Question. Let $A$ be an integral domain and $\tilde{A}$ be its integral closure in the field of fractions $K$. Assume that $\tilde{A}$ is a finitely generated $A$-module. I want to prove that if $\tilde{A}$ is flat over $A$, then $A$ is integrally closed.

I thought the following fact would be useful:

Fact: Let $A$ be an integral domain and $K$ be its field of fractions. Also let $B$ be a finitely generated $A$-submodule of $K$. Then $B$ is flat iff $B$ is locally free of rank $1$.

By the above fact, I think we may assume that $\tilde{A}$ is locally free of rank $1$, i.e., $\tilde{A}_{\mathfrak{p}}$ is free of rank $1$ over $A_\mathfrak{p}$ for every prime ideal $\mathfrak{p}$ of $A$. However, I don't think that this would immediately imply that $A=B$ but I don't know how to use the fact that $\tilde{A}$ is the integral closure of $A$.


Let $A\subseteq B$ be an extension of integral domains such that $B_{\mathfrak p}=x_{\mathfrak p}A_{\mathfrak p}$ for every prime ideal $\mathfrak p$ of $A$ and some element $x_{\mathfrak p}\in B_{\mathfrak p}$. Then $x_{\mathfrak p}$ is invertible in $B_{\mathfrak p}$ and therefore $B_{\mathfrak p}=A_{\mathfrak p}$. One then gets

$B\subseteq\bigcap\limits_{\mathfrak p} B_{\mathfrak p} =\bigcap\limits_{\mathfrak p} A_{\mathfrak p}=A$.


One can use the following:

  1. If $A\subset B$ is an integral extension and $B$ is flat over $A$, then $B$ is faithfully flat over $A$.

  2. If $A\subset B$ are integral domains with the same field of fractions and $B$ is faithfully flat over $A$, then $A=B$. (Matsumura, Commutative Ring Theory, Exercise 7.2)

Remark. This argument shows that the condition "$\tilde{A}$ is a finitely generated $A$-module" is superfluous.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.