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Question. Prove that $\prod\limits_{\mathbb N}\mathbb Z$ is not a free $\mathbb Z$-module.

I have searched over MSE and found answers here and here. The question I want to ask is, whether there is an alternative proof for this fact, as the solution provided in this question used the fact that free $\mathbb Z$-module has no non-trivial divisible submodule to force a contradiction in cardinality. In particular, in both questions above it is mentioned that counting the cardinality of $\mathrm{Hom}_{\mathbb Z}(\prod\limits_{\mathbb N}\mathbb Z,\mathbb Z)$ can also lead to a contradiction if $\prod\limits_{\mathbb N}\mathbb Z$ is free, so I also would like to ask how to prove the Question in this way. Any help is appreciated.

Thanks in advance..

josephz
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  • Doesn't the accepted answer already mention a theorem of Specker? – Hagen von Eitzen Nov 01 '17 at 16:26
  • I think Specker's original proof (published in 1950 in Portugaliae Math., vol. 9, pp. 131-140) is different from the proofs on MSE. – Andreas Blass Nov 01 '17 at 17:09
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    Here is the exercise in Dummit&Foote that walks through the proof (I find it very easy to understand): https://i.stack.imgur.com/eL5Si.png This is from pg. 358 in the 3rd edition. – Steve D Nov 01 '17 at 17:42
  • @HagenvonEitzen Yes, but only knowing "a theorem of Specker" I cannot find the exact source of it. – josephz Nov 02 '17 at 01:05
  • @AndreasBlass Thank you! I will check this source.. – josephz Nov 02 '17 at 01:11
  • @SteveD Thank you! It is indeed easy to understand, yet is also rather alike one of the answers mentioned above... – josephz Nov 02 '17 at 01:26

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