Ideal class group of a one-dimensional Noetherian domain

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.

Since Matt proved (1) and (2), I'll prove the rest.

(3) Let $RI^+(A)$ be the set of regular ideals of $A$. Clearly $RI^+(A)$ is an ordered commutative monoid with mulitiplications of ideals. Let $RI^+(B)$ be the set of ideals of $B$ which are relatively prime to $\mathfrak{f}$. $RI^+(B)$ is also an ordered commutative monoid. By (1) and (2), $RI^+(A)$ is canonically isomorphic to $RI^+(B)$ as an ordered commutative monoid. Since $B$ is a Dedekind domain, (3) follows immediately.

(4) follows immediately from (3) and the following lemma.

Lemma 1 Let $P$ be a maximal ideal of $A$. $P$ is invertible if and only if $P$ is regular.

Proof: Suppose P is regular. By this, $A_P$ is integrally closed. Since $A_P$ is integrally closed, Noetherian and of dimension 1, it is a discrete valuation ring. Hence $PA_P$ is principal. Let $Q$ be a maximal ideal such that $Q \neq P$. Since $P$ is not contained in $Q$, $PA_Q = A_Q$. Hence $PA_Q$ is also principal. Since $A$ is Noetherian, $P$ is finitely generated over $A$. Hence P is invertible by this.

Suppose conversely P is invertible. By this, $PA_P$ is principal. Hence $A_P$ is a discrete valuation ring(e.g Atiyah-MacDonald). Hence $A_P$ is integrally closed. By this, $P$ is regular. QED

Lemma 2 Let $A$ be a commutative Noetherian ring. Let $I$ be a proper ideal of $A$ such that $dim A/I = 0$. Then $A/I$ is canonically isomorphic to $\prod_P A_P/IA_P$, where $P$ runs over all the maximal ideals of $A$ such that $I \subset P$.

This is well knowm.

Lemma 3 Let $A$ be a Noetherian domain of dimension 1. Let $I$ be a non-zero proper ideal of $A$. Then $(A/I)^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (A_\mathfrak{p}/IA_\mathfrak{p})^*$ as abelian groups, where $\mathfrak{p}$ runs over all the maximal ideals of $A$ such that $I \subset \mathfrak{p}$.

This follows immediately from Lemma 2.

Lemma 4 Let $A, K, B, \mathfrak{f}$ be as in the title question. Then $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ as abelian groups, where $\mathfrak{p}$ runs over all the maximal ideal of $A$.

Proof: By Lemma 3, $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{P}} (B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P})^*$ as an abelian group, where $\mathfrak{P}$ runs over all the maximal ideal of $B$ such that $\mathfrak{f} \subset \mathfrak{P}$. If $\mathfrak{p}$ is a regular prime ideal of $A$, $A_\mathfrak{p}$ is integrally closed by this.Hence $B_\mathfrak{p} = A_\mathfrak{p}$. Since $\mathfrak{f}A_\mathfrak{p} = A_\mathfrak{p}$, $B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p} = A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p} = 0$. Hence we only need to consider $\mathfrak{p}$ such that $\mathfrak{f} \subset \mathfrak{p}$. It's easy to see that $B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p}$ is canonically isomorphic to $\prod B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P}$, where $\mathfrak{P}$ runs over all the maximal ideals of $B$ lying over $\mathfrak{p}$. Hence $(B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ is canonically isomorphic to $(\bigoplus B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P})^*$, where $\mathfrak{P}$ runs over all the maximal ideals of $B$ lying over $\mathfrak{p}$. QED

Lemma 5 Let $B$ be an integral domain. Let $A$ be a subring of $B$ such that $B$ is integral over $A$. Let $I$ be an ideal of $A$. Let $\mathfrak{p}$ be a prime ideal of $A$ such that $I \subset \mathfrak{p}$. Let $B_\mathfrak{p}$ be the localization of $B$ with respect the multiplicative subset $A - \mathfrak{p}$. Let $f:B_\mathfrak{p} \rightarrow B_\mathfrak{p}/IB_\mathfrak{p}$ be the canonical homomorphism. $f$ induces a group homomorphism $g: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/IB_\mathfrak{p})^*$. Then $g$ is surjective.

