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1.66 (1965)

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Suppose that $T$ is a periodic abelian group, and $\mathfrak{m}$ an uncountable cardinal number. Does there always exist an abelian torsion-free group $U(T, \mathfrak{m})$ of cardinality $\mathfrak{m}$ with the following property: for any abelian torsion-free group $A$ of cardinality $\leqslant \mathfrak{m}$, the equality $\text{Ext}(A, T) = 0$ holds if and only if $A$ is embeddable in $U(T, \mathfrak{m})$?

Contributor: L. Ya. Kulikov

Progress

No, not always. There is a model of ZFC in which for a certain class of cardinals $\mathfrak{m}$ the answer is negative (S. Shelah, L. Strüngmann, J. London Math. Soc., 67, no. 3 (2003), 626–642). On the other hand, under the assumption of Gödel’s constructivity hypothesis ($V = L$) the answer is affirmative for any cardinal if $T$ has only finitely many non-trivial bounded $p$-components (L. Strüngmann, Ill. J. Math., 46, no. 2 (2002), 477–490).

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