An "existence contra partition of unity" statement for integer matrices?

Question

While reading a blog post 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.

Answer 1

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).