Basis of the module $Z^{n}$

A student who is taking linear algebra asked me the following problem.

If we consider the field $R$ restricted to $Z$ , and create a “vector space” on $Z$ . How do we know if $v, u \in Z^{2}$ “spans” $Z^{2}$ ?

Formally, what can we say about $v$ and $u$ if for every $w \in Z^{2}$ , there exist $n, m \in Z$ , such that $n v + m u = w$ .

We can generalize it and put it in terms of modules, as $Z$ is only a ring but not a field.

Theorem1

$v_{1}, \dots, v_{n}$ is a basis for the module $Z^{n}$ iff the matrix $M$ formed by the vectors is a unimodular matrix.

Proof

$\Rightarrow$ If $det (M) = 0$ , then $v_{1}, \dots, v_{n}$ are not linearly independent. If $∣ det (M) ∣ \geq 2$ , then the parallelepiped formed by $v_{1}, \dots, v_{n}$ has volume $\geq 2$ . If there is any integer point not on the corners of the parallelepiped, then that point can’t be written as linear combination of $v_{1}, \dots, v_{n}$ . Notice that it must contain some lattice points not on the corners of the parallelepiped. One can see why by consider a large box that contain volume of $m$ such parallelepiped, but contain at least $2 m$ lattice points.

This shows if $M$ is not unimodular, then $v_{1}, \dots, v_{n}$ can’t be a basis.

Alternative proof: $M$ is not unimodular then $M^{- 1}$ contain a non-integer entry. This shows there exist a $b$ , such that the solution $x$ to $M x = b$ contain a non-integer entry. (proposed by Thao Do)

$\Leftarrow$ $∣ det (M) ∣ = 1$ implies it has a inverse over $Z$ , thus $M x = b$ for any $b \in Z^{n}$ always has a solution.

Posted by Chao Xu on 2012-08-30.

Tags: math.

Basis of the module Zn

Basis of the module $Z^{n}$