Preliminary Remarks

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

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

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

$\displaystyle \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

$\displaystyle \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

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


$\displaystyle A = \left(
\begin{matrix}1&1\ \sqrt{2}&-\sqrt{2}

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.

William Stein 2004-05-06