# 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 $v,u∈Z_{2}$ “spans” $Z_{2}$?

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

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

$v_{1},…,v_{n}$ is a basis for the module $Z_{n}$ iff the matrix $M$ formed by the vectors is a unimodular matrix.

$⇒$ If $det(M)=0$, then $v_{1},…,v_{n}$ are not linearly independent. If $∣det(M)∣≥2$, then the parallelepiped formed by $v_{1},…,v_{n}$ has volume $≥2$. If there is any integer point not on the corners of the parallelepiped, then that point can’t be written as linear combination of $v_{1},…,v_{n}$. 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 $2m$ lattice points.

This shows if $M$ is not unimodular, then $v_{1},…,v_{n}$ 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 $Mx=b$ contain a non-integer entry. (proposed by Thao Do)

$⇐$ $∣det(M)∣=1$ implies it has a inverse over $Z$, thus $Mx=b$ for any $b∈Z_{n}$ always has a solution.