Direct Products of Free Modules Need Not be Free (10.3.24)

Dummit and Foote Abstract Algebra, section 10.3, exercise 24:

For each $i \in \mathbb{Z}^+$ let $M_i=\mathbb{Z}$ be free and let $M=\prod_{i \in \mathbb{Z}^+}M_i$. Let $N$ be the restricted direct product of $M$ considered as a submodule of $M$ by componentwise multiplication by $\mathbb{Z}$. Assume $M$ is free with basis $B$.

(a) Prove $N$ is countable.
(b) Prove there exists countable $B_1 \subseteq B$ such that $N \subseteq N_1 = RB_1$. Prove $N_1$ is countable.
(c) Prove $\overline{M}=M/N_1$ is free. Deduce that for any nonzero $\overline{x} \in \overline{M}$ there are only a finite number of integers $k$ such that $\overline{x}=k\overline{m}$ for some $\overline{m} \in \overline{M}$.
(d) Let $S=\{(b_1,...)~|~b_i = \pm i!\}$, and prove $S$ is uncountable. Deduce there exists $s\in S$ with $s \not \in N_1$.
(e) Prove there exists $s \in S$ with $s \not \in N_1$ such that for every positive integer $k$ there exists $\overline{m} \in \overline{M}$ such that $\overline{s}=k\overline{m}$, a contradiction by (c).

Proof: (a) Recall that countable unions of countable sets are countable. We have $$N=\bigcup_{i∈\mathbb{Z}^+}N_i'$$ where $N_i'=\{x~|~x∈N \land \text{last}(x)=i\}$ for $\text{last}(x)$ being the position of the last nonzero coordinate of $x$, is a countable union, so that it suffices to show $N_i'$ is countable. We have $$N_i'=\bigcup_{j∈\mathbb{Z}^++\{0\}}N_{i,j}''$$ where $N_{i,j}''=\{x~|~x∈N_i'' \land \sum_{y∈x} |y| = j\}$ is also a countable union, so we must prove $N_{i,j}''$ is countable; in fact, since for every $y∈x∈N_{i,j}''$ we must have $-j≤y≤j$ implying $|N_{i,j}''| < (2j+1)^i$ is finite, and as such is countable.

In general, this implies that any restricted direct product of copies of $\mathbb{Z}$ will be countable.

(b) Let $$B_1=\bigcup_{x∈N}V_x$$ where $V_x$ is the set of elements from $B$ involved in the nonzero terms of the unique sum representation of $x$. Since each $V_x$ is finite and $N$ is countable, $B_1$ is countable and clearly $N \subseteq N_1 = RB_1$. Since $N_1$ has $B_1=\{a_1,...\}$ as a free basis (uniqueness of sums by the freeness of $M$), we have $N_1 ≅ Ra_1 \oplus ...$ by $n=z_1a_1+... \mapsto (z_1,...)$ and in this fashion $N_1$ is the restricted direct product of a number of copies of $\mathbb{Z}$ and by the note at the end of (a) $N_1$ is thus countable.

(c) We claim $\overline{B \setminus B_1}$ is a free basis for $\overline{M}$. We have $\overline{B}$ generates $\overline{M}$, and the terms involving elements from $B_1$ may be removed as they are within $N_1$, so that $\overline{B \setminus B_1}$ generates $\overline{M}$. Further, if $\overline{z_1c_1+...+z_kc_k}=\overline{z_1'c_1+...+z_k'c_k}$ as sums with $c_i∈B \setminus B_1$, then for some $n∈N-1$ we have $z_1c_1+...+z_kc_k+n=z_1c_1+...+z_kc_k+z_1''a_1+...+z_n''a_n=z_1'c_1+...+z_k'c_k$ and now $z_1'c_1+...+z_k'c_k-z_1c_1-...-z_kc_k=z_1''a_1+...+z_n''a_n$. Since the terms $a_i$ and $c_i$ don't coincide, by the freeness of $M$ we can conclude $z_1c_1+...+z_kc_k=z_1'c_1+...+z_k'c_k$ and again by freeness that $z_i=z_i'$.

Since such $k$ manifest as the common divisors of the coefficients in $\mathbb{Z}$ of the elements from $\overline{B \setminus B_1}$, and there are only a finite number of divisors for any integer, we have there are only a finite number of $k$ per $\overline{x}$.

(d) Since there are two choices per coordinate of an element of $b_i$, this is equivalent to proving $\prod^\infty \{0,1\}$ is uncountable, which is standard by Cantor's diagonal argument. Since $S$ is now uncountable, $S \subseteq N_1$ would allow a natural surjection $\mathbb{Z} → S$ by the countability of $N_1$; therefore, such $s∈S$ and $s∉N_1$ exists.

(e) Let $s=(b_1,...)$ be as in (d) and choose arbitrary $k∈\mathbb{Z}$. We have $$\overline{(b_1,...)}=\overline{(0,...,b_k,...)+(b_1,...,b_{k-1},0,...)}=$$ $$\overline{(0,...,b_k,...)}=k\overline{(0,...,b_k',...)}$$ as $k \mid b_i=\pm i!$ for all $i \geq k$.$~\square$