# Factor ring by a regular ideal of a one-dimensional Noetherian domain

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.

#### Solutions Collecting From Web of "Factor ring by a regular ideal of a one-dimensional Noetherian domain"

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 $a(IB)\subset I$).