While reading a [blog post](http://sbseminar.wordpress.com/2007/12/14/the-nullstellensatz-and-partitions-of-unity/) on partitions of unity at the Secret Blogging Seminar the following question came into my mind.


Let $n$ be a positive integer and let $B\_1$ and $B\_2$ be $n \times n$ matrices with integer entries. Is it true that exactly one of the following two statements is true?


1. There is a vector $v \in \mathbb{Q}^n \backslash \mathbb{Z}^n$ such that both $B\_1v$ and $B\_2v$ are in $\mathbb{Z}^n$.
2. There are matrices $A\_1$ and $A\_2$ with integer entries such that $A\_1B\_1+A\_2B\_2=I$.


Here, $I$ denotes the $n \times n$ identity matrix. The case $n=1$ is Bézout's identity.



An "existence contra partition of unity" statement for integer matrices?
I believe that what you say is true. I'll sketch an argument.


Let f:Zn ---> Z2n be the map of free Z-modules given by the matrices B1, B2 put in column (i.e. the direct sum of the morphisms given by B1 and B2). Now we rephrase conditions (1) and (2) in a slightly more abstract way:


* (1) fails to hold if, and only if there exists p:Z2n ---> Zn such that, together with f, fit in a short exact sequence


0 ---> Zn ---> Z2n ---> Zn ---> 0 (\*)


Indeed, the failure of (1) means that any v in Qn such f(v) in Z2n must be integral (i.e. v in Zn). In particular, this implies that f is injective. Moreover, take w in Z2n representing a nonzero torsion element in the cokernel of f. As w represents a torsion element, Nw belongs to the image of f for some big enough positive integer N, so there is v in Zn such that f(v) = Nw. But now f(1/N v) = w, and this means, by the failure of (1), that 1/N v is integral, so w is in the image of f and the cokernel of f has no torsion. As a finitely generated torsion-free Z-module is free, we get an exact sequence like (\*) above. This argument can easily be reversed, to show the equivalence between the existence of this exact sequence and the failure of (1). 


* (2) holds if, and only if there exists a morphism of Z-modules r:Z2n ----> Zn such that rf = id.


Let r be represented by a matrix (A1,A2). Then gf has matrix A1B1 + A2B2, and gf = id if, and only if (2) holds.


Now, the proof of what you asked for is easy. (1) fails if, and only if we can form the exact sequence (\*), but such an exact sequence is always split because Z^n is projective, so we can form such exact sequence if, and only if there exists a splitting r:Z2n ----> Zn, which is exactly condition (2).


