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Let $A$ be an integral domain, $K$ its field of fractions. Let $M$ be a fractional ideal of $A$. I'd like to prove that $M$ is invertible if and only if $MA_P$ is a principal fractional ideal of $A_P$ for every maximal ideal $P$ of $A$.

EDIT As Georges Elencwajg pointed out, it seems that we need to assume $M$ is finitely generated to prove if part.

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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 7:08
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2 Answers 2

up vote 4 down vote accepted

Reminder
A non zero fractional ideal is invertible if and only if it is projective , and then it has rank one.

a) If $M$ is invertible it is projective and non-zero. Hence $M_P$ is also non-zero and projective of rank one over $A_P$, hence free of rank one because $A_P$ is local and thus a principal fractional ideal for $A_P$.

b) Here I assume that $M$ is finitely generated over $A$.
If $M_P$ is principal fractional it certainly is free of rank one, hence projective of rank one .
Since this is true for all maximal $P$ , the module $M$ is (non-zero) projective: this is where I use that $M$ is finitely generated .
Hence $M$ is an invertible fractional ideal of $A$, according to the Reminder.

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As for b), You don't need to assume that $M$ has a finite presentation? –  Makoto Kato Jun 25 '12 at 10:38
    
You seem to be a very precise person Makoto: I like that! But, no, what I wrote is correct. It is true that finite generation does not suffice in general and that you must require finite presentation. But the subtle point is that here you have the supplementary hypothesis that the rank of $M_P$ at each $P$ is constant, namely $1$. And this is then sufficient: Bourbaki, Comm.Alg., Chap.II,§5,Theorem 2. –  Georges Elencwajg Jun 25 '12 at 10:50
    
Yes, I remember. Thanks. –  Makoto Kato Jun 25 '12 at 10:53
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I'll prove the title assertions (Proposition 1 and Proposition 2) using the following lemmas.

Lemma 1 Let $A$ be an integral domain. Let $M$ be an invertible fractional ideal of A. Then $M$ is finitely generated as an $A$-module.

Proof: There exists a fractional ideal $N$ of $A$ such that $MN = A$. Hence there exist $x_i \in M, y_i \in N, i = 1, ..., n$ such that $1 = \sum x_iy_i$. Hence, for every $x \in M$, $x = \sum x_i(xy_i)$. Since $xy_i \in A$, $M$ is generated by $x_1, ..., x_n$ over $A$. QED

Lemma 2 Let $A$ be an integral domain. Let $M$ be an invertible fractional ideal of $A$. Then $M$ is projective as an $A$-module.

Proof: There exists a fractional ideal $N$ of $A$ such that $MN = A$. Hence there exist $x_i \in M, y_i \in N, i = 1, ..., n$ such that $1 = \sum x_iy_i$. For each i, define A-homomorphism $f_i: M \rightarrow A$ by $f_i(x) = y_ix$. Since $x = \sum x_i(y_ix)$ for every $x \in M$, $x = \sum f_i(x)x_i$. As shown in the proof of Lemma 1, $M$ is generated by $x_1, ..., x_n$ over $A$. Let $L$ be a free $A$-module with basis $e_1, ..., e_n$. Define $A$-homomorphism $p: L \rightarrow M$ by $p(e_i) = x_i$ for each $i$. Define $A$-homomorphism $s: M \rightarrow L$ by $s(x) = \sum f_i(x)e_i$. Let $K = Ker(p)$. We get an exact sequence: $0 \rightarrow K \rightarrow L \rightarrow M \rightarrow 0$. Since $ps = 1_M$, this sequence splits. Hence $M$ is projective. QED

Lemma 3 Let $A$ be a local ring. Let $M$ be a finitely generated projective $A$-module. Then $M$ is a free $A$-module of finite rank.

Proof: Let $\mathfrak{m}$ be the maximal ideal of $A$. Let $k = A/\mathfrak{m}$. Since $M \otimes k$ is a free k-module of finite rank, there exists a free $A$-module $L$ of finite rank and a surjective homomorphism $f: L \rightarrow M$ such that $f \otimes 1_k: L \otimes k \rightarrow M \otimes k$ is an isomorphism. Let $K = Ker(f)$. We get an exact sequence: $0 \rightarrow K \rightarrow L \rightarrow M \rightarrow 0$

Since $M$ is projective, this sequence splits. Hence the following sequence is exact. $0 \rightarrow K \otimes k \rightarrow L \otimes k \rightarrow M \otimes k \rightarrow 0$

Since $f \otimes 1_k: L \otimes k \rightarrow M \otimes k$ is an isomorphism, $K \otimes k = 0$. Since $K$ is a direct summand of $L$, $K$ is a finitely generated $A$-module. Hence $K = 0$ by Nakayama's lemma. QED

Lemma 4 Let $A$ and $B$ be commutative rings. Let $f: A \rightarrow B$ be a homomorphism. Let $M$ be a projective $A$-module. Then $M \otimes_A B$ is projective as a $B$-module.

