Coprime cyclic quotients over a PID split by the Chinese remainder map

Coprimality means $(a,b)=R$, so choose $r,s\in R$ with $ra+sb=1$.

The map $\mathrm{\Phi }:R\mathrm{/}(ab)\to R\mathrm{/}(a)\oplus R\mathrm{/}(b)$ given by $x+(ab)\mapsto (x+(a),x+(b))$ is well defined. If both residues vanish, $a\mid x$ and $b\mid x$; multiplying the Bezout identity appropriately gives $ab\mid x$, so $\mathrm{\Phi }$ is injective. Given residues $u+(a)$ and $v+(b)$, the element $usb+vra$ maps to them, so $\mathrm{\Phi }$ is surjective and is an $R$-module isomorphism.

Repeatedly apply step 2.1 to a finite pairwise coprime family; the product of any subfamily remains coprime to the next factor. This gives the displayed finite direct sum, with the empty and singleton conventions stated above and with unit factors contributing zero quotients.