An (essentially) the same question has been asked and answered here Definition of graded rings, but the user who posted the answer has not been here for a while, and I am confused by the answer, so I post a new question here.
A graded is a ring $S$ together with a family $(S_{d})_{d\geq 0}$ of subgroups of the additive group of $S$, such that $S=\bigoplus_{d\geq 0}S_{d}$ and $S_{e}S_{d}\subseteq S_{e+d}$ for all $e,d\geq 0$.
I want to show that
$S_{0}$ is a subring of $S$.
It is clear from the definition that $S_{0}\leq S$, so we only need to show $S_{0}$ is closed under multiplication and contains the multiplicative identity $1$.
The closure under multiplication follows immeidately from $S_{0}S_{0}\subseteq S_{0+0}=S_{0}$.
However, I got stuck in why $1\in S_{0}$. The answer referred above goes as follows. Write $1=\sum_{d\geq 0}x_{d}$, where $x_{d}=x_{d}(1)$ is uniquely determined by $1$, $x_{d}\in S_{d}$, and only finitely many of $x_{d}$'s are nonzero.
Fix $n\in\mathbb{Z}_{\geq 0}$. Then, $$x_{n}=x_{n}1=\sum_{d\geq 0}x_{d}x_{n}=x_{0}x_{n}+x_{1}x_{n}+\cdots.$$
So far so good, but then the answer argues that "by comparing degree", we can conclude that $x_{n}=x_{0}x_{n}.$
I don't understand why this follows. I understand by saying "comparing degree", the answer might actually mean the following. Everything here is a homogenous element of degree $\geq n$, i.e. $x_{n}\in S_{n}$, $x_{n}x_{0}\in S_{n}$, $x_{1}x_{n}\in S_{n+1}$, $x_{2}x_{n}\in S_{n+2},\cdots$.
So in the RHS of the equation, only $x_{0}x_{n}$ is of degree $n$ (which equals to the degree of $x_{n}$), and other summands are of degree $\geq n+1$. But why does this imply that we must have $x_{0}x_{n}=x_{n}$? Equivalently, why does this imply $x_{d}x_{n}=0$ for all $d\geq 1$?
I know that $S_{i}\cap S_{j}=0$ for $i\neq j$, but why does this prevent a homogeneous element of degree $n$ from being written as a sum of homogeneous elements of different degrees?
Thank you!
You know much more than just $S_i\cap S_j=0$. You know $S$ is the direct sum of the $S_i$. This by definition means that every element of $S$ is uniquely written as a sum $\sum a_i$ where $a_i\in S_i$ (and all but finitely many of them are $0$). In particular, this means the unique way to write $x_n$ as such a sum is taking $a_n=x_n$ and $a_i=0$ for $i\neq n$. Since $x_n=\sum_d x_dx_n$ gives another such way of representing $x_n$ (with $a_i=0$ for $i<n$ and $a_i=x_{i-n}x_n$ for $i\geq n$) it must be the same, i.e. we must have $x_n=x_0x_n$ and $x_dx_n=0$ for all $d>0$.