I am familiar with the proof of the fact that if $K/F$ is a finite extension, then |Aut$(K/F)$| $\leq [K:F]$. I am wondering if the following proof works in the case of a separable finite extension:
By the primitive element theorem, $K = F(\alpha)$ for some $\alpha \in \overline{F}$. Then, any $\sigma \in$ Aut$(K/F)$ is completely determined by where it sends $\alpha$. If $m_\alpha$ denotes the minimal polynomial of $\alpha$ over $F$, then $\sigma(\alpha)$ is a root of $m_\alpha$. Thus, there are $\deg(m_\alpha) = [F(\alpha):F]$ possibilities for $\sigma(\alpha)$ and so |Aut$(F(\alpha)/F)$| $\leq [F(\alpha):F]$.
Does this proof make sense?
Thank you
Once again $K/F$ is separable iff $|Hom_{F-algebra}(K,\overline{K})| = [K:F]$,
In general write $K/F$ as a tower of simple extensions $F_n[x_{n+1}]/(f_{n+1})/F_n$, multiply the number of distinct roots of the $f_n$, what is called the separable degree $[K:F]_s$, you'll have $|Hom_{F-algebra}(K,\overline{K})|=[K:F]_s$.
Then $Aut(K/F)=Hom_{F-algebra}(K,K)$ and $Hom_{F-algebra}(K,\overline{K})=Hom_{F-algebra}(K,K)$ iff $K/F$ is normal.