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Let $L$ be a subfield of $\Complex$
Definition 1
- A real embedding of $L$ is an injective field homomorphism $$ \sigma\colon L \hookrightarrow \Reals $$
- A (non-real) complex embedding of $L$ is an injective field homomorphism $$ \tau\colon L \hookrightarrow \Complex $$ such that $\tau(L)\nsubseteq \Reals$
- We denote $\Sigma_L$ the set of all embeddings, real and complex, of $L$ in $\Complex$ (note that all of them must fix $\Rats$ since they are field homomorphisms).
Note that if $\sigma$ is a real embedding then $\bar{\sigma}=\sigma$ where $\overline{\cdot}$ denotes the complex conjugation automorphism: $$ \overline{\cdot}\colon \Complex \to \Complex,\quad \overline{(a+bi)}=a-bi$$ On the other hand, if $\tau$ is a complex embedding, then $\bar{\tau}$ is another complex embedding, so the complex embeddings always come in pairs $\{\tau,\bar{\tau}\}$
Let $K\subseteq L$ be another subfield of $\Complex$ Moreover, assume that $[L:K]$ is finite (this is the dimension of $L$ as a vector space over $K$ . We are interested in the embeddings of $L$ that fix $K$ pointwise, i.e. embeddings $\psi\colon L \hookrightarrow \Complex$ such that $$\psi(k)=k,\quad \forall k\in K$$
Theorem 1 For any embedding $\psi$ of $K$ in $\Complex$ there are exactly $[L:K]$ embeddings of $L$ such that they extend $\psi$ In other words, if $\varphi$ is one of them, then $$\varphi(k)=\psi(k),\quad \forall k\in K$$ Thus, by taking $\psi=\operatorname{Id}_K$ there are exactly $[L:K]$ embeddings of $L$ which fix $K$ pointwise.
Hence, by the theorem, we know that the order of $\Sigma_L$ is $[L:\Rats]$ The number $[L:\Rats]$ is usually decomposed as $$[L:\Rats]=r_1+2r_2$$ where $r_1$ is the number of embeddings which are real, and $2r_2$ is the number of embeddings which are complex (non-real). Notice that by the remark above this number is always even, so $r_2$ is an integer.
Remark: Let $\psi$ be an embedding of $L$ in $\Complex$ Since $\psi$ is injective, we have $\psi(L)\cong L$ so we can regard $\psi$ as an automorphism of $L$ When $L/\mathbb{Q}$ is a Galois extension, we can prove that $\Sigma_L\cong \operatorname{Gal}(L/\Rats)$ and hence proving in a different way the fact that $$\mid \Sigma_L \mid = [L:\Rats]= \mid \operatorname{Gal}(L/\Rats)\mid$$
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