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

Solution 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.