Prove $\theta:S\to E^G$ is $R$-module isomorphism

Question

I'm currently studying the paper Galois Theory and Galois Cohomology of Commutative Rings by S.U. Chase, D.K. Harrison and Alex Rosenberg. I'm trying to understand step by step the proof of Theorem 1.3, and there's a detail in (d)$\implies$(e) that I don't understand. Since there are useless context in the proof where I don't have issues with, I'll try to formulate a more general question that will solve my specific problem while being easier to understand:

Let $S$ be a commutative ring, $G$ a finite group of automorphisms of $S$, and $R=S^G$ (subring of $S$ whose elements are left fixed by every element of $G$). Let $E$ be the $S$-algebra of all functions from $G$ to $S$ under pointwise addition and multiplication.

Let $G$ act on $E$ by setting, being $\sigma,\tau\in G$, $s\in S$, $v\in E$: $$(\sigma v)(\tau)=\sigma(v(\sigma^{-1}\tau)).$$ Now, $E^G$ (elements of $E$ left fixed by every element of $G$) is easily seen to be the $G$-homomorphisms of $G$ to $S$ and thus the map $\theta:S\to E^G$ defined by $\theta(s)(\sigma)=\sigma(s)$ is an $R$-module isomorphism.

I've succesfully understood that $E^G$ is made of the $G$-homomorphisms from $G$ to $S$. My question is about the proof if $\theta$ being $R$-module isomorphism. I think understand it's $R$-module homomorphism (due to the fact that $R=\{r\in S \mid \sigma(r)=r, \forall\sigma\in G\}$). However, I am unable to prove it's an isomorphism. I managed to prove it's injection, but can't do the same surjectivity. How can I prove it's a surjection?

Any help or hint will be appreciated, thanks in advance.

Answer 1

Notice that $\sigma \in G$, so it is an automorphism of $S$. This means that $\sigma$ is invertible (for composition). Injectivity is then easy to prove.

Let consider $f \in E^G$. To show surjectivity we need to find $ s\in S$ such that $\theta (s) = f$. Since $f : G \rightarrow S$ consider $s = f(1)$ where $1 \in G$ is the identity element.

Then

$$ \theta(s)(\sigma) = \sigma(s) = \sigma(f(1)) = \sigma(f(\sigma^{-1}\sigma)) = (\sigma\cdot_G f) (\sigma)=f(\sigma)$$

where $\cdot_G$ denotes the action of $G$ on $E$. The above equations hold for all $\sigma \in G$ so we have $\theta(s) = f$.