How to show the set $\operatorname{Hom}_K(L,\bar{K})$ of all $K$-embeddings of $L$ is partitioned into $m$ equivalence classes of $d$ elements each?

Question

Let $L|K$ be a finite separable extension. Denote the algebraic closure of $K$ by $\bar K$.

$\forall x\in L$, denote $d=[L:K(x)]$ and $m=[K(x):K]$.

How to show the set $\operatorname{Hom}_K(L,\bar{K})$ of all $K$-embeddings of $L$ is partitioned by the equivalence relation $$\sigma\sim \tau \Longleftrightarrow \sigma x=\tau x$$ into $m$ equivalence classes of $d$ elements each?

Answer 1

I guess because the degree of the minimal polynomial $p(x)$ of $x$ over $K$ is $m$, then $p(x)$ has $m$ different roots $a_1,\cdots,a_m$ in $\bar K$, we can let $\sigma x=a_i(i=1,\cdots,m)$. But How to show there exist $d$ different $K$-embeddings sending $x$ to $a_i$ for each $i$?

Answer 2

One way to think of this via Galois theory, is that if you take $F$ to be the Galois closure of $L$ over $K$, then

1. $Hom_K(L, \overline{K}) \cong Hom_K(L, F) \cong Gal(F/K) / Gal(F/L)$
2. The partition you are talking about is really the map $$ Hom_K(L, F) \stackrel{res}{\longrightarrow} Hom_K(K(x), F)$$ So if you think from the perspective of group theory, it's really the quotient map $$ Gal(F/K) / Gal(F/L) \to Gal(F/K) / Gal(F/K(x))$$ So the size of each equivalence class being equal really comes from all cosets have the same size.