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TauCeti.AlgebraicGeometry.EllipticCurve.Affine.Point.Index.Two

Points with singular reduction: the index two cases #

Let v be a valuation on a field F, with valuation ring O, maximal ideal 𝔪 and residue field k, and let W be a Weierstrass curve over F with an integral model W_O over O whose reduction is singular at the origin, that is a₃, a₄, a₆ ∈ 𝔪. 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 computes the index at the reduction symbols III and III*, where c_v = 2, and shows that it divides 2 when the tangent quadratic at the singular point has no root in the residue field. Over a perfect residue field this is a nonsplit node, where c_v is 1 or 2.

The points of W(F) outside E₀(F) are the affine points (x, y) with x, y ∈ 𝔪 (WeierstrassCurve.Affine.some_notMem_nonsingularReduction_iff). Let P = (x₁, y₁) and Q = (x₂, y₂) be two of them with P + Q ≠ O, let ℓ be the slope of the line through them, and suppose that P + Q = (x₃, y₃) lies outside E₀(F) as well. Then ℓ is integral, since ℓ² + a₁ ℓ − a₂ = x₁ + x₂ + x₃, and comparing the coefficients of X in the factorisation of the Weierstrass cubic restricted to the line, with ν = y₁ − ℓ x₁ (WeierstrassCurve.Affine.exists_chord_of_add_notMem_nonsingularReduction),

a₄ − a₃ ℓ − (2 ℓ + a₁) ν = x₁ x₂ + x₁ x₃ + x₂ x₃ ∈ 𝔪².

So the residue of ℓ is a root of the tangent quadratic T² + a₁ T − a₂. When that quadratic has no root in k this is impossible, and at the symbol III, where ϖ ∣ b₂, ϖ ∣ a₃, ϖ ∣ a₄, ϖ² ∣ a₆ and ϖ³ ∤ b₈ for a generator ϖ of 𝔪, the root is double, so 2 ℓ + a₁ ∈ 𝔪 and a₄ ≡ a₃ ℓ modulo ϖ². Then

b₈ = b₂ a₆ − (a₄ − a₃ ℓ) (a₄ + a₃ ℓ + a₁ a₃) − a₃² (ℓ² + a₁ ℓ − a₂)

is divisible by ϖ³, which is excluded. In both cases the sum of two points outside E₀(F) lies in E₀(F), and the index divides 2.

At the symbol III over a Henselian valuation ring with perfect residue field, a point with singular reduction exists, so the index is exactly 2. Writing x = ϖ X and y = ϖ s X, where s lifts the double root of the tangent quadratic, the Weierstrass equation divided by ϖ² becomes a cubic in X whose reduction is linear with unit coefficient a₄ / ϖ − s a₃ / ϖ: this is again the condition ϖ³ ∤ b₈. Its leading coefficient ϖ is not a unit, so the root is lifted by Hensel's lemma for polynomials that are not monic (Polynomial.exists_isRoot_sub_mem).

At the symbol III*, where ϖ ∣ a₁, ϖ² ∣ a₂, ϖ³ ∣ a₃, ϖ³ ∣ a₄, ϖ⁵ ∣ a₆ and ϖ⁴ ∤ a₄, comparing powers of ϖ in the Weierstrass equation shows that a point with singular reduction has ϖ² ∣ x and ϖ³ ∣ y. If P, Q and P + Q all had singular reduction, then ℓ ∈ 𝔪 because ℓ² + a₁ ℓ − a₂ = x₁ + x₂ + x₃, and then ϖ⁴ would divide every term of a₄ = x₁ x₂ + x₁ x₃ + x₂ x₃ + a₃ ℓ + (2 ℓ + a₁) ν, which is excluded. So again the index divides 2. Over a Henselian valuation ring, with any residue field, the point (ϖ² X, 0) has singular reduction, where X is a root of the Weierstrass cubic x³ + a₂ x² + a₄ x + a₆ at x = ϖ² X, divided by ϖ⁵, whose reduction is linear with the unit coefficient a₄ / ϖ³.

