Points with singular reduction at the symbols I₀*, IV* and II*

Let v be a valuation on a field F, with valuation ring O, maximal ideal 𝔪 = (ϖ) and residue field k, and let W be a Weierstrass curve over F with an integral model W_O over O. The points with nonsingular reduction form the subgroup E₀(F) of W(F) (WeierstrassCurve.Affine.nonsingularReduction). When F is complete for a discrete valuation with perfect residue field and the equation is minimal, the index [E(F) : E₀(F)] is the local index c_v that Tate's algorithm computes (Silverman, Advanced Topics, IV.9, Table 4.1). This file shows that E₀(F) = W(F), so that the index is one, at three of the symbols that Tate's algorithm reaches from the normal form ϖ ∣ a₁, a₂, ϖ² ∣ a₃, a₄, ϖ³ ∣ a₆ of its Step 6:

• at I₀*, when the residue cubic T³ + (a₂/ϖ) T² + (a₄/ϖ²) T + a₆/ϖ³ has no root in k;
• at IV*, when the residue quadratic T² + (a₃/ϖ²) T − a₆/ϖ⁴ has no root in k;
• at II*, always.

These are the cases in which Table 4.1 reports c_v = 1 at these symbols. In general Tate's algorithm reports c_v = 1 + #{roots of the cubic in k} at I₀* and c_v = 3 at IV* when the quadratic has a root in k.

The reduction of W_O is singular at the origin, so a point with singular reduction is an integral solution (x, y) with x, y ∈ 𝔪 (WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_forall_mem_maximalIdeal_not_equation). Comparing powers of ϖ in the Weierstrass equation rules such solutions out. At I₀*, writing x = ϖ X and y = ϖ Y, the equation divided by ϖ² forces ϖ ∣ Y, and then divided by ϖ³ forces X to reduce to a root of the residue cubic. From the normal form of Step 8, where ϖ ∣ a₁, ϖ² ∣ a₂, a₃, ϖ³ ∣ a₄ and ϖ⁴ ∣ a₆, the same comparison gives x, y ∈ 𝔪², and y / ϖ² reduces to a root of the residue quadratic. At II* that quadratic is T², so y ∈ 𝔪³, and the equation then forces ϖ⁶ ∣ a₆, which the stopping test for II* excludes.

Main results

• WeierstrassCurve.sq_dvd_and_dvd_cubic_of_equation: over a domain with a prime ϖ, on the normal form of Step 6 of Tate's algorithm, a solution (ϖ X, y) has ϖ² ∣ y, and X is a root of the residue cubic modulo ϖ.
• WeierstrassCurve.pow_six_dvd_a₆_of_equation: over a domain with a prime ϖ, a solution with both coordinates divisible by ϖ on the normal form of Step 10 of Tate's algorithm gives ϖ⁶ ∣ a₆. On the normal form of Step 8 such a solution gives a root of the residue quadratic modulo ϖ (WeierstrassCurve.exists_eq_sq_mul_of_equation), and on the normal form of Tate's algorithm at III* it has ϖ² ∣ x and ϖ³ ∣ y (WeierstrassCurve.sq_dvd_and_pow_three_dvd_of_equation).
• WeierstrassCurve.dvd_quadratic_of_equation_odd: on the normal form of an odd stage k of the Iₙ* subprocedure of Step 7, a solution (ϖ^(k+2) X, ϖ^(k+2) Y) makes Y a root of the residue quadratic modulo ϖ. WeierstrassCurve.dvd_quadratic_of_equation_even is the analogue at an even stage, where X is a root of the residue quadratic modulo ϖ.
• WeierstrassCurve.dvd_and_dvd_add_of_equation, WeierstrassCurve.dvd_of_equation_sq_mul, WeierstrassCurve.dvd_and_dvd_of_equation_pow and WeierstrassCurve.pow_dvd_and_pow_dvd_of_equation: coordinate divisibility on the normal forms of the Iₙ* subprocedure over a domain with a prime uniformiser.
• WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal: at an additive symbol, if no integral solution has both coordinates divisible by ϖ, then E₀(F) = W(F).
• WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal_IStar_zero: at I₀*, if the residue cubic has no root in k, then E₀(F) = W(F).
• WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal_IVStar: at IV*, if the residue quadratic has no root in k, then E₀(F) = W(F).
• WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal_IIStar: at II*, E₀(F) = W(F).

