# Ideal class group 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 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.

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Just a note: this $\mathfrak f$ is often called the conductor [führer in German] of the extension of rings. – Dylan Moreland Jun 15 '12 at 22:57
To someone who down voted, may I ask the reason to improve my question? – Makoto Kato Jun 15 '12 at 23:24
This is a weird question. When you say you found those results, what do you mean? You worked out some examples and this is a conjecture? Or you proved them and you are not sure your proof is right? You are looking for a reference? – M Turgeon Jun 15 '12 at 23:56
I edited my question to make it clearer. – Makoto Kato Jun 16 '12 at 0:13
@Makoto I'd still want to know where do these questions come from. At least question (1) was asked yesterday, though it seems to me it was without the regularity condition on the ideal. Are these questions from some book, paper, work...? – DonAntonio Jun 16 '12 at 1:49

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.

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Dear Matt, could you explain why the kernel of $M/IM \to N/IN$ is annihlated by $\mathfrak f$? I guess I'm missing something here. – Makoto Kato Jun 17 '12 at 7:51
@MakotoKato: Dear Makoto, Actually, I should have had $\mathfrak f^2$ here, rather than just $\mathfrak f$. I have edited the answer to sketch a proof. I've used (a part of) the Tor long exact sequence for tensoring with $A/I$, since this is probably the most natural approach, but I'm sure you could check it directly just by unwinding the snake lemma computations that underly the existence of that long-exact sequence. Regards, – Matt E Jun 18 '12 at 6:55
I think, by borrowing your idea, the injectivity of $A/I \to B/IB$ can also be proved as follows. Since $\mathfrak{f}BI \subset I$, $(IB \cap A)/I$ is annihilated by both $I$ and $\mathfrak{f}$. Hence $I = IB \cap A$. – Makoto Kato Jun 18 '12 at 10:35

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.

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Please stop deleting your posts. That is highly frowned upon. – Bill Dubuque Jul 19 '12 at 22:10
@BillDubuque Some people are objecting me answering my questions. So I think I better delete such answers. – Makoto Kato Jul 19 '12 at 23:02
No, that is not what folks are objecting to, as has been explained on meta. Your posts add great value to the site. Please don't destroy them. We can solve any remaining meta-problems. But there is no rush. Let's get back to mathematics! – Bill Dubuque Jul 19 '12 at 23:12
@BillDubuque Some people just don't let me concentrate on mathematics. – Makoto Kato Jul 19 '12 at 23:53