TITLE: Prove by using diagonalization
QUESTION [5 upvotes]: Can anyone give me some hints on how to prove this question?
Q:  Use diagonalization to prove that if $A \subset B$ are lattices then $[B:A ]=\frac{\Delta(A)}{\Delta(B)}$.
Added:
Definition:  A lattice $A$ in the plane $\mathbb{R}^2$ is generated or spanned by a set $B$ if every element of $A$ can be written as an integer combination of elements of $B$.

REPLY [4 votes]: Since the original question is rather vague, I make explicit here 
what definitions I am working with. A  lattice  is a free abelian subgroup
of ${\mathbb Z}^d$ that has a basis $\cal B$ of ${\mathbb R}^d$ as a set of free generators. I call $\cal B$ a lattice basis for $A$. The index $\Delta (A)$
of a lattice $A$ the volume of a fundamental parallepiped.
Let $(a',b')=((a'_1,a'_2, \ldots ,a'_d)(b'_1,b'_2, \ldots ,b'_d))$ be a pair where $a'$ is a lattice basis for $A$ and $b'$ is a lattice basis for $B$. By the so-called
diagonalization algorithm, we can find a sequence of 
elementary moves that transforms $(a’,b’)$ into another pair 
$(a,b)=((a_1,a_2, \ldots ,a_d)(b_1,b_2, \ldots ,b_d))$ where $a$ is a lattice basis for $A$, $b$ is a lattice basis for $B$, and for each index $k$, we have
$a_k=m_kb_k$ where $m_k$ is a nonnegative integer. Then
$$
\Delta(B)=\prod_{k=1}^{n} ||b_k|| , \Delta(A)=\prod_{k=1}^{n} ||m_kb_k||,
[B:A]= \prod_{k=1}^{n} m_k
$$
and the desired equality follows.