References

• J. Silverman, The Arithmetic of Elliptic Curves, VII.6.
• J. H. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, GTM 151, IV.9.
• J. Tate, Algorithm for determining the type of a singular fibre in an elliptic pencil, in Modular Functions of One Variable IV, LNM 476 (1975), 33–52.

Solutions with coordinates divisible by ϖ

theorem WeierstrassCurve.dvd_and_dvd_add_of_equation {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {π : R} (hπ : Prime π) (h₁ : π ∣ W.a₁) {A₂ : R} (h₂ : W.a₂ = π * A₂) (h₃ : π ^ 2 ∣ W.a₃) (h₄ : π ^ 3 ∣ W.a₄) (h₆ : π ^ 4 ∣ W.a₆) {X Y : R} (hX : ¬π ∣ X) (h : W.toAffine.Equation (π * X) (π * Y)) : π ∣ Y ∧ π ∣ X + A₂

On the normal form π ∣ a₁, a₂ = π A₂, π² ∣ a₃, π³ ∣ a₄, π⁴ ∣ a₆ of Step 7 of Tate's algorithm, a solution (π X, π Y) with π ∤ X has π ∣ Y and X ≡ −A₂ modulo π: the residue of X is the simple root of the residue cubic T² (T + A₂).

theorem WeierstrassCurve.dvd_of_equation_sq_mul {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {π : R} (hπ : Prime π) (h₃ : π ^ 2 ∣ W.a₃) (h₄ : π ∣ W.a₄) (h₆ : π ^ 3 ∣ W.a₆) {X Y : R} (h : W.toAffine.Equation (π ^ 2 * X) (π * Y)) : π ∣ Y

If π² ∣ a₃, π ∣ a₄ and π³ ∣ a₆, a solution (π² X, π Y) has π ∣ Y.

theorem WeierstrassCurve.dvd_and_dvd_of_equation_pow {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {π : R} (hπ : Prime π) (i : ℕ) (h₁ : π ∣ W.a₁) {A₂ : R} (h₂ : W.a₂ = π * A₂) (hA₂ : ¬π ∣ A₂) (h₃ : π ^ (i + 3) ∣ W.a₃) (h₄ : π ^ (i + 4) ∣ W.a₄) (h₆ : π ^ (2 * i + 6) ∣ W.a₆) {X Y : R} (h : W.toAffine.Equation (π ^ (i + 2) * X) (π ^ (i + 2) * Y)) : π ∣ X ∧ π ∣ Y

One stage of the Iₙ* subprocedure, on points. With π ∣ a₁, a₂ = π A₂, π ∤ A₂, and π^(j+1) ∣ a₃, π^(j+2) ∣ a₄, π^(2j+2) ∣ a₆ for j = i + 2, a solution (π^j X, π^j Y) has π ∣ X and π ∣ Y: the equation divided by π^(2j) forces π ∣ Y, and then divided by π^(2j+1) it forces π ∣ A₂ X².

theorem WeierstrassCurve.pow_dvd_and_pow_dvd_of_equation {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {π : R} (hπ : Prime π) {k : ℕ} (h₁ : π ∣ W.a₁) {A₂ : R} (h₂ : W.a₂ = π * A₂) (hA₂ : ¬π ∣ A₂) (h₃ : π ^ (k + 2) ∣ W.a₃) (h₄ : π ^ (k + 3) ∣ W.a₄) (h₆ : π ^ (2 * k + 4) ∣ W.a₆) {x y : R} (h : W.toAffine.Equation x y) (hx : π ^ 2 ∣ x) (hy : π ^ 2 ∣ y) : π ^ (k + 2) ∣ x ∧ π ^ (k + 2) ∣ y

The Iₙ* subprocedure, on points. With π ∣ a₁, a₂ = π A₂, π ∤ A₂, π^(k+2) ∣ a₃, π^(k+3) ∣ a₄ and π^(2k+4) ∣ a₆, a solution (x, y) with π² ∣ x and π² ∣ y has π^(k+2) ∣ x and π^(k+2) ∣ y.

