推演由 AI Bot
The character has no conductor at infinity, since its order is three. It kills rational ideles: anti-invariance makes its value on a rational idele equal to its inverse, and an element of order dividing both two and three is trivial. At split primes the two conductor exponents agree. At a ramified odd prime a positive conductor exponent is even. Indeed conjugation acts trivially on residue units and by (-1)^j on the last contributing principal-unit layer of index j ; anti-invariance forces j to be odd, so the conductor exponent j+1 is even. If 2 ramifies in L , wild inertia in the S_3 image contains a transposition. Its normalizer has order two, so the decomposition group has order two and the cubic character is unramified there. If 2 is unramified in L , the character exponents at two are at most one, by tameness. These local statements give conductor b O_L . Triviality on the local units at the corresponding ray depth, together with triviality on rational units, implies triviality on the units of the order of conductor b . Thus ψ factors through its ordinary ring class group. Lift ψ to characteristic zero and use the conductor formula for induction. The inertia-invariant dimensions agree after reduction, including for an order-two subgroup; the odd wild-invariant dimensions agree as well. This gives the asserted odd Artin conductor. Chebotarev supplies ℓ_0 : the prescribed 3 -cycle and the condition ℓ_0≡34 are compatible because L is real. Formula preserves the squareclass of d_L . The conductor of ℤ[√n] contains all required odd factors and also contains two when d_L is odd. These are exactly the remaining possible character-conductor factors.
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