I'm looking for the most simple example of a quasicoherent sheaf $\mathcal{F}$ over a scheme $X$ (preferably affine for simplicity) which has a free stalk $\mathcal{F}_x$ at a point $x \in X$ and yet for every open neighborhood $x \in U$ the restricted sheaf $\mathcal{F}|_{U}$ isn't free.
By semicontinuity theorem $\mathcal{F}$ must jump down rank at $x$. I'm trying to figure out how badly behaved these jumps are.
In terms of modules I'm looking for a module $M$ over $A$ which has a free localization $M_{\mathfrak{p}}$ but s.t. for every $f \notin \mathfrak p\subset A$ the localization $M_f$ isn't free.
Bonus points: Find such a finite module over a noetherian ring!
Let $R=\mathbb F_2^{\mathbb N}$ the ring of sequences over $\mathbb F_2$, $I=\mathbb F_2^{(\mathbb N)}$ the ideal all sequences of finite support, and $M=I\oplus R/I$.
Since every $R$-module is flat from the short exact sequence $0\to I\to R\to R/I\to 0$ we have that either $I_P$ is free (of rank one) and $(R/I)_P=0$, or vice versa. Thus $M_P$ is free (of rank one) for every prime $P$.
There is a prime $P$ such that $M_f$ is not free for any $f\notin P$. Suppose the contrary, and denote by $f_P$ an element not in $P$ such that $M_{f_P}$ is free. Then $1=\sum_{i=1}^nr_if_{P_i}$ with $r_i\in R$. Since $M_P$ is a finitely generated $R_P$-module for any prime ideal $P$ we get that $M_{{f_P}_i}$ is a finitely generated $R_{{f_P}_i}$-module for all $i=1,\dots,n$. This shows that $M$ is a finitely generated $R$-module. But this is false since $I$ is not finitely generated.