7
$\begingroup$

If $V$ is an $n$-dimensional real vector space, a lattice in $V$ is a subgroup of the form $\Gamma=\mathbb{Z}v_1+\dots+\mathbb{Z}v_m$ where $v_1,\dots,v_m\in V$ are are $\mathbb{R}$-linearly independent. The lattice is complete if $m=n$.

Theorem: A subgroup $\Gamma\subset V$ is a lattice iff it is discrete.

This is proven in Neukirch's Algebraic Number Theory but there is a step I don't understand. Here's his proof:

enter image description here

In the last step, why does $q\Gamma\subset \Gamma_0$?

UPDATE: Also, in the line just before that, why does that end the proof that $(\Gamma:\Gamma_0)$ is finite? I see it proves that there are finite $\mu_i$, but why finite $\gamma_i$?

1 Answers 1

6

The additive group $\Gamma/\Gamma_0$ has order $q$, so for every $\gamma+\Gamma_0\in\Gamma/\Gamma_0$, $q(\gamma+\Gamma_0)=0+\Gamma_0=\Gamma_0$. But $q(\gamma+\Gamma_0)=q\gamma+\Gamma_0$, so $q\gamma+\Gamma_0=\Gamma_0$, and therefore $q\gamma\in\Gamma_0$. Since $\gamma\in\Gamma$ was arbitrary, $q\Gamma\subseteq\Gamma_0$.

  • 3
    @user46225: Because the map $\gamma_i\to\mu_i$ is a bijection. Suppose that $\mu_i=\mu_j$; then $\gamma_i-\gamma_j\in\Gamma_0$, and therefore $i=j$, since the $\gamma_i$’s were all from different cosets of $\Gamma_0$.2012-11-29