Viewing $ \O_K$ as a Lattice in a Real Vector Space

Let $ K$ be a number field of degree $ n$. By the primitive element theorem, $ K = \mathbf{Q}(\alpha)$ for some $ \alpha$, so we can write $ K\cong
\mathbf{Q}[x]/(f)$, where $ f\in\mathbf{Q}[x]$ is the minimal polynomial of $ \alpha$. Because $ \mathbf{C}$ is algebraically closed and $ f$ is irreducible, it has exactly $ n=[K:\mathbf{Q}]$ complex roots. Each of these roots $ z\in\mathbf{C}$ induces a homomorphism $ \mathbf{Q}[x] \to \mathbf{C}$ given by $ x\mapsto z$, whose kernel is the ideal $ (f)$. Thus we obtain $ n$ embeddings of $ K\cong
\mathbf{Q}[x]/(f)$ into  $ \mathbf{C}$:

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

Example 6.1.1   We compute the embeddings listed above for $ K=\mathbf{Q}(\sqrt[3]{2})$.
sage: K = QQ[2^(1/3)]; K
Number Field in a with defining polynomial x^3 - 2
sage: K.complex_embeddings()
[Ring morphism: ...
  Defn: a |--> -0.629960524947 - 1.09112363597*I,
 Ring morphism: ...
  Defn: a |--> -0.629960524947 + 1.09112363597*I,
 Ring morphism: ...
  Defn: a |--> 1.25992104989]

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

Lemma 6.1.2   Suppose $ L\subset \mathbf{R}^n$ is a subgroup of the vector space  $ \mathbf{R}^n$. Then the induced topology on $ L$ is discrete if and only if for every $ H>0$ the set

$\displaystyle X_H = \{v \in L : \max\{\vert v_1\vert,\ldots, \vert v_n\vert\} \leq H \}

is finite.

Proof. If $ L$ is not discrete, then there is a point $ x \in L$ such that for every $ \varepsilon >0$ there is $ y\in L$ such that $ 0 < \vert x-y\vert<\varepsilon $. By choosing smaller and smaller  $ \varepsilon $, we find infinitely many elements $ x-y\in L$ all of whose coordinates are smaller than $ 1$. The set $ X_1$ is thus not finite. Thus if the sets $ X_H$ are all finite, $ L$ must be discrete.

Next assume that $ L$ is discrete and let $ H>0$ be any positive number. Then for every $ x\in X_H$ there is an open ball $ B_x$ that contains $ x$ but no other element of $ L$. Since $ X_H$ is closed and bounded, it is compact, so the open covering $ \cup B_x$ of $ X_H$ has a finite subcover, which implies that $ X_H$ is finite, as claimed. $ \qedsymbol$

Lemma 6.1.3   If $ L$ if a free abelian group that is discrete in a finite-dimensional real vector space $ V$ and $ \mathbf{R}{}L=V$, then the rank of $ L$ equals the dimension of $ V$.

Proof. Let $ x_1,\ldots, x_m \in L$ be an $ \mathbf{R}$-vector space basis for $ \mathbf{R}{}L$, and consider the $ \mathbf {Z}$-submodule $ M=\mathbf{Z}x_1 + \cdots + \mathbf{Z}
x_m$ of $ L$. If the quotient $ L/M$ is infinite, then there are infinitely many distinct elements of $ L$ that all lie in a fundamental domain for $ M$, so Lemma 6.1.2 implies that $ L$ is not discrete. This is a contradiction, so $ L/M$ is finite, and the rank of $ L$ is $ m=\dim(\mathbf{R}{}L)$, as claimed. $ \qedsymbol$

Proposition 6.1.4   The $ \mathbf{R}$-vector space  $ V=\mathbf{R}\sigma(K)$ spanned by the image $ \sigma(K)$ of $ 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 nonzero elements of $ \O_K$ are nonzero integers.

Since $ \sigma(\O_K)$ is discrete in $ \mathbf{C}^n$, Lemma 6.1.3 implies that $ \dim(V)$ equals the rank of $ \sigma(\O_K)$. Since $ \sigma$ is injective, $ \dim(V)$ is the rank of $ \O_K$, which equals $ n$ by Proposition 2.4.5. $ \qedsymbol$

William Stein 2012-09-24