Is there any relation between the number of distinct embeddings of a number field into $\mathbb C$ and the number of $\mathbb Q$ automorphisms of it?

Question

Suppose that $K$ is a finite extension of $\mathbb Q$, say of degree $n$. By the primitive element theorem, $K=\mathbb Q(\alpha)$. Then $\alpha$ has $n$ conjugates and we correspondingly get $n$ embeddings of $K$ into $\mathbb C$. But I believe that all these embeddings need to be have the same images in $\mathbb C$ (as an example, $\mathbb Q(\sqrt 2)$ and $\mathbb Q(-\sqrt 2)$ are really the same field). Is there any relation between the number of embeddings with distinct images and the order of the group $Aut(K/\mathbb Q)$? On working a few examples, it seems that if $|Aut(K/\mathbb Q)|=a$ and there are $b$ embeddings with distinct images in $\mathbb C$, then $ab=n$, the degree of the extension. Is this true? If so, how can I prove it? I don't seem to be having much success. I'd appreciate some help.

Answer 1

For an algebraic extension $K/\mathbb{Q}$ of degree $n$ there are $n$ embeddings of $K$ into $\mathbb{C}$, see (The number of) embeddings of an algebraic extension of $\mathbb{Q}$ into $\mathbb{C}$. For example, let $K=\mathbb{Q}(\sqrt[4]{2})$. Then $a=|Aut_{\mathbb{Q}}(K)|=2$, $b=4$ and $n=4$. Hence $ab\neq n$.

Answer 2

Yes, of course there is, and this is one of the main results in fields extensions-Galois Theory.

Theorem: If $\;K/F\;$ is a fields extension of degree $\;n\;$ and $\;S,\,F\subset S\subset K\;$ is the separable closure of this extension, then there are $\;[S:F]\;$different embeddings $\;K\mapsto\overline{\Bbb F}\;$ , when this last is any algebraically closed field containing $\;F\;$ ( we can take the algebraic closure of $\;F\;$) .

If we assume the extension $\;K/F\;$ is finite and separable, then the number of embeddings equals $\;n=[K:F]\;$ iff $\;K/F\;$ is normal, which would mean the extension $\;K/F\;$ is Galois. In this last case, any such embedding is an automorphism of $\;K/F\;$ , meaning $\;\sigma K=K\;$ for any embedding $\;\sigma:K\to\overline F\;$, and $\;\sigma f=f\;,\;\;\forall\,f\in F\;$ .

Answer 3

Maybe an example is appropriate. Let $p$ be a polynomial of degree $3$ with rational coefficients and one real root $r$ and two complex (conjugate) roots ($c$ and $\overline{c}$). The splitting field $F$ of $p$ is an extensioin of degree $6$ and the Galois group of $p$ is the symmetric group $S_3$, the group that permutes the three roots. The Galois group consists of field automorphisms of $F$. On the other hand the field $M$ generated by the root $r$ is a subfield of $F$ is of degree $3$ and is not normal. The only elements of the Galois group that leave this field invariant (and so define automorphisms on $M$) are thos that permute the other two root. This makes that $M$ has two other different embeddings in $F$ that are images of automorphisms of $F$.