The cases c_v = 1 are WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_Δ_notMem, WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_not_splits and WeierstrassCurve.Affine.nonsingularReduction_eq_top_of_a₆_notMem.

Main results #

References #

The slope of a chord through two points with singular reduction whose sum has singular reduction. At a reduction singular at the origin, the slope ℓ = W.slope x₁ x₂ y₁ y₂ of the line through the two points (x₁, y₁) and (x₂, y₂) is integral and its residue is a root of the tangent quadratic T² + a₁ T − a₂, and with ν = y₁ − ℓ x₁ ∈ 𝔪 the coefficient a₄ − a₃ ℓ − (2 ℓ + a₁) ν of the Weierstrass cubic restricted to the line lies in 𝔪².

A tangent quadratic with no root #

When the tangent quadratic has no root, the sum of two points with singular reduction has nonsingular reduction. Here the reduction of the integral model is singular at the origin and its tangent quadratic T² + a₁ T − a₂ has no root in the residue field. When the residue field is perfect, or b₂ is a unit, the singular point is then a node whose tangent directions are not rational, and for a discrete valuation, on a minimal equation, this is nonsplit multiplicative reduction. Over an imperfect residue field of characteristic 2 the singular point may instead be a cusp.

When the tangent quadratic has no root, [E(F) : E₀(F)] divides 2. Here the reduction of the integral model is singular at the origin and its tangent quadratic T² + a₁ T − a₂ has no root in the residue field. When the residue field is perfect, or b₂ is a unit, this is a nonsplit node; for a discrete valuation, on a minimal equation with nonsplit multiplicative reduction of type Iₙ, Tate's algorithm reports the local index gcd(2, n).

The reduction symbol III #

At the reduction symbol III, the sum of two points with singular reduction has nonsingular reduction. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III: ϖ ∣ b₂, ϖ ∣ a₃, ϖ ∣ a₄, ϖ² ∣ a₆ and ϖ³ ∤ b₈.

At the reduction symbol III, [E(F) : E₀(F)] divides 2. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III.

At the reduction symbol III, some point has singular reduction, when the valuation ring is Henselian, for instance complete, and its residue field is perfect. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III. The point is (ϖ X, ϖ s X), where s lifts the double root of the tangent quadratic T² + a₁ T − a₂ and X is a root of the Weierstrass equation along the line y = s x, divided by ϖ²: a cubic whose reduction is linear with unit coefficient.

At the reduction symbol III, [E(F) : E₀(F)] = 2, when the valuation ring is Henselian, for instance complete, and its residue field is perfect. This is the local index c_v = 2 that Tate's algorithm reports at III. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III.

The reduction symbol III* #

At the reduction symbol III*, the sum of two points with singular reduction has nonsingular reduction. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III*: ϖ ∣ a₁, ϖ² ∣ a₂, ϖ³ ∣ a₃, ϖ³ ∣ a₄, ϖ⁵ ∣ a₆ and ϖ⁴ ∤ a₄.

At the reduction symbol III*, [E(F) : E₀(F)] divides 2. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III*.

At the reduction symbol III*, some point has singular reduction, when the valuation ring is Henselian, for instance complete. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III*. The point is (ϖ² X, 0), where X is a root of x³ + a₂ x² + a₄ x + a₆ at x = ϖ² X, divided by ϖ⁵: a cubic whose reduction is linear with the unit coefficient a₄ / ϖ³. No hypothesis on the residue field is needed.

At the reduction symbol III*, [E(F) : E₀(F)] = 2, when the valuation ring is Henselian, for instance complete. This is the local index c_v = 2 that Tate's algorithm reports at III*. Here ϖ generates the maximal ideal of the valuation ring, and Tate's algorithm, run on the integral model with uniformiser ϖ, stops with the symbol III*.