Kourovka Notebook Problem 21.13
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Definitions and mathematical statement
Let $\omega=\{0,1,2,\ldots\}$ and let $\mathbb Z^\omega=\prod_{i\in\omega}\mathbb Z$ be the unrestricted product, with coordinatewise addition. Its elements are all integer sequences, without a finite-support condition. For an abelian subgroup $A\le\mathbb Z^\omega$, define its algebraic dual by \[ A^*=\operatorname{Hom}_{\mathbb Z}(A,\mathbb Z), \qquad (f+g)(a)=f(a)+g(a). \] Homomorphisms are abstract additive maps. In particular this is the integer-valued algebraic dual, not a dual defined using the circle group or a topology.Write $\mathfrak c=2^{\aleph_0}$ for the cardinality of the continuum. Does there exist a subgroup $A\le\mathbb Z^\omega$ for which $A^*$ is free abelian of rank $\kappa>\mathfrak c$? Freeness means that some family $(f_\alpha)_{\alpha<\kappa}$ is a basis: every member of $A^*$ has a unique expression as a finite integer linear combination of these homomorphisms. Equivalently, $A^*\cong\bigoplus_{\alpha<\kappa}\mathbb Z$.
The known continuum-rank case motivates the strict inequality; it would not answer the question. The subgroup $A$ itself is not required to be free, countable, or of finite index. Since $|A|\le\mathfrak c$, there are at most $\aleph_0^{\mathfrak c}=2^{\mathfrak c}$ maps $A\to\mathbb Z$, giving the upper bound $\kappa\le2^{\mathfrak c}$. Attaining that upper bound is not required: any free dual rank strictly above $\mathfrak c$ would suffice.
Short English statement
Can a subgroup of the countable product of integers have an integer-valued algebraic dual that is free abelian of rank above the continuum?Sources
- S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2522 (KOU-21.13), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
- S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.13. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.
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