Confused about part of Milne's notes on the Galois theory of étale algebras over a field

Question

Let $F$ be a perfect field, let $F^c$ be an algebraic closure and let $G=\text{Gal}(F^c/F).$

To each finite set $S$ with continuous $G$-action, Milne associates the $F$-algebra

$$A(S):=\{f:S \to F^c \; | \; f(\sigma\cdot s)=\sigma(f(s)) \;\; \forall s \in S \; \; \forall \sigma \in G\}$$

of $G$-equivariant maps $S \to F^c.$

He then claims that this is an étale $F$-algebra.

It is clear to me that this algebra is reduced but not clear that it is finite-dimensional over $F.$

Confusion: For example, if $G$ acts transitively on $S,$ then surely $A(S)$ is in bijection with $F^c$ and this isn't finite-dimensional over $F$...

Answer 1

To begin with, the functor $A$ from the category of finite sets with a continuous $G$-action into the category of $F$-algebras takes coproducts to products, i.e. $A(S\sqcup T)=A(S)\times A(T)$ holds for any finite $G$-sets $S$ and $T$. This is not hard to show. It is therefore sufficient to consider a transitive $G$-set $S$, noting that every finite $G$-set decomposes into a finite coproduct of such transitive $G$-sets.

Now fix any point $s\in S$ and let $H\subset G$ be the stabilizer of $s$, which means that we find $S\cong G/H$ as $G$-sets by the orbit-stabilizer theorem. Observe that $H$ is a closed subgroup of finite index in $G$. Hence $A(S)\cong A(G/H)$ and I claim that the latter is isomorphic as an $F$-algebra to the fixed field of $F^c$ by $H$ by virtue of the $F$-algebra homomorphism $$A(G/H)\rightarrow (F^c)^H,~f\mapsto f(H).$$ This is not hard to show, I'll let you work out the details by yourself. Finally, infinite Galois theory tells us that the right-hand side is a finite subextension of $F^c/F$ and therefore étale by separability. Hence $A(S)$ is étale over $F$, which proves the claim.

Answer 2

I assume $F$ is finite. Since $S$ is finite, $G/H$ is finite where $H$ is the kernel of the homomorphism $G\rightarrow Aut(S)$. This implies that the image of every element of $A(S)$ is contained in $L$ where $L$ is the subfield of $F^c$ fixed by $H$. Since $L$ is finite, the result follows.