Lemma 10.29.372EDF5773FAC

Proof. A subgroup $G$ of $\mathbb R^ s$ is dense unless there is a nonzero linear form $\xi $ on $\mathbb R^ s$ with $\xi (G)\subseteq \mathbb Z$: the closure of $G$ is the sum of a linear subspace $V$ and a discrete subgroup of a complement of $V$, and if $V\neq \mathbb R^ s$ one can take $\xi $ vanishing on $V$ and integral on that discrete subgroup. Suppose $\xi (\sigma _{\mathbb R}(a))=\sum _ j\xi _ j\sigma _ j(a)\in \mathbb Z$ for every $a\in \mathcal O_ K$. Since $\mathcal O_ K$ spans $K$ over $\mathbb Q$, the map $a\mapsto \sum _ j\xi _ j\sigma _ j(a)$ is a $\mathbb Q$-linear map $K\to \mathbb Q$. The trace form of $K$ is nondegenerate, so there is $b\in K$ with

where $\sigma $ runs over all $s+2t$ embeddings of $K$ into $\mathbb C$. Distinct field embeddings are linearly independent over $\mathbb C$, by Dedekind's lemma on characters. Comparing the coefficients of a complex embedding $\sigma _{s+1}$, which occurs only on the right, gives $\sigma _{s+1}(b)=0$. Hence $b=0$, and then $\xi _ j=\sigma _ j(b)=0$ for every $j$. $\square$