I'm studying the paper Galois Theory and Galois Cohomology of Commutative Rings by S. U. Chase, D. K. Harrison and Alex Rosenberg. A couple days ago I posted this question where I wanted to prove that the map $j$ defined as below was $S$-module homomorphism. I may be redundant but here are the definitions of the paper (almost the same as the linked post ones).
"$S$ 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 $1$. "
So I went directly to prove $j$ is $S$-module homomorphism without having really understood what the nature of $D$ as an $R$-algebra is. The paper states $D$ has $R$-algebra structure with the product defined above, but unfortunately for me its proof is left as an exercise to the reader. So I've been trying to prove but got stuck. The work I've done is the following:
- At first I notice that $S$ is $R$-module. This is trivial since $S$ is a ring and $R$ is an $S$ subring, having its $R$-module structure defined by the inclusion $(r,s)\mapsto \iota(r)s$ (the properties required are verified due to $S$ being commutative ring).
- Secondly, I noticed how $\text{Aut}(S)$ is also $R$-module (with pointwise product and the inclusion as above for $S$), hence $G$ is so.
- Thirdly I attempted to prove $D$ is $R$-module (this is required for it to be an $R$-algebra later). The paper states that $D$ is in fact a free $S$-module with generators $u_\sigma$ ($\sigma$ in $G$), but I don't know how to prove it and don't really get what those $u_\sigma$ are like, so I'll approach this taking in consideration the elements of $D$ are of the form $\sum_is_i\sigma_i$ (finite sums where $s_i\in S$ and $\sigma_i\in G$). I proved it's $R$-module under the action $\left(r,\sum_is_i\sigma_i\right)\mapsto\left(\sum_i(\iota(r)s_i)\sigma_i\right)$. Then I proceed like this (using that $S$ and $G$ both are $R$-modules as stated above): $$\text{ (I) } \boxed{r\left(r'\sum s_i\sigma_i\right)}=r\sum(\iota(r')s_i)\sigma_i=\sum(\iota(r)(\iota(r')s_i))\sigma_i=\sum(\iota(rr')s_i)\sigma_i=\boxed{(rr')\sum s_i\sigma_i}$$ $$\text{ (II) }\boxed{(r+r')\sum s_i\sigma_i} = \sum (\iota(r+r')s_i)\sigma_i =\sum (\iota(r)s_i+\iota(r')s_i)\sigma_i= \sum((\iota(r)s_i)\sigma_i+(\iota(r's_i))\sigma_i) =$$ $$=\sum(\iota(r)s_i)\sigma_i + \sum(\iota(r')s_i)\sigma_i = \boxed{r\sum s_i\sigma_i + r'\sum s_i\sigma_i}$$ $$\text{ (III) }\boxed{r\left( \sum s_i\sigma_i + \sum t_i\tau_i \right)} = r\sum \bar s_i\bar\sigma_i = \sum (\iota(r)\bar s_i)\bar\sigma_i=\sum (\iota(r)s_i)\sigma_i + \sum (\iota(r)t_i)\tau_i=\boxed{r\sum s_i\sigma_i + r\sum t_i\tau_i}$$ So with this I've proven that $D$ is an $R$-module with the action described.
- Finally, the last step is proving $D$ is $R$-algebra with the multiplication given by the paper. In order for this to be true, then $D$ must verify: it's $R$-module (proven above), and that its multiplication is bilinear, associative and has identity element. I guess this would be way easier to prove taking in consideration the basis of $D$ as a free $S$-module with elements $u_\sigma$ ($\sigma\in G$). My problem is that I don't get what these are. So here goes my first question: Is $u_\sigma$ "the same" as $\sigma$? If not, what is $u_\sigma$ then? I've been able to prove the existence of the identity element for multiplication, it's obviously $u_1$: $$(u_1)(su_\sigma)=1(s)u_{1\sigma}=su_\sigma, \phantom{a}\forall s\in S, \sigma\in G.$$ I also proved associativity: $$((su_\sigma)(tu_\tau))(wu_\omega)=(s\sigma(t)u_{\sigma\tau})(wu_\omega)=s\sigma(t)\sigma(\tau(w))u_{\sigma\tau\omega}=s\sigma(t\tau(w))u_{\sigma\tau\omega}=$$ $$= (su_\sigma)(t\tau(w)u_{\tau\omega})=(su_\sigma)((tu_\tau)(wu_\omega)).$$ But I did not have the same luck proving bilinearity, to be said, proving that $a(b+c)=ab+ac$, $(a+b)c=ac+dc$ and $(ra)b=a(rb)=r(ab)$, $\forall a,b,c\in D, r\in R$. So my second question is how can I prove this last property.
To sum up my questions are: What are those $u_\sigma$ like? Are they "the same" as $\sigma$? How can I prove the bilinearity of the multiplication? Any help or hint will be appreciated, thanks in advance.
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.