Explicit Kummer Theory for Quadratic Fields

Kummer Theory for Multiquadratic or Quartic Cyclic Number Fields

“…β e q q such that p, q are odd prime divisors of N and the integers e ∈ {0, 1, 2, 3}, e p ∈ {0, 2}, and e q ∈ {0, 1, 2} satisfy the following conditions: ) generate a cyclic subextension of L 2 /K (respectively, L p /K) of degree dividing 4, and of degree dividing 2 if p ≡ 5 mod 8. By taking products of these roots, we get an extension of K of degree 4 unless all roots generate extensions of degree at most 2, so we may conclude with a counting argument as in the proof of [7,Theorem 11].…”

“…is called the -adelic failure. By [7,Remark 17] we may compute at once all numbers C( n , n ) where is a prime number and n 1, so we only need to provide formulas for the -adelic failure B(M, n ).…”

“…In [7,12] we have described a finite procedure for the computation of the --adelic failure over Q and over quadratic number fields. Now we consider number fields which are either multiquadratic or quartic cyclic, and we provide an explicit finite procedure to compute the -adelic failure B(M, n ) for all prime numbers , all n 1, and for all M 1 such that n divides M , see Sections 8 and 9.…”

“…β only when n + x t + 3 (and t ∈ {2, 3}) and of the form ζ n+x 3 √ β only when n + x 7 (and t = 3). P r o o f. For t ∈ {1, 2} we refer to [7,Theorem 12], so suppose that t = 3. Moreover, the proof of that result also covers the case h 2 and…”

“…P r o o f. If K( √ α)/Q is Galois, then it is abelian because its Galois group has order 8 and it has a quotient isomorphic to Z/4Z. For the second equivalence we may reason as in the proof of [7,Lemma 4], where we consider α • σ(α) instead of N K/Q (α).…”