For each positive integer $i$ let $M_i$ be the free $\Bbb Z$-module $\Bbb Z$, and let $M$ be the direct product $\prod _{i \in \Bbb Z^+} M_i$. Each element of $M$ can be written uniquely in the form $(a_1,a_2,a_3,...)$ $a_i \in \Bbb Z$ for all $i$ Let $N$ be the submodule of $M$ consisting of all such tuples with only finitely many nonzero $a_i$. Assume $M$ is a free $\Bbb Z$-module with basis $B$.
i) Show that $N$ is countable.
ii) Show that $\exists B_1$ countable subset of $B$ s.t $N$ is contained in the submodule, $N_1$, generated by $B_1$. Show also that $N_1$ is countable.
iii) Let $\overline M= M/N_1$. Show that $\overline M$ is a free $\Bbb Z$-module. Deduce that if $\overline x$ is any nonzero element of $\overline M$ then there are only finitely many distinct positive integers $k$ s.t $\overline x=k \overline m$ for some $m \in M$ (depending on $k$).
iv) Let $S= \{(b_1,b_2,b_3,.....)| b_i=\pm i! \forall i\}$. Prove that $S$ is uncountable. Deduce that there is some $s \in S$ with $s \notin N_1$.
v) Show that the assumption $M$ is free leads to a contradiction: by iv) we may choose $s \in S$ with $s \notin N_1$. Show that for each positive integer $k$ there is some $m \in M$ with $\bar s= k \bar m$, contrary to iii).
I have proved i), iii). Showed $\overline M$ is a $\Bbb Z$ module. But having problems to show the remaining parts ii), ii), v). Can someone help me?
iii) Note that we still assume that $M$ is free with basis $B$. Let $B_2=\overline{(B\setminus B_1)}$. Then $\overline M$ is free with basis $B_2$. Indeed given any map $B_2\to A$, we can lift it to a map $B\setminus B_1\to A$, extend this by zero to a map $B\to A$, hence get a homomorphism $M\to A$ that vanishes on $B_1$ and hence also on $N_1$, hence this factors over a homomorphism $\bar M\to A$. Check that this is unique (use that $b\mapsto b+N_1$ is a bijection between $B\setminus B_1$ and $B_2$).
Let $\bar x\in\overline M$ be nonzero. Then in the basis expansion $x=\sum_{b\in B}a_bb$, some coefficient $a_b$ with $b\in B\setminus B_1$ is nonzero. If $\bar x=k\overline{m}$, then $k$ must be one of the finitely many divisors of $a_b$ (because the $b$-coefficient of $x-km$ is zero).
iv) By the sign choices, the elements of $S$ are in bijection with $\mathcal P(\mathbb N)$, which is uncountable. As $N_1$ is countable, $S$ cannot be a subset of $N_1$.
v) Given $k$, note that $s\equiv s'$ where $s'$ is the same as $s$ except that has the first $k-1$ components are set to zero. Then $s'$ can be divided by $k$ componentwise.