Does the size of the automorphism group divide the separable degree
问题内容
Let $E/K$ be a finite field extension, $G=\operatorname{Aut}_K(E)$ be the group of automorphisms fixing elements of $K$, and $E^G$ be the fixed field of $E$ under $G$.
We know $E/E^G/K$ and so $$[E:K]=[E:E^G][E^G:K]=|G|\,[E^G:K].$$
Additionally, $[E:K]=[E:K]_s[E:K]_i,$ where $[E:K]_s$ denote the separable degree and $[E:K]_i$ inseparable one. Consequently, $$|G|\,[E^G:K]=[E:K]_s[E:K]_i.$$
If $E/K$ is normal, then $\frac{[E:K]_s}{|G|}=1$, if $E/K$ is separable, then $$\frac{[E:K]_s}{|G|}=[E^G:K].$$ While $|G|\le[E:K]_s$ always holds, is it sure that $|G|$ divides $[E:K]_s$? If the answer is no, $E/K$ must be inseparable and not normal. If the answer is no, would you give me a counterexample?
回答 (1)
The argument is correct. The only point that should be stated explicitly is the use of the multiplicativity of separable degrees in towers, namely $$ [E:K]_s = [E:E^G]_s [E^G:K]_s. $$ Since $E/E^G$ is Galois, we have $$ [E:E^G]_s = [E:E^G] = \lvert\operatorname{Aut}_K(E)\rvert. $$ Therefore, $$ [E:K]_s = \lvert\operatorname{Aut}_K(E)\rvert \, [E^G:K]_s, $$ which immediately implies $$ \lvert\operatorname{Aut}_K(E)\rvert \mid [E:K]_s. $$ With this standard result, the proof is complete.