to the memory of professor makoto ishidaFor quartic fields with a quadratic subfield, explicit lower bounds of the regulators are given in terms of the discriminant. Further, these bounds are shown to be in a sense best possible by constructing infinitely many quartic fields with explicit fundamental units whose regulators approach our lower bounds as their discriminants become large. The fields constructed are new families of non-galois fields given by explicit quartic polynomials.
All number fields we consider are in the complex number field. The symbol (S} denotes a multiplicative group generated by S.For a finite extension k/Q, let E be the group of units of k, and E be the group generated by all units o proper subfields of k together with roots of unity in k. We define the group H of relative units of k by H= { e E N/,(z) is a root of unity ior a proper subfield k' o k}.Let us consider the problem to construct E with the help of E. It is interesting to utilize H together with E when (E "E)= / oo.Hasse [2] has treated such a case when k is a real cyclic quartic number field. We are going to treat the case when k is a non-galois quartic (resp. sextic) number field having a quadratic subfield (resp. a quadratic and a cubic subfields). Then the galois closure o k/Q is a dihedral extension o degree 8 or 12 over Q. We restrict .our investigation on such extensions.From now on, we assume n= 2 or 3. Let L/Q be a galois extension of degree 4n with the galois group G (, } = () 1.The invariant subfield of the subgroup (r} (resp. (ar}, (an}) is denoted by K (resp. F,/2), and the maximal abelian subfield by A. Then K and F are non-galois number fields o degree 2n which we are going to study.The quadratic subfield of K (resp. F) is denoted by K. (resp. F.). When n=3, the cubic subfield o both K and F is denoted by K. The quartic field A is the composite field of K. and F. which contains another quadratic subfield A. Note that A= 9 when n=2.It is easy to show the following, which is in Nagell [6] when n= 2. Proposition 1. When LR=f2, we have E:=E and EF=E.Therefore we treat the two cases: CaseI: LR=K. CaseII: LcR. Taking into account that all roots of unity of L is contained in the quartic subfield A, we take and fix a generator (resp. , p) o the group of roots o unity o A (resp. A., F).1. Type of EK and EF. A typical example o K and F are a pure number field of degree 2n. The method, which is used in Stender [8],
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