The Volume of $\O_K$

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 we define the volume of $V/\sigma(\O_K)$ to be $\vert\det(A)\vert$.

Example 6.1.5   The ring $\O_K = \mathbf{Z}[i]$ of integers of $K=\mathbf{Q}(i)$ has $\mathbf {Z}$-basis $w_1=1$, $w_2=i$. 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 $V/\sigma(\O_K)$ is $2\sqrt{2}$.

As the above example illustrates, the volume $V/\sigma(\O_K)$ need not be an integer.