Lemma 48 ([Scalar residual order]). For an odd prime q N , let b_q∈ℤ20 be the Γ -Frobenius exponent at one of the places over q , and put g_q=2^v₂(b_q) . Then ord_{tL_} alg (C(t)) =4∑_q N; q≠2 g_q.

Let w^h=(α) as ideals, α=(a+b√-d)/2 , a^2+db^2=2^ h+2 . Both a,b are odd: otherwise α would have the same reduction at both primes over 2, impossible. Each odd prime of b has (2/q)=1 (here q denotes that prime), by the norm equation and h odd; hence b≡±18 . For h≥3 , a^2≡916 , and the unit α at w is congruent to a modulo 8. For h=1 we have d=7 , and the unit root of X^2±X+2 is again ±38 . Thus -1,2,α give a basis of local squareclasses there. They are also a basis of global squareclasses with only dyadic support (odd h ). With trivial scalar 𝔽_2 and no odd places yet allowed, the strict problem is acyclic: restriction to w is an isomorphism in degree one by that basis and in degree zero on constants, so use the above amplitude and Euler characteristic. On either individual residual constituent with scalar twist Ψ_t , now adding the two places over each odd q N gives singular blocks by the unramified inflation comparison. Each has determinant over 𝔽_2[[t]] of order g_q=2^ v_2(b_q) in t , b_q∈ℤ_2 the exponent at one place over q . Indeed Frobenius-minus-one on the scalar inertia H^1 coefficient is (1+t)^ ±b_q -1 (the cardinality twist reduces to one). Extension of the two constituents is only needed at full support. There are two places over each q and two residual constituents, so summing these orders gives. Determinant multiplicativity for the residual filtration allows the two constituents to form a nonsplit extension.