Factoring 1 + π^k through a prime-to-3 automorphism #
For odd k = 2r+1, the isogeny 1 + π^k factors as (1 + π) ∘ G with
G = a - bπ, where a + 2b = 1, a - b = (-2)^r and a² + 2b² = m, and
H = a + bπ satisfies H ∘ G = G ∘ H = [m]. Whenever [m] is invertible on
the group — which on E(K) follows from an annihilator N coprime to m via
Bézout — G is bijective.
That isolates the entire 3-primary obstruction into the single factor 1 + π,
and it is all abstract: this file is a calculation in a ℤ-module equipped
with an endomorphism π satisfying π² = [-2]. No field, curve, Frobenius
map or algebraic closure appears.
The n-fold iterate of pi.
Equations
- KasamiCyclicAdditive.Isogeny.piIter pi 0 x✝ = x✝
- KasamiCyclicAdditive.Isogeny.piIter pi n.succ x✝ = pi (KasamiCyclicAdditive.Isogeny.piIter pi n x✝)
Instances For
a - b*pi.
Equations
- KasamiCyclicAdditive.Isogeny.gMap pi a b x = a • x - b • pi x
Instances For
a + b*pi.
Equations
- KasamiCyclicAdditive.Isogeny.hMap pi a b x = a • x + b • pi x
Instances For
Arithmetic coefficients for the factorisation.
For k = 2*r+1 and 2^k+1 = 3*m, there are integers a,b
such that
a + 2b = 1,
a - b = (-2)^r,
a^2 + 2b^2 = m.
Explicitly one may take
b = (1 - (-2)^r)/3,
a = (1 + 2*(-2)^r)/3.
The factorization (1 + π) ∘ G = 1 + π^(2r+1), with no invertibility
hypothesis. Stated separately because it is needed over the algebraic closure,
where the group has no finite annihilator and G need not be bijective.
Invertibility of G = a - b*pi.
Assume pi² = [-2] on an abelian group, and put H = a + b*pi. The norm
identity a² + 2b² = m gives HG = GH = [m], so if every element of the group
is killed by an integer N coprime to m, multiplication by m is invertible
by Bezout and hence G is bijective.
This needs neither r nor the coefficient identities: the factorization
(1 + pi) G = 1 + pi^(2r+1) is the separate statement gMap_factor.