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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 $f$?

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    $\begingroup$ Write $K = \mathbb Q(\alpha)$. Elements of the Galois group and embeddings are all determined by their action on $\alpha$. Moreover, any embedding must take $\alpha$ to another root of the minimal polynomial of $\alpha$ in $\mathbb C$. $\endgroup$
    – Mathmo123
    May 21, 2020 at 6:18

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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 $E$.)

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  • $\begingroup$ I haven't heard of $G$-equivariant bijections before –– what are these? And as a follow-up, what would be a canonical element of $E$? $\endgroup$
    – Azhao17
    May 20, 2020 at 18:54
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    $\begingroup$ 1. When you have two sets $X, Y$ with group actions of $G$ on each, then a map $f : X \to Y$ is said to be $G$-equivariant when it commutes with the action of $G$ : $f(g*x) = g*f(x)$ for all $g \in G, x \in X$. Here, we have $X = G$ with the action given by right multiplication: $g*h = hg$, and $Y = E$ with the action given by precomposition: $g * e = e \circ g$. A $G$-equivariant bijection $f : G \to E$ is determined by the image of $1 \in G$. So producing a $G$-equivariant bijection is easier than producing any bijection: you only need to pick one element of $E$, and this fixes $f$. $\endgroup$
    – Bart Michels
    May 22, 2020 at 10:46
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    $\begingroup$ 2. In general, there is no canonical embedding $e : K \to \mathbb C$. For example, if I define $K$ to be the splitting field of $x^3+4x^2-2$, what is a canonical embedding $K \to \mathbb C$? There is none. As I say in my answer, you do have a canonical embedding if $K$ is defined as a subfield of $\mathbb C$, in which case you can take $e$ to be the inclusion map. For example, when you define $K = \mathbb Q(\sqrt 2)$, where $\sqrt 2$ is defined as being the positive element of $\mathbb R$ whose square is $2$, then $K \subset \mathbb C$ (simply by definition of $\mathbb Q(\ldots)$). $\endgroup$
    – Bart Michels
    May 22, 2020 at 10:51
  • $\begingroup$ Thank you for the clear answers! So what I understand then is that if $K$ is a number field, it is a subset of $\mathbb{C}$, and so there exists this $G$-equivariant bijection $G \to E$ (where $G = \text{Gal}(K/\mathbb{Q})$) where $1 \in G$ is sent to $1 \in E$. $\endgroup$
    – Azhao17
    May 22, 2020 at 17:59
  • $\begingroup$ $K$ is not necessarily a subfield of $\mathbb C$, but yes, there always exist $G$-equivariant bijections. If $K$ is a subfield, then yes, there exists a $G$-equivariant bijection that sends $1 \in G$ to the inclusion $\in E$. $\endgroup$
    – Bart Michels
    May 24, 2020 at 13:15

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