While reading the notes from Tom Marley about graded rings and modules I came across the following statement:
"Define M(n) (read "M twisted by n") to be equal to M as an R-module, but with it´s grading defined by $M(n)_k = M_{n+k}$.
(For example, if $M = R(-3)$ then $1 \in M_3$)"
My question here is why 1 is not in $M_{-3}$, since we know that $1 \in R_0$ for the graded ring R. By following the definition this would imply: $1 \in M_{-3+0}= M_{-3}$ And not $1\in M_3$
$M_3=R(-3)_3=R_{-3+3}=R_0\ni 1$