Two questions about graded rings From Wikipedia, a graded ring is a ring $R$ that can be decomposed as the direct sum
$$
R=\bigoplus_{n\geq0}R_n
$$
of additive subgroups such that $R_mR_n\subset R_{m+n}$ for all $m,n\geq 0$. Wiki then claims that $R_0$ is a subring of $R$ but does not prove this. They then claim that "by definition of direct sum", every element $a$ of $R$ can be written uniquely as
$$
a=a_0+a_1+\ldots+a_n
$$
where each $a_i\in R_i$. Here are my questions:

*

*How do we prove $R_0$ is a subring of $R$? I have searched many internet sources and I cannot seem to find a proof of this fact. We know it is an additive subgroup, so it must be closed under subtraction. But the subring test requires closure under multiplication, and $1\in R_0$ as well. How can we verify these?

*I don't understand what they mean with "by definition of direct sum". I checked Wikipedia's definition of the direct sum of abelian groups, and it is simply the subgroup of the direct product such that all but finitely many of the entries are zero. I don't see how this yields the uniqueness property. I have, however, seen this uniqueness property when working with direct sums of vector subspaces, but with the additional assumption that the subspaces satisfy
$$(U_1+U_2+\ldots+\hat{U_i}+\ldots+U_n)\cap U_i=\{0\}.$$
That is, they are "mutually disjoint" in some sense (I know that's not what this property is called, I'm just not sure what else to call it). Anyway, without this assumption on the $R_i$, how do we know every element $a\in R$ can be written uniquely as above?

 A: As mentioned in a comment, $R_0$ being a subring of $R$ follows from how $R_0R_0\subseteq R_{0+0}$.
As for your question on direct sums, it might be helpful to look at the finite analogue.
Given two rings $R$ and $S$, the direct sum $R\oplus S$ will coincide with (as in, it is isomorphic to) the direct product $R\times S$.
The latter is a cartesian product of rings, so the elements of $R\times S$ are of the form $(r,s)$ for $r\in R$ and $s\in S$, with addition and multiplication done component-wise.
With this notation, saying that an element of $R\times S$ is "in $R$" means that the $S$ component is zero. In other words, $R$ can be identified with the subring $\{(r,0)\mid r\in R\}$ of $R\times S$, and similarly for $S$.
Now, saying that an element of $R\times S$ can be written uniquely as a sum of an element of $R$ with an element of $S$ is to say that $(x,y)=(r,0)+(0,s)$ forces $x=r$ and $y=s$, which follows by computing the sum on the right-hand side.
In the infinite setting, this is no longer true for products only because infinite sums don't make sense in rings. For example, you can't decompose $(1,1,1,\dots)\in\prod_{i=0}^\infty\mathbb Z$ into a sum where each summand lies in one of the $\mathbb Z$'s because such a sum would have to be infinite.
If we restrict our attention to the subring of $\prod_{i=0}^\infty\mathbb Z$ where