FormalConjectures/Wikipedia/ResolutionOfSingularities.lean

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7 declarations, 4 contain sorry, 0 declared as axioms

Resolution of singularities

A variety $X$ over a field $k$ admits a resolution of singularities if there is a smooth $k$-variety $Y$ and a proper birational morphism $Y \to X$. Hironaka proved that every variety over a field of characteristic zero admits one. In positive characteristic this is known over a perfect field in dimension at most three, and is open from dimension four on. Perfectness of $k$ cannot be dropped from the statement in this form: over an imperfect field it fails already in dimension zero, see exists_not_hasResolution_of_not_perfectField below. Over an arbitrary field one asks instead that $Y$ be regular.

References:

Imports (1)
  1. def AlgebraicGeometry.Scheme.HasResolution Documented declaration: HasResolution line 67 source

    Source documentation

    A scheme X over a base S with structure morphism sX : X ⟶ S admits a resolution of singularities if there is an integral scheme Y, smooth over S, together with a proper morphism f : Y ⟶ X that is birational over X, i.e. that restricts to an isomorphism between dense open subschemes of Y and of X.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: AlgebraicGeometry.Scheme, Quiver.Hom

    Proof uses: Exists, And, AlgebraicGeometry.IsIntegral, AlgebraicGeometry.IsProper, AlgebraicGeometry.Smooth, CategoryTheory.CategoryStruct.comp, AlgebraicGeometry.Scheme.BirationalOver, CategoryTheory.CategoryStruct.id

    7 instances, coercions and auxiliary constants left out.

    def Scheme.HasResolution {S X : Scheme.{u}} (sX : X ⟶ S) : Prop :=
    …
  2. theorem AlgebraicGeometry.Scheme.hasResolution_of_smooth A scheme that is already smooth over the base is its own resolution line 73 source

    Source documentation

    A scheme that is already smooth over the base is its own resolution.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: AlgebraicGeometry.Scheme, Quiver.Hom, AlgebraicGeometry.IsIntegral, AlgebraicGeometry.Smooth, AlgebraicGeometry.Scheme.HasResolution

    Proof uses: Exists.intro, Exists, And, AlgebraicGeometry.IsProper, CategoryTheory.CategoryStruct.comp, AlgebraicGeometry.Scheme.BirationalOver, CategoryTheory.CategoryStruct.id, And.intro, inferInstance, Eq.mpr, id, Eq, congrArg, CategoryTheory.Category.id_comp, AlgebraicGeometry.Scheme.BirationalOver.refl

    12 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem Scheme.hasResolution_of_smooth {S X : Scheme.{u}} (sX : X ⟶ S) [IsIntegral X]
        [Smooth sX] : Scheme.HasResolution sX :=
    …

    Mentions: AlgebraicGeometry.Scheme.HasResolution

  3. theorem AlgebraicGeometry.Scheme.surjective_of_birationalOver_id A proper birational morphism onto a scheme is surjective line 83 source

    Source documentation

    A proper birational morphism onto a scheme is surjective. In particular a resolution of singularities is onto, which is what rules out the inclusion of the smooth locus of X as a trivial solution.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: AlgebraicGeometry.Scheme, Quiver.Hom, AlgebraicGeometry.IsProper, AlgebraicGeometry.Scheme.BirationalOver, CategoryTheory.CategoryStruct.id, Function.Surjective, CommRingCat, AlgebraicGeometry.SheafedSpace.toPresheafedSpace, AlgebraicGeometry.LocallyRingedSpace.toSheafedSpace, AlgebraicGeometry.Scheme.toLocallyRingedSpace, TopCat, ContinuousMap, CategoryTheory.ConcreteCategory.hom, AlgebraicGeometry.PresheafedSpace.Hom.base, AlgebraicGeometry.LocallyRingedSpace.Hom.toHom, AlgebraicGeometry.Scheme.Hom.toLRSHom'

