Tate's algorithm: the triple-root branch

Let R be a discrete valuation ring with uniformiser ϖ and perfect residue field k. Step 6 of Tate's algorithm (WeierstrassCurve.exists_variableChange_dvd_a₁_a₂_a₃_a₄_a₆) brings an equation with additive reduction to the normal form ϖ ∣ a₁, a₂, ϖ² ∣ a₃, a₄, ϖ³ ∣ a₆, and reads off it the residue cubic

P(T) = T³ + (a₂/ϖ) T² + (a₄/ϖ²) T + a₆/ϖ³ over k.

This file treats the branch of the algorithm in which P has a triple root (Steps 8 to 10):

• Step 8: translate x ↦ x + r so that the triple root is T = 0. Then ϖ² ∣ a₂, ϖ³ ∣ a₄ and ϖ⁴ ∣ a₆. If the quadratic Y² + (a₃/ϖ²) Y − a₆/ϖ⁴ has distinct roots, that is if ϖ⁵ ∤ b₆, the reduction symbol is IV*.
• Step 9: otherwise translate y ↦ y + t so that its double root is Y = 0. Then ϖ³ ∣ a₃ and ϖ⁵ ∣ a₆. If ϖ⁴ ∤ a₄, the reduction symbol is III*.
• Step 10: otherwise, if ϖ⁶ ∤ a₆, the reduction symbol is II*.

If all three tests fail, then ϖⁱ ∣ aᵢ for every i and the equation was not minimal (Step 11, WeierstrassCurve.not_isMinimal_baseChange_of_pow_dvd). So on a minimal equation this branch always stops at Step 8, 9 or 10.

The triple-root condition is stated on the coefficients of W, without dividing by ϖ: by TauCeti.Polynomial.exists_cubic_eq_X_sub_C_pow_three_iff, a monic cubic T³ + b T² + c T + d over the perfect field k is a cube exactly when b² = 3c, c² = 3bd and bc = 9d. For P these conditions read ϖ³ ∣ a₂² − 3 a₄, ϖ⁵ ∣ a₄² − 3 a₂ a₆ and ϖ⁴ ∣ a₂ a₄ − 9 a₆. Perfectness of k enters twice: to find the triple root, a cube root in characteristic 3, and to complete the square in Step 9, a square root in characteristic 2.

Main results

• Irreducible.exists_dvd_of_residue_cubic_eq_pow_three: over a perfect residue field, a cubic whose residue is a cube has a lifted triple root ρ.
• WeierstrassCurve.exists_variableChange_pow_dvd_a₂_a₄_a₆: Step 8 of Tate's algorithm. If the residue cubic has a triple root, a translation x ↦ x + r over R produces ϖ ∣ a₁, ϖ² ∣ a₂, a₃, ϖ³ ∣ a₄ and ϖ⁴ ∣ a₆.
• WeierstrassCurve.exists_variableChange_pow_dvd_a₃_a₆: Step 9 of Tate's algorithm. If moreover ϖ⁵ ∣ b₆, a translation y ↦ y + t over R produces in addition ϖ³ ∣ a₃ and ϖ⁵ ∣ a₆.
• WeierstrassCurve.exists_variableChange_not_pow_dvd_b₆_or_a₄_or_a₆: the triple-root branch terminates. On a minimal equation with nonzero discriminant whose residue cubic has a triple root, a change of variables over R reaches the normal form of Step 8 and one of ϖ⁵ ∤ b₆ (Step 8, IV*), ϖ⁴ ∤ a₄ (Step 9, III*) or ϖ⁶ ∤ a₆ (Step 10, II*) holds, the last two in the normal form of Step 9.

References

• J. H. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, GTM 151, IV.9, Steps 8 to 10 of Tate's algorithm.
• 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.

theorem Irreducible.exists_dvd_of_residue_cubic_eq_pow_three {R : Type u_1} [CommRing R] [IsDomain R] [IsDiscreteValuationRing R] {ϖ : R} [PerfectField (IsLocalRing.ResidueField R)] (hϖ : Irreducible ϖ) {a₂ a₄ a₆ A₂ A₄ A₆ : R} (hA₂ : a₂ = ϖ * A₂) (hA₄ : a₄ = ϖ ^ 2 * A₄) (hA₆ : a₆ = ϖ ^ 3 * A₆) (hb : ϖ ^ 3 ∣ a₂ ^ 2 - 3 * a₄) (hc : ϖ ^ 5 ∣ a₄ ^ 2 - 3 * a₂ * a₆) (hd : ϖ ^ 4 ∣ a₂ * a₄ - 9 * a₆) : ∃ (ρ : R), ϖ ∣ A₂ + 3 * ρ ∧ ϖ ∣ A₄ + 2 * ρ * A₂ + 3 * ρ ^ 2 ∧ ϖ ∣ A₆ + ρ * A₄ + ρ ^ 2 * A₂ + ρ ^ 3

