Tate's algorithm: the stopping conditions, and termination #
Tate's algorithm reads the reduction symbol of a minimal Weierstrass equation over a discrete
valuation ring R with uniformiser ϖ and perfect residue field k. It alternates changes of
variables over R with divisibility tests, and stops as soon as one test succeeds. This file
records the test at which the algorithm stops as a predicate on an integral equation,
WeierstrassCurve.IsTateTerminal W ϖ s, and proves that the algorithm terminates: every minimal
equation is carried by a change of variables over R to an equation on which one of the stopping
tests succeeds.
The stopping conditions are stated on the coefficients of W without dividing by ϖ. With the
residue cubic P(T) = T³ + (a₂/ϖ) T² + (a₄/ϖ²) T + a₆/ϖ³ of the Step 6 normal form, they are:
I₀:ϖ ∤ Δ(Step 1);Iₙforn ≥ 1: the reduction is singular at the origin (ϖ ∣ a₃, a₄, a₆),ϖ ∤ b₂, andv(Δ) = n(Step 2);II,III,IV: the reduction is singular at the origin,ϖ ∣ b₂, and respectivelyϖ² ∤ a₆;ϖ² ∣ a₆andϖ³ ∤ b₈;ϖ² ∣ a₆,ϖ³ ∣ b₈andϖ³ ∤ b₆(Steps 3 to 5);I₀*: the Step 6 normal formϖ ∣ a₁, a₂,ϖ² ∣ a₃, a₄,ϖ³ ∣ a₆, andPhas nonzero discriminant, that isϖ⁷ ∤ disc(T³ + a₂ T² + a₄ T + a₆)(WeierstrassCurve.pow_seven_dvd_discr_iff);Iₙ*forn ≥ 1: the stopping stagenof the subprocedure of Step 7 (WeierstrassCurve.IsTateOddStage,WeierstrassCurve.IsTateEvenStage);IV*,III*,II*: the normal forms and tests of Steps 8, 9 and 10.
The equation must be minimal for the algorithm to stop: on a non-minimal equation the triple-root branch may reach Step 11, which rescales the equation and starts again. Minimality is used only there.
The algorithmic Ogg exponent v(Δ) − m + 1 of a symbol with m components
(TauCeti.ReductionSymbol.algorithmicOggExponent) is 0 at I₀ and 1 at Iₙ, since there
v(Δ) = n; at every additive symbol the stopping conditions force v(Δ) > m, so the exponent is
at least 2.
Main definitions #
WeierstrassCurve.IsTateTerminal: Tate's algorithm, run onW, stops with outputswithout a further change of variables.
Main results #
WeierstrassCurve.exists_isTateTerminal: Tate's algorithm terminates. Over a discrete valuation ring with perfect residue field, a minimal equation with nonzero discriminant is carried by a change of variables overRto an equation on which the algorithm stops.WeierstrassCurve.IsTateTerminal.isTateOddStage_zero_of_IStar: every positiveIₙ*stopping condition implies the stage-one normal form.WeierstrassCurve.IsTateTerminal.isTateOddStage_of_IStar_two_mul_add_oneandWeierstrassCurve.IsTateTerminal.isTateEvenStage_of_IStar_two_mul_add_two: the odd and even stopping conditions supply their respective stage and discriminant tests.WeierstrassCurve.isTateTerminal_map_iff: the stopping conditions transfer along a ring map that reflects divisibility by powers of the uniformiser.WeierstrassCurve.IsTateTerminal.addVal_Δ_eq: at a symbolIₙ, the discriminant has valuationn.WeierstrassCurve.IsTateTerminal.pow_componentCount_add_one_dvd_Δ: at an additive symbol withmcomponents,ϖ^(m+1) ∣ Δ.WeierstrassCurve.IsTateTerminal.dvd_b₂_and_dvd_a₃_and_dvd_a₄_and_dvd_a₆andWeierstrassCurve.IsTateTerminal.dvd_c₄: at an additive symbol,ϖdividesb₂,a₃,a₄,a₆and hencec₄.WeierstrassCurve.IsTateTerminal.ne_zero: at a symbol other thanI₀, the uniformiserϖis nonzero.WeierstrassCurve.IsTateTerminal.prime_and_not_dvd_of_IVStar: atIV*, witha₃ = ϖ² A₃anda₆ = ϖ⁴ A₆, the uniformiser is prime andϖ ∤ A₃² + 4 A₆.WeierstrassCurve.IsTateTerminal.dvd_Δ_iff: the algorithm stops at a symbol other thanI₀exactly whenϖ ∣ Δ.WeierstrassCurve.IsTateTerminal.two_le_algorithmicOggExponent: at an additive symbol, the algorithmic Ogg exponent is at least2.
