TITLE: Basis for $\mathbb{Z}^2$
QUESTION [17 upvotes]: Let $x = (a, b), y = (c, d) \in \mathbb{Z}^2$. What is the condition on $a, b, c, d$ so that ${x, y}$ is a basis?

My answer: $ad\neq bc$ and $gcd(a, c) = gcd(b, d) = 1$. 
The first condition ensures that they aren't the same vector; the second ensures that we can actually "get" all of the integer values/lattice points. 
Is this correct? 
Thanks.

REPLY [2 votes]: 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 $\gcd(a,c) = \gcd(b,d) = 1$.