The residue cubic step of Tate's algorithm. Let a₂ = ϖ A₂, a₄ = ϖ² A₄ and a₆ = ϖ³ A₆ satisfy the triple-root conditions ϖ³ ∣ a₂² − 3 a₄, ϖ⁵ ∣ a₄² − 3 a₂ a₆ and ϖ⁴ ∣ a₂ a₄ − 9 a₆. Over a perfect residue field the cubic T³ + A₂ T² + A₄ T + A₆ is then a cube modulo ϖ, so some ρ ∈ R lifts its triple root: ϖ divides the coefficients A₂ + 3 ρ, A₄ + 2 ρ A₂ + 3 ρ² and A₆ + ρ A₄ + ρ² A₂ + ρ³ of the shifted cubic P(T + ρ).

theorem WeierstrassCurve.exists_variableChange_pow_dvd_a₂_a₄_a₆ {R : Type u_1} [CommRing R] [IsDomain R] [IsDiscreteValuationRing R] {ϖ : R} [PerfectField (IsLocalRing.ResidueField R)] (hϖ : Irreducible ϖ) (W : WeierstrassCurve R) (h₁ : ϖ ∣ W.a₁) (h₃ : ϖ ^ 2 ∣ W.a₃) (h₄ : ϖ ^ 2 ∣ W.a₄) (h₆ : ϖ ^ 3 ∣ W.a₆) (hb : ϖ ^ 3 ∣ W.a₂ ^ 2 - 3 * W.a₄) (hc : ϖ ^ 5 ∣ W.a₄ ^ 2 - 3 * W.a₂ * W.a₆) (hd : ϖ ^ 4 ∣ W.a₂ * W.a₄ - 9 * W.a₆) : ∃ (r : R), ϖ ∣ ({ u := 1, r := r, s := 0, t := 0 } • W).a₁ ∧ ϖ ^ 2 ∣ ({ u := 1, r := r, s := 0, t := 0 } • W).a₂ ∧ ϖ ^ 2 ∣ ({ u := 1, r := r, s := 0, t := 0 } • W).a₃ ∧ ϖ ^ 3 ∣ ({ u := 1, r := r, s := 0, t := 0 } • W).a₄ ∧ ϖ ^ 4 ∣ ({ u := 1, r := r, s := 0, t := 0 } • W).a₆

Tate's algorithm, Step 8. Over a discrete valuation ring with uniformiser ϖ and perfect residue field, let W be an equation in the normal form of Step 6, ϖ ∣ a₁, ϖ² ∣ a₃, a₄, ϖ³ ∣ a₆, whose residue cubic T³ + (a₂/ϖ) T² + (a₄/ϖ²) T + a₆/ϖ³ has a triple root: ϖ³ ∣ a₂² − 3 a₄, ϖ⁵ ∣ a₄² − 3 a₂ a₆ and ϖ⁴ ∣ a₂ a₄ − 9 a₆ (TauCeti.Polynomial.exists_cubic_eq_X_sub_C_pow_three_iff). Then a translation x ↦ x + r with r ∈ R moves the triple root to 0: the new equation has ϖ ∣ a₁, ϖ² ∣ a₂, ϖ² ∣ a₃, ϖ³ ∣ a₄ and ϖ⁴ ∣ a₆. (The condition ϖ ∣ a₂ of the Step 6 normal form follows from the others.)

