Volume of fundamental domain in lattice of $\mathbb{Z}^2$ I'm trying to prove exercise 6.1 on Algebraic number theory by Stewart and Tall and essentially it states:
If $L$ is a lattice in $\mathbb{R^2}$ with basis $u, v\in \mathbb{Z^2}$, $T$ its fundamental domain, i.e.
$$T=\{xu+yv: 0\leq x,y <1\}$$
and $n$ is the number of points of $\mathbb{Z}^2$ that are in $T$ then
$$\operatorname{vol}(T)=n$$
I know Pick's theorem but the exercise is stated without any knowledge on that topic and I don't know how to prove it just knowing the definitions I gave. I've tried to prove it inductively on $n$ but I can't even prove the base case and I don't know how to relate in any way the volume and $n$. Any ideas?
 A: This exercise does not seem to be correct:  Take for example T = \Z^2, the standard integer lattice in \R^2.  By your definition of fundamental domain, the only lattice point in T would be the origin so n=1, but of course vol(T)=1 \not= n-2.
A: (I'm posting this as a separate answer to avoid cofusion with my previous reply).
The corrected form of your question is a simple consequence of Hermit normal form of integer matrices, aka row reduction.
Fix a basis $(u1, u2), (v1,v2)\in \mathbf Z^2$ of your sublattice, and consider the matrix
$
M = \bigl(
\begin{smallmatrix} 
  u1 & u2 \\
  v1 & v2 
\end{smallmatrix}
\bigr)
$.
To say that your sublattice has index n is to say that
$
\det(M)=n
$.
By row reduction over $\mathbf Z$, which is equivalent to mutiplying $M$ on the left by a matrix in $SL_2(\mathbf Z)$, we can turn $M$ into its unique Hermite normal form, i.e. a matrix of the form
$
H = \bigl(
\begin{smallmatrix} 
  a & b \\
  0 & d
\end{smallmatrix}
\bigr)
$
where $a, b, d\in \mathbf Z$ with $a|d$, and of course $ad=n$.  That means
$(a,b), (0,d)$ is a basis of your sublattice, and your exercise follows immediately (draw a picture).
