Primes corresponding to embeddings of a number field

Question

Let $k$ be a number field. Define a prime of $k$ to be an equivalence class of absolute values on $k$. If $\sigma:k\hookrightarrow \mathbb{C}$ is an embedding of $k$ into the complex numbers then we know that $|a|=|\sigma(a)|$ defines an absolute value on $k$.

My question is:

Why is there exactly one prime for each real embedding and for each pair of complex embeddings of $k$?

How can we show this?

Answer 1

There is the so called "Extension Theorem".

Theorem: Let $\mid \;\mid_v$ be an absolute value on a field $K$ and $L$ an algebraic extension. Denote by $K_v$ the completion of $K$ w.r.t $\mid \;\mid_v$ and by $\bar K_v$ its algebraic closure. Note that $\mid\;\mid_v$ extends uniquely to $\bar K_v$, because $K$ is dense in $K_v$, which is complete w.r.t. $\mid\;\mid_v$. Then the following hold:
(1) Every extension of $\mid\;\mid_v$ to $L$ is of the form $|x|_{\tau v}=|\tau x|_v$ for some $K$-embedding $\tau :L\to \bar K_v$
(2) Two extensions $\mid\;\mid_{\tau v}$ and $\mid\;\mid_{\tau'v}$ are equivalent if and only if $\tau$ and $\tau'$ are conjugate over $K_v$, i.e. there is a $\sigma \in Gal(\bar K_v/K_v)$ such that $\sigma \circ \tau =\tau'$.

Apply this to $K=\Bbb Q, L=k$ with the usual absolute value, then $K_v=\Bbb R, \bar K_v=\Bbb C$. Therefore two embeddings $\tau, \tau'$ are conjugate if and only if they are equal or complex conjugates of each other, because $Gal(\Bbb C/\Bbb R)=\{id, \sigma\}$, where $\sigma$ is complex conjugation. Hence, every infinite prime of a number field is either a pair of complex conjugate embeddings or a real embedding (which is its own complex conjugate).

Answer 2

Fix a generator $a \in k$, so that $k = \mathbb Q(a)$. Let $f \in \mathbb Q[X]$ be its minimal polynomial, and let $$ \alpha_1, \dots, \alpha_n \in \mathbb C $$ be the complex roots of $f$; then any embedding $\sigma \colon k \to \mathbb C$ is determined by where it sends $a$, which is one of the $\alpha_i$'s.

Suppose we have two embeddings $\sigma_1, \sigma_2$, which map $a$ to $\alpha_1$ and $\alpha_2$ respectively, such that $|\sigma_1(t) | = |\sigma_2(t)|$ for all $t \in k$.

Then in particular for all $x, y \in \mathbb Q$, we have $$ |x + y \alpha_1 |^2 = |x + y \alpha_2|^2,$$ which holds if and only if $$Re(\alpha_1) = Re(\alpha_2) \qquad \text{ and} \qquad Im(\alpha_1) = \pm Im(\alpha_2). $$

This means that either $\alpha_1 = \alpha_2$ or that $\alpha_1 = \overline{\alpha_2}$; in the first case the embeddings are the same, in the second they are conjugate.