theorem WeierstrassCurve.exists_variableChange_pow_dvd_a₃_a₆ {R : Type u_1} [CommRing R] [IsDomain R] [IsDiscreteValuationRing R] {ϖ : R} [PerfectField (IsLocalRing.ResidueField R)] (hϖ : Irreducible ϖ) (W : WeierstrassCurve R) (h₁ : ϖ ∣ W.a₁) (h₂ : ϖ ^ 2 ∣ W.a₂) (h₄ : ϖ ^ 3 ∣ W.a₄) (h₆ : ϖ ^ 4 ∣ W.a₆) (hb₆ : ϖ ^ 5 ∣ W.b₆) : ∃ (t : R), ϖ ∣ ({ u := 1, r := 0, s := 0, t := t } • W).a₁ ∧ ϖ ^ 2 ∣ ({ u := 1, r := 0, s := 0, t := t } • W).a₂ ∧ ϖ ^ 3 ∣ ({ u := 1, r := 0, s := 0, t := t } • W).a₃ ∧ ϖ ^ 3 ∣ ({ u := 1, r := 0, s := 0, t := t } • W).a₄ ∧ ϖ ^ 5 ∣ ({ u := 1, r := 0, s := 0, t := t } • W).a₆

Tate's algorithm, Step 9. Over a discrete valuation ring with uniformiser ϖ and perfect residue field, let W be an equation in the normal form of Step 8, ϖ ∣ a₁, ϖ² ∣ a₂, ϖ³ ∣ a₄, ϖ⁴ ∣ a₆, on which Step 8 does not stop: the quadratic Y² + (a₃/ϖ²) Y − a₆/ϖ⁴ has a double root, that is ϖ⁵ ∣ b₆. Then a translation y ↦ y + t with t ∈ R moves the double root to 0: the new equation has ϖ ∣ a₁, ϖ² ∣ a₂, ϖ³ ∣ a₃, ϖ³ ∣ a₄ and ϖ⁵ ∣ a₆. (The condition ϖ² ∣ a₃ of the Step 8 normal form follows from b₆ = a₃² + 4 a₆.)

theorem WeierstrassCurve.exists_variableChange_not_pow_dvd_b₆_or_a₄_or_a₆ {R : Type u_1} [CommRing R] [IsDomain R] [IsDiscreteValuationRing R] {ϖ : R} {K : Type u_2} [Field K] [Algebra R K] [IsFractionRing R K] [PerfectField (IsLocalRing.ResidueField R)] (hϖ : Irreducible ϖ) (W : WeierstrassCurve R) (hΔ : W.Δ ≠ 0) [IsMinimal R (W.baseChange K)] (h₁ : ϖ ∣ W.a₁) (h₃ : ϖ ^ 2 ∣ W.a₃) (h₄ : ϖ ^ 2 ∣ W.a₄) (h₆ : ϖ ^ 3 ∣ W.a₆) (hb : ϖ ^ 3 ∣ W.a₂ ^ 2 - 3 * W.a₄) (hc : ϖ ^ 5 ∣ W.a₄ ^ 2 - 3 * W.a₂ * W.a₆) (hd : ϖ ^ 4 ∣ W.a₂ * W.a₄ - 9 * W.a₆) : ∃ (r : R) (t : R), ϖ ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₁ ∧ ϖ ^ 2 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₂ ∧ ϖ ^ 2 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₃ ∧ ϖ ^ 3 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₄ ∧ ϖ ^ 4 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₆ ∧ (¬ϖ ^ 5 ∣ ({ u := 1, r := r, s := 0, t := t } • W).b₆ ∨ ϖ ^ 3 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₃ ∧ ϖ ^ 5 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₆ ∧ (¬ϖ ^ 4 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₄ ∨ ¬ϖ ^ 6 ∣ ({ u := 1, r := r, s := 0, t := t } • W).a₆))

The triple-root branch of Tate's algorithm terminates. Over a discrete valuation ring with uniformiser ϖ, perfect residue field and fraction field K, let W be an equation with nonzero discriminant, minimal over K, in the normal form of Step 6 (ϖ ∣ a₁, ϖ² ∣ a₃, a₄, ϖ³ ∣ a₆) and whose residue cubic has a triple root (ϖ³ ∣ a₂² − 3 a₄, ϖ⁵ ∣ a₄² − 3 a₂ a₆, ϖ⁴ ∣ a₂ a₄ − 9 a₆). Then a change of variables x ↦ x + r, y ↦ y + t over R produces an equation in the normal form of Step 8 (ϖ ∣ a₁, ϖ² ∣ a₂, a₃, ϖ³ ∣ a₄, ϖ⁴ ∣ a₆) on which the algorithm stops at Step 8 (ϖ⁵ ∤ b₆, symbol IV*), or reaches the normal form of Step 9 (ϖ³ ∣ a₃, ϖ⁵ ∣ a₆) and stops at Step 9 (ϖ⁴ ∤ a₄, symbol III*) or at Step 10 (ϖ⁶ ∤ a₆, symbol II*).