Abstract.
In {\it Almost Free Modules, Set-theoretic Methods}, Eklof and Mekler [5,p. 455, Problem 12] raised the question about the existence of dual abelian groups G which are not isomorphic to \({\mathbb Z} \oplus G\). Recall that G is a dual group if \(G \cong D^*\) for some group D with \(D^* ={\rm Hom}(D,{\mathbb Z})\). The existence of such groups is not obvious because dual groups are subgroups of cartesian products \({\mathbb Z}^{D}\) and therefore have very many homomorphisms into \({\mathbb Z}\). If \(\pi\) is such a homomorphism arising from a projection of the cartesian product, then \(D^*\cong{\rm ker} \pi \oplus{\mathbb Z}\). In all `classical cases' of groups {\it D} of infinite rank it turns out that \(D^* \cong{\rm ker} \pi\). Is this always the case? Also note that reflexive groups G in the sense of H. Bass are dual groups because by definition the evaluation map \(\sigma: G \longrightarrow G^{**}\) is an isomorphism, hence G is the dual of \(G^*\). Assuming the diamond axiom for \(\aleph_1 (\Diam_{\aleph_1})\) we will construct a reflexive torsion-free abelian group of cardinality \(\aleph_1\) which is not isomorphic to \({\mathbb Z} \oplus G\). The result is formulated for modules over countable principal ideal domains which are not field.
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Received July 1, 1999; in final form January 26, 2000 / Published online April 12, 2001
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Göbel, R., Shelah, S. Some Nasty reflexive groups. Math Z 237, 547–559 (2001). https://doi.org/10.1007/PL00004879
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DOI: https://doi.org/10.1007/PL00004879