ − y³) has the origin as its only ℚ-point: −3y² and 6z⁵ are partial derivatives of z⁶ − y³.