Module and group ring: definitions and notations I apologize in advance for the stupid questions and the bad English, but I've started studying math a few months ago. I've some problems with the definitions of a group ring, modules, and their notations.
I've found in a textbook, the expression "$R[G]$-module" (Robinson textbook) (where $G$ denotes a group and $R$ a ring). How should I read it? I mean: I know that the expression $A$-module indicates a module over a ring $A$ and the expression $R[G]$ indicates the group ring of $G$ over $R$; so when I find "$R[G]$-module" should I take $A=R[G]$ (because if $R$ is a ring then $R$ is always an $R$-module on itself, right?)?
Moreover, I read that a group ring $R[G]$ is a free module and at the same time a ring, constructed from any given ring $R$ and any given group $G$. Is there a relationship between $R[G]$ (considered as a free module) and an $R$-module consisting of an abelian group $(G,+)$ and an operation $R×G\rightarrow G$? Thanks in advance to everyone!
 A: Don't worry about asking stupid questions, especially if they're about definitions. Math is hard, and a big part of that is that definitions can be tricky, and often appear unmotivated. Group rings and their modules give a lot of people a lot of problems, and there are a few math professors at my institution who would probably be very unhappy if you asked them questions about group rings out of the blue. You're learning, and it's OK to be confused.
Alright, I'll get off my soapbox and start answering your question now. That said, I'm not sure exactly what your background is, so please forgive me if I start with material that you already know. 

Let's fix some ring $R$ (which, in practice, is often $\mathbb{Z}$ or a field $k$),
and a group $G$ (which does not have to be abelian).
The group ring $R[G]$ is the ring of polynomials with coefficients in $R$ and variables coming from $G$ (which explains the notation). However, we define
the multiplication of two polynomials using the multiplication of our group.
As an example, consider $\mathbb{Z}[C_3]$ where $C_3 = \langle g ~|~ g^3 = 1 \rangle$ is the cyclic group with 3 elements. Members of $\mathbb{Z}[C_3]$ look like:
$a + bg + cg^2$
Addition is exactly what you expect: 
$(a + bg + cg^2) + (a' + b'g + c'g^2) = (a + a') + (b + b')g + (c + c')g^2$
Multiplication is exactly what you expect too:
$(a + bg + cg^2)(a' + b'g + c'g^2) = a a' + a b' g + a c' g^2 + b a' g + b b' g^2 + b c' g^3 + c a' g^2 + c b' g^3 + c c' g^4$
But since our coefficients form a group, we can reduce this further, to:
$(a a' + b c' + c b') + (a b' + b a' + c c') g + (a c' + b b' + c a') g^2$

As for the first question you actually asked, you are exactly right! An $R[G]$-module is a module whose ring is $R[G]$. 
Precisely, an $R[G]$-module $(M,+)$ is an abelian group which admits scalar multiplication by elements of $R[G]$.
Of course, we can describe this structure in a different way, making explicit reference to both $R$ and $G$. An $R[G]$ module $(M,+)$ is an $R$-module with a distinguished $G$ action $\varphi : G \to \text{Aut}(M)$. Then $r \cdot m$ is well defined because $M$ is an $R$-module, and $g \cdot m$ is well defined because $G$ acts on $M$. Then a polynomial does what is has to do in order to be linear. As a quick example:
$(r_1 g_1 + r_2 g_2) \cdot m = r_1 \cdot (g_1 \cdot m) + r_2 \cdot (g_2 \cdot m)$

As for what you mean by $R[G]$ being "free", there are two interpretations I can think of:
$R[G]$ is a free module over $R$, with one generator for each element of $G$.
$R[G]$ is free with respect to maps out of $G$. That is:
For every ring homomorphism $\varphi : R \to S$, for every group homomorphism $h : G \to S^\times$ (the group of units in $S$), there is a unique ring homomorphism $\psi : R[G] \to S$ such that $\psi(r) = \phi(r)$ and $\psi(g) = h(g)$ for every $r \in R$ and $g \in G$.
Another way to phrase this is the "group ring" functor $R[-]$ is left adjoint to the "group of units" functor $(-)^\times$, if you're familiar with the categorical language.

For the last question, you ask if $R[G]$ has anything to do with $G$ being an $R$ module. It does not. $G$ is allowed to be any group, in particular nonabelian ones, in the construction of a group ring. However for $G$ to be an $R$-module, $G$ must be abelian.

If I missed part of your question, feel free to comment and I'll update my answer.
I hope this helps ^_^
