The Deuring normal form of an elliptic curve #
For a : R, the Weierstrass curve in Deuring normal form is
Y² + aXY + Y = X³,
that is, a₁ = a, a₃ = 1 and a₂ = a₄ = a₆ = 0 (WeierstrassCurve.deuringNormalForm). Its
invariants are
c₄ = a (a³ - 24), c₆ = -a⁶ + 36a³ - 216, Δ = a³ - 27, j = a³ (a³ - 24)³ / (a³ - 27),
so it is elliptic exactly when a³ - 27 is a unit. The point (0, 0) lies on it, with horizontal
tangent Y = 0 meeting the curve there to order three: (0, 0) is a flex, a point of order 3,
and the 3-division polynomial Ψ₃ vanishes at 0.
Conversely, a point at which Ψ₃ vanishes can be moved to the origin in this shape (Silverman,
AEC, Appendix A, Proposition 1.3). Let (x, y) be a point of W with Ψ₃(x) = 0, and suppose
that ψ₂(x, y) = 2y + a₁x + a₃ is a cube u³ of a unit. The translation (r, t) = (x, y) moves
the point to the origin (a₆ = 0), the shear s with s ψ₂(x, y) = a₄ + 2a₂x - a₁y + 3x² makes
the tangent line there horizontal (a₄ = 0), and the scaling by u makes a₃ = 1. The coefficient
a₂ then vanishes because a₂ ψ₂(x, y)² = Ψ₃(x) after the translation and shear
(WeierstrassCurve.a₂_mul_ψ₂_sq_eq_eval_Ψ₃).
Away from characteristic 3 every elliptic curve takes this form after a finite separable
extension of the base field: adjoin a root x of the separable polynomial Ψ₃, a matching y,
and a cube root of ψ₂(x, y), which is nonzero because Ψ₃ and Ψ₂Sq have no common root. This
holds in characteristic 2 too, where the Legendre form y² = x (x - 1) (x - λ) of
TauCeti/AlgebraicGeometry/EllipticCurve/Legendre.lean is not available. Since j is an explicit
rational function of a, the Deuring form ties the reduction of an elliptic curve to its
j-invariant in residue characteristic 2, as the Legendre form does away from it: this is how
the criterion WeierstrassCurve.hasPotentialGoodReduction_of_isInteger_j for potential good
reduction is proved in residue characteristic 2.
Main definitions #
WeierstrassCurve.deuringNormalForm a: the curveY² + aXY + Y = X³.
Main results #
WeierstrassCurve.deuringNormalForm_ΔandWeierstrassCurve.deuringNormalForm_j: its discriminant andj-invariant.WeierstrassCurve.exists_variableChange_smul_eq_deuringNormalForm: a curve with a point at whichΨ₃vanishes andψ₂is a cube of a unit is a change of variables away from Deuring normal form.WeierstrassCurve.exists_variableChange_smul_baseChange_eq_deuringNormalForm: away from characteristic3, an elliptic curve takes Deuring normal form over a finite separable extension.
References #
- J. Silverman, The Arithmetic of Elliptic Curves, Appendix A, Proposition 1.3, and VII.5.5.
The Deuring normal form #
The Weierstrass curve Y² + aXY + Y = X³ in Deuring normal form: a₁ = a, a₃ = 1 and
a₂ = a₄ = a₆ = 0.
Instances For
The discriminant of the Deuring normal form Y² + aXY + Y = X³ is a³ - 27.
The Deuring normal form Y² + aXY + Y = X³ is elliptic exactly when a³ - 27 is a unit.
The j-invariant of the Deuring normal form Y² + aXY + Y = X³ is
a³ (a³ - 24)³ / (a³ - 27).
The point (0, 0) lies on the Deuring normal form.
On the Deuring normal form, the 3-division polynomial Ψ₃ vanishes at 0: the origin is a
point of order 3.
Changes of variables to Deuring normal form #
Deuring normal form at a point of order 3. Let (x, y) be a point of W at which the
3-division polynomial Ψ₃ vanishes, and u a unit with u³ = ψ₂(x, y) = 2y + a₁x + a₃. Then a
change of variables carries W to a curve Y² + aXY + Y = X³ in Deuring normal form.
Deuring normal form over a finite separable extension. Away from characteristic 3, an
elliptic curve over K becomes isomorphic to a curve Y² + aXY + Y = X³ over a finite separable
extension L of K: one obtained by adjoining a root x of the 3-division polynomial Ψ₃, a
point (x, y) of the curve above it, and a cube root of 2y + a₁x + a₃.