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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 $[B:A ]=\frac{\Delta(A)}{\Delta(B)}$.


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 $B$.

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This will take a few definitions for non experts in lattices. –  1015 Mar 20 '13 at 5:11
I suppose lattice is a free abelian group in $\,\Bbb R^n\,$ containing a basis for the vector space $\,\Bbb R^n_{\Bbb R}\,$ which is also a set of free generators...and perhaps $\,\Delta A\,$ is the volume of a (the) fundamental paralleliped determined by the lattice...? –  DonAntonio Mar 20 '13 at 5:17
that is correct –  user67258 Mar 20 '13 at 5:20
Then please edit these clarifications into your question. People shouldn't have to comb through the comments to understand the question. –  Gerry Myerson Mar 20 '13 at 5:31
It is unclear whether the question is about lattices in general or only lattices in ${\mathbb R}^2$. –  Ewan Delanoy Mar 22 '13 at 10:24

1 Answer 1

up vote 4 down vote accepted

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.

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