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I came across the following assertion in the Wikipedia article about totally imaginary number fields.

Let $K/\mathbb{Q}$ be an algebraic number field that is Galois over $\mathbb{Q}$. Then $K$ is totally real or it is totally imaginary.

Now, since I don't have any references for this result I was trying to prove it myself, and my supposed proof is the following.

What I think is that I basically only need the fact that $K/\mathbb{Q}$ is a normal extension, and then I use the following condition, that is equivalent to normality. Since $K/\mathbb{Q}$ is normal then every embedding $\sigma: K \hookrightarrow \overline{\mathbb{Q}}$ is an automorphism of $K$, which means in particular that $\sigma(K) = K$. Then since we either have $K \subset \mathbb{R}$ or else $K \cap (\mathbb{C} \setminus \mathbb{R} ) \neq \emptyset$ then this shows that $K$ is either totally real or totally imaginary.

Now my questions are if my proof is correct or if I'm missing something and if someone can provide me some references where totally real, totally imaginary and CM-fields are treated at least is some detail or where I can find some basic properties of these types of number fields.

Thank you very much.

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    @MattE Dear Matt, thank you very much. That's a very interesting way of looking at it.2012-04-16

1 Answers 1

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Your proof is correct and here's a variant.

Suppose $\Bbb Q\subset K$ Galois with $[K:\Bbb Q]=n$. Then $\text{Gal}(K/\Bbb Q)$ is the group of automorphisms of $K$ and consists of $n$ elements. It acts by composition on the set of embeddings $K\hookrightarrow\Bbb C$.

Suppose that $K$ admits a real embedding $\phi:K\rightarrow\Bbb R$ and consider its Galois orbit. It consists of the embeddings $ K\stackrel\sigma\longrightarrow K\stackrel\phi\longrightarrow\Bbb R $ as $\sigma\in\text{Gal}(K,\Bbb Q)$. Since $\phi$ is injective they are all different. Thus the orbit consists of $n$ embeddings.

But we know that $n$ is also the number of all embeddings. Therefore $\text{Gal}(K/\Bbb Q)$ acts transitively on the set of embeddings and they are all real.

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    Tha$n$k you very much for this answer. It's nice to see different ways to prove the statement in my question.2012-04-16