I read that the group $\prod_{n\in{\mathbb N}}\mathbb Z$ is not a free abelian group. Can someone point me to a proof of this fact?
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1Can you update your question with the definition of a free Abelian group and try if the axioms fit? – gt6989b Apr 29 '21 at 14:05
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1Do you know the difference between $\bigoplus_{n\in\mathbb N}\mathbb Z$ and $\prod_{n\in\mathbb N}\mathbb Z$? – Mathmo123 Apr 29 '21 at 14:05
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@Mathmo123: Could you try to answer the question? – Nandor Apr 29 '21 at 14:11
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I don't think there is a trivial proof. Here is one answer: (free $\mathbb{Z}$-module is the same thing as free abelian group) https://math.stackexchange.com/questions/320444/why-isnt-an-infinite-direct-product-of-copies-of-bbb-z-a-free-module – Mark Apr 29 '21 at 14:18
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2Does this answer your question? Why isn't an infinite direct product of copies of $\Bbb Z$ a free module? – Jendrik Stelzner Apr 29 '21 at 14:23
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Yes it does, thanks! – Nandor Apr 29 '21 at 14:26
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5The question could have been closed as a duplicate. But, as often happens, I do not understand why it was closed for "lack of context". Did all of the people who voted to close know the answer? If not, why are they voting to close - other people might be interested in the answer. – Derek Holt Apr 29 '21 at 14:27
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incidentally, this is the "theorem of Specker" referred to in the answer linked above. as a corollary, Specker's result shows that, rather remarkably, there are only $\aleph_0$ many group homomorphisms from $\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$! – Atticus Stonestrom Apr 29 '21 at 14:35