Torsion-free module ($T(M)=0$)

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.

Apply clearing denominators relative to an independent module subset to find $0\ne a$ with $aM$ lying in a finite free module. Multiplication by $a$ is injective on a torsion-free module, giving the required embedding. A submodule need not itself be free over a general integral domain.

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.

A finitely generated module $M$ over a commutative ring $A$ is reflexive when its evaluation homomorphism $M\to M^{\ast \ast }$ to the double dual module is an isomorphism.

Let $A$ be a Noetherian ring that is an integral domain. Ifis exact, $N$ is a finitely generated free module, and $P$ is a torsion-free module, then $M$ is a reflexive module. In particular, the dual of every finitely generated $A$-module is reflexive.