Research proof index

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.

The classical formula for the quadratic twist away from characteristic 2. Suppose char K ≠ 2, so that after completing the square we may assume E has the form y² = x³ + a₂x² + a₄x + a₆, and suppose L = K(α) where α² = d is a nonsquare in K (every separable quadratic extension arises this way when char K ≠ 2). Then the quadratic twist of E...

Twisting the Galois action. The Galois action on the points of the quadratic twist is the Galois action on the points of E, twisted by the quadratic character of L/K: for σ ∈ Aut(M/K), transporting the action of σ on Eᴸ(M) through Eᴸ(M) ≅ E(M) gives χ(σ) times the action of σ on E(M). Taking M to be a separable closure of K, this...