# Galois Group of $K/\mathbb Q$ and Embeddings $K\to\mathbb C$

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 $$f$$?

• 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 $\mathbb C$. May 21, 2020 at 6:18

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 $$E$$.)

• I haven't heard of $G$-equivariant bijections before –– what are these? And as a follow-up, what would be a canonical element of $E$? May 20, 2020 at 18:54
• 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 $f$. May 22, 2020 at 10:46
• 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 $\mathbb Q(\ldots)$). May 22, 2020 at 10:51
• 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 $1 \in E$. May 22, 2020 at 17:59
• $K$ 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 $\in E$. May 24, 2020 at 13:15