# Fields whose embeddings into the complex numbers are invariant under complex conjugation

Is there a general notion/description of fields $K$ such that the image of any embedding $K \hookrightarrow \mathbb{C}$ is invariant under complex conjugation, thus inducing an involution on $K$ which is independent on the chosen embedding into $\mathbb{C}$ but gives complex conjugation when choosing one?

Examples of such fields are $\mathbb{Q}$, CM-fields like cyclotomic fields, but also $\mathbb{R}$ (which is not a number field, I want to include this case!).

In the book "Unitary Reflection Groups" by Lehrer-Taylor on page 20 (see http://books.google.com/books?id=7QSFEnNh7WkC&lpg=PP1&pg=PA20#v=onepage&q&f=false) it is somehow mentioned that any abelian number field has this property. Is this correct? I doubt this but would be happy if so.

I would like to use this to define the notion of inner products for vector spaces for fields different from $\mathbb{C}$ and $\mathbb{R}$. This is also the context in which Lehrer-Taylor use this. I think it's not a good idea to just take a subfield of $\mathbb{C}$ which is invariant under complex conjugation since then all notions depend on the chosen embedding. Any ideas on how to properly do this are welcome, too.

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Doesn't every number field which is galois over $\mathbb Q$ have this property? In this case there are as many complex embeddings as there are elements of the Galois group, hence every complex embedding is obtained if you fix one and apply elements of the Galois group to it. So all complex embeddings have the same image. – Anton May 22 '13 at 18:13
I do not think that it is true for $\mathbb R$: they are other subfields of $\mathbb C$ isomorphic to $\mathbb R$. Those fields cannot be invariant under complex conjugation, since $\mathbb R$ has no non-trivial automorphism. In fact, I think that $K$ must be algebraic over $\mathbb Q$. – Pierre Simon May 22 '13 at 19:04
As for your abelianness question, every abelian field is either totally real or CM, hence the answer is definitely yes. – Filippo Alberto Edoardo May 22 '13 at 20:12
@anton: No, if the complex conjugation is not in the center then the choice of embedding into $\mathbf{C}$ will make a difference. The CM number fields and totally real number fields are precisely the ones for which all embeddings into $\mathbf{C}$ are stable under complex conjugation with induced effect on the field independent of the embedding. – user29283 May 23 '13 at 1:43

The only fields with this property are algebraic over $\mathbb{Q}$. They are either totally real or a quadratic imaginary extension of a totally real field (but not necessarily finite).
Proof: Note that if a field $L$ satisfies your condition, then any subfield $K$ such that $L/K$ is algebraic also satisfies the condition (embeddings into alg. closed fields can always be extended to embeddings of a given finite extension).
Now suppose there was a field transcendental over $\mathbb{Q}$ which satisfied this property. Then by taking a transcendence basis, the field $\mathbb{Q}(\lbrace x_i\rbrace_{i\in I})$ would have this property, where the $x_i$ are mutually transcendental and $I$ is a non-empty set of cardinality $\leq 2^{\aleph_0}$.
However this last field clearly has both real embeddings and non-real embeddings into $\mathbb{C}$, so could not satisfy your condition.
This proves that any such field has to be algebraic over $\mathbb{Q}$. Then it basically follows by definition that the field is either totally real or a quadratic imaginary extension (note that there is a maximal such field, call it $\mathbb{Q}^{cm}$; it is the subfield of $\overline{\mathbb{Q}}$ fixed by the subgroup $[c,G_\mathbb{Q}]\subset G_\mathbb{Q}$, where $c$ is some choice of complex conjugation).