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# real and complex embeddings

Let $L$ be a subfield of $\mathbb{C}$.

###### Definition 1.

1. A

*real embedding*of $L$ is an injective field homomorphism$\sigma\colon L\hookrightarrow\mathbb{R}$ 2. A (non-real)

*complex embedding*of $L$ is an injective field homomorphism$\tau\colon L\hookrightarrow\mathbb{C}$ such that $\tau(L)\nsubseteq\mathbb{R}$.

3. We denote $\Sigma_{L}$ the set of all embeddings, real and complex, of $L$ in $\mathbb{C}$ (note that all of them must fix $\mathbb{Q}$, 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:

$\overline{\cdot}\colon\mathbb{C}\to\mathbb{C},\quad\overline{(a+bi)}=a-bi$ |

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 $\mathbb{C}$. 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 $\psi\colon L\hookrightarrow\mathbb{C}$ such that

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

###### Theorem 1.

For any embedding $\psi$ of $K$ in $\mathbb{C}$, 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 $[L:\mathbb{Q}]$. 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). 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 $\mathbb{C}$. 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 $\Sigma_{L}\cong\operatorname{Gal}(L/\mathbb{Q})$, and hence proving in a different way the fact that

$\mid\Sigma_{L}\mid=[L:\mathbb{Q}]=\mid\operatorname{Gal}(L/\mathbb{Q})\mid$ |

## Mathematics Subject Classification

12D99*no label found*

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