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.


1 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).

  • $\begingroup$ Very good solution. Thank you. By the way, the statement is also true (and much simpler) if we replace Q/Z by some field k in the following sense. It is an exercise in linear algebra to show that if B<sub>1</sub> and B<sub>2</sub> are nxn matrices with entries in k then either they have a common eigenvector with eigenvalue 0 or there exist nxn matrices A<sub>1</sub> and A<sub>2</sub> with entries in k such that A<sub>1</sub>B<sub>1</sub>+A<sub>2</sub>B<sub>2</sub>=I. $\endgroup$ Commented Oct 17, 2009 at 19:35

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