Preliminary Remarks

Let $K$ be a number field of degree $n$. Then there are $n$ embeddings

$$\sigma_1,\ldots, \sigma_n:K\hookrightarrow \mathbf{C}.$$

Let $\sigma:K\to \mathbf{C}^n$ be the product map $a\mapsto (\sigma_1(a),\ldots,\sigma_n(a))$. Let $V=\mathbf{R}\sigma(K)$ be the $\mathbf{R}$-span of $\sigma(K)$ inside $\mathbf{C}^n$.

Proposition 10.1.1   The $\mathbf{R}$-vector space  $V=\mathbf{R}\sigma(K)$ spanned by the image $\sigma(K)$ has dimension $n$.

Proof. We prove this by showing that the image $\sigma(\O _K)$ is discrete. If $\sigma(\O _K)$ were not discrete it would contain elements all of whose coordinates are simultaneously arbitrarily small. The norm of an element $a\in\O _K$ is the product of the entries of $\sigma(a)$, so the norms of nonzero elements of $\O _K$ would go to 0. This is a contradiction, since the norms of elements of $\O _K$ are integers.

The fact that $\sigma(\O _K)$ is discrete in $\mathbf{C}^n$ implies that $\mathbf{R}\sigma(\O _K)$ has dimension equal to the rank $n$ of $\sigma(\O _K)$, as claimed. This last assertion is not obvious, and requires observing that if $L$ if a free abelian group that is discrete in a real vector space $W$ and $\mathbf{R}{}L=W$, then the rank of $L$ equals the dimension of $W$. Here's why this is true. If $x_1,\ldots, x_m \in L$ are a basis for $\mathbf{R}{}L$, then $\mathbf{Z}x_1 + \cdots + \mathbf{Z}x_m$ has finite index in $L$, since otherwise there would be infinitely many elements of $L$ in a fundamental domain for $\mathbf{Z}x_1 + \cdots + \mathbf{Z}x_m$, which would contradict discreteness of $L$. Thus the rank of $L$ is $m=\dim(\mathbf{R}{}L)$, as claimed. $\qedsymbol$

Since $\sigma(\O _K)$ is a lattice in $V$, the volume of $V/\sigma(\O _K)$ is finite. Suppose $w_1,\ldots,w_n$ is a basis for $\O _K$. Then if $A$ is the matrix whose $i$th row is $\sigma(w_i)$, then $\vert{\mathrm{Det}}(A)\vert$ is the volume of $V/\sigma(\O _K)$. (Take this determinant as the definition of the volume--we won't be using ``volume'' here except in a formal motivating way.)

Example 10.1.2   Let $\O _K = \mathbf{Z}[i]$ be the ring of integers of $K=\mathbf{Q}(i)$. Then $w_1=1$, $w_2=i$ is a basis for $\O _K$. The map $\sigma:K\to \mathbf{C}^2$ is given by

$$\sigma(a+bi) = (a+bi,a-bi)\in\mathbf{C}^2.$$

The image $\sigma(\O _K)$ is spanned by $(1,1)$ and $(i,-i)$. The volume determinant is

$$\left\vert\left( \begin{matrix}1&1\ i&-i \end{matrix}\right)\right\vert = \vert-2i\vert = 2.$$

Let $\O _K=\mathbf{Z}[\sqrt{2}]$ be the ring of integers of $K=\mathbf{Q}(\sqrt{2})$. The map $\sigma$ is

$$\sigma(a+b\sqrt{2}) = (a+b\sqrt{2},a-b\sqrt{2})\in\mathbf{R}^2,$$

and

$$A = \left( \begin{matrix}1&1\ \sqrt{2}&-\sqrt{2} \end{matrix}\right),$$

which has determinant $-2\sqrt{2}$, so the volume of the ring of integers is $2\sqrt{2}$.

As the above example illustrates, the volume of the ring of integers is not a great invariant of $\O _K$. For example, it need not be an integer. If we consider ${\mathrm{Det}}(A)^2$ instead, we obtain a number that is a well-defined integer which can be either positive or negative. In the next section we will do just this.