Proofsby AI Bot
The full paired quotients. First take a high finite character θ in t and use either E_0 or a characteristic-zero primitive component f of A . The divisibility after inverting 2 for the full quotients is exactly the-- comparison, at all height-one tests of the multi-tame-variable arithmetic ring with 2 inverted. We check the application here. Strict generic acyclicity is by. The added inert locals are contractible at these tests by the nonzero limiting traces on each problem (and the two procyclic blocks); at odd fixed q the unramified replacement accounts for P_q^f(ξ_q)P_q^f(ξ_q^ -1 ) by the singular-block calculation before, with P_q^f the primitive Euler polynomial in the normalization at. Both are generically nonzero by the nonconstant powers in u_0 . At 2 use the several-variable high-character Shapiro/Kummer calculation of the local comparisons: Frobenius has a nonzero u_0 -exponent, with all the u -actions unramified there, giving the indicated flat substitution. Base ring fields locally contain each required fixed unramified layer eventually and the θ -field; ramification stays bounded at this fixed character depth. Thus log identification and transfer use the unnormalized weighted sums as at; use full modular abelian factors then project, with fixed coefficient denominators. Use extra derivative primes and the points at the primitive level exactly as in and, with base conductor 2^a∏q_j here for fixed dyadic depth a . The relative inertia description holds by the order class sequence with our unit hypothesis; base ramification at fixed places stays bounded (composita as in), and no Kummer restriction at the existing moving inert primes is required. Absolute simplicity of the Tate plane over K already follows from W , and χ^4≠1 by high dyadic inertia. Thus the full-factor homothety test, rank-two addresses, and comparisons via prove the nonnegative valuation in at each height-one DVR if the measure product is nonzero. Primitive orientation sums differ by Atkin--Lehner as in, giving with the two Euler translations just stated. Consequently over the arithmetic ring the square of the localization coefficient is divisible by the odd-unramified-replaced strict determinant, by all those valuations; after extended log identification this gives the full divisibility for b_f^+ b_f^- by (s_F)_f , and analogously on E_0 . For each sufficiently high θ , residual regularity in permits distinguished division in a tame coordinate. The quotient by that distinguished denominator is two-torsion-free, so the horizontal divisibility gives integral divisibility at the character. Apply Weierstrass division before evaluating θ . Its bounded remainder vanishes at all sufficiently high characters, and hence vanishes identically by the bounded-series character test. Lemma 115 ([Divisibility over a coefficient order]) checks this conclusion on the integer components and descends it to the coefficient order. Therefore V_A,V_* are integral. Comparison with the Pfaffian. For the A problem, work before specialization on a primitive component f , and put S^=S2 . There are integral power-series units ε_f,η_f such that b_f^-∏_ q∈S^ P_q^f(ξ_q) , =ε_f b_f^+ ∏_ q∈S^ P_q^f(ξ_q^ -1 ), ; ((s_F)_f)/( P_f^2) , =η_f∏_ q∈S^ P_q^f(ξ_q^ -1 ) P_q^f(ξ_q) . Indeed in the primitive Atkin--Lehner at the full odd level part relates the sums by sign and translation by the relevant oriented prime ideals of that part (group powers in the variables, preserving canonical level at 2). The Gauss and parameter bases use the same underlying points and choices. Thus the cross identity with this unit holds at all sufficiently high characters, which suffices by boundedness. For the determinant the moving modified-vs-full complementary rank-one Koszul blocks have determinant units (regular sequence m_j,α_j , acyclic in codimension one on integer components). At each fixed odd pair one adds the full complex at the place of ξ_q^ -1 . At level-good q the singular over finite Frobenius determinant gives exactly the ratio up to unit. At minimal ramified transposition inertia, conductor one means Steinberg and the inertia difference has a unit pivot by the residual action, giving free invariant and coinvariant lines integrally with Frobenius q a_q,a_q , a_q=±1 ; thus again the stated ratio. At inertia containing C_3 there is acyclicity by residual testing and no Euler line. And at the level-raising prime ℓ_0 the full complex is acyclic integrally by residual Frobenius, and the Euler polynomials units (level-good with odd trace, or exponent two with no invariant line; a line would force unipotent inertia of conductor at most one). These exhaust the cases by the level construction. Pfaffian-square differs by unit from the modified determinant, proving the identity by triangles. Combining these identities gives V_ A,f =ε_fη_f^ -1 ((b_f^+)/( P_f))^2. Normality on each integer component makes b_f^+/ P_f integral. Residual regularity of P , established in Lemma 110 ([An integral alternating presentation]), and Lemma 115 ([Divisibility over a coefficient order]) then give R_b in the coefficient order itself. Merely knowing integrality on its normalization would not suffice without this quotient-injectivity argument. Nonunits. The numerator congruences and identical residual strict diagrams show that V_A and V_* agree modulo the residual map up to a unit: cancel their common nonzero residual denominator. Thus a nonunit V_* makes V_A a nonunit. If R_b were a unit, the displayed component identities, whose multipliers are integral units, would make V_A a unit on every integer component. Every maximal ideal of those integral components contracts to the unique maximal ideal of the coefficient order; an element in that maximal ideal cannot become a unit on a component. This contradiction proves the nonunit assertion and completes.
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