I am having trouble to prove that $\mathbb{Z}^{(\mathbb{N})}$ is isomorphic as a $\mathbb{Z}$-module to $\text{Hom}_\mathbb{Z}(\mathbb{Z}^\mathbb{N},\mathbb{Z})$, where the isomorphism $\varphi$ is given by $\varphi(e_n)=\text{pr}_n$.

I have read in this other link that the key step to prove this is the fact that any morphism from $\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$ that vanishes in $\mathbb{Z}^{(\mathbb{N})}$ must be the zero morphism, and I am able to prove this fact, but I do not know how to use it to prove that $\varphi$ is an epimorphism.

  • $\begingroup$ There is one more thing you need to prove : that if $f: \mathbb Z^{\mathbb N} \to \mathbb Z$ is a morphism, then it is only nonzero on finitely many $e_i$. This is not super easy $\endgroup$ – Maxime Ramzi Jun 11 '19 at 14:04
  • $\begingroup$ That was indeed my idea, to show that it is non zero on finitely many $e_i$, but I could not solve it. Any other hint, or solution would be appreciated. $\endgroup$ – AMarchionna Jun 11 '19 at 14:08
  • $\begingroup$ It's not that easy. You have to assume it's not the case, then reduce to the case where all of them are nonzero to simplify, then let $a_k =f(e_k)$ and create a sequence $(b_k)$ defined by induction with some considerations on $a_k$ such that looking at $f((b_k))$ will be a contradiction. This hint is probably not enough, so I'll add some words about it when I have the time $\endgroup$ – Maxime Ramzi Jun 11 '19 at 14:24
  • $\begingroup$ Just a note about what I found, trying to find something about homomorphisms $\mathbb Z^{\mathbb N}\rightarrow\mathbb Z$ being non-zero only on finitely many $e_i$'s: This is a theorem due to Specker from 1950 (E. Specker, Additive Gruppen von Folgen ganzer Zahlen). In more modern terminology this would be formulated as "$\mathbb Z$ is slender group" (it is exactly the definition of being slender, see e.g. wiki article on that). The group $\mathbb Z^{\mathbb N}$ is called Baer–Specker group. $\endgroup$ – OnDragi Jun 13 '19 at 14:41
  • $\begingroup$ Similar question is here with links to proofs of the original claim in your question, for example here $\endgroup$ – OnDragi Jun 13 '19 at 15:25

I will post a proof that uses the idea provided by Max. I hope this is correct.

Consider a $\mathbb{Z}$-module morphism $f:\mathbb{Z}^\mathbb{N} \rightarrow \mathbb{Z}$, and suppose that $f(e_n) \neq 0$ for infinitely many $n$. Without loss of generality, we can assume that $f(e_n)>0$ for infinitely many $n$. By considering the projections onto those coordinates, we may assume that $f(e_n)>0$ for all $n$.

Let's consider a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that:

  • $f(2^{\alpha_n}e_n)>2\sum_{i=1}^{n-1}f(2^{\alpha_i}e_i)$
  • $\alpha_{n+1} > \alpha_n$

Let $a_n = 2^{\alpha_n}$, and consider $X=f((a_n)_{n\in\mathbb{N}})$. Then, for every $n\in \mathbb{N}$:

$X \equiv \sum_{i=1}^{n-1}f(a_ie_i) \pmod{a_n}$

Let's take $k$ sufficiently large such that $\sum_{i=1}^{k-1}f(a_ie_i) > |X|$, $a_k - \sum_{i=1}^{k-1}f(a_ie_i) > |X|$ and $a_k > |X|$, and this is possible because $f(e_n)>0$ for all $n$ and the first condition imposed to the sequence.

Then, considering $X$ modulo $a_k$:

$X \equiv \sum_{i=1}^{k-1}f(a_ie_i) \pmod{a_k}$

But because $k$ is large enough, we obtain that $X = \sum_{i=1}^{k-1}f(a_ie_i)$, or $X = \sum_{i=1}^{n-1}f(a_ie_i) - a_k$, but this contradicts our choice for $k$.

Hence our initial assumption was wrong.

This means that $f=a_1\text{pr}_{x_1}+...+a_n\text{pr}_{x_n}$ in $\mathbb{Z}^{(\mathbb{N})}$, but using the fact that if a $\mathbb{Z}$ module morphsim $g:\mathbb{Z}^\mathbb{N} \rightarrow \mathbb{Z}$ that vanishes in $\mathbb{Z}^{(\mathbb{N})}$ must vanish everywhere, we obtain that $f=a_1\text{pr}_{x_1}+...+a_n\text{pr}_{x_n}$ everywhere, as desired.

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  • $\begingroup$ Seems essentially correct to me, just few details: 1) $k$ and $n$ in the formulas should be the same letter; 2) I think that as the first condition for the sequence you want $2^{\alpha_n}>2\sum_{i=1}^{n-1}f(2^{\alpha_i}e_i)$, so that you have $a_k - \sum_{i=1}^{k-1}f(2^{\alpha_i}e_i)$ increasing, right? $\endgroup$ – OnDragi Jun 14 '19 at 8:54
  • $\begingroup$ Yes you are right, I need to fix the formulas. Yes, I use that condition to guarantee the expression to be strictly increasing. $\endgroup$ – AMarchionna Jun 14 '19 at 11:36

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