Abstract. We investigate effective properties of uncountable free abelian groups. We show that identifying free abelian groups and constructing bases for such groups is often computationally hard, depending on the cardinality. For example, we show, under the assumption V " L, that there is a first-order definable free abelian group with no first-order definable basis.
IntroductionHow complicated is it to find a basis of a free abelian group? Can it be done recursively, as we do when building bases for vector spaces? Here by a basis we mean a subset which is both linearly independent and spans the whole group (with integer, rather than rational coefficients). The difficulty is that unlike vector spaces, free abelian groups can contain maximal linearly independent subsets which are not bases. For countable groups, there is a strengthening of linear independence, originally used by Pontryagin [28], which allows us to recover a recursive construction. This notion generalises p-independence, which is widely used in the study of torsion-free abelian groups. Recall that a subgroup H of a torsion-free abelian group G is pure if G X QH " H; that is, if for all n P Z and all h P H, if n divides h in G then it also divides it in H. Definition 1.1. Let G be a torsion-free abelian group. A subset A Ď G is Pindependent if it is linearly independent and its span is a pure subgroup of G.Note that any subset of a P -independent set is also P -independent. Let Z ω " À kPN Z denote the countably generated free abelian group. The following is implicit in Pontryagin's work, and is stated in the following way, for example, in Downey and Melnikov's [5] (who generalised it to completely decomposable groups).Again, to be specific, SpanpBq is the set of elements of G of the form ř m i a i where a i P B and m i P Z; B is a basis of Z ω if it is linearly independent and spans Z ω , if and only if Z ω " À bPB Zb. Of course every basis of Z ω must be Pindependent. Proposition 1.2 tells us that a basis for Z ω can be built recursively, repeatedly extending finite P -independent subsets while ensuring that the next element of the group (in some arbitrary ω-enumeration of the elements of the group) belongs to the span of the basis that we are building. Can this process be mimicked when we are given an uncountable free abelian group? We know that there is no important difference between countable and uncountable vector spaces. A basis for a vector space can be built by transfinite recursion, extending as usual at successor steps and taking unions at limit stages. Searching the literature, we found no such construction for uncountable free abelian groups. The purpose of this paper is to show that in most cases, such a construction cannot be performed. One key point is that Proposition 1.2 heavily relies on the fact that B is finite.1 A recursive construction can get stuck at a limit stage: we can find elements a 1 , a 2 , . . . , of a free abelian group G such that each finite initial segment ta 1 , a 2 , . . . , a n u can be extended to a bas...