Prove $j:D\to \text{Hom}_R(S,S)$ is left $S$-module homomorphism

Question

I'm studing the paper Galois Theory and Galois Cohomology of Commutative Rings by S. U. Chase, D. K. Harrison and Alex Rosenberg. I'm trying to read it carefully and understanding the definitions (which are a lot), and there's a moment in pages 3 and 4 where it's stated:

"$S$ is a commutative ring, G is a finite group of ring automorphisms of $S$, and $R=S^G$, the subring of $S$ consisting of all elements of $S$ left fixed by every element of $G$. "

"Let $D=D(S,G)$ denote the trivial crossed product of $S$ with $G$. This means that $D$ is a free $S$-module with generators $u_\sigma$ ($\sigma$ in $G$), with $R$-algebra structure defined by the formula $$(su_\sigma)(tu_\tau)=s\sigma(t)u_{\sigma\tau}\phantom{a},\phantom{a}(s,t\in S; \sigma,\tau\in G).$$ The identity of $D$ is $u_1$, and we shall denote it by the symbol $1$. There is an $R$-algebra homomorphism $j:D\to\text{Hom}_R(S,S)$ defined by $j(su_\sigma)(x)=s\sigma(x)$ for $s,x$ in $S$ and $\sigma$ in $G$. $j$ is also a left $S$-module homomorphism, where the $S$-module structure on $\text{Hom}_R(S,S)$ arises from the $S$-module structure of the covariant argument. "

I've been trying to prove the statement "$j$ is also a left $S$-module homomorphism", but there is one point I'm not able to do and for the other I would like to check if they're correct.

My work:

1. $D$ is left $S$-module: I think this one's trivial by the definition of $D$.

2. $\text{Hom}_R(S,S)$ is left $S$-module: This is my main problem. The paper says it comes from the $S$-module structure of the covariant argument, but I'm unable to find what that is. So I tried proving it by hand, to be said, proving that $r,s\in S$, $f\in\text{Hom}_R(S,S)$ implies that $r(sf)=(rs)f$ and $(r+s)f=rf+sf$. I guess it must be trivial since $f(x)\in S$ $\forall x\in S$ and given that $S$ is a commutative ring then those properties must be verified (Correct me if I'm wrong). I'm probably wrong with this (I did not use the fact that $f$ is a left $R$-module homomorphism), so this is where my main question is (I'll ask it at the end again).

3. Preserves module structures: I didn't have problems here I think. It was pretty straight forward that $$j\left((su_\sigma)(tu_\tau)\right)=j(s\sigma(t)u_{\sigma\tau})=s\sigma(t)(\sigma\tau)=(s\sigma)(t\tau)=j(su_\sigma)j(tu\tau),$$ $$j(rsu_\sigma)=rs\sigma=r(s\sigma)=rj(su_\sigma).$$

My question:

I think parts 1 and 3 of my proof are correct, but I would like confirmation of it. Part 2 is the problematic one, how can I prove $\text{Hom}_R(S,S)$ is left $S$-module? Is it trivial as I thought?

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

Answer 1

$\newcommand{\Hom}{\operatorname{Hom}}$ I will phrase things a bit more generally, and just consider $S$ as a ring extension of $R$.

As for 1., yes, the action on $D$ is $s.(s' u_\sigma)= (ss')u_\sigma$.

For 2, we have the following. The $S$-action on $\Hom_R(S, S)$ being induced by the covariant argument means the following. For $f \in \Hom_R(S, S)$ and $s \in S$, we define $s.f$ as the endomorphism of the abelian group $S$ given by $(s.f)(x) = s f(x)$. (note that your equations $r(sf) = (rs)f$ etc don't make sense since you have not defined what the symbol $sf$ means.)

Now you have to show that $s.f$ is actually not just a morphism of abelian groups, but also intertwines the $R$-action. In terms of the inclusion $i \colon R \to S$, the relevant $R$-action on $S$ is of course given by $r.s = i(r)s$. Then $(s.f)(r.x) = sf(i(r)x) = s i(r) f(x) = i(r) s f(x) = r. ((s.f)(x))$, where both the fact that $f$ is in $\Hom_R(S, S)$ and that $i(R)$ is in the center of $S$ were used.

This establishes $s.f \in \Hom_R(S,S)$.

And for the last point, you are almost correct, but you have to show that actually the functions $j((s u_\sigma) (t u_\tau))$ and $j(s u_\sigma) \circ j(t u_\tau)$ as well as $j(s s' u_\sigma)$ and $s . j(s' u_\sigma)$ agree, and I feel you have not really done that, because you have not actually compared the functions. Instead, to be 100% precise, the computation should probably look something like \begin{align} j((s u_\sigma) (t u_\tau)) (x) &= j( s \sigma(t) u_{\sigma \tau}) (x) = s \sigma(t) \sigma (\tau (x)) = s \sigma ( t \tau (x)) \\ &= j(s u_\sigma) ( t \tau(x)) = j(s u_\sigma) \circ j(t u_\tau)(x) \end{align}

This might seem like a nitpicky distinction, but it's not.