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For a finitely generated module over a principal ideal domain, its invariant factors are nonzero nonunits $d_i$ in a decomposition $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, ordered by $d_i\mid d_{i+1}$. They are unique up to multiplication by units and describe the torsion submodule. Over $F[t]$, making $t$ act as a linear operator recovers the invariant factors of a linear operator used in its rational canonical form.

For a module $M$ over an integral domain, the quotient $M/T(M)$ by its torsion submodule is torsion-free. Every homomorphism from $M$ to a torsion-free module factors uniquely through $M/T(M)$ by the universal property of a quotient module.

Let $T$ consist of all torsion elements of a module in $M$. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$. Since $R$ is an integral domain, $rs\ne 0$. Also $r(am)=0$ for every scalar $a\in R$. Thus $T$ is the torsion submodule.

If $r(m+T)=0$ in $M/T$ with $r\ne 0$, then $rm\in T$, so some $s\ne 0$ satisfies $srm=0$. The domain property again gives $sr\ne 0$, hence $m\in T$. Therefore $M/T$ is a torsion-free module.

Finally, any submodule $N$ with the stated properties satisfies $N\subseteq T$. Conversely, if $m\in T$, its coset in $M/N$ is torsion and therefore zero, so $m\in N$. Hence $\boxed{N=T}$ is the unique choice.

The structure theorem for finitely generated modules over a principal ideal domain applies because a Euclidean domain is a principal ideal domain. For a finitely generated $R$-module,where the $d_i$ are nonzero nonunits. The free rank $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. Equivalently, the torsion submodule decomposes into cyclic modules $R/(P^a)$ for irreducible $P$, with uniquely determined prime-power elementary factors.

Let the torsion submodule of the module over a ring $M$ beThis is a submodule. Indeed, if $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs\ne 0$ 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: if $0\ne r\in R$ and $r(m+T(M))=0$, then $rm\in T(M)$, so $srm=0$ for some $s\ne 0$. Since $sr\ne 0$, 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 $r\ne 0$, then $rf(m)=f(rm)=0$, so torsion-freeness gives $f(m)=0$. Hence $T(M)\subseteq \ker f$, and the universal property of a quotient module gives a unique homomorphismwith $\boxed{f=f_0\circ q}$. Thus $M/T(M)$ is the maximal torsion-free quotient of $M$.

For a module $M$ over an integral domain $R$, a torsion element is an $m\in M$ with $rm=0$ for some $0\ne r\in R$. This includes zero. Equivalently, its annihilator ${\mathrm{Ann}}_R(m)=\{r\in R:rm=0\}$ is a nonzero ideal. The torsion elements of a module form the torsion submodule.

For $M=\mathbb{Z}/6\mathbb{Z}$, computed annihilators areFor these three elements, the additive order of a group element is respectively $6$, $3$, and $2$: these are the smallest positive integers killing them. Torsion over $\mathbb{Z}$ agrees exactly with finite additive order of a group element. Over a general integral domain, torsion concerns the chosen scalars instead; the polynomial module from a linear operator gives a torsion example whose nonzero elements have infinite additive order of a group element.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

For a module $M$ over an integral domain $R$, its torsion submodule isThis is a submodule. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$ and $rs\ne 0$; also $r(am)=a(rm)=0$ for every $a\in R$. The integral domain hypothesis supplies both the nonzero product and commutativity. These are the conventions of ^[ref].

A mixed example is the direct sum $M={\mathbb{Z}}^2\oplus \mathbb{Z}/6\mathbb{Z}$. If a nonzero integer kills $(a,b,\overline{c})$, then $a=b=0$. Conversely, $6$ kills every $(0,0,\overline{c})$. ThusThe torsion submodule isolates the finite-order part while the quotient module keeps the independent integer directions. In general $M/T(M)$ is a torsion-free module: if $r(m+T(M))=0$ with $r\ne 0$, some $s\ne 0$ kills $rm$, so $sr\ne 0$ kills $m$.

The hypothesis matters. In the ring $R=\mathbb{Z}/6\mathbb{Z}$ viewed as a module over itself, the elements $\overline{2}$ and $\overline{3}$ are killed by nonzero scalars $\overline{3}$ and $\overline{2}$, but their sum $\overline{5}$ is a unit and is killed by no nonzero scalar. Thus using arbitrary nonzero scalars in a ring with zero divisors need not give a submodule. A common broader convention uses regular elements of a ring as annihilators; the integral domain convention here avoids that ambiguity.