Wikipedia states that every free abelian group is slender. Where can I find a proof?
If this is not trivial, then I will also need a reference to use in my paper.
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$\begingroup$ Welcome to MathOverflow. Your question is a good one, but needs improvement for this forum. Use the Help Center in the help menu to find out more about asking questions on MathOverflow. In particular, it is good to show more of what you tried (e.g. textbooks, colleagues, other websites besides Wikipedia) in researching before asking here. You might change the focus of your question into a request for references, once you've shown that other efforts on your part have failed. For this question, I might try books on abelian groups, perhaps by Fuchs or by Kaplansky. $\endgroup$– The Masked AvengerCommented Nov 15, 2014 at 18:36
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4$\begingroup$ @TheMaskedAvenger I think this question is fine, and that you could have left "books by Fuchs or Kaplansky" as an answer $\endgroup$– Yemon ChoiCommented Nov 15, 2014 at 18:37
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$\begingroup$ the short paper by Nunke is available; it appears everything relevant happened from the late 1950's to the early 1960's, various authors projecteuclid.org/euclid.bams/1183524151 $\endgroup$– Will JagyCommented Nov 15, 2014 at 18:37
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$\begingroup$ @WillJagy Why not leave this as an answer? $\endgroup$– Yemon ChoiCommented Nov 15, 2014 at 18:39
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$\begingroup$ @Yemon, alright. $\endgroup$– Will JagyCommented Nov 15, 2014 at 18:39
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2 Answers
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This follows fairly straightforwardly from the fact that every map from $A=\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$ factors through a finite subproduct (let me call this "Specker's theorem"). Suppose $F$ is a free abelian group and $f:A\to F$ is a homomorphism. Since a subgroup of a free abelian group is free, we may assume $f$ is surjective. If $F$ is finitely generated, the conclusion follows from Specker's theorem, so we may assume $F$ is infinitely generated. Since $F$ is free, the map $f$ splits and so $F$ is a direct summand of $A$. But there are uncountably many homomorphisms $F\to\mathbb{Z}$, and hence uncountably many homomorphisms $A\to \mathbb{Z}$. This contradicts Specker's theorem.
answered Nov 15, 2014 at 21:26
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The short article by Nunke is available, with its own bibliography. It seems everything relevant happened from the late 1950's to the early 1960's.
I'm looking at Kap's book Infinite Abelian Groups. I do not see the word slender. However, Kap admired Baer, and one of nine references by Baer is Die Torsionsuntergruppe... , plus Baer is thanked in the Introduction. So I think it is a matter of figuring out how the concept "slender" is discussed here.
Hmmm; originally 1954, but second edition 1968. Should also check Fuchs, who has more than one edition.
answered Nov 15, 2014 at 18:47
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$\begingroup$ Sorry, this is not quite an answer... this might be, or might not be, a useful direction to look for an answer. $\endgroup$ Commented Nov 15, 2014 at 18:58
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1$\begingroup$ @AlexanderGelbukh, would you prefer that people not contribute what they have? I expect your kind of reaction on MSE, where most questions are homework and people complain if they cannot turn in an answer word for word as their own work. $\endgroup$ Commented Nov 15, 2014 at 19:03
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3$\begingroup$ Thank you very much for your contribution! Still I think advising on useful direction to look for an answer should be made as a comment, not as an answer. Then we can upvote the comment if it proved to be useful. An answer is as a proof as a theorem: you don't say "PROOF. I think one could look for a proof along these lines... QED" $\endgroup$ Commented Nov 15, 2014 at 19:06
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1$\begingroup$ @AlexanderGelbukh it is the best response so far, and many members who might have better ones have likely not seen your question yet. I recommend withholding judgment for a while. $\endgroup$ Commented Nov 15, 2014 at 19:06
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5$\begingroup$ @TheMaskedAvenger, also Yemon asked me to post it as an answer. Note that Alexander and Irina are probably slightly different people, but likely relatives. $\endgroup$ Commented Nov 15, 2014 at 19:09