This question is motivated by my most spectacular  [answer](https://mathoverflow.net/questions/10239/is-it-true-that-as-z-modules-the-polynomial-ring-and-the-power-series-ring-over/10240#10240)  on MO (: 


Let $A$ be a module over $\mathbb Z$. $A$ is said to be torsion-free if $na=0$ implies $n=0$ or $a=0$ for any $n\in \mathbb Z, a\in A$. $A$ is torsionless if the map $\phi: A \to A^{\*\*}$ is injective (here ${}^\*$ means $\text{Hom}\_{\mathbb Z}(A,\mathbb Z)$).


If $A$ is finite, then torsion-free and torsionless are equivalent. In general, it is not hard to see that being torsionless implies torsion-free. On the other hand, $A=\mathbb Q$ is torsion-free but not torsionless since $A^\*=0$. But the question and answers quoted above (which shows that for $A=\mathbb Z[x]$, $\phi$ is an isomorphism) raised the following:


Question: If $A$ is a finite $\mathbb Z[x\_1,...,x\_d]$-module, are being torsion-free and torsionless equivalent? 



Torsion-free and torsionless abelian groups
If by finite you mean finitely presented, then the answer is no. For instance, Let $A = \mathbb{Z}[x]/(2x-1) = \mathbb{Z}[\frac12]$. Then, like $\mathbb{Q}$, $\text{Hom}(A,\mathbb{Z}) = 0$.


