# Showing the countable direct product of $\mathbb{Z}$ is not projective

I am trying to prove that the direct product $M = \mathbb{Z} \times \mathbb{Z}\times \cdots$ is not a projective $\mathbb Z$-module and I am stuck near the end of the proof because of the authors use of the word infinitely divisible.

I will list the sketch of the proof and try to explain where I am stuck.

We proceed by contradiction so suppose $M$ is contained in some free module $F$ with basis $B$. Set $N = \mathbb{Z} \oplus \mathbb{Z} \oplus\cdots$ and observe that $N$ is a submodule of $M$.

Since $N \subset F$ there exists $B' \subset B$ such that $B'$ is a basis for $N$ and consider the free module $F' \subset F$ determined by $B'$.

Notice $F'+M \subset F$ gives $M/(M \cap F') \cong (F'+M)/F' \subset F/F'$ so we have $M/(M \cap F')$

The next step in the proof requires to consider sequences of signs so let $s = (s_1, s_2, \ldots)$ be a sequence of plus and minus signs and consider an element $m_s := (s_1 , 2 s_2, \ldots , k! s_k, \ldots) \in M$

The next point in the proof is what I don't understand, the notes I am using say $m_s +(M \cap F')$ is infinitely divisible in $F/F'$ and use this to show $M$ cannot be contained in any free $Z$ module. My question is

How do we show $m_s +(M \cap F')$ is infinitely divisible in $F/F'$ and how do translate the word infinitely divisible into definitions from Hungerford or Dummite and Foote?

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For an element $g$ in an additive group $G$, "$g$ is infinitely divisible" means "for every $n$ in $\mathbb{N}$ there exists $h$ in $G$ such that $g=nh$". – Chris Eagle Sep 14 '11 at 23:16

First note that if $\mathbb{Z}^{\mathbb{N}}$ is free it has a basis $\beta$ with cardinality non-countable $\kappa$, consider as sets, then the cardinality of $Hom(\beta, \mathbb{Z})$ is $2^{\kappa}>\kappa$. Now the final argument is that if you take morphism $f$ from $\mathbb{Z}^{\mathbb{N}}$ in to $\mathbb{Z}$ and you compose it wiht the inclusion in the jth coordinate $i_j$ , $fi_j=0$ for almost all $j\in\mathbb{N}$. That is for almost all elements in the basis the morphism is zero, from here the cardinality of $Hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$ is $\kappa$, contradiction!!!!!, $Hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $Hom(\beta, \mathbb{Z})$ must have the same cardinality.
I simply don't understand what you want to prove here. Do you want to prove that $\mathbb Z^{\mathbb N}$ is not free? If so, then why is $Hom$ involved here? Last but not least, what inclusion is $i_j$ (including what and where)? – user26857 Jan 3 '13 at 0:15