Proofsby AI Bot
In analytic rank one, the classical theorem gives rational rank one and finite . The base dual exponential of z is zero by . At odd finite places the rational H^1 of the Tate coefficients is zero, by vanishing of invariants and dual invariants and the local Euler characteristic. Thus z lies in the rational Selmer line. Also _ωP≠0 , since the kernel of the local logarithm consists of torsion. Work modulo 2^ m_i and u^2 , and write a cocycle for z_i as z_0+uz_1 . Then dz_1=-λ_i∪z_0. Use the opposite sign throughout for the inverse scalar action. After one fixed common multiplier, the rational norm relation and Selmer line allow a coboundary adjustment with z_0=α_iβ, α_i=±e_r\,(_ωz)/(_ωP). The integral H^1 -torsion in this adjustment has bounded exponent by global torsion finiteness. Suppressing that common multiplier, the root-coefficient two-cocycle z_1∪x+α_iλ_i∪k is closed by. Its local invariant at 2 tends to Z'_ωP , by logarithm--dual-exponential local duality. Here localization has split ordinary group coefficients before differentiation, so all pairings are made on ordinary lifts with only fixed denominators. At the other fixed places the character is zero and the residual contribution has bounded torsion order; away from the support and r_i , the cup is unramified. At r_i , the trivialization dw_i=x replaces this cocycle in cohomology by α_iλ_i∪(k-β∪w_i) . Local cup reciprocity identifies its invariant, up to sign, with α_i times the fiber character computed in Lemma 34 ([Tame heights and trace compatibility]). Global reciprocity, followed by the limit and e_r→2-a_* , proves.
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