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Is the dual of the product of infinite cyclic groups a free abelian group ?  

By a theorem of Specker, the group $Hom(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$$\mathrm{Hom}(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this generalizes to all cardinals:

Question: Is it true that $Hom(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$$\mathrm{Hom}(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$ is a free abelian group for each cardinal $\kappa$ ?

As the answer to such questions sometimes depends on the underlying set theory, I included the "set-theory" tag.

Is the dual of the product of infinite cyclic groups a free abelian group ?  

By a theorem of Specker, the group $Hom(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this generalizes to all cardinals:

Question: Is it true that $Hom(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$ is a free abelian group for each cardinal $\kappa$ ?

As the answer to such questions sometimes depends on the underlying set theory, I included the "set-theory" tag.

Is the dual of the product of infinite cyclic groups a free abelian group ?

By a theorem of Specker, the group $\mathrm{Hom}(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this generalizes to all cardinals:

Question: Is it true that $\mathrm{Hom}(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$ is a free abelian group for each cardinal $\kappa$ ?

As the answer to such questions sometimes depends on the underlying set theory, I included the "set-theory" tag.

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Ralph
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Is the dual of the product of infinite cyclic groups a free abelian group ?

By a theorem of Specker, the group $Hom(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this generalizes to all cardinals:

Question: Is it true that $Hom(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$ is a free abelian group for each cardinal $\kappa$ ?

As the answer to such questions sometimes depends on the underlying set theory, I included the "set-theory" tag.