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Let $A$ be a one-dimensional Noetherian domain. Let $K$ be its field of fractions. Let $B$ be the integral closure of $A$ in $K$. Suppose $B$ is a finitely generated $A$-module. It is well-known that $B$ is a Dedekind domain. Let $\mathfrak{f} = \{a \in A; aB \subset A\}$. Let $I$ be an ideal of $A$. If $I + \mathfrak{f} = A$, we call $I$ regular. I came up with the following proposition.

Proposition Let $I$ be a regular ideal of $A$. Then the canonical homomorphism $A/I \rightarrow B/IB$ is an isomorphism.

Outline of my proof I used the result of this question.

My question How do you prove the proposition? I would like to know other proofs based on different ideas from mine. I welcome you to provide as many different proofs as possible. I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.

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I noticed that someone serially upvoted for my questions. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. –  Makoto Kato Nov 27 '13 at 7:07

1 Answer 1

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Since $I + \mathfrak{f} = A$ one can write $1=\alpha+a$ with $\alpha\in I$ and $aB\subset A$. Then, for $b\in B$ we have $b=\alpha b+ab$, so $b-ab\in IB$, $ab\in A$, proving that the canonical $A/I\to B/IB$ is surjective. In order to show that it is injective we have to prove that $IB\cap A=I$: let $x\in IB\cap A$; then $x=\alpha x+ax$ with $\alpha x\in I$ (since $\alpha \in I$ and $x\in A$) and $ax\in I$ (since $aB\subset A$ implies $a(IB)\subset I$).

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This is nice. Thanks. –  Makoto Kato Nov 7 '13 at 11:00
    
I accepted your answer. However, it does not mean the question is over. I always welcome anyone who would post other proofs of the proposition. –  Makoto Kato Nov 7 '13 at 22:51

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