The output of the algorithm is not shown here to be independent of the choices made along the way.
References #
- 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.
The stopping conditions of Tate's algorithm. W.IsTateTerminal ϖ s says that Tate's
algorithm, run on the integral equation W with uniformiser ϖ, stops with output s without a
further change of variables. The conditions are those listed in the module docstring; for Iₙ*
the stage n = 2k + 1 or n = 2k + 2 is the normal form IsTateOddStage ϖ k or
IsTateEvenStage ϖ k of the subprocedure of Step 7.
Instances For
Both branches of the stopping test for Iₙ* satisfy the normal form of stage 1 of the
subprocedure: ϖ ∣ a₁, ϖ ∥ a₂, ϖ² ∣ a₃, ϖ³ ∣ a₄ and ϖ⁴ ∣ a₆.
The stopping test at I*_{2k+1} is the odd stage k with ϖ^(2k+5) ∤ b₆.
The stopping test at I*_{2k+2} is the even stage k with
ϖ^(2k+7) ∤ a₄² − 4 a₂ a₆.
The stopping conditions are read along a ring map that reflects divisibility by powers of
the uniformiser. If f ϖ ^ n ∣ f x ↔ ϖ ^ n ∣ x for all n and x, then Tate's algorithm stops
on W.map f with uniformiser f ϖ exactly where it stops on W with uniformiser ϖ. For
instance the cast from ℤ to a discrete valuation ring with uniformiser a prime p reflects
divisibility by powers of p (Nat.Prime.pow_dvd_intCast_iff), so the symbol of an integral
equation over ℤ can be computed over ℤ.
At a stopping symbol other than Iₙ with m components, ϖ^(m+1) divides the
discriminant. This is the bound that makes the algorithmic Ogg exponent at least 2 there.
At a stopping symbol other than Iₙ, the equation is in the additive normal form of Step 2 of
Tate's algorithm: ϖ divides b₂, a₃, a₄ and a₆.
At a stopping symbol other than Iₙ, ϖ divides c₄. In the additive normal form ϖ
divides b₂ and b₄, and c₄ = b₂² − 24 b₄.
At a stopping symbol other than I₀, the uniformiser is nonzero. If ϖ = 0, the
divisibility conditions force the relevant coefficients to vanish, and then the stopping test
¬ 0 ^ k ∣ _ fails.
At the symbol IV*, the residue quadratic has distinct roots. If (ϖ) is a prime ideal,
a₃ = ϖ² A₃ and a₆ = ϖ⁴ A₆, then the test ϖ⁵ ∤ b₆ = ϖ⁴ (A₃² + 4 A₆) says that ϖ is prime
and does not divide the discriminant A₃² + 4 A₆ of T² + A₃ T − A₆.
Tate's algorithm stops at a symbol other than I₀ exactly when ϖ divides the
discriminant.
The I₀* test is the residue-cubic test. If a₂ = ϖ A₂, a₄ = ϖ² A₄ and a₆ = ϖ³ A₆, then
ϖ⁷ divides the discriminant of T³ + a₂ T² + a₄ T + a₆ exactly when the residue cubic
T³ + A₂ T² + A₄ T + A₆ over the residue field has zero discriminant.
At the symbol Iₙ, the discriminant has valuation n.
The algorithmic Ogg exponent at an additive symbol is at least 2. If Tate's algorithm
stops on W with a symbol other than Iₙ, and n is the valuation of the discriminant, then
n − m + 1 ≥ 2, where m is the component count of the symbol.
Termination #
Tate's algorithm terminates. Over a discrete valuation ring with uniformiser ϖ and
perfect residue field, a Weierstrass equation over R with nonzero discriminant that is minimal
over the fraction field K is carried by a change of variables over R to an equation on which
one of the stopping tests of Tate's algorithm succeeds.