# Number of embeddings in algebraic closure

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?

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Use the primitive element theorem. – Qiaochu Yuan Jun 9 '12 at 23:31
For the second question, apply the first result to $L$ as an extension of $K$ and then to $L$ as an extension of $K(x)$. – Qiaochu Yuan Jun 9 '12 at 23:49
Where exactly are you in that page 11 in Neukirch's book? I've the 1999 edition of the book and in page 11 he talks about discriminant, proposition 2.8 ...so where are you? – DonAntonio Jun 10 '12 at 2:12

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.