# The group of invertible fractional ideals of a Noetherian domain of dimension 1

Let $A$ be a Noetherian domain of dimension 1. Let $I(A)$ be the group of invertible fractional ideals of $A$. Let $P$ be a maximal ideal of $A$. Let $I(A_P)$ be the group of invertible fractional ideals of $A_P$. Let $I$ be an invertible fractional ideal of $A$. $IA_P$ is an invertible fractional ideal of $A_P$. It's easy to see that $IA_P = A_P$ except for finitely many $P$s. Hence we get a homomorphism $f\colon I(A) \rightarrow \bigoplus_P I(A_P)$, where $P$ runs over all the maximal ideals of $A$. I need to prove that $f$ is an isomorphism. So my question is: Is $f$ an isomorphism?

Perhaps this question is a special case of Proposition (21.9.4) of EGA IV-4, p.285. But I think it's desirable to have a direct proof.

This question is related to this.

EDIT[Jul 3, 2012] Martin Brandenburg Nothing new? Perhaps you are right(this is not MathOverflow), but I hadn't known the proof. Could you please show us your proof or the reference?

EDIT It is well-known that an invertible fractional ideals of a semilocal domain is principal(see Bourbaki, Algebre Commutative, Chapitre II $\S$ 5, or this for a local domain). Hence $I(A_P)$ is the group of prinicipal fractional ideals of $A_P$.

EDIT[Jun/25/2012] Since there has been no answer so far, I'll try to prove the above result. I need several lemmas most of which are well-known, but I will prove them for the readers' convenience.

Lemma 1 Let $A$ be a commutative ring. Let $S$ be a multiplicative subset of $A$. Let $M$ be an $A$-module. Let $N$ and $L$ be $A$-submodules of $M$. Then $(N \cap L)_S = N_S \cap L_S$ as submodules of $M_S$.

Proof: The following sequence is exact. $0 \rightarrow N \cap L \rightarrow M \rightarrow M/N \oplus M/L$

Hence we get an exact sequence: $0 \rightarrow (N \cap L)_S \rightarrow M_S \rightarrow M_S/N_S \oplus M_S/L_S$

Hence $(N \cap L)_S = N_S \cap L_S$ QED

Lemma 2 Let $A$ be a Noetherian commutative ring. Let $\mathfrak{p}$ and $\mathfrak{q}$ be prime ideals of $A$. Suppose $\mathfrak{q}$ is not contained $\mathfrak{p}$. Let $\mathfrak{Q}$ be a $\mathfrak{q}$-primary ideal. Then $\mathfrak{Q}A_\mathfrak{p} = A_\mathfrak{p}$.

Proof: There exists $n \gt 0$ such that $\mathfrak{q}^n \subset \mathfrak{Q}$. Since $\mathfrak{q}$ is not contained $\mathfrak{p}$, $\mathfrak{q}A_\mathfrak{p} = A_\mathfrak{p}$. Hence $\mathfrak{q}^nA_\mathfrak{p} = A_\mathfrak{p}$. Hence $\mathfrak{Q}A_\mathfrak{p} = A_\mathfrak{p}$. QED

Lemma 3 Let $A$ be a Noetherian commutative ring. Let $\mathfrak{m}_1, ..., \mathfrak{m}_n$ be distinct maximal ideals of $A$. For each $i$, let $\mathfrak{q}_i$ be an ($\mathfrak{m}_iA_{\mathfrak{m}_i}$)-primary ideal of $A_{\mathfrak{m}_i}$. Let $\mathfrak{Q}_i$ be the inverse image of $\mathfrak{q}_i$ by the canonical homomorphism $A \rightarrow A_{\mathfrak{m}_i}$. Let $I = \mathfrak{Q}_1 \cap ... \cap \mathfrak{Q}_n$. Then $IA_{\mathfrak{m}_i} = \mathfrak{q}_i$ for each $i$.

Proof: By Lemma 1, $IA_{\mathfrak{m}_i} = \mathfrak{Q}_1A_{\mathfrak{m}_i} \cap ... \cap \mathfrak{Q}_nA_{\mathfrak{m}_i}$. By Lemma 2, $\mathfrak{Q}_jA_{\mathfrak{m}_i} = A_{\mathfrak{m}_i}$ for $i \neq j$. Hence $IA_{\mathfrak{m}_i} = \mathfrak{Q}_iA_{\mathfrak{m}_i} = \mathfrak{q}_i$. QED

Lemma 4 Let $A$ be a commutative ring. Let $M$ be an $A$-module. Let $N$ and $L$ be $A$-submodules of $M$. Suppose $N_\mathfrak{m} = L_\mathfrak{m}$ for every maximal ideal $\mathfrak{m}$ of $A$. Then $N = L$.

