Regulator of number fields doesn't vanish The regulator of a number field $K$ is usually presented at the beginning of books on algebraic number theory, alongside the class number group, Dirichlet unit theorem...
But the only proof for the fact that $R_K\neq0$ usually involves L-functions and the class number formula.
Is there an "elementary" proof of this? (by that I mean, one that doesn't involve L-functions)
 A: Dirichlet's unit theorem explicitly states that if $|\cdot |_1, \ldots, |\cdot |_r$ are the real absolute values and $|\cdot |_{r+1},\ldots, |\cdot|_{r+s}$ are the complex ones, then the log map
$$l:\begin{cases}\mathcal{O}_K\to \Bbb R^{r+s} \\ l(\alpha) = (\log |\alpha|_1,\ldots , \log |\alpha|_r, \log|\alpha|_{r+1},\ldots, \log |\alpha|_{r+s})\end{cases}$$
with kernel $\mathcal{O}_K^\times$ is of maximal rank in the trace-$0$ subspace $\left\{\mathbf{x}\in\mathbb{R}^{r+s} : \sum_i x_i = 0\right\}$.
This means that it is a lattice, which by definition means that $\mathcal{O}_K$ has $r+s-1$ $\Bbb R$-linearly independent vectors within it. But then $R_K$ is defined as the volume of the parallelopiped spanned by a $\Bbb Z$-basis for this lattice, which--having maximal rank--implies that the volume is non-zero just by linear algebra since it is the absolute value of the determinant of the matrix formed by these vectors. Since a determinant is non-zero iff the vectors are $\Bbb R$-linearly dependent, and we know this is not so by Dirichlet, the regular does not vanish.
