Abelian Extensions of Arbitrary Fields

Let k be an Hilbertian field, i.e. a field for which Hilbert's irreducibility theorem holds (cf. [1, 5]). It is obvious that the degree of the algebraic closure k of k is infinite with respect to k. It is not obvious that the same is true for the maximal p-extension of k, p a prime number. Let A be a finite abelian group. The question whether there exists a Galoisian extension l/k with Galois group A is, classically. known to be solvable if there exists a finite group G, and a surjective homomorphism G ~ A , such that the following condition is satisfied. Suppose M is a faithful k[G]-module, and let Sk(M ) denote its symmetric algebra over k. The group G acts upon Sk(M ) and on its field of quotients k(M) in a natural way. Then the condition is that the subfield k(M) G of k(M) of all G-invariants is a purely transcendental field extension of k (of. [6, 5]). This applies in particular to the case G = A, and M is the group ring k[A]. In that case we denote k(M) ~ by ka. Let k be an arbitrary field. Recently, the second named author [4-] gave necessary and sufficient conditions in order that, for given k and A, the extension k~/k is purely transcendental, as follows. To check the pure transcendency of kA one has to look at a finite set of Dedekind domains Dq(A)=Z[~q~a)], where the positive integer q(A) runs through a finite subset of Z and (q~A~ is a primitive q(A)-th root of unity. Then one can determine in every Dq(A) an ideal Iq~A~ with the property: k A is purely transcendental over k if and only if the two following conditions are satisfied: (i) every ideal lq~A) is a principal ideal, (ii) if 2" is the highest power of 2 dividing the exponent of A and if the characteristic of k is not equal to 2, then the extension k((2,)/k has cyclic Galois group. This leads to