Proof: Let $x \in N$. Let $I = \{a \in A; ax \in L\}$. Suppose $I \neq A$. Then there exist a maximal ideal $\mathfrak{m}$ such that $I \subset \mathfrak{m}$. Since $x/1 ∈ N_\mathfrak{m} = L_\mathfrak{m}$, there exists $s \in A - \mathfrak{m}$ such that $sx \in L$. Hence $s \in I$. This is a contradiction. Hence $I = A$. Hence $x \in L$. Therefore $N \subset L$. Similarly $L \subset N$. QED

Lemma 5[this lemma is unnecessary, but I leave it here] Let $A$ be an integral domain. Let $K$ be its field of fractions. Then $A = \cap A_\mathfrak{m}$, where $\mathfrak{m}$ runs over all the maximal ideals of $A$ and each $A_\mathfrak{m}$ is regarded as a subring of $K$.

Proof: Let x $\in \cap A_\mathfrak{m}$. Let $I = \{a \in A; ax \in A\}$. Suppose $I \neq A$. Then there exists a maximal ideal $\mathfrak{m}$ such that $I \subset \mathfrak{m}$. Since $x/1 \in A_\mathfrak{m}$, there exists $s \in A - \mathfrak{m}$ such that $sx \in A$. Hence $s \in I$. This is a contradiction. Hence $I = A$. Hence $x \in A$. Therefore $\cap A_\mathfrak{m} \subset A$. The other inclusion is clear. QED

Proposition 1 Let $A$ be a Noetherian domain of dimension 1. The above homomorphism $f\colon I(A) \rightarrow \bigoplus_P I(A_P)$ is injective.

Proof: Let $K$ be the field of fractions of $A$. Let $M$ be an invertible fractional ideal of $A$. Suppose $f(M) = 0$. Hence $M_P = A_P$ for all the maximal ideals $P$ of $A$. Since $M$ and $A$ are $A$-submodules of $K$, by Lemma 4, $M = A$. QED

Proposition 2 Let $A$ be a Noetherian domain of dimension 1. The above homomorphism $f\colon I(A) \rightarrow \bigoplus_P I(A_P)$ is surjective.

Proof: Let $x = (x_P)$ be an element of $\bigoplus_P I(A_P)$. Since $I(A_P)$ is the group of principal ideals of $A_P$ for each P, $x_P = (a_P/b_P)A_P$ for some $a_P, b_P \in A$. $(a_P/b_P)A_P = A_P$ for all but finitely many $Ps$. If $(a_P/b_P)A_P = A_P$, we can assume $a_P = b_P = 1$. For each $P$, let $I(P) = A \cap a_PA_P$. If $a_PA_P \neq A_P$, $a_PA_P$ is $(PA_P)$-primary ideal. Let $I = \cap I(P)$. By Lemma 3, $IA_P = a_PA_P$ for all $P$. Since $I$ is locally principal, $I$ is invertible(see Bourbaki, Algebre Commutative, Chapitre II $\S$ 5). Similarly, for each $P$, let $J(P) = A \cap b_PA_P$. Let $J = \cap J(P)$. Since $J$ is locally principal, $J$ is invertible. Let $M = IJ^{-1}$. Then $M_P = (a_P/b_P)A_P$ for each $P$. Hence $f$ is surjective. QED

• Erm, this seems to be an excerpt from a commutative algebra textbook. Nothing new. And where is the question?! – Martin Brandenburg Jun 24 '12 at 14:55
• @MakotoKato, if you think you have a proof, you should consider posting it as an answer to your question. There is nothing wrong with answering your own question (and even accepting that answer). If you think there is a gap in your proposed proof (and I don't see one), you should make it clear what you are worried about. Also, in case you don't know, "\bigoplus" is the tex command for $\bigoplus$, the standard notation for direct sum (rather than $\sum$, which I don't think is used enough to have a standard meaning, but I normally see it used to mean something different than "direct sum"). – B R Jun 24 '12 at 15:56
• @BR Thanks for the advice. – Makoto Kato Jun 24 '12 at 16:04
• @MakotoKato, also, "\oplus" gives the smaller version $\oplus$ when you want to write $A\oplus B$, instead of $\bigoplus_p A_p$. – B R Jun 24 '12 at 16:58
• @BR Thanks, again. – Makoto Kato Jun 24 '12 at 22:04