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 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.
Are the following assertions true?
If yes, how do you prove them?
(1) Let $I$ be a regular ideal.
Then $I = IB \cap A$.
(2) Let $\mathfrak{I}$ be an ideal of B such that $\mathfrak{I} + \mathfrak{f} = B$.
Let $I = \mathfrak{I} \cap A$.
Then $I$ is regular and $IB = \mathfrak{I}$.
(3) A regular ideal is uniquely decomposed as a product of regular prime ideals.
(4) A regular ideal is invertible.
(5) Let $I(A)$ be the group of invertible fractional ideals of $A$.
Let $P(A)$ be the group of principal ideals of $A$.
Let $RI(A)$ be the group of regular fractional ideals of $A$.
Let $RP(A)$ be the group of regular principal ideals of $A$.
Then $RI(A)/RP(A)$ is isomorphic to $I(A)/P(A)$.
EDIT[Jun 26, 2012]
(6) There exists the following exact sequence of abelian groups.
$0 \rightarrow B^*/A^* \rightarrow (B/\mathfrak{f})^*/(A/\mathfrak{f})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$
In paticular,
$[I(A) : P(A)] = [I(B) : P(B)][(B/\mathfrak{f})^* : (A/\mathfrak{f})^*]/[B^* : A^*]$.
EDIT[Jun 27, 2012]
The converse of (4) is false.
Let $\alpha$ be a nonzero element of a non-regular maximal ideal $P$ of $A$.
Then $\alpha A$ is invertible, but not regular.
EDIT
My motivation came from the theory of binary quadratic forms over the ring of rational integers. It has close relationship with the ideal theory of orders of quadratic number fields. I got some idea from the books of Hilbert and Neukirch on algebraic number theory.
EDIT
I think you need this.
EDIT
Let K be an algebraic number field.
Let $A$ be its order(a subring of K which is a finitely generated $\mathbb{Z}$-module and contains a $\mathbb{Q}$-basis of $K$).
Let $B$ be the ring of algebraic integers in K.
Usually $B$ is hard to be determined while $A$ is easily found.
For example, let $\theta$ be an algebraic integer which generates $K$.
Then $A = \mathbb{Z}[\theta]$ is an order of $K$.
In this case, the prime decomposition of a regular ideal of $A$ can be rather easily calculated than in $B$.
By the above results, we can get information of prime decompositions of ideals of $B$ which are prime to $\mathfrak{f}$.
EDIT[Jun 28, 2012]
I think Hilbert(or someone else before him) proved (1), (2), (3), (4) and a modified version of (6) (using $RI(A)/RP(A)$ instead of $I(A)/P(A)$).
However, I think (5) is non-trivial.
To prove (5), I needed to prove (6) by a method whose basic idea I borrowed from Neukirch's book on algebraic number theory.
EDIT[Nov 26, 2013]
I asked alternative proofs of (5) here in MathOverflow because I think my proof is detoured.
Best Answer
Let $M \to N$ be a morphism of $A$-modules, whose kernel and cokernel are both annihilated by an ideal $\mathfrak f$ of $A$. Suppose that $I$ is another ideal of $A$. Then the cokernel of $M/IM \to N/IN$ is annihlated by both $I$ and $\mathfrak f$, while the kernel of this map is annihilated by $I+\mathfrak f^2$ (and so in particular by $(I+\mathfrak f)^2$). In particular, if $I + \mathfrak f = A$, the kernel and cokernel of $M/IM \to N/IN$ vanish, i.e. this map is an isomorphism.
[Let $M'$ and $M''$ be the kernel and image of $M \to N$, so that there is a short exact sequence $0 \to M' \to M \to M'' \to 0.$ Let $N''$ denote the cokernel of $M \to N$, so that there is an a short exact sequence $0 \to M'' \to N \to N'' \to 0$. We then obtain exact sequences $$ M'/IM' \to M/IM \to M''/IM'' \to 0$$ and $$Tor_1^A(A/I,N'') \to M''/IM'' \to N/IN \to N''/IN'' \to 0.$$ Considering these, we see that the cokernel of $M/IM \to N/IN$ is $N''/IN'',$ and that it is annihilated by $I + \mathfrak f$, while the kernel of $M/IM \to N/IN$ is an extension of a submodule of $Tor_1^A(A/I,N'')$ by a quotient of $M'/IM'$, and so is annihilated by $(I + \mathfrak f)^2$. It is also clearly annihilated by $I$, and so is annihilated by $I + \mathfrak f^2$.]
Applying this to your set-up, we see that $A/I \to B/IB$ is an isomorphism if $I$ is a regular ideal. This proves (1).
If $\mathfrak I$ is an ideal of $B$ such that $\mathfrak I + \mathfrak f = B,$ then since $\mathfrak f \subset A$, one immediately checks that $A \cap \mathfrak I + \mathfrak f = A$. Thus $I := A \cap \mathfrak I$ is regular. Now by (1), the map $A/I \to B/IB$ is an isomorphism. On the other hand, composing this with surjection $B/IB \to B/\mathfrak I$ (induced by the evident inclusion of $IB$ in $\mathfrak I$), we obtain $A/I \to B/\mathfrak I$, which is injective, by the definition of $I$. Thus $B/IB \to B/\mathfrak I$ is injective as well as surjective, hence is an isomorphism, hence $I B = \mathfrak I$. This proves (2).
Perhaps, with these in hand, you can try to prove some of the rest yourself.