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.
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.
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
