real and complex embeddings

Let $L$ be a subfield of $ℂ$.

Definition 1.

1. A real embedding of $L$ is an injective fieldhomomorphism

   $$\sigma :L\hookrightarrow ℝ$$

2. A (non-real) complex embedding of $L$ is an injectivefield homomorphism

   $$\tau :L\hookrightarrow ℂ$$

   such that $\tau (L)⊈ℝ$.

3. We denote ${\mathrm{\Sigma }}_L$ the set of all embeddings, real andcomplex, of $L$ in $ℂ$ (note that all of them must fix$\mathbb{Q}$, since they are field homomorphisms).

Note that if $\sigma$ is a real embedding then$\overline{\sigma }=\sigma$, where $\overline{\cdot }$ denotes thecomplex conjugation automorphism:

$$\overline{\cdot }:ℂ\to ℂ,\overline{(a+bi)}=a-bi$$

On the otherhand, if $\tau$ is a complex embedding, then $\overline{\tau }$ isanother complex embedding, so the complex embeddings always comein pairs $\{\tau ,\overline{\tau }\}$.

Let $K\subseteq L$ be another subfield of $ℂ$. Moreover,assume that $[L:K]$ is finite (this is the dimension of $L$ as avector space over $K$). We are interested in the embeddings of $L$that fix $K$ pointwise, i.e. embeddings $\psi :L\hookrightarrow ℂ$ such that

$$\psi (k)=k,\forall k\in K$$

Theorem 1.

For any embedding $\psi$ of $K$ in $\mathrm{C}$, there are exactly$\mathrm{[}L\mathrm{:}K\mathrm{]}$ embeddings of $L$ such that they extend $\psi$. In otherwords, if $\phi$ is one of them, then

$$\phi (k)=\psi (k),\forall k\in K$$

Thus, by taking $\psi \mathrm{=}{\mathrm{Id}}_K$, there are exactly$\mathrm{[}L\mathrm{:}K\mathrm{]}$ embeddings of $L$ which fix $K$ pointwise.

Hence, by the theorem, we know that the order of ${\mathrm{\Sigma }}_L$ is$[L:\mathbb{Q}]$. The number $[L:\mathbb{Q}]$ is usually decomposed as

$$[L:\mathbb{Q}]=r_1+2r_2$$

where $r_1$ is the number of embeddings which are real, and $2r_2$is the number of embeddings which are complex (non-real). Noticethat by the remark above this number is always even, so $r_2$ isan integer.

Remark: Let $\psi$ be an embedding of $L$ in $ℂ$.Since $\psi$ is injective, we have $\psi (L)\cong L$, so we canregard $\psi$ as an automorphism of $L$. When $L/\mathbb{Q}$ is aGalois extension, we can prove that ${\mathrm{\Sigma }}_L\cong \mathrm{Gal}(L/\mathbb{Q})$, and hence proving in a different waythe fact that

$$\mid {\mathrm{\Sigma }}_L\mid =[L:\mathbb{Q}]=\mid \mathrm{Gal}(L/\mathbb{Q})\mid$$