Cyclic extensions without relative integral bases

Abstract

Let K be an algebraic number field and o the ring of algebraic integers in K. If o is a principal ideal domain (p.i.d.) then any finite extension A/K has an integral basis over o (i.e., the ring of integers D = ?)(A) of A is a free o-module). The converse of this was shown by Mann [5 ]. More precisely, he proved that if o is not a p.i.d., there is a quadratic extension A/K which has no integral basis over o. Thus o is a p.i.d. if and only if every quadratic extension of K has an integral basis. One can also show ([7] or the corollary below) that if K contains a primitive cube root of 1, then o is a p.i.d. if and only if every cyclic extension of degree 3 has an integral basis. However, the analogous theorem with 3 replaced by a prime p >3 is false. The problem considered here is the following. Given a finite group G of order n and an algebraic number field K, consider all normal extensions A/K with Galois group isomorphic to G. What are the omodule types of the Z?(A) for these extensions? In particular, when are all the Z(A) free? In Theorems 1 and 2 we answer these questions in the case that G is cyclic of order n and K contains the nth roots of unity. A finitely generated torsion free o-module M of a given o-rank is characterized by its Steinitz class C(M) = CD(M) which is an o-ideal class of K. Specifically, M-o(r-l) (DJ where r is the o-rank of M, 0(r-1) is a free o-module of rank r1, and J is any ideal in the class C(M). If A/K is a finite extension, let Z=Z(Z(A)/o) be the discriminant ideal and let A =A(A/K) be the discriminant of a basis of A/K. It was shown by Artin that the ideal (Z/(A))1I2 is an o-ideal lying in CD(Z(A)). (For proofs of the above remarks, see Artin [ ] or Fr6hlich [2] and [3].) DEFINITION. If I is an odd prime, let d(l) = (1-1)/2, and letd(2) = 1. We define, for any integer n, d(n) =g.c.d. { d(l) I I is a prime divisor of n}.