A not free $\mathbb{Z}$-module

Question

From this post I see that $\mathbb{Z}^{\mathbb{N}}$ is not a free abelian group. Someone can explain me why?

Thanks a lot!

Answer 1

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.

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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.