A not free $\mathbb{Z}$-module From this post I see that $\mathbb{Z}^{\mathbb{N}}$ is not a free abelian group. Someone can explain me why?
Thanks a lot!
 A: The answer https://mathoverflow.net/a/10249 to 
Is it true that, as $\mathbb{Z}$-modules, the polynomial ring and the power series ring over integers are dual to each other? on MathOverflow shows that
$$
\operatorname{Hom}(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})
$$ 
is isomorphic to $\mathbb{Z}^{(\mathbb{N})}$.
If $\mathbb{Z}^{\mathbb{N}}$ were free it would be isomorphic to $\mathbb{Z}^{(X)}$ for some infinite set $X$, so we'd have
$$
\mathbb{Z}^{(\mathbb{N})}\cong
\operatorname{Hom}(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})\cong
\operatorname{Hom}(\mathbb{Z}^{(X)},\mathbb{Z})\cong
\mathbb{Z}^X
$$
which is false because the first group is countable, whereas the final one isn't.

There are different proofs of the result above. An abelian group $L$ is called slender if, for every homomorphism $f\colon \mathbb{Z}^{\mathbb{N}}\to L$, $f(e_n)=0$ for all but a finite number of $n\in\mathbb{N}$ (where $e_n$ is the function which is $1$ on $n$ and $0$ elsewhere).
A theorem by Nunke (Acta Sci. Math. Szeged, 23 (1962)) characterizes slender groups: 

An abelian group $L$ is slender if and only if
  
  
*
  
*$L$ is reduced and torsionfree and
  
*$L$ has no subgroup isomorphic to $\mathbb{Z}^{\mathbb{N}}$ or to the group of $p$-adic integers for any prime $p$.
  

(I'm pretty sure you can find a proof of Nunke's theorem in Fuchs's Infinite Abelian Groups.)
In particular $\mathbb{Z}^{(X)}$ is slender for any countable set $X$. A property of slender groups is the following.

If $L$ is a slender group, $(G_n)$ is a countable family of torsionfree abelian groups and $f\colon \prod_n G_n\to L$ is a homomorphism, then
  
  
*
  
*$f(G_n)=0$ for all but a finite number of $n\in\mathbb{N}$ (the canonical embedding of $G_n$ in the product is considered)
  
*$f$ is zero if and only if it is zero on $\bigoplus_n G_n$ (considered as a subgroup of the product).
  

A consequence of this is that for every homomorphism $f\colon\mathbb{Z}^{\mathbb{N}}\to L$ a slender group, the image of $f$ is finitely generated.
So, since $\mathbb{Z}^{(\mathbb{N})}$ is slender, no homomorphism $\mathbb{Z}^{\mathbb{N}}\to\mathbb{Z}^{(\mathbb{N})}$ is surjective, which proves $\mathbb{Z}^{\mathbb{N}}$ is not free.
