extend trace and norm from a number field $K$ to $ K \otimes_{\mathbb Q} \mathbb R $ I read somewhere that we can extend the trace and the norm of a number field $K$ to the commutative algebra $ V=K \otimes_{\mathbb Q} \mathbb R$.
Before state exactly my question, let me write the notation and the setting:
Let $K$ be a number field of degree $n=[K: \mathbb Q]$ over $\mathbb Q$, and let $\mathcal{O}_K$ be its ring of integers. Fix an integral ideal $ \mathfrak n$ of $\mathcal{O}_K$ and a $\mathbb Z$-basis $\omega_1, \omega_2, \dots, \omega_n$ of $\mathfrak n$. Through the canonical embedding $ \displaystyle K \to K \otimes_{\mathbb Q} \mathbb R=:V$ we obtain that $\omega_1, \omega_2, \dots ,\omega_n$ is an $\mathbb R$-basis of $V$.  Let now $K_i$ be the completion of $K$ with respect to the embedding $ \sigma_i, \hspace{0.1in} i=1,2,\dots, n_1 + n_2 $, so we have $K_i = \mathbb R $ for $ i \leq n_1 $ and $ K_i= \mathbb C $ for $i>n_1$. Hence we can consider $ \displaystyle  V= K \otimes_{\mathbb Q} \mathbb R  \simeq \bigoplus_{i=1}^{n_1 + n_2} K_i $ which is an $n$-dimensional commutative $\mathbb R$-algebra. Using now the canonical embedding $ K \hookrightarrow V, \hspace{0.1in} \alpha \mapsto \bigoplus_i \sigma_i (\alpha) $ we can identify $K$ with its image in $V$ and hence $\mathfrak n$ forms a lattice in $V$. Therefore, we may consider $V$ to be the set $$V= \{r_1 \omega_1+ r_2 \omega_2 + \cdots + r_n \omega_n: r_i \in \mathbb R\}$$ Using the above identifications we now extend the trace and norm maps on the number field $K$ to $V$ as follows $$ \boxed{ \text{Tr} (x)= x^{(1)} + x^{(2)} + \cdots + x^{(r)} + 2\text{Re}(x^{(r+1)}) + 2\text{Re}(x^{(r+2)}) + \cdots + 2\text{Re}(x^{(r+s)})} $$ and $$ \boxed{ \text{Nm}(x)= x^{(1)} x^{(2)} \cdots x^{(r)} |x^{(r+1)}|^2 |x^{(r+2)}|^2 \cdots |x^{(r+s)}|^2 }$$

Question: How do we obtain these two formulas ?

Thank you in advance.
 A: This answer was already pointed out in the comments.
Suppose $K$ is a field and $L$ is a finite $K$-algebra, meaning $\dim_K L < \infty$. Then for any $a\in L$, we have a $K$-linear map map $L_a(x): L \rightarrow L, x \mapsto ax$. By definition ${\rm{Tr}}_{L/K}(a)$ and ${\rm{Nm}}_{L/K}(a)$ are the trace and determinant of $L_a$ as a linear transformation of $K$-vector spaces. They can be computed by taking a $K$-basis for $L$, writing the matrix of $L_a$, and taking its trace and determinant.
In fact it's not necessary that $K$ be a field. It's enough that $L$ is free over $K$, so a $K$-basis of $L$ exists.
Now if $R$ is any $K$-algebra, and $L$ is free over $K$, then $L\otimes_K R$ is a free algebra over $K\otimes_K R = R$ of the same rank. In fact the same basis of $L$ over $K$ will be an $R$-basis of $L\otimes_K R$ over $R$, under the inclusion $L \hookrightarrow L \otimes_K R, y \mapsto y \otimes 1.$ Moreover, if $a\in L$, then the matrix of $L_a$ as a map of $K$-algebras is the same as the matrix of $L_{a\otimes 1}$ as a map of $R$-algebras. Therefore ${\rm Tr}_{L/K}(a) = {\rm Tr}_{L\otimes_K R/R}(a\otimes 1)$. This shows the inclusion $L \hookrightarrow L \otimes_K R$ really does extend trace and norm the way you want.
Now as for the particular formulas. If $K$ is a number field, and $\Phi\subset {\rm Hom}(K,\mathbb{C})$ is such that $\Phi \cup \overline{\Phi} = {\rm{Hom}}(K,\mathbb{C})$ and $\Phi \cap \overline{\Phi} = {\rm Hom}(K,\mathbb{R})$,  then $K\otimes \mathbb{R} \rightarrow \bigoplus \mathbb{R}^r\oplus \mathbb{C}^s$ is the isomorphism in your question. Then under this isomorphism, using the standard basis of $\mathbb{R}^r\oplus \mathbb{C}^s$, the matrix of $L_a$ is almost diagonal: it has as $2\times 2$ entry for each copy of $\mathbb{C}$. Computing the trace and determinant of this matrix gives the formula you want.
