Friday, 18 July 2008

gr.group theory - dual of Z^I for uncountable I

With regard to Mariano's answer, I believe some clarification is in order. A closely related question was asked by Michael Barr and answered by user Ralph here. In brief, the homomorphism named in Martin's question is in fact an isomorphism, provided that $I$ has cardinality less than the first measurable cardinal.



Shelah and Strüngmann (accessible here) refer to this result as well, using the same source given by Ralph, just before Definition 2.1:




For generalizations to products of larger cardinalities and the resulting definition of slenderness for abelian groups we refer to [EM] or [F1]




where [EM] is the text by Eklof and Mekler. It seems that Shelah and Strüngmann are talking about something slightly different: homomorphisms out of free complete products (but using a notation which could unfortunately suggest direct products).

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