theorem WeierstrassCurve.sq_dvd_and_dvd_cubic_of_equation {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {ϖ : R} (hϖ : Prime ϖ) (h₁ : ϖ ∣ W.a₁) {A₂ A₄ A₆ : R} (h₂ : W.a₂ = ϖ * A₂) (h₃ : ϖ ^ 2 ∣ W.a₃) (h₄ : W.a₄ = ϖ ^ 2 * A₄) (h₆ : W.a₆ = ϖ ^ 3 * A₆) {X y : R} (h : W.toAffine.Equation (ϖ * X) y) : ϖ ^ 2 ∣ y ∧ ϖ ∣ X ^ 3 + A₂ * X ^ 2 + A₄ * X + A₆

On the normal form of Step 6 of Tate's algorithm, a solution (ϖ X, y) has ϖ² ∣ y, and X is a root of the residue cubic modulo ϖ. Over a domain with a prime ϖ, suppose that ϖ ∣ a₁, ϖ² ∣ a₃, a₂ = ϖ A₂, a₄ = ϖ² A₄ and a₆ = ϖ³ A₆; the residue cubic is T³ + A₂ T² + A₄ T + A₆.

theorem WeierstrassCurve.dvd_quadratic_of_equation_odd {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {ϖ : R} (hϖ : ϖ ≠ 0) {k : ℕ} (h₁ : ϖ ∣ W.a₁) (h₂ : ϖ ∣ W.a₂) {A₃ A₆ : R} (h₃ : W.a₃ = ϖ ^ (k + 2) * A₃) (h₄ : ϖ ^ (k + 3) ∣ W.a₄) (h₆ : W.a₆ = ϖ ^ (2 * k + 4) * A₆) {X Y : R} (h : W.toAffine.Equation (ϖ ^ (k + 2) * X) (ϖ ^ (k + 2) * Y)) : ϖ ∣ Y ^ 2 + A₃ * Y - A₆

On the normal form of an odd stage of the Iₙ* subprocedure of Step 7 of Tate's algorithm, a solution (ϖ^(k+2) X, ϖ^(k+2) Y) makes Y a root of the residue quadratic modulo ϖ. Over a domain with ϖ ≠ 0, suppose that ϖ ∣ a₁, ϖ ∣ a₂, a₃ = ϖ^(k+2) A₃, ϖ^(k+3) ∣ a₄ and a₆ = ϖ^(2k+4) A₆. Then ϖ ∣ Y² + A₃ Y − A₆. At k = 0 this covers the normal form of Step 8.

theorem WeierstrassCurve.dvd_quadratic_of_equation_even {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {ϖ : R} (hϖ : Prime ϖ) {k : ℕ} (h₁ : ϖ ∣ W.a₁) {A₂ A₄ A₆ : R} (h₂ : W.a₂ = ϖ * A₂) (h₃ : ϖ ^ (k + 3) ∣ W.a₃) (h₄ : W.a₄ = ϖ ^ (k + 3) * A₄) (h₆ : W.a₆ = ϖ ^ (2 * k + 5) * A₆) {X Y : R} (h : W.toAffine.Equation (ϖ ^ (k + 2) * X) (ϖ ^ (k + 2) * Y)) : ϖ ∣ A₂ * X ^ 2 + A₄ * X + A₆

On the normal form of an even stage of the Iₙ* subprocedure of Step 7 of Tate's algorithm, a solution (ϖ^(k+2) X, ϖ^(k+2) Y) makes X a root of the residue quadratic modulo ϖ. Over a domain with a prime ϖ, suppose that ϖ ∣ a₁, a₂ = ϖ A₂, ϖ^(k+3) ∣ a₃, a₄ = ϖ^(k+3) A₄ and a₆ = ϖ^(2k+5) A₆. Then ϖ ∣ A₂ X² + A₄ X + A₆.

theorem WeierstrassCurve.exists_eq_sq_mul_of_equation {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {ϖ : R} (hϖ : Prime ϖ) (h₁ : ϖ ∣ W.a₁) (h₂ : ϖ ^ 2 ∣ W.a₂) {A₃ A₆ : R} (h₃ : W.a₃ = ϖ ^ 2 * A₃) (h₄ : ϖ ^ 3 ∣ W.a₄) (h₆ : W.a₆ = ϖ ^ 4 * A₆) {x y : R} (h : W.toAffine.Equation x y) (hx : ϖ ∣ x) (hy : ϖ ∣ y) : ∃ (X : R) (Y : R), x = ϖ ^ 2 * X ∧ y = ϖ ^ 2 * Y ∧ ϖ ∣ Y ^ 2 + A₃ * Y - A₆

