I am trying to understand the simple graded modules over a graded ring $R$ (all the gradings over $\mathbb{Z}$).
I know that there exists a bijection between the simple graded $R$-modules and simple modules of $R_0$. But now I wonder: If $S$ is a simple graded module then is $S(n)$, the module defined as $$ S(n)_t:=S_{n+t}, $$ still a simple module?
If I consider the category of graded modules, in which morphisms are graded maps of degree zero, I have that $S(n) \ncong S$ in general as I can not find an isomorphism, so I do not see how the bijection can hold.
Thanks for your help!
For any graded module $S$ and any $n\in\mathbb{Z}$, there is a bijection between the set of graded submodules of $S$ and the set of graded submodules of $S(n)$: given a graded submodule $T\subseteq S$, send it to the graded submodule $T(n)\subseteq S(n)$. Since a graded module is simple iff it has exactly two graded submodules, we conclude that $S$ is simple iff $S(n)$ is simple.
As for the assertion that "there exists a bijection between the simple graded $R$-modules and simple modules of $R_0$", this is simply not true in most cases. The only statement I can think of that is close to this and is true is that if $R_n=0$ for all $n<0$, then there is a bijection between the simple $R_0$-modules and the simple graded $R$-modules $S$ such that $S_0\neq 0$ (namely, every simple graded $R$-module is nonzero in only one degree, and in that degree it is a simple $R_0$-module).