Invariants of Group and Subspace (questions from a paper) I am reading the following paper:invariants of finite groups and their applications to combinatorics  (p.3)
https://pdfs.semanticscholar.org/bb23/fbe9718a7ccb11e0a625d43c431c14338428.pdf 
Or the following part:  

My questions:  


*

*What is the relation between $R^G$ and $T$? (Consider $C = R^G \cap T$, then $C$ is an invariant subspace in $T$ which contradicts that "T, $G$ acts irreducibly".)   

*Why is that called "$G$-module isomorphism"?  (I think $f\in R$, so it should be called $R$-module isomorphism.)    

*Why $R^G\cdot R^G_\chi \in R^G_\chi$? 
 A: I try to give short answers to the questions, then there will be some comments.
(1)
$R^G$ is the sum of all "$T$'s", of all irreducible representations for the trivial character, so
$$R^G = R^G_{\text{trivial character }\varepsilon} \ .$$
(The notation for $R^G_\chi$ is done so that it extends the usual $R^G$ notation for the fixed set of an action of a group $G$ on a set $R$. Here $R$ has moreover linear structure.)
(2)
It is an isomorphism in the category of $G$--($R$-)modules. These are $R$-modules with a supplementary action of the group $G$. Equivalently, $R[G]$-modules. Here, $R[G]$ is the group ring associated to the (commutative) ring $R$, and the (possibly non-commutative) group $G$. The $R$-structure is obvious. Multiplication by $f$ is an $R$-morphism (of $R$-modules) in a trivial way. But it is moreover compatible with the $G$-action, this is the new information. So $T$ is an $R[G]$-morphism. (It hurts to type it like that, most time people write $G$-morphism only.)
(3)
Let us start with $f\in R^G$, $F\in R^G_\chi$, and a character $\chi$ of $G$, $\chi:G\to \Bbb C$, so that for all $g\in G$ we have 
$$
\begin{aligned}
g.f &=\varepsilon(g)\cdot f = 1\cdot f= f\ ,\\
g.F &=\chi(g)\cdot F\ .
\\[2mm]
&\text{Then:}\\[2mm]
g(f\cdot F)
&=(g.f)\cdot(g.F)\\
&=f\cdot(\chi(g)\cdot F)\\
&=\chi(g)\cdot (fF)\ .
\end{aligned}
$$
So $fF$ transforms by the action of a $g\in G$ (factorizing) through $\chi$ (applied on $g$).
Comments:
An example makes things easier to digest.
Let the symmetric group $G=S(3)$ act on $R=\Bbb C[x_1,x_2,x_3]$ by permuting the indices of $x_1,x_2,x_3$. For instance, the transposition $(12)$ acts on $7x_1+8x_2+9x_3$ mapping it to  $7x_2+8x_1+9x_3$.
It is a result of invariant theory that a polynomial $f$ is in $R^G$ iff it is an algebraic (polynomial) expression in the elementary symmetric polynomials $e_1=x_1+x_2+x_3$, $e_2=x_1x_2+x_2x_3+x_3x_1$, $e_3=x_1x_2x_3$.
Now let us consider also $F= (x_1-x_2)(x_2-x_3)(x_3-x_1)$.
Then $G$ acts trivially, i.e. via the trivial character on typo, edited $f=e_1^3(e_1e_2+e_3)$. (Just a messy example.) Explicitly, let us fix a permutation $g\in G$, then:
$$
\begin{aligned}
&g.(\ e_1^3(e_1e_2+e_3)\ )\\
&\qquad =g.\Big(\ (x_1+x_2+x_3)((x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+x_1x_2x_3\ \Big)\\
&\qquad =\Big(\ (x_{g1}+x_{g2}+x_{g3})((x_{g1}+x_{g2}+x_{g3})(x_{g1}x_{g2}+x_{g2}x_{g3}+x_{g3}x_{g1})+x_{g1}x_{g2}x_{g3}\ \Big)\\
&\qquad =g.\Big(\ (x_1+x_2+x_3)((x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+x_1x_2x_3\ \Big)\\
&\qquad =(x_1+x_2+x_3)((x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+x_1x_2x_3\\
&\qquad =e_1^3(e_1e_2+e_3)\ ,\\[2mm]
&\varepsilon(g)\cdot(\ e_1^3(e_1e_2+e_3)\ )\\
&\qquad =1\cdot(\ e_1^3(e_1e_2+e_3)\ )\\
&\qquad = e_1^3(e_1e_2+e_3)\ .
\end{aligned}
$$
It may be that i used the opposite action on the indices.
Also, acts on $F$ via $\text{sign}$, the sign character. For example, consider $g=(12)$, the transposition of $1$, $2$, then we compare
$$
\begin{aligned}
F&=(x_1-x_2)(x_2-x_3)(x_3-x_1)\\
(12).F
&=(12).\Big(\ (x_1-x_2)(x_2-x_3)(x_3-x_1)\ \Big)\\
&=\phantom{(12).}\Big(\ (x_2-x_1)(x_1-x_3)(x_3-x_2)\ \Big)\\
&=(-1)^3\cdot F=(-1)\cdot F=\text{sign}((12))\cdot F\ .
\end{aligned}
$$
To understand the meaning of (3) in this example, we can try to see how does $G$ act on $fF$... 
A: 1.$R^G$ is one of the $R^G_\chi$, specifically it's $R^G_\epsilon$  ($\epsilon$ is the trivial character). Hence $R^G$ is a direct sum of some of the $T$'s.
Your proof using $C=R^G\cap T$ is correct but has the wrong conclusion : there's no contradiction here, it just implies $R^G\cap T = T$ or $0$, in other words for any $T$, $T\subset R^G$ or $T\cap R^G=0$, and this makes perfect sense when you see the definitions.


*It's also an $R$-module. However since $T$ and $fT$ are $G$-invariant, they are also $G$-modules and this map is also a $G$-module; and this is what's important for the rest (I think that it should be $f\in R^G$ though, not $f\in R$, otherwise $fT$ is not $G$-invariant and so not a $G$-module)

*Well precisely what came before : $T\to fT$ is a $G$-morphism between irreducible $G$-spaces so must be $0$ or an isomorphism. If it's $0$, then who cares, we get $fT\subset T$; but if it's an isomorphism then this means that the character of $fT$ is the same as that of $T$ : in any case, if $T\subset R^G_\chi$, $fT\subset R^G_\chi$.
Therefore multiplication by an element of $R^G$ (this confirms the correction I mentioned in 2.) leaves $R_\chi^G$ stable : this is precisely what $R^GR_\chi^G \subset R^G_\chi$ means
Note : in 3., the context probably yields that $fT\neq 0$ but theres not sufficient context in the picture to be sure of that
