Number of embeddings in algebraic closure

Question

I'm having trouble following the details of discussion on pages 9 and 10 of Neukirch's algebraic number theory book.

Suppose $L$ is a separable extension of $K$ with degree n. Consider the set of embeddings of $L$ into $\bar K$, the algebraic closure of $K$, that fix $K$ (K-embeddings). Why are there $n$ embeddings in this set?

EDIT: Also, consider some element $x\in K$. Let $d$ be the degree of $L$ over $K(x)$ and $m$ be the degree of $K(x)$ over $K$. Why are the $K$ embeddings of $L$ partitioned by the equivalence relation

$$\sigma\sim\tau\ \Leftrightarrow\ \sigma x = \tau x$$

into $m$ equivalence relations of $d$ elements?

Answer 1

The idea behind the proof is that for a field $K$ and an element $\alpha \in \bar{K}$, the roots of the minimal polynomial of $\alpha \in \bar{K}$ are exactly the conjugates of $\alpha$ over $K$. Then taking $L = K(\alpha)$ each conjugate of $\alpha$ defines a unique embedding from $L$ to $\bar{K}$. Since $[L: K] = n$, there are $n$ distinct embeddings.

For the full details of this proof look at Lemma 5.17 and Theorem 5.18 of this.