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 $a(IB)\subset I$).