If any, what is the relation between $\text{Gal}(K/\mathbb{Q})$ and the set of embeddings (say $E$) of $K \to \mathbb{C}$? I ask this for two reasons:
(1) The orders of $E$ and $\text{Gal}(K/\mathbb{Q})$ are identical and equal to $[K: \mathbb{Q}]$.
(2) The embeddings $e \in E$ permute $K$ in the sense that for a minimal element $\alpha \in K$ with minimum polynomial $f$ over $\mathbb{Q}$, $f(e(\alpha)) = 0$, and so $e$ represents a permutation of the roots of $f$, which brings to mind the action of the Galois group of the separable extension $\mathbb{Q}[x]/(f(x)) \simeq \mathbb{Q}[\alpha]$.
Is there an easy isomorphism between $E$ and $\text{Gal}(K/\mathbb{Q})$? Or is perhaps $K$ the splitting field of $f$?
Best Answer
One remark: the orders $|E|$ and $|\operatorname{Gal}(K/\mathbb Q)|$ are equal only when $K/\mathbb Q$ is a Galois extension.
Under this hypothesis, $E$ is a homogeneous set (or principal homogeneous space, or $G$-torsor) for $G = \operatorname{Gal}(K/\mathbb Q)$, where the Galois group acts by precomposition. That is, $G$ acts simply transitively on $E$.
A $G$-equivariant bijection $G \to E$ is determined by the image of $1 \in G$. Unless there is a canonical element of $E$, there is no canonical bijection $G \to E$. (When $K \subset \mathbb C$, one can take the inclusion map, which gives a canonical element of $E$.)