## Background

I'm reading the article "Topological Representations of Algebras" by Arens, Kaplansky. In the proof of Theorem 6.1 we have the following situation: $X$ is a Stone space, $X_\alpha$ is a family of closed subsets, $K$ is a field with an algebraic closure $M$ and separable closure $L \subseteq M$, $L_\alpha$ are the intermediate fields of $M/L$, $R$ is the ring of continuous functions $X \to K^{alg}$ which map $X_\alpha$ to $K_\alpha$, $G = Aut(L/K)$ is the absolute Galois group of $K$, which may be identified with $Aut(M/K)$. Now, why does $G$ act on $R$ via $(g f)(x) = g f(x)$?

## Question

Why is this action well-defined? Don't we need that $L_\alpha / K $ is normal for that? But I doubt that this is true for every algebraic extension of $L$.

There are several other problems in the proof, which arise from the fact that the isomorphism with the ring of study with $R$ is not canonical, but perhaps I'm just missing something ...

Edit: I wonder why nobody answers ... is it too easy? :-)