Choose a finite presentation $F_1\to F_0\to M\to0$ with $F_0,F_1$ finite free. Applying $\operatorname{Hom}_A(-,A)$ gives
$$
0\longrightarrow M^*\longrightarrow F_0^*
\longrightarrow F_1^*.
$$
Let $N=F_0^*$ and let $P$ be the image in $F_1^*$. Then
$$
\boxed{0\longrightarrow M^*\longrightarrow N
\longrightarrow P\longrightarrow0}
$$
is exact, $N$ is finite free, and $P$ is torsion-free because it is a submodule of the free module $F_1^*$ over the domain $A$.

The <reflexive-module second-syzygy criterion> says that the kernel of a map from a finite free module to a torsion-free module over a Noetherian domain is reflexive. Applying it to this sequence shows that $M^*$ is reflexive. Concretely, after localizing at the fraction field, every functional on $(M^*)^*$ represented generically by an element of $N$ has no denominator: torsion-freeness of $P$ forces its image to vanish already over $A$. Thus the natural evaluation map $M^*\to M^{***}$ is an isomorphism.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 101 / 4 / a / Solution

Let $A$ be a <Noetherian ring> that is an <integral domain>. If
$$
0\longrightarrow M\longrightarrow N\longrightarrow P\longrightarrow0
$$
is <exact sequence>[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.

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