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  $\mathbf{C}$:

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

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

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

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. $\qedsymbol$