Proof: Since $A_\mathfrak{p}$ is a local ring and $B_\mathfrak{p}$ is integtral over $A_\mathfrak{p}$, every maximal ideal $\mathfrak{Q}$ of $B_\mathfrak{p}$ lies over $\mathfrak{p}A_\mathfrak{p}$. Hence $\mathfrak{Q}$ = $\mathfrak{P}B_\mathfrak{p}$, where $\mathfrak{P}$ is a maximal ideal of $B$ lying over $\mathfrak{p}$. Since $I \subset \mathfrak{p}$, $I \subset \mathfrak{P}$. Hence $IB_\mathfrak{p} \subset \mathfrak{Q}$. Let $x \in B_\mathfrak{p}$. Suppose $f(x)$ is invertible. Then $f(x)$ is not contained in any maximal ideal of $B_\mathfrak{p}/IB_\mathfrak{p}$. Suppose $x$ is not invertible. $x$ is contained in a maximal ideal of $B_\mathfrak{p}$. This is a contradiction. QED

Lemma 6 Let $A, K, B, \mathfrak{f}$ be as in the title question. Let $\mathfrak{p}$ be a prime ideal of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$. Since $\mathfrak{f}$ is an ideal of both $A$ and $B$, $\mathfrak{f} = \mathfrak{f}A = \mathfrak{f}B$. Hence $\mathfrak{f}A_\mathfrak{p} = \mathfrak{f}B_\mathfrak{p}$. Hence $A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p} \subset B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p}$. Hence $(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^* \subset (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$. We claim $(B_\mathfrak{p})^*/(A_\mathfrak{p})^*$ is isomorphic to $(B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$.

Proof: By lemma 5, $g: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ is surjective. Let $\pi: (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ be the canonical homomorphism. Let $h: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ be $\pi g$. Let $x \in (B_\mathfrak{p})^*$. Suppose $h(x) = 0$. Then $g(x) \in (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$. Hence thers exists $y \in A_\mathfrak{p}$ such that $x \equiv y$ (mod $\mathfrak{f}B_\mathfrak{p}$). Since $\mathfrak{f}B_\mathfrak{p} = \mathfrak{f}A_\mathfrak{p}$, $x \in A_\mathfrak{p}$. Since $g(x) \in (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$, $x \in (A_\mathfrak{p})^*$. Hence Ker$(h) = (A_\mathfrak{p})^*$. QED

(6) Let $A, K, B, \mathfrak{f}$ be as in the title question. There exists the following exact sequence of abelian groups.

$0 \rightarrow B^*/A^* \rightarrow (B/\mathfrak{f})^*/(A/\mathfrak{f})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$

Proof: By this, there exists the following exact sequence of abelian groups.

$0 \rightarrow B^*/A^* \rightarrow \bigoplus_{\mathfrak{p}} (B_{\mathfrak{p}})^*/(A_{\mathfrak{p}})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$

Here, $\mathfrak{p}$ runs over all the maximal ideals of $A$.

If $\mathfrak{p}$ is a regular prime ideal of $A$, $A_\mathfrak{p}$ is integrally closed. Hence $B_\mathfrak{p} = A_\mathfrak{p}$. Hence, in $\bigoplus_{\mathfrak{p}} (B_{\mathfrak{p}})^*/(A_{\mathfrak{p}})^*$, it suffices to consider only $\mathfrak{p}$ such that $\mathfrak{f} \subset \mathfrak{p}$.

By Lemma 3, $(A/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_\mathfrak{p} (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ as an abelian group, where $\mathfrak{p}$ runs over all the maximal ideals of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.

By Lemma 4, $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ as an abelian group, where $\mathfrak{p}$ runs over all the maximal ideal of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.

Now, by Lemma 6, we are done. QED

Lemma 7 Let $A, K, B, \mathfrak{f}$ be as in the title question. Let $\phi: I(A)/P(A) \rightarrow I(B)/P(B)$ be the canonical homomorphism. Let $C \in$ Ker($\phi$). Then $C$ contains an ideal of the form $A \cap \beta B$, where $\beta$ is an element of $B$ such that $\beta B + \mathfrak{f} = B$.

This follows immediately from (6).

(5) Let $I$ be an invertible ideal of $A$. Since $B$ is a Dedekind domain, by this, there exist an ideal $\mathfrak{J}$ of $B$ and $\gamma \in K$ such that $\mathfrak{J} + \mathfrak{f} = B$ and $IB = \mathfrak{J}\gamma$. Let $J = A \cap \mathfrak{J}$. By (2), $J$ is regular and $JB = \mathfrak{J}$. By (4), $J$ is invertible. Since $IB = \mathfrak{J}\gamma = J\gamma B$, $IJ^{-1}B = \gamma B$. By Lemma 7, there exists $\beta \in B$ such that $\beta B + \mathfrak{f} = B$ and $IJ^{-1} \equiv A \cap \beta B$ mod($P(A)$). Hence $I \equiv J(A \cap \beta B)$ mod($P(A)$). Since $J$ and $A \cap \beta B$ are regular, we are done.