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Let $A$ be a Noetherian domain of dimension 1. Let $K$ be its field of fractions. Let $B$ be the integral closure of $A$ in $K$. Suppose $B$ is finitely generated as an $A$-module. It is well-known that $B$ is a Dedekind domain. Let $I(A)$ be the group of invertible fractional ideals of $A$. Let $P(A)$ be the group of principal fractional ideals of $A$. Similarly we define $I(B)$ and $P(B)$.

Then there exists the following exact sequence of abelian groups(Neukirch, Algebraic number theory p.78).

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

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

Since we use this result to prove this, it'd be nice that we have the proof here(I don't understand well Neukirich's proof).

EDIT Since someone wonders what my question is(though I think it is obvious), I state it more clearly: How do you prove it?

EDIT[July 11, 2012] May I ask the reason for the downvote so that I could improve my question?

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Makoto Kato
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  • Thank you for the clarification. It would be best if you included a summary (or scan of the relevant pages) of Neukirch's proof, so people know what approach you are confused by and can either explain it in more detail, or offer alternative proofs. – Zev Chonoles Jun 26 '12 at 02:34
  • I forgot why I didn't understand well Neukirch's proof. Personally this does not matter anymore because I came up with my own proof. – Makoto Kato Jun 26 '12 at 02:53
  • I noticed that someone serially upvoted for my questions. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. – Makoto Kato Nov 27 '13 at 07:07

1 Answers1

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Lemma 1 Let $A$ be a Noetherian domain of dimension 1. Let $K$ be its field of fractions. Then there exists the following exact sequence of abelian groups.

$0 \rightarrow K^*/A^* \rightarrow \bigoplus K^*/(A_{\mathfrak{p}})^* \rightarrow I(A)/P(A) \rightarrow 0$

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

Proof: By this, $I(A)$ is canonically isomorphic to $\bigoplus I(A_{\mathfrak{p}})$. By this, $I(A_{\mathfrak{p}})$ is the group of principal fractional ideals of $A_{\mathfrak{p}}$. Hence $I(A_{\mathfrak{p}})$ is canonically isomorphic to $K^*/(A_{\mathfrak{p}})^*$.

On the other hand, $P(A)$ is canonically isomorphic to $K^*/A^*$. QED

Lemma 2 Let $A$ be a Noetherian domain of dimension 1. Let $K$ be its field of fractions. Let $B$ be the integral closure of $A$ in $K$. Suppose $B$ is finitely generated as an $A$-module. Then there exists the following exact sequence of abelian groups.

$0 \rightarrow K^*/B^* \rightarrow \bigoplus K^*/(B_{\mathfrak{p}})^* \rightarrow I(B)/P(B) \rightarrow 0$

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

Proof: This follows immediately from the proposition of this. QED

The proof of the exactness of the title sequence There exists a canonical morphism from the exact sequence of Lemma 1 to that of Lemma 2. The exactness of the title sequence follows immediately by snake lemma. QED

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