Past exam of the rings and modules course of the University of Oxford / 2023 / 2 / b

For $j\in {\mathbb{N}}_0$, let $e_j\in B$ have $e_j(j)=1$ and $e_j(k)=0$ for $k\ne j$. If $f,g\in F$ and $z\in \mathbb{Z}$, then $f+g$ can be nonzero only where $f$ or $g$ is nonzero, and $zf$ can be nonzero only where $f$ is nonzero. These are finite sets. Since $0\in F$, this proves that $F$ is a submodule.

Every $f\in F$ has the finite expressionEvaluating at coordinate $k$ recovers the coefficient of $e_k$, so the expression is unique. Thus $\{e_j:j\in {\mathbb{N}}_0\}$ is a basis of a module for $F$, andis a free module. By contrast, the ambient module of integer sequences $B={\prod }_{j\in {\mathbb{N}}_0}\mathbb{Z}$ allows arbitrary integer sequences.

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 $\mathbb{Z}$.

ConsiderThe map sending a sequence of signs $({\epsilon }_j)$ to $({\epsilon }_j{2}^j)$ is a bijection from $\{-1,1{\}}^{{\mathbb{N}}_0}$ onto $W$. This is an uncountable set by the Cantor diagonal argument: given a proposed list $w_0,w_1,\dots$ of its elements, define $w(j)=-w_j(j)$. Then $w\in W$ differs from $w_j$ at coordinate $j$ for every $j$.

Since $\ker \pi$ is countable, $W$ cannot be contained in it. Consequently there exists

For every $i\in {\mathbb{N}}_0$, truncate $w$ before coordinate $i$:This $f_i$ lies in $F$. For $j<i$, $(w-f_i)(j)=0$. For $j\ge i$, $(w-f_i)(j)=w(j)=\pm 2^j$, which is divisible by $2^i$. In particular the integer sequencebelongs to $B$ and satisfies $w=f_i+2^i{g}_i$.

Apply the module homomorphism $\pi$. Since $f_i\in F\subseteq \ker \pi$,Fixing any coordinate $j$, the integer $\pi (w)(j)$ is therefore divisible by $2^i$ for every $i$. The only such integer is zero: if $a\ne 0$, choose $i$ with $2^i>|a|$, which rules out $a=2^{i}b$ for any nonzero integer $b$. Thus every coordinate of $\pi (w)$ vanishes, contradicting part (b)(iii).

It follows thatThe decisive property is vanishing of infinitely divisible integer sequences, namely ${\bigcap }_{i\ge 0}{2}^{i}B=\{0\}$.