= Solution
Let the <torsion submodule> of the <module over a ring> $M$ be
$$
T(M)=\{m\in M:rm=0\text{ for some }0\ne r\in R\}.
$$
This is a <submodule>. Indeed, if $rm=0$ and $sn=0$ with $r,s\ne0$, then $rs\ne0$ because the scalar ring is an <integral domain>, and $rs(m+n)=0$; scalar multiples and additive inverses are handled similarly.
Take the <quotient module> $M_0=M/T(M)$ and let $q:M\to M_0$ be the quotient <R-module homomorphism>. It is <torsion-free module>[torsion-free]: if $0\ne r\in R$ and $r(m+T(M))=0$, then $rm\in T(M)$, so $srm=0$ for some $s\ne0$. Since $sr\ne0$, this puts $m$ in $T(M)$ and therefore $m+T(M)=0$.
Now let $N$ be torsion-free and let $f:M\to N$ be an $R$-module homomorphism. If $m\in T(M)$ and $rm=0$ for $r\ne0$, then $r f(m)=f(rm)=0$, so torsion-freeness gives $f(m)=0$. Hence $T(M)\subseteq\ker f$, and the <universal property of a quotient module> gives a unique homomorphism
$$
f_0:M/T(M)\longrightarrow N,
\qquad f_0(m+T(M))=f(m),
$$
with $\boxed{f=f_0\circ q}$. Thus $M/T(M)$ is the <maximal torsion-free quotient> of $M$.
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