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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.

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2 Answers 2

If $M$ is an $A$-module and $N$ is a $B$-module, then $Hom_A(B,M)$ is naturally a $B$-module (via the $B$-module structure on $B$ itself), and evaluation at $1 \in B$ gives an $A$-linear morphism $Hom_A(B,M) \to M$. This induces a morphism $Hom_B(N,Hom_A(B,M)) \to Hom_A(N,M)$, which you can easily check is an isomorphism.


Applying the preceding adjunction shows that $$Hom_A(Hom_A(B,A),A) = Hom_B(Hom_A(B,A),Hom_A(B,A)) = End_B(Hom_A(B,A)),$$ and so your question becomes the question of why the natural map $B \to End_B(Hom_A(B,A))$ is an isomorphism.

Now $B$ is finite over $A$, hence $Hom_A(B,A)$ is finite over $A$, thus also over $B$, and hence $End_B(Hom_A(B,A))$ is a finite $B$-algebra. We will show that it equals $B$ by showing that it embeds into the fraction field of $B$ (and then appealing to the integral closedness of $B$).

If we write $K$ and $L$ for the fraction fields of $A$ and $B$ respectively, then $K\otimes_A B = L$, and so $Hom_A(B,A) \hookrightarrow Hom_A(B,K) = Hom_K(L,K)$. The last of these spaces is one-dimensional over $L$, and hence $End_L(Hom_K(L,K)) = L.$

From the results of the preceding paragraph, we can draw some conclusions:

  1. The natural map $Hom_A(B,A) \to L\otimes_B Hom_A(B,A)$ is injective, i.e. $Hom_A(B,A)$ is torsion-free over $B$, and hence $End_B(Hom_A(B,A))$ is torsion-free over $B$ as well, i.e. the natural map $End_B(Hom_A(B,A)) \to L\otimes_B End_B(Hom_A(B,A))$ is also injective.

  2. $L\otimes_K Hom_A(B,A) = Hom_K(L,K)$, and hence $L\otimes_B End_B(Hom_A(B,A)) = End_L(Hom_K(L,K)) = L.$

From 1. and 2. we see that $End_B(Hom_A(B,A))$ is a subalgebra of $L$ containing $B$. It is finite over $B$ (as noted above) and so equals $B$.

This completes the argument. Note that I didn't use the assumption that $A$ is integrally closed, so either the statement is true without this hypothesis, or I made a mistake. I wouldn't rule out the latter possibility!

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@navigetor23: Dear Navigetor, I looked over it again; I don't see the mistake. Regards, –  Matt E Oct 29 '12 at 3:01

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:

  1. $M$ is reflexive;
  2. $M$ is torsion-free and equals the intersection of its localizations at the prime ideals of height $1$;
  3. $M$ satisfies Serre's condition $(S_2)$ and its support is $\mathrm{Spec}(R)$;
  4. $M$ 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 $B=\bigcap B_{\mathfrak{p}}$.

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