Proof: Let $N$ be a $B$-module. $N$ can be regarded as an $A$-module via $f$. $Hom_B(M \otimes_A B, N)$ is canonically isomorphic to $Hom_A(M, N)$. This isomorphism is functorial in $N$. Since $Hom_A(M, -)$ is an exact functor, $Hom_B(M \otimes_A B, -)$ is exact. Hence $M \otimes_A B$ is projective as a $B$-module. QED

Lemma 5 Let $A$ be an integral domain. Let $K$ be the field of fractions of $A$. Let $M$ and $N$ be $A$-submodules of $K$. Let $MN$ be the $A$-submodule of $K$ generated by the set {$xy; x \in M, y \in N$}. Let $M^{-1} = \{x \in K; xM \subset A\}$. Suppose $MN = A$. Then $N = M^{-1}$.

Proof: Since $N \subset M^{-1}$, $MN \subset MM^{-1} \subset A$. Since $MN = A$, $MM^{-1} = A$. Multiplying the both sides of $MN = A$ by $M^{-1}$, we get $M^{-1}MN = M^{-1}$. Hence $N = M^{-1}$. QED

Lemma 6 Let $A$ be an integral domain. Let $K$ be the field of fractions of $A$. Let $M$ be finitely generated $A$-submodule of $K$. Let $M^{-1} = \{x \in K; xM \subset A\}$. Let $P$ be a prime ideal of $A$. Let $(M_P)^{-1} = \{x \in K; xM_P \subset A_P\}$. Then $(M^{-1})_P = (M_P)^{-1}$.

Proof: Let $x \in M^{-1}$. Since $xM \subset A$, $xM_P \subset A_P$. Hence $x \in (M_P)^{-1}$. Hence $M^{-1} \subset (M_P)^{-1}$. Hence $(M^{-1})_P \subset (M_P)^{-1}$.

Let $x_1, ..., x_n$ be generators of $M$ as an $A$-module. Let $y \in (M_P)^{-1}$. Then $yx_i \in A_P$ for $i = 1, ..., n$. Hense there exists $s \in A - P$ such that $syx_i \in A$ for $i = 1, ..., n$. Since $sy \in M^{-1}$, $y \in (M^{-1})_P$. Hence $(M_P)^{-1} \subset (M^{-1})_P$ QED

Proposition 1 Let $A$ be an integral domain, $K$ its field of fractions. Let $M$ be an invertible fractional ideal of $A$. Let $P$ be a prime ideal of $A$. Then $M_P$ is a principal fractional ideal.

Proof: By Lemma 1, $M$ is finitely generated as an $A$-module. Hence $M_P$ is finitely generated as an $A_P$-module. By Lemma 2, $M$ is projective as an $A$-module. Hence by Lemma 4, $M_P$ is projective as an $A_P$-module. Therefore, by Lemma 3, $M_P$ is a free $A_P$-module of finite rank. Since $M_P \neq 0$ and an $A_P$-basis of $M_P$ is linearly independent over $K$, $M_P$ is a free $A_P$-module of rank 1. Hence $M_P$ is a principal fractional ideal. QED

Proposition 2 Let $A$ be an integral domain, $K$ its field of fractions. Let $M$ be a finitely generated fractional ideal of $A$. Suppose $M_P$ is a principal fractional ideal of $A_P$ for every maximal ideal $P$ of $A$. Then $M$ is invertible.

Proof: Let $M^{-1} = \{x \in K; xM \subset A\}$. Let $P$ be a maximal ideal of $A$. Hence by Lemma 6, $(M^{-1})_P = (M_P)^{-1}$. Hence, $(MM^{-1})_P = (M_P)(M^{-1})_P = (M_P)(M_P)^{-1}$ Since $M_P$ is a principal, $M_P$ is invertible. Hence by Lemma 5, $(M_P)(M_P)^{-1} = A_P$. Hence $(MM^{-1})_P = A_P$. Hence by Lemma 4 of this question, $MM^{-1} = A$. QED

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Hi Makoto. What are you doing: expanding my proof to give full details of prerequisites? That would not be a bad idea. –  Georges Elencwajg Jun 25 '12 at 10:24
    
Hi, Georges Yes, I'm trying to write a detailed proof of the title theorem. Perhaps I may need some of the lemmas elsewhere later. –  Makoto Kato Jun 25 '12 at 10:50
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This is an excellent and very useful initiative , also for other users, Makoto. Best wishes! –  Georges Elencwajg Jun 25 '12 at 10:56
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