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The module of integer sequences consists of all functions ${\mathbb{N}}_0\to \mathbb{Z}$, with pointwise addition and integer scalar multiplication. Its submodule of finitely supported sequences is the direct sum of modules ${⨁}_{j\ge 0}\mathbb{Z}$, a free module with coordinate basis of a module $(e_j)$. The full product is not a free module.

To prove this, suppose it had a basis of a module $E$. The countable basis support lemma places every finitely supported sequence in a submodule $C$ generated by a countable subset of $E$. The basis partition gives a module direct sum lemma supplies a projection $\pi$ onto its complementary direct summand, with kernel $C$. By countability of countably generated abelian groups, $C$ is countable. The Cantor diagonal argument makes the family of all sequences $w(j)=\pm 2^j$ uncountable, so choose one with $\pi (w)\ne 0$. Truncating $w$ before coordinate $i$ writes $w=f_i+2^i{g}_i$, where $f_i$ is finitely supported and $g_i$ is an integer sequence. Thus $\pi (w)=2^i\pi (g_i)$ for every $i$. The vanishing of infinitely divisible integer sequences forces $\pi (w)=0$, a contradiction.

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.