Research proof index

If the root of the reduced node polynomial P̄ (assumed irreducible) satisfies a monic quadratic relation X² - t·X + n over the residue field, then comparing with the defining relation of P̄ (aeval_root_nodePoly_map) and using the linear independence of 1 and the root (AdjoinRoot.eq_zero_of_mul_root_add_eq_zero) yields the relations `φc₄·t + φ(...

Twisting flips the square class (residue characteristic ≠ 2). Combining the split criterion nodePoly_map_splits_iff_isSquare with the coefficient scaling of the quadratic twist (c₄_quadraticTwistOf, c₆_quadraticTwistOf), the node polynomial of W.quadraticTwistOf t n splits over a field k of characteristic ≠ 2 exactly when D · (-c₄ c₆) is...

If the residues of (t', n') satisfy the trace and norm relations cut out by the node polynomial, then the node polynomial of the quadratic twist of the integral model by (t', n') splits over the residue field: the key identity φc₄ · φ(t'² - 4n') = -φc₆ (residue_c₄_mul_residue_eq_neg_c₆) reduces this to a square-class computation for residue charac...

The residue characteristic 2 case of nodePoly_quadraticTwistOf_map_splits_of_residue: the Artin–Schreier split condition (nodePoly_map_splits_iff_of_two_eq_zero) holds with z = 0, because φ κ_W = 0. Indeed κ_W = D³κ - D²·n·a₁²·c₄ (kappa_quadraticTwistOf), and φκ = -φc₄·φn (hB), φa₁ = -φt' (hA), φD = φt'² (as 4 = 0), so `φκ_W = -φ...

If j(E) ∉ {0, 1728} (so that the only automorphisms of E are ±1) then the quadratic twist is not isomorphic to E over K: twisting by L/K is a nontrivial operation. This can fail for j ∈ {0, 1728}: e.g. for E : y² = x³ + x of j-invariant 1728 over K = ℚ(i), the quadratic twist by any L = K(d^{1/2}) with d ∈ (K^×)⁴ ∖ (K^×)² is is...

If the residue of an integral element θ of S does not come from the residue field of R, then θ does not come from K either: an element of K integral over the integrally closed R lies in R, and residues are compatible.