Suppose $L$ is a field and $H$ is a finite subgroup of $Aut(L)$.
Then $[L:L^H]=|H|$
I've seen a proof of this that uses the Rank-Nullity theorem but it's rather cumbersome and difficult to remember the large number of steps required.
Is there a simpler proof of this that perhaps doesn't use the Rank-Nullity theorem?
Every $\alpha$ has a finite orbit under H $\{\sigma_1(\alpha),\ldots,\sigma_r(\alpha)\}$. Then the polynomial $$\prod (x-\sigma_i(\alpha)) $$ Is in $L^H$. This shows that the extension is separable, normal and every element has degree lower than the cardinality of $H$. So $[L:L^H] $ is finite and given that $H<Gal(L/L^H) $ you have the equality.