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The double dual module is obtained by dualizing the dual module again. For a left module $M$ over a possibly noncommutative ring $R$, the correct sided formula isIt is again a left module. The evaluation homomorphism sends $m$ to the functional $f\mapsto f(m)$; right linearity of this functional follows from $(f\cdot r)(m)=f(m)r$.

Over a commutative ring, both Hom operations can be written with subscript $R$. Evaluation is an isomorphism for finite projective modules, while a reflexive module is defined by requiring it to be an isomorphism. It need not even be injective in general: the double dual of $\mathbb{Z}/n\mathbb{Z}$ is zero when $n>1$.

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.