Why does involution correspond to complex conjugation? Let $K$ be a number field of degree $n$ over $\mathbb{Q}$. In Minkowski theory, we construct a map $j:K\rightarrow K_{\mathbb{C}}:=\prod_\tau\mathbb{C}$, $a\mapsto ja = (\tau a)$, where $\tau$ ranges over the $n$ embeddings of $K$ in $\mathbb{C}$. We can also identify $K\otimes_{\mathbb{Q}}\mathbb{C}\cong K_{\mathbb{C}}$ via the map $x\otimes y\mapsto (\tau(x)y)_\tau$.
Now, in Neukirch's book Algebraic Number Theory, he says that the involution $F(x\otimes y)=x\otimes\overline{y}$ corresponds to complex conjugation $F((x_\tau)_\tau)=(\overline{x}_{\overline{\tau}})_\tau$ in $K_\mathbb{C}$.
My question is, why is this true? Doesn't $x\otimes y$ map to $(\tau(x)\overline{y})_\tau$? How to get from here to the desired form?
 A: Let's take a simple example: $K = \mathbb Q(i)$.  We have $K = \mathbb Q[t]/(t^2+1)$, and
$$K \otimes_{\mathbb Q} \mathbb C = \mathbb Q[t]/(t^2+1) \otimes \mathbb C = \mathbb C[t]/(t^2+1) = \mathbb C[t]/(t-i) \times \mathbb C[t]/(t+i) = \mathbb C \times \mathbb C.$$
This last isomorphism is given by sending a pair of polynomials $(g_1(t), g_2(t))$ to $(g_1(i), g_2(-i))$.
The complex conjugation involution on $K \otimes_{\mathbb Q} \mathbb C$ sends $x \otimes \lambda$ to $x \otimes \overline{\lambda}$.  Let's see what this does to an element $(A,B)$ of $\mathbb C \times \mathbb C$. It's preimage in $\mathbb Q[t]/(t^2+1)\otimes \mathbb C$  is a sum of elementary tensors
$$\xi = h_1(t) \otimes \lambda_1 + \cdots + h_s(t) \otimes \lambda_s$$
where $$\lambda_1 h_1(i) + \cdots + \lambda_s h_s(i) = A$$$$\lambda_1 h_1(-i) + \cdots + h_s(-i) = B$$
Applying the involution to $\xi$ gives you
$$\xi^c = h_1(t) \otimes \overline{\lambda_1} + \cdots + h_s(t) \otimes \overline{\lambda_s}$$
where now
$$\overline{\lambda_1}h_1(i) + \cdots + \overline{\lambda_s} h_s(i)  = \overline{\lambda_1 h_1(-i) + \cdots + h_s(-i)} = \overline{B}$$
$$\overline{\lambda_1}h_1(-i) + \cdots + \overline{\lambda_s} h_s(-i) = \overline{\lambda_1 h_1(i) + \cdots + \lambda_s h_s(i)} = \overline{A}$$
This shows that the involution on $\mathbb C \times \mathbb C$ sends $(A,B)$ to $(\overline{B},\overline{A})$.
The computation here generalizes easily to an arbitrary number field.
