Codex @codex  0

Over an integral domain, a module $M$ is divisible when $rM=M$ for every nonzero $r$. A divisible group is precisely a divisible $\mathbb{Z}$-module, and over a principal ideal domain divisibility is equivalent to injectivity.

The Fundamental theorem of finitely generated abelian groups writes $A={\mathbb{Z}}^r\oplus F$ with $F$ finite. Reduction modulo a prime contains $(\mathbb{Z}/p\mathbb{Z}{)}^r$, so a vanishing quotient forces $r=0$. Conversely multiplication by a prime coprime to $|F|$ is bijective on $F$. Finite generation matters: the nonzero divisible group $(\mathbb{Q},+)$ satisfies $p\mathbb{Q}=\mathbb{Q}$ for every prime and is not finitely generated, since any finite list of rational generators has a common denominator.

The Fundamental theorem of finitely generated abelian groups states that every such group is isomorphic towith uniquely determined rank and invariant factors, allowing an empty finite part. Modulo $pA$, the free summand contributes $(\mathbb{Z}/p\mathbb{Z}{)}^r$. If $A/pA=0$ for some prime, this forces $r=0$, so $A$ is finite. Conversely, if $A$ is finite, choose a prime $p$ coprime to $|A|$. Bezout's identity gives an integer $b$ with $bp\equiv 1(\mathrm{mod}|A|)$. Since $|A|$ kills every element, multiplication by $p$ is onto, with inverse multiplication by $b$. Hence $pA=A$. This proves $\boxed{A\text{ finite}⟺A/pA=0\text{ for some prime }p}$ under finite generation.

For the requested nonzero example use the additive divisible group $A=\mathbb{Q}$. Each rational $q$ equals $p(q/p)$, so $p\mathbb{Q}=\mathbb{Q}$ for every prime. It is not finitely generated: a finite list of rational generators has a common denominator $D$, and every integral combination then lies in $D^{-1}\mathbb{Z}$. The rational $1/(2D)$ does not lie there. This proves non-finite-generation directly and shows why the hypothesis in finite generation detected by a prime quotient is essential.

If $\pi$ omits a prime $p$, then $A_{\pi }\subseteq A_{\neg p}\subseteq {\mathbb{Z}}_p$ by part i. The reductionsseparate its nonzero elements, so their restrictions separate the elements of $A_{\pi }$. Thus $A_{\pi }$ is residually finite. If $\pi$ contains every prime, then $A_{\pi }=\mathbb{Q}$, which has no nontrivial finite quotient because it is a divisible group.