Proof. Take a nondyadic prime q of F with s(Fq) = 2 (there must be one since s(F ) = 2). We show that the order d of the ideal class [q] in the ideal class group of F is even. For suppose d is odd. Then q = (a) is a principal ideal and the number a is a q-adic prime times a q-adic square. Since −1 is not a local square at q, we have (−1, a)q = −1. On the other hand, we claim that (−1, a)p = 1 for all the remaining primes p of F. For a dyadic prime p this is obvious since −1 is a local square at p (the local dyadic levels are all equal to 1). When p is a finite nondyadic prime and p 6= q, then a is a local unit at p (by our choice of a), hence again the Hilbert symbol is trivial. Since s(F ) = 2, there are no real infinite primes, and at complex infinite primes any Hilbert symbol is trivial. This proves our claim, contradicting Hilbert reciprocity law. Hence d must be even, and so also the ideal class number of F is even. If K is any number field Witt equivalent to F, then K and F have the same Witt equivalence invariant. Hence, if F satisfies Conner’s level conditions, so does K, and, as has been already proved, it has even class number.