Infinite primes of a number field

Question

Let $K$ be a number field. I know that to each real and to each complex conjugate pair of embeddings of $K$ there corresponds exactly one prime (equivalence class of absolute values) of $K$. How do I show that distinct embeddings give rise to distinct primes?

Many thanks!

Answer 1

Since nobody answered yet, let me have a try. But be careful: This is just an idea. I didn't check it in all details so there is no guarantee for correctness.

You already know that each complex embedding of $K$ gives an archimedean absolute value. If we find a way to assign an embedding to each (class of) archimedean absolute value, you can easily show that both mappings are inverse to each other, and the correspondence turns out to be bijective. So let $|\cdot|$ be an arbitrary archimedean absolute value.

1. Show that the restriction of $|\cdot|$ on $\mathbb{Q}$ is equivalent to the usual absolute value. So without loss of generality we may assume that they coincide on $\mathbb{Q}$.
2. The completions of $\mathbb{Q}$ and $K$ with respect to $|\cdot|$ yield an injective (canonical) homomorphism $\mathbb{R} = \widehat{\mathbb{Q}} \stackrel{\varphi}{\to} \widehat{K}$. Show that it is algebraic.
3. By the fundamental theorem of algebra there are only two possibilities: $\varphi$ must either be an isomorphism or $\varphi$ defines a field extension of degree $2$ in which case $\widehat{K} \cong \mathbb{C}$. The latter isomorphism is not canonical, but there are again only two possibilities (compatible with $\varphi$): If we take a root $x \in \widehat{K}$ of the polynomial $X^2+1$ we can either assign $x \mapsto i$ or $x \mapsto -i$.

Conclusion: For each archimedean absolute value $|\cdot|$ on $K$ we get a canonical embedding $K \to \mathbb{R}$ or a canonical pair of embeddings $K \to \mathbb{C}$.