Finding the volume of $\mathcal{O}$

Question

Let $K$ be a number field, $r_1$ denotes the number of real embeddings and $2r_2$ denotes so that $r_1+2r_2 = n = [K:\mathbb{Q}]$. Define the ring homomorphism, canonical imbedding of $K$,
$\sigma:K \to \mathbb{R}^{r_1}\times\mathbb{C}^{r_2}$ as $$\sigma(x) = (\sigma_1(x), \dots, \sigma_{r_1}(x),\dots, \sigma_{r_1+r_2}(x))$$ .

We only consider the 'first' $r_2$ complex embeddings since the rest are conjugate of the previous ones.

So, there is this lemma that I do not know how to start, what to think:

Lemma.If M is a free $\mathbb{Z}$-module of $K$ of rank $n$, and if $(x_i)_{1\leq i \leq n}$ is a $\mathbb{Z}-$base of $M$, then $\sigma(M)$ is a lattice in $\mathbb{R}^n$, whose volume is $$v(\sigma(M))=2^{-r_2}|\det_{1\leq i,j \leq n}(\sigma_i(x_j))|$$

The determinant on the right hand side is $\sqrt{\Delta_K}$ as far as I know. However, I appreciate any help to understand the proof of this lemma.

Edit: The lemma does not say about anything about $\mathcal{O}$, but it is a free $\mathbb{Z}$-module of rank $n$. So the lemma is more general, $\mathcal{O}$ is a specific case but I wrote the title as this way. You can either take a $\mathbb{Z}$-base for $\mathcal{O}$ and continue or work on any free $ \mathbb{Z}$-module of rank $n$.

Answer 1

You say that you don't understand the proof of the lemma without writing which proof you are talking about. Here is a proof I know, I left a few details for you to check: Let $\alpha_1,\ldots, \alpha_n$ be a $\mathbb{Z}$-basis of $M$. Then $\sigma(M)$ is spanned by $\sigma(\alpha_1),\ldots,\sigma(\alpha_n)$ over $\mathbb{Z}$. Let $A=(\sigma(\alpha_1),\ldots,\sigma(\alpha_n))$ be the matrix with $i$-th column given by $\sigma(\alpha_i)$, where you split up the complex number $\sigma_j(\alpha_i)$ into its real and imaginary part for $j=r_1+1,\ldots,r_1+r_2$. If you do some row operations, you will see that the determinant of $A$ is equal to $-2^{-r_2} \operatorname{det}(\sigma_i(\alpha_j))$. Since the trace form is non-degenerate, this will be non-zero, so $A$ defines an isomorphism $\mathbb{R}^n \cong \mathbb{R}^n$ taking $[0,1]^n$ to the fundamental parallelepiped of $\sigma(\alpha_1),\ldots,\sigma(\alpha_n)$. This gives you the desired formula for the volume of the lattice $\sigma(M)$.

Answer 2

As in my earlier comment, I think the issue here is not about rings of algebraic integers, but just about lattices $\Lambda$ in $\mathbb R^n$, meaning discrete subgroups $\Lambda\subset \mathbb R^n$ such that the quotient $\mathbb R^n/\Lambda$ is compact.

Although Minkowski did not quite argue in such terms, in contemporary terms we have a fundamental (measure/integral-theory) theorem, sometimes attributed to A. Weil (from his book on integration in topological groups and applications): for example, with abelian topological group $A$ and discrete subgroup $\Lambda\subset A$, with Haar measure $da$ on $A$, there is a unique invariant measure $d\dot{a}$ on $A/\Lambda$ such that, for all $f\in C^o_c(A)$, $$ \int_A f(a)\;da \;=\; \int_{A/\Lambda} \sum_{\lambda\in\Lambda} f(\dot{a}+\lambda)\;d\dot{a} $$ With respect to this measure, the total measure of $A/\Lambda$ is the "co-volume" of $\Lambda$. Yes, it depends on the Haar measure on $A$. For $A=\mathbb R^n$, we have a standard Haar measure.

(The detail about powers of $2$ is about normalization of measures on $\mathbb C$...)