Let $R$ be a commutative ring and let $M$ be a finitely generated $R$-module such that the $R_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$ is free of rank $1$ for every prime ideal $\mathfrak{p}$. Can we conclude that $M$ is locally free of rank $1$?
This is true when $M$ is also assumed to be finitely presented (in particular, there is no problem when $R$ is noetherian). But what happens in general?
Edit: I've found a proof, see below.
More generally, let $X$ be a locally ringed space and $M$ be an $\mathcal{O}_X$-module of finite type such that $M_x$ is free of rank $1$ for every $x \in X$. I claim that $M$ is locally free of rank $1$.
Proof. We may work locally on $X$ and hence assume that $M$ is generated by global sections $m_1,\dotsc,m_n$. For every $x \in X$ then $(m_1)_x,\dotsc,(m_n)_x$ is a generating system of $M_x$. Since $M_x \cong \mathcal{O}_{X,x}$ is local, it follows that there is some $1 \leq i \leq n$ such that $(m_i)_x$ already generates $M_x$. Write each $(m_j)_x$ as a multiple of $(m_i)_x$. Then, there is some open neighborhood $U$ of $x$ such that each $m_j|_U$ is a multiple of $m_i|_U$. It follows that $M|_U$ is generated by $m_i|_U$. Since $M$ is of finite type, we have $\mathrm{Ann}(M)_y=\mathrm{Ann}(M_y)=0$ for every $y \in Y$, hence $\mathrm{Ann}(M)=0$. Therefore, $M|_U$ is freely generated by $m_i|_U$. $\square$