Example of a module which is free at an isolated point 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!
 A: 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.
A: I'm surprised no one has mentioned the following very simple example.  Let $A=\mathbb{Z}$ and let $M=\mathbb{Q}$.  Then $M$ is free at the generic point of $\operatorname{Spec} A$, but is not free in any open neighborhood.  Or, if you want $x$ to be a closed point, you can instead take $M=\mathbb{Z}_{(p)}$ for some prime $p\in\mathbb{Z}$.
More generally, if $X$ is any scheme and $x\in X$ is any point such that every neighborhood of $x$ contains a point which is not a generalization of $x$, then the local ring at $x$ will be free at $x$ but not in any neighborhood of $x$ (it can't be free in any neighborhood because its fiber is $1$-dimensional at $x$ but $0$-dimensional at any point which is not a generalization of $x$).