    Proof uses: AlgebraicGeometry.Scheme.PartialIso, AlgebraicGeometry.Scheme.PartialIso.IsOver, AlgebraicGeometry.Scheme.PartialIso.source, Eq, AlgebraicGeometry.Scheme.Opens.ι, AlgebraicGeometry.Scheme.PartialIso.target, CategoryTheory.Iso.hom, AlgebraicGeometry.Scheme.PartialIso.iso, id, Eq.mp, CategoryTheory.CategoryStruct.comp, TopCat.Hom.hom, congrArg, congrFun', ContinuousMap.comp, AlgebraicGeometry.PresheafedSpace.Hom, AlgebraicGeometry.LocallyRingedSpace.Hom, CategoryTheory.Category.comp_id, Eq.symm, Dense and 24 more

    37 instances, coercions and auxiliary constants left out.

    Tactic trace: 11 steps

    Open to load the checked before-and-after goal states.

    theorem Scheme.surjective_of_birationalOver_id {X Y : Scheme.{u}} (f : Y ⟶ X) [IsProper f]
        (hf : Scheme.BirationalOver f (𝟙 X)) : Function.Surjective f.base := by
    …
  4. theorem AlgebraicGeometry.resolution_of_singularities Resolution of singularities in positive characteristic sorry line 105 source

    Source documentation

    Resolution of singularities in positive characteristic. Let $k$ be a perfect field of characteristic $p > 0$ and let $X$ be an integral scheme that is separated and of finite type over $k$. Then there is an integral scheme $Y$ that is smooth over $k$ together with a proper birational morphism $Y \to X$.

    This is open from dimension four on; see [Hau2010]. Perfectness of $k$ is needed for the conclusion as stated; see exists_not_hasResolution_of_not_perfectField.

    depends on sorry what this means

    Statement uses: Field, PerfectField, Nat, Fact, Nat.Prime, CharP, AlgebraicGeometry.Scheme, Quiver.Hom, AlgebraicGeometry.Spec, CommRingCat.of, AlgebraicGeometry.IsIntegral, AlgebraicGeometry.LocallyOfFiniteType, AlgebraicGeometry.QuasiCompact, AlgebraicGeometry.IsSeparated, AlgebraicGeometry.Scheme.HasResolution

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous, OfNat.ofNat

    17 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem resolution_of_singularities (k : Type u) [Field k] [PerfectField k] (p : ℕ)
        [Fact p.Prime] [CharP k p] {X : Scheme.{u}} (sX : X ⟶ Spec (.of k)) [IsIntegral X]
        [LocallyOfFiniteType sX] [QuasiCompact sX] [IsSeparated sX] :
        Scheme.HasResolution sX := by
    …

    Mentions: AlgebraicGeometry.Scheme.HasResolution

  5. theorem AlgebraicGeometry.resolution_of_singularities_of_charZero Hironaka's theorem sorry line 117 source

    Source documentation

    Hironaka's theorem. Every integral scheme that is separated and of finite type over a field of characteristic zero admits a resolution of singularities. This is the main theorem of [Hir1964]; see [Kol2007], Theorem 36, for a statement in the generality used here.

    depends on sorry what this means

    Statement uses: Field, CharZero, AlgebraicGeometry.Scheme, Quiver.Hom, AlgebraicGeometry.Spec, CommRingCat.of, AlgebraicGeometry.IsIntegral, AlgebraicGeometry.LocallyOfFiniteType, AlgebraicGeometry.QuasiCompact, AlgebraicGeometry.IsSeparated, AlgebraicGeometry.Scheme.HasResolution

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous, OfNat.ofNat, Nat

    17 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem resolution_of_singularities_of_charZero (k : Type u) [Field k] [CharZero k]
        {X : Scheme.{u}} (sX : X ⟶ Spec (.of k)) [IsIntegral X] [LocallyOfFiniteType sX]
        [QuasiCompact sX] [IsSeparated sX] :
        Scheme.HasResolution sX := by
    …

    Mentions: AlgebraicGeometry.Scheme.HasResolution

  6. theorem AlgebraicGeometry.resolution_of_singularities_of_topologicalKrullDim_le_three Resolution of singularities holds in dimension at most three over any perfect field sorry line 134 source

