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Trying to answer this question, I encountered the following question, the answer to which should be known but it is hard to Google, so I did not find it.

Let $G=\prod_{n\in\mathbb N}\mathbb Z_n$ be a full product of finite cyclic groups $\mathbb Z_n=\mathbb Z/n\mathbb Z$. Exists there a homomorphism $f:G\to\mathbb Z$ such that $f((1,1,\dots))=1$?

According to this question, my question should be equivalent to the following.

Is the group $\{(m,m,\dots)\in G: m\in\mathbb Z\}$ a direct summand of the group $G$?

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Alex Ravsky
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  • Interesting question! Just a thought: by composing with the projection $\prod_{n \in \mathbb{Z}{>0}} \mathbb{Z} \to \prod{n \in \mathbb{Z}{>0}} \mathbb{Z}/n\mathbb{Z}$ we can instead ask if there is a homomorphism $\prod{n \in \mathbb{Z}_{>0}} \mathbb{Z} \to \mathbb{Z}$ with the above properties. – Viktor Vaughn Dec 10 '14 at 08:57
  • Does $\mathbb{N}$ include $0$? $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$ and $\mathbb{Z}/\mathbb{Z} = {0}$. Or does it start at $2$? – KDuck Dec 10 '14 at 08:58
  • @SpamIAm A homomorphism from $f:\prod_{n\in\mathbb N}\mathbb Z$ to $\mathbb Z$ such that $f((1,1,\dots))=1$ exists - it is a projection to 1st coordinate. – Alex Ravsky Dec 10 '14 at 09:02
  • @Krahapp No, $\mathbb N$ does not include $0$. The factor $\mathbb Z_1$ can be skipped. – Alex Ravsky Dec 10 '14 at 09:04
  • Hm, but that homomorphism doesn't descend to the quotient. I guess we also need to have $\prod_{n \in \mathbb{Z}_{>0}} n \mathbb{Z}$ contained in the kernel of the homomorphism. – Viktor Vaughn Dec 10 '14 at 09:06

1 Answers1

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The only homomorphisms $\varphi:\prod_{n \in \mathbb{N}} \mathbb{Z} \to \mathbb{Z}$ are the obvious ones $$\varphi\left(a_1,a_2,\dots\right)=\sum_{n\in\mathbb{N}}a_nb_n,$$ where $\left(b_1,b_2,\dots\right)$ is a sequence of integers with $b_n=0$ for all but finitely many $n$. In particular, $\varphi$ is determined by its restriction to $\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$. For a proof, see for example https://mathoverflow.net/questions/10239/is-it-true-that-as-z-modules-the-polynomial-ring-and-the-power-series-ring-ove/10249#10249.

But the obvious map $\theta:\prod_{n \in \mathbb{N}} \mathbb{Z} \to \prod_{n \in \mathbb{N}} \mathbb{Z}/n\mathbb{Z}$ sends every element of $\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ to a torsion element, which must therefore be in the kernel of any homomorphism $\alpha:\prod_{n \in\mathbb{N}} \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}$. So by the theorem above, $\alpha\circ\theta=0$ and so $\alpha=0$.

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Jeremy Rickard
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