TITLE: Why is this module reflexive?
QUESTION [6 upvotes]: Let $A\subset B$ be two integrally closed Noetherian domains with $B$ finitely generated as an $A$-module. Then $B$ is reflexive. 

Could you explain me why, please?
Reflexive means that the natural map from $B$ to $B^{**}$, where $B^{**}=\mathrm{Hom}_A(\mathrm{Hom}_A(B,A),A)$, is an isomorphism.

REPLY [2 votes]: For finitely generated modules over noetherian integral domains there are several equivalent conditions for reflexivity which can be very useful in order to solve this type of problems. Let me borrow them from this topic:

Let $M$ be a finitely generated module over an integrally closed noetherian domain $R$. TFAE:

$M$ is reflexive;
$M$ is torsion-free and equals the intersection of its localizations at the prime ideals of height $1$;
$M$ satisfies Serre's condition $(S_2)$ and its support is $\mathrm{Spec}(R)$;
$M$ is the dual of a f.g. module.


For our problem we shall use $2.$ Obviously $B$ is a torsion-free $A$-module. It remains to prove that $B$ is the intersection of its localizations at the prime ideals of $A$ of height $1$. One knows that $B$ is the intersection of its localizations at the prime ideals of height $1$. Now take $x\in \bigcap B_{\mathfrak{p}}$, where $\mathfrak{p}\subset A$ is a height $1$ prime. Let $P\subset B$ a prime ideal of height $1$. Since the extension $A\subset B$ is integral, we have that the height of $\mathfrak{p}=P\cap A$ is least $1$. But we also know that $A$ is integrally closed and $B$ is a domain. This shows that the going-down theorem applies and therefore the height of $\mathfrak{p}$ is $1$. Note now that $B_{\mathfrak{p}}\subset B_P$ and therefore $x\in B_P$. We have proved that $\bigcap B_{\mathfrak{p}}\subset \bigcap B_P=B$ and then $B=\bigcap B_{\mathfrak{p}}$.