# Why is $S_{0}$ a subring of the graded ring $S=\bigoplus_{d\geq 0}S_{d}$.

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 $$i and $$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$$.