Prove $D$ is $R$-algebra

Solution 1:

It might be helpful to think about a related, easier construction, which is when the $G$ action on $S$ is trivial, and so $R=S$. So then, our algebra $D$ has a basis $u_\sigma$ (as an S module), and the multiplication is given by: $$u_\sigma\cdot u_\tau:=u_{\sigma \tau}$$

And in this case, our "scalars" $S$ don't interact in any way with the $u_\sigma$. Explicitly, when we multiply $s_1 u_\sigma$ and $s_2 u_\tau$, we just get $s_1s_2 u_{\sigma\tau}$, and extend by linearity.

This is called the group ring, usually denoted $S[G]$, and first understanding why this is an $S$ algebra addresses some of your confusion. In particular, $S$ sits inside $S[G]$ as $s\cdot u_1$, and we can check this forms an $S$ algebra structure. As a concrete example, one should try showing that this group ring $S[C_n]$ for the cyclic group $C_n$ is isomorphic to the ring $S[x]/\langle x^n-1 \rangle$.

Once you are more comfortable with this usual construction (which is itself very relevant in your setting of Galois theory), then one can more profitably think about this twisted version. In the twisted version (that you described), the scalars $S$ actually interact with the group elements, by the formula you gave, and the way I like to think about this is that when they "pass through each other", the group elements speak to the scalars, and do their action to them. One needs to verify this works, and describes a valid algebra structure still, but its not too tricky once the usual group ring is more familiar.

If you've seen some group theory, it can be helpful to look at the definition of a semidirect product of groups here, we get a valid ring structure for the same reason the semidirect product of groups yields a valid group.

To address the more conceptual question of what the $u_\sigma$ are like, we can think of this process of taking a group $G$ and spitting out a ring $S[G]$ (in the untwisted group ring version) as making a ring that is a "best approximation" to the group $G$, in a precise sense. To make this precise is probably going a bit further afield, but the distinguished basis $u_\sigma$ of $S[G]$ (or $D$) can be interpreted as remnants of this universal process lurking in the background that turns groups into rings, in an efficient way.