Number of embeddings in algebraic closure I'm having trouble following the details of the 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 $x\in L$. 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 classes of $d$ elements each?

I'll provide a slightly different solution to your second question. Assuming that $\#\text{Hom}(L,\bar K)=n$, let $x=x_1,\dots,x_m\in \bar K$ be the conjugates of $x$ (i.e., the roots in $\bar K$ of the minimal polynomial $m_{x,K}(t)$ for $x$ over $K$). Recall that each of the $m$ isomorphisms $K(x)\rightarrow K(x_j)$ extend to $d$ isomorphisms $L=K(x,\theta)\rightarrow K(x_j,\theta_k)$ where $\theta\in \bar K$ is a primitive element for $L$ over $K(x)$ (i.e. $L=K(x,\theta)$) and  $\theta=\theta_1,\dots,\theta_d$ are its conjugates. (This is Theorem 8 in section 13.1 of Dummit and Foote.) This accounts for all $n=md$ embeddings $L\rightarrow \bar K$ via $L=K(x,\theta)\rightarrow  K(x_j,\theta_k)\rightarrow\bar K$, where the last map is inclusion. Thus, what we find is that for each $1\leq j \leq m$ there are $d$ embeddings $\sigma:L\rightarrow \bar K$ with $\sigma: x\mapsto x_j$. Hence there are $m$ equivalence classes, each with $d$ elements. 
