A finitely generated PID module is its torsion submodule direct-summed with a finite free module

In [L1], every cyclic quotient $R\mathrm{/}(a_i)$ is torsion, and every torsion element has zero component in the free summand because a free module over a domain is torsion-free. Thus the direct sum of the cyclic quotients is exactly $\mathrm{Tor}(M)$.

Substituting that identification into the invariant-factor decomposition gives $M\cong \mathrm{Tor}(M)\oplus R^r$. It includes pure torsion when $r=0$, pure free modules when $\mathrm{Tor}(M)=0$, and the zero module when both vanish.

The torsion submodule is canonical because it is defined by $M$ alone, while a free complement is not. Suppose $a\in R$ is a nonzero nonunit and put $M=R\oplus R\mathrm{/}(a)$. Since $R$ is a domain, an element $(x,y+(a))$ killed by some nonzero scalar has $x=0$, and $a$ kills every $(0,y+(a))$; hence $\mathrm{Tor}(M)=0\oplus R\mathrm{/}(a)$. Both $C_1=R(1,0+(a))$ and $C_2=R(1,1+(a))$ are free of rank one, because $r(1,\ast )=0$ forces $r=0$, and each meets $\mathrm{Tor}(M)$ only in $0$ while $(x,y+(a))=(0,(y-x)+(a))+x(1,1+(a))$ shows $\mathrm{Tor}(M)+C_2=M$ and likewise for $C_1$. They are distinct: $(1,1+(a))\in C_2$ lies in $C_1$ only if $a\mid 1$, contrary to $a$ being a nonunit.