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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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Your proof is correct. – Kevin Apr 15 '12 at 17:48
@GregMartin "Totally imaginary" is pretty standard, I think, by analogy with "totally real" -- what's the problem? – David Loeffler Apr 15 '12 at 19:10
If you like, you can also use the primitive element theorem. Let $K=\mathbb{Q}(\alpha)$. Then the minimal polynomial of $\alpha$ either has a real root, or it doesn't. – Brett Frankel Apr 15 '12 at 19:14
Dear Adrian, Here is one way to think of this, which may be helpful: a standard fact, which you quite likely know, is that in a Galois extension of number fields, if $\mathfrak p$ is any prime ideal downstairs, and $\mathfrak q$ is any prime ideal upstairs dividing $\mathfrak p$, then the ramification and intertial indices of $\mathfrak q$ depend only on $\mathfrak p$ (i.e. they are the same for all primes above $\mathfrak p$). Well, the same is true for archimedean primes (where there is no inertia, but there is ramification: real over real or complex over complex is unramified, but ... – Matt E Apr 16 '12 at 5:00
... complex over real is ramified). In particular, when the bottom field is $\mathbb Q$, it has just one archimedean prime, and so if $K$ is Galois over $\mathbb Q$, then every archimedean prime of $K$ (all of which lie over the unique such prime of $\mathbb Q$) is either real or complex, i.e. $K$ is either totally real or totally complex (the latter being my preferred term for what you call "totally imaginary" --- but that latter term is also fine). Regards, – Matt E Apr 16 '12 at 5:02
up vote 6 down vote accepted

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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Thank you very much for this answer. It's nice to see different ways to prove the statement in my question. – Adrián Barquero Apr 16 '12 at 13:00
You are welcome! – Andrea Mori Apr 16 '12 at 22:02

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