Since this subject is full of misunderstandings (see [here](https://math.stackexchange.com/q/94422/3217), [here](https://math.stackexchange.com/q/800913/3217), [here](https://math.stackexchange.com/questions/2713073/locally-free-module), and [here](https://math.stackexchange.com/q/1905409/3217)) let us fix a precise terminology.  

Let $A$ be a commutative ring and $P$ an $A$-module.


I) We'll say that $P$ is a **locally free module of rank one** or is **invertible** if $P$ is finitely generated, projective and of rank one in the sense that for every prime ideal $\mathfrak p$ of $A$ the localized $A\_\mathfrak p$- module $P\_\mathfrak p$ (which is free by projectiveness) is of dimension $1$.  

These modules correspond bijectively, by a well known result of Serre in FAC, to locally free sheaves $\tilde P$ of rank $1$ on $\operatorname {Spec}A$, also known as invertible sheaves. This is one motivation for the above terminology.  

Another justification for the terminology "invertible" is that these modules are exactly those for which the canonical evaluation map $ P^\*\otimes\_AP\to A$ is an isomorphism. 


II) If $B\supset A$ is an overring of $A$ and $P\subset B$ is an $A$-module, we'll say that it is **concretely invertible** with respect to $B$ if $P.(A:P)\_B=A$.  

[As is standard $(A:P)\_B$ denotes the set of elements $b\in B$ such that $bP\subset A$]  

Lam proves in his [Lectures on Modules and Rings](https://books.google.fr/books?id=r9VoYbk-8c4C), that these concretely invertible modules are invertible. What about the converse? 


**Question:**  

**Is an invertible $A$-module $P$ isomorphic as an $A$-module to a concretely invertible module $P'\subset B$ with respect to a suitable overring $B\supset A$?**


**Remarks**  

a) Denote by $\operatorname {Quot} A$ the total quotient ring of $A$ obtained by inverting the regular (=not zero-divisors) of $A$, so that $A\hookrightarrow \operatorname {Quot} A$ is injective.  

Then a submodule $P\subset \operatorname {Quot}A$ is invertible if and only if it is concretely invertible.  

b) The answer is "yes" if $A$ is an integral domain: we can take $P'$ sitting inside $B=\operatorname {Frac}A$.  

c) The answer is "yes" if $A$ is semi-local, since then $P$ is free of rank $1$: see [here](https://math.stackexchange.com/q/3469106/3217).  

d) The answer is "yes" if $A$ is noetherian: in Eisenbud's [Commutative Algebra](https://books.google.fr/books/about/Commutative_Algebra.html?id=Fm_yPgZBucMC&redir_esc=y), page 253, Theorem 11.6 b., it is proven that every invertible module $P$ over a **noetherian** ring $A$ is isomorphic to a concretely invertible submodule $P'\subset \operatorname {Quot} A$ of its total quotient ring.  

e) Whatever the answer to the question is, it is definitely not true that we can always find the required $P'$ inside the total quotient ring $B=\operatorname {Quot} A$.  

Lam gives a counter-example in his [Lectures on Modules and Rings](https://books.google.fr/books?id=r9VoYbk-8c4C), Example (2.22)(A), page 37. 



Is every locally free module of rank $1$ over a commutative ring concretely invertible?
The answer is yes. Recall that given an invertible $A$-module $P$ and $n \in \mathbf{Z}$ there is an invertible $A$-module $P^{\otimes n}$ such that $P^{\otimes 0} = A$, $P^{\otimes 1} = P$, and $P^{\otimes n} \otimes\_A P^{\otimes m} = P^{\otimes n + m}$. Set $B = \bigoplus\_{n \in \mathbf{Z}} P^{\otimes n}$; this is a commutative $\mathbf{Z}$-graded $A$-algebra (details omitted). Then $P \subset B$ in degree $1$ and $(P : A)\_B = P^{\otimes -1}$ sitting in degree $-1$ and we have $P \cdot P^{\otimes -1} = A$.


Remark. The spectrum of $B$ is the $\mathbf{G}\_m$-torsor over $\text{Spec}(A)$ corresponding to $P$.


