>>184 より (a, b, c)T = (c, -b, a) だから
Ψ(fT) = [([c, (b + √D)/2], sign(c))]

I = [a, (-b + √D)/2]
J = [c, (b + √D)/2]
θ = (-b + √D)/2
とおく。

θ'I = [a(-b - √D)/2, ac]
= a[(-b - √D)/2, c]
= a[c, (b + √D)/2]
= aJ
よって
I = (a/θ')J

N(θ') = ac だから
N(a/θ') = a/c

Ψ(fT) = [((a/θ')[c, (b + √D)/2], sign(c)sign(N(a/θ')))]
= [([a, (-b + √D)/2], sign(c)sign(a/c))]
= [([a, (-b + √D)/2], sign(a))]
= Ψ(f)