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

Question

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.

Answer 1

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).