On the normal form of Tate's algorithm at the symbol IV*, where ϖ ∣ a₁, ϖ² ∣ a₂, a₃ = ϖ² A₃, ϖ³ ∣ a₄ and a₆ = ϖ⁴ A₆, a solution with both coordinates divisible by the prime ϖ is (ϖ² X, ϖ² Y) with Y² + A₃ Y − A₆ divisible by ϖ.

theorem WeierstrassCurve.sq_dvd_and_pow_three_dvd_of_equation {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {ϖ : R} (hϖ : Prime ϖ) (h₁ : ϖ ∣ W.a₁) (h₂ : ϖ ^ 2 ∣ W.a₂) (h₃ : ϖ ^ 3 ∣ W.a₃) (h₄ : ϖ ^ 3 ∣ W.a₄) (h₆ : ϖ ^ 5 ∣ W.a₆) {x y : R} (h : W.toAffine.Equation x y) (hx : ϖ ∣ x) (hy : ϖ ∣ y) : ϖ ^ 2 ∣ x ∧ ϖ ^ 3 ∣ y

On the normal form of Tate's algorithm at the symbol III*, where ϖ ∣ a₁, ϖ² ∣ a₂, ϖ³ ∣ a₃, ϖ³ ∣ a₄ and ϖ⁵ ∣ a₆, a solution with both coordinates divisible by the prime ϖ has ϖ² ∣ x and ϖ³ ∣ y.

theorem WeierstrassCurve.pow_six_dvd_a₆_of_equation {R : Type u_1} [CommRing R] [IsDomain R] (W : WeierstrassCurve R) {ϖ : R} (hϖ : Prime ϖ) (h₁ : ϖ ∣ W.a₁) (h₂ : ϖ ^ 2 ∣ W.a₂) (h₃ : ϖ ^ 3 ∣ W.a₃) (h₄ : ϖ ^ 4 ∣ W.a₄) (h₆ : ϖ ^ 5 ∣ W.a₆) {x y : R} (h : W.toAffine.Equation x y) (hx : ϖ ∣ x) (hy : ϖ ∣ y) : ϖ ^ 6 ∣ W.a₆

On the normal form of Step 10 of Tate's algorithm, a solution divisible by ϖ forces ϖ⁶ ∣ a₆. Over a domain with a prime ϖ, suppose that ϖ ∣ a₁, ϖ² ∣ a₂, ϖ³ ∣ a₃, ϖ⁴ ∣ a₄ and ϖ⁵ ∣ a₆. If (x, y) is a solution with ϖ ∣ x and ϖ ∣ y, then ϖ⁶ ∣ a₆. The stopping test for II* asks for ϖ⁶ ∤ a₆, so there such a solution does not exist.

Every point has nonsingular reduction

theorem WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal {F : Type u_1} {Γ₀ : Type u_2} [Field F] [LinearOrderedCommGroupWithZero Γ₀] (v : Valuation F Γ₀) {W : Affine F} [IsIntegral (↥v.valuationSubring) W] [DecidableEq F] {ϖ : ↥v.valuationSubring} (hϖ : IsLocalRing.maximalIdeal ↥v.valuationSubring = Ideal.span {ϖ}) {s : TauCeti.ReductionSymbol} (hs : (integralModel (↥v.valuationSubring) W).IsTateTerminal ϖ s) (hsI : ∀ (n : ℕ), s ≠ TauCeti.ReductionSymbol.I n) (h : Prime ϖ → ∀ (x y : ↥v.valuationSubring), ϖ ∣ x → ϖ ∣ y → ¬(integralModel (↥v.valuationSubring) W).toAffine.Equation x y) : nonsingularReduction v W = ⊤

If Tate's algorithm stops at an additive symbol on the integral model with a generator ϖ of the maximal ideal, and no integral solution has both coordinates divisible by ϖ, then every point has nonsingular reduction. The stopping test forces ϖ ≠ 0, so ϖ is prime, which the hypothesis on solutions may use.

