Volume of fundamental domain in lattice of $\mathbb{Z}^2$

Question

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?

Answer 1

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.

Answer 2

(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).