Let $A=K[[X\_1,\dots,X\_n]]$ where $K$ is a field. Let $M$ be a finitely generated torsion-free $A$-module, such that


1. for all $k$, the $A[1/X\_k]$-module $M[1/X\_k]$ is free of rank $d$;
2. for every $i \neq j$, we have $M = M[1/X\_i] \cap M[1/X\_j]$.


Does this imply that $M$ is free? 


It certainly does if $n=1$ (easy), and also if $n=2$ (reduce $M$ modulo $X$), but things seem trickier if $n \geq 3$. 


This question comes up when trying to prove that some $(\varphi,\Gamma)$-modules over rings in several variables are free.



A criterion for freeness over a local ring
No, this is false as soon as $n ≥ 3$. A second syzygy $M$ of the residue field $K$ gives a counterexample: each $M[1/X\_i]$ is projective, hence free, and it is reflexive, so the second condition is satisfied. On the other hand the projective dimension of $K$ as a module is $n$, so $M$ can not be free.