theorem WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal_IStar_zero {F : Type u_1} {Γ₀ : Type u_2} [Field F] [LinearOrderedCommGroupWithZero Γ₀] (v : Valuation F Γ₀) {W : Affine F} [IsIntegral (↥v.valuationSubring) W] [DecidableEq F] {ϖ : ↥v.valuationSubring} (hϖ : IsLocalRing.maximalIdeal ↥v.valuationSubring = Ideal.span {ϖ}) (h : (integralModel (↥v.valuationSubring) W).IsTateTerminal ϖ (TauCeti.ReductionSymbol.IStar 0)) {A₂ A₄ A₆ : ↥v.valuationSubring} (h₂ : (integralModel (↥v.valuationSubring) W).a₂ = ϖ * A₂) (h₄ : (integralModel (↥v.valuationSubring) W).a₄ = ϖ ^ 2 * A₄) (h₆ : (integralModel (↥v.valuationSubring) W).a₆ = ϖ ^ 3 * A₆) (hroot : ∀ (t : IsLocalRing.ResidueField ↥v.valuationSubring), t ^ 3 + (IsLocalRing.residue ↥v.valuationSubring) A₂ * t ^ 2 + (IsLocalRing.residue ↥v.valuationSubring) A₄ * t + (IsLocalRing.residue ↥v.valuationSubring) A₆ ≠ 0) : nonsingularReduction v W = ⊤

At the symbol I₀* with no rational root of the residue cubic, every point has nonsingular reduction: E₀(F) = W(F). Here ϖ generates the maximal ideal, Tate's algorithm stops at I₀* on the integral model, and with a₂ = ϖ A₂, a₄ = ϖ² A₄ and a₆ = ϖ³ A₆ the residue cubic T³ + A₂ T² + A₄ T + A₆ has no root in the residue field.

theorem WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal_IVStar {F : Type u_1} {Γ₀ : Type u_2} [Field F] [LinearOrderedCommGroupWithZero Γ₀] (v : Valuation F Γ₀) {W : Affine F} [IsIntegral (↥v.valuationSubring) W] [DecidableEq F] {ϖ : ↥v.valuationSubring} (hϖ : IsLocalRing.maximalIdeal ↥v.valuationSubring = Ideal.span {ϖ}) (h : (integralModel (↥v.valuationSubring) W).IsTateTerminal ϖ TauCeti.ReductionSymbol.IVStar) {A₃ A₆ : ↥v.valuationSubring} (h₃ : (integralModel (↥v.valuationSubring) W).a₃ = ϖ ^ 2 * A₃) (h₆ : (integralModel (↥v.valuationSubring) W).a₆ = ϖ ^ 4 * A₆) (hroot : ∀ (t : IsLocalRing.ResidueField ↥v.valuationSubring), t ^ 2 + (IsLocalRing.residue ↥v.valuationSubring) A₃ * t - (IsLocalRing.residue ↥v.valuationSubring) A₆ ≠ 0) : nonsingularReduction v W = ⊤

At the symbol IV* with no rational root of the residue quadratic, every point has nonsingular reduction: E₀(F) = W(F). Here ϖ generates the maximal ideal, Tate's algorithm stops at IV* on the integral model, and with a₃ = ϖ² A₃ and a₆ = ϖ⁴ A₆ the residue quadratic T² + A₃ T − A₆ has no root in the residue field.

theorem WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_isTateTerminal_IIStar {F : Type u_1} {Γ₀ : Type u_2} [Field F] [LinearOrderedCommGroupWithZero Γ₀] (v : Valuation F Γ₀) {W : Affine F} [IsIntegral (↥v.valuationSubring) W] [DecidableEq F] {ϖ : ↥v.valuationSubring} (hϖ : IsLocalRing.maximalIdeal ↥v.valuationSubring = Ideal.span {ϖ}) (h : (integralModel (↥v.valuationSubring) W).IsTateTerminal ϖ TauCeti.ReductionSymbol.IIStar) : nonsingularReduction v W = ⊤

At the symbol II*, every point has nonsingular reduction: E₀(F) = W(F). Here ϖ generates the maximal ideal and Tate's algorithm stops at II* on the integral model.