On the 2-class Field Tower of a Quadratic Number Field

Let k = k be a quadratic number field with discriminant ∆. For n ≥ 0, we define fields k inductively by taking k as the compositum of all unramified quadratic extensions of k that are central over k. Then k(∞) = ⋃∞ n=0 k (n,2) is the 2-class field tower of k. In the following, we call k the n central 2-step. The structure of the Galois group Gal (k/k) of the first central 2-step is determined by the principal genus theorem. We want to show (Theorem 1 and 3), that Gal (k/k) is determined completely by the values of the Legendre symbols (ap ), where a | ∆ and p | ∆. Known results in this direction are the theorem of L. Rédei and H. Reichardt [4] on the invariants of the class group of k divisible by 4, as well as a sufficient condition by A. Fröhlich [2] for the class field tower of k to be non-abelian. Moreover we investigate in which cases the 2-class field tower terminates after the first and second central 2-step, i.e., when k(∞) = k and k(∞) = k, respectively. According to A. Fröhlich [2], we have k(∞) 6= k if ∆ has more than three prime divisors. In the case where ∆ has two prime divisors, the theorem of Rédei and Reichardt gives rise to a necessary and sufficient condition for k(∞) 6= k. In this paper, we give a necessary and sufficient condition for k(∞) 6= k in the remaining case where ∆ has three prime divisors (Theorem 4) and show that the 2-class field tower terminates after the second central 2-step if Gal (k/k) is the quaternion group (Theorem 5). In part II of this paper we shall prove that k(∞) 6= k if ∆ has more than four prime divisors. If ∆ has exactly four primes divisors, then the 2-class field tower stops for some groups Gal (k/k) of order 32 after the second central 2-step. In all other cases we have k(∞) 6= k. I would like to thank Professor H. Reichardt for some very valuable suggestions.