Basis for $\mathbb{Z}^2$

Solution 1:

This has a beautiful geometrical interpretation. Note $\rm\, x,y\, $ is a $\rm\,\mathbb Z$-basis of $\rm\, \mathbb Z^2\, $ iff $\rm\, \mathbb Z^2\, $ is tiled by the fundamental parallelogram $\rm P $ with sides $\rm\,x,y.\, $ But this is true iff the only lattice points that are inside $\rm P $ or on the boundary of $\rm\,P\,$ are its vertices. However, by Pick's area formula, this is true iff

$$\rm\ area\ P =\text{ #interior_points } + \frac{1}2\text{ #boundary_points}- 1\, =\, 0 + \frac{4}2 - 1\, =\, 1\qquad$$

But by basic analytic geometry $\rm\, area\ P\, =\, |\det(x,y)|.\,$ Therefore, combining the two, we conclude that $\rm\, x,y\,$ is a $\rm\,\mathbb Z$-basis of $\rm\, \mathbb Z^2\! \iff |\det(x,y)| = 1.$

In fact it deserves to be much better known that Pick originally applied his area formula in a similar way to give a beautiful geometric proof of the Bezout linear representation of the gcd.

Solution 2:

If that is a basis, then you can write $e_1=(1,0)$ and $e_2=(0,1)$ in terms of the basis using integer coefficients. This implies that the matrix determined by the basis in invertible in $\mathbb Z$. That is your condition. It can be expressed neatly using determinants.

Solution 3:

For $ (a, b) $ and $ (c, d) $ to be a basis, they must be linearly independent. In other words,
$\det \begin{bmatrix} a &b \\ c& d \end{bmatrix} $ must be invertible. Over a field, this would imply that $\det(A) \neq 0 $, but since we are in $\mathbb{Z}$, we require $\det(A) = \pm 1$ so that each entry in $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d &-b \\ -c& a \end{bmatrix}$ is an integer.

Note that $ad - bc = 1$ implies that $\gcd(a,c) = \gcd(b,d) = 1$.