How to prove algebraic integers of degree $n$ number field form a lattice in $\mathbb{R}^n$?

Question

Let $K$ be a degree $n$ number field. I know that the ring of integers $O_K$. Let $\sigma_1, ..., \sigma_r$ be the $r$ real embeddings and $\tau_i$ and $\tau'_i$ be the pairs of complex embeddings $(1 \leq i\leq s)$. Let $w_1, ..., w_n$ be the integral basis of $O_K$. We can then view $O_K$ as a lattice in $\mathbb{R}^n$ via the map $$ j: K \rightarrow \mathbb{R}^r \times \mathbb{C}^s \cong \mathbb{R}^n $$ and so $$ j(O_K) = \{ \sum a_i j(w_j) : a_i \in \mathbb{Z} \}. $$ My question is how do we know that the vectors $j(w_i)$'s are linear independent over $\mathbb{R}$ (to form a lattice in $\mathbb{R}^n$)? Thank you very much!

Answer 1

Hint: Show that the matrix $A$ whose $i$th row is $$ (\sigma_1(w_i),\ldots\sigma_r(w_i), \Re{(\sigma_{r+1}(w_i)}),\Im{(\sigma_{r+1}(w_i)}), \ldots) $$ has non-zero determinant. You should be able to relate the determinant of $A$ to the determinant of the matrix $B$ whose $i$th row is $$ (\sigma_1(w_i),\ldots\sigma_r(w_i),{\sigma_{r+1}(w_i)},\overline{{\sigma_{r+1}(w_i)}}, \ldots,\overline{{\sigma_{r+s}(w_i)}}) $$ using column operations. Since $\text{det}(B)^2=\text{disc}(w_1,\ldots,w_n)\neq 0$, the result for $\text{det}(A)$ follows.

See page 79 of Milne from which the above was taken for details.

Answer 2

As an alternative to the solution by sharding4, observe that the map $j$ is precisely the composition $K\to K\otimes_{\mathbb Q} \mathbb R\to \mathbb R^r \times \mathbb C^s$ where the right homomorphism is an isomorphism and the left one is an inclusion. Now the image of your integral basis is precisely the image of the $w_i\otimes 1$ under the right isomorphism. But a basis is still a basis after extending scalars, hence the $w_i\otimes 1$ are linearly independent over $\mathbb R$, hence so are your $j(w_i)$.