= Torsion-free module
{title2=$T(M)=0$}
{wiki}
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 $q/2\notin\mathbb Zq$ for $q\ne0$. The finite-generation hypothesis cannot be dropped. See https://kconrad.math.uconn.edu/blurbs/linmultialg/modulesoverPID.pdf[Keith Conrad's Modules over a PID, Corollary 2.6].
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