    Source documentation

    Resolution of singularities holds in dimension at most three over any perfect field. In characteristic zero this is [Hir1964]. In positive characteristic, dimension at most two is [Lip1978] and dimension three is due to Cossart and Piltant: [CP2008] and [CP2009] prove it for quasi-projective varieties over a field that is differentially finite over a perfect subfield, and [CP2019], Theorem 1.1, removes both restrictions, resolving every reduced separated Noetherian quasi-excellent scheme of dimension at most three in any characteristic. That theorem gives an everywhere regular source, which over a perfect field is the same as a smooth one. The problem is open in positive characteristic from dimension four on.

    depends on sorry what this means

    Statement uses: Field, PerfectField, AlgebraicGeometry.Scheme, Quiver.Hom, AlgebraicGeometry.Spec, CommRingCat.of, AlgebraicGeometry.IsIntegral, AlgebraicGeometry.LocallyOfFiniteType, AlgebraicGeometry.QuasiCompact, AlgebraicGeometry.IsSeparated, LE.le, WithBot, ENat, topologicalKrullDim, CommRingCat, AlgebraicGeometry.SheafedSpace.toPresheafedSpace, AlgebraicGeometry.LocallyRingedSpace.toSheafedSpace, AlgebraicGeometry.Scheme.toLocallyRingedSpace, OfNat.ofNat, Nat and 1 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    34 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem resolution_of_singularities_of_topologicalKrullDim_le_three (k : Type u) [Field k]
        [PerfectField k] {X : Scheme.{u}} (sX : X ⟶ Spec (.of k)) [IsIntegral X]
        [LocallyOfFiniteType sX] [QuasiCompact sX] [IsSeparated sX]
        (hX : topologicalKrullDim X ≤ 3) :
        Scheme.HasResolution sX := by
    …

    Mentions: AlgebraicGeometry.Scheme.HasResolution

  7. theorem AlgebraicGeometry.exists_not_hasResolution_of_not_perfectField Resolution of singularities fails over imperfect fields sorry line 156 source

    Source documentation

    Resolution of singularities fails over imperfect fields. If the field $k$ is not perfect, then there is an integral scheme $X$ of dimension zero that is separated and of finite type over $k$ and has no resolution of singularities: no integral scheme smooth over $k$ admits a proper birational morphism to $X$.

    A witness: $k$ has characteristic $p > 0$ and there is some $a \in k \setminus k^p$. Take $X = \operatorname{Spec} k(a^{1/p})$. Every integral scheme birational to $X$ has function field $k(a^{1/p})$, which is not separable over $k$, whereas the function field of an integral scheme smooth over $k$ is separable over $k$.

    So the hypothesis PerfectField k cannot be dropped from resolution_of_singularities or from resolution_of_singularities_of_topologicalKrullDim_le_three.

    depends on sorry what this means

    Statement uses: Field, Not, PerfectField, Exists, AlgebraicGeometry.Scheme, Quiver.Hom, AlgebraicGeometry.Spec, CommRingCat.of, And, AlgebraicGeometry.IsIntegral, AlgebraicGeometry.LocallyOfFiniteType, AlgebraicGeometry.QuasiCompact, AlgebraicGeometry.IsSeparated, Eq, WithBot, ENat, topologicalKrullDim, CommRingCat, AlgebraicGeometry.SheafedSpace.toPresheafedSpace, AlgebraicGeometry.LocallyRingedSpace.toSheafedSpace and 3 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous, Nat

    29 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem exists_not_hasResolution_of_not_perfectField (k : Type u) [Field k]
        (hk : ¬ PerfectField k) :
        ∃ (X : Scheme.{u}) (sX : X ⟶ Spec (.of k)), IsIntegral X ∧ LocallyOfFiniteType sX ∧
          QuasiCompact sX ∧ IsSeparated sX ∧ topologicalKrullDim X = 0 ∧
          ¬ Scheme.HasResolution sX := by
    …

    Mentions: AlgebraicGeometry.Scheme.HasResolution