# Isomorphism between Z-modules

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.

• 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 – Maxime Ramzi Jun 11 '19 at 14:04
• 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. – AMarchionna Jun 11 '19 at 14:08
• 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 – Maxime Ramzi Jun 11 '19 at 14:24
• 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. – OnDragi Jun 13 '19 at 14:41
• Similar question is here with links to proofs of the original claim in your question, for example here – 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.

• 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? – OnDragi Jun 14 '19 at 8:54
• Yes you are right, I need to fix the formulas. Yes, I use that condition to guarantee the expression to be strictly increasing. – AMarchionna Jun 14 '19 at 11:36