Embedding the ring of algebraic integers into $\mathbb{R}^n$ (Serge Lang Algebra Exercise 7.4) 
Let $L$ be a finite extension of $\mathbb{Q}$ and let $\mathfrak{o}_L$ be the ring of algebraic integers in $L$. Let $\sigma_1,\dots,\sigma_n$ be the distinct embeddings of $L$ into the complex numbers. Embedded $\mathfrak{o}_L$ into a Euclidean space by the map
$$
\alpha \mapsto (\sigma_1\alpha,\sigma_2\alpha,\dots,\sigma_n\alpha).
$$
Show that in any bounded region of space there is only a finite number of elements of $\mathfrak{o}_L$.

I'm not asking about how to show the discreteness of $\mathfrak{o}_L$, as stated in the last sentence. My question is, does the map
$$
\alpha \mapsto (\sigma_1\alpha,\sigma_2\alpha,\dots,\sigma_n\alpha)
$$
really maps $\mathfrak{o}_L$ into $\mathbb{R}^n$? More precisely, is it guaranteed that $\sigma_k\alpha \in \mathbb{R}$ (instead of $\mathbb{C} \cong \mathbb{R}^2$)? I highly doubt this as I can't even find it working in $\mathbb{Q}(i)/\mathbb{Q}$ with $\mathfrak{o}_L = \mathbb{Z}[i]$. However, Lang did mean the map is $\mathfrak{o}_L \to \mathbb{R}^n$, because later he asked the reader to use another exercise about additive subgroups in $\mathbb{R}^n$ which shows that $\mathfrak{o}_L$ is free with $\le n$ generators, not $2n$.
 A: There is a difference between real embeddings and complex embeddings. Notice that if $\sigma$ is an embedding from $L$ to $\mathbb{C}$, then $\overline{\sigma}$ is also an embedding and $\sigma=\overline{\sigma}$ if and only if $\sigma$ is a real embedding.
If we denote $\sigma_{1},\dots,\sigma_{r_{1}}$ for those real embeddings $L\rightarrow \mathbb{R}$ and $\sigma_{r_{1}+1},\overline{\sigma_{r_{1}+1}},\dots,\sigma_{r_{1}+r_{2}},\overline{\sigma_{r_{1}+r_{2}}}$ for complex embeddings. Then the map should be $$\alpha\mapsto (\sigma_{1}(\alpha),\dots,\sigma_{r_{1}}(\alpha),\operatorname{Re}\sigma_{r_{1}+1}(\alpha),\operatorname{Im}\sigma_{r_{1}+1}(\alpha),\dots,\operatorname{Re}\sigma_{r_{1}+r_{2}}(\alpha),\operatorname{Im}\sigma_{r_{1}+r_{2}}(\alpha)),$$
or $$\alpha\mapsto (\sigma_{1}(\alpha),\dots,\sigma_{r_{1}}(\alpha),\sigma_{r_{1}+1}(\alpha),\dots,\sigma_{r_{1}+r_{2}}(\alpha))$$
if we view $\mathbb{C}$ as $\mathbb{R}^{2}$. In this way we can show that $\mathfrak{o}_{L}$ is a lattice of rank $r_{1}+2r_{2}=n$.
A: Of course, many of the $\sigma_i$ will have image in $\mathbb{C}$ and not just $\mathbb{R}$, but $\mathbb{C}^n$ is also a Euclidean space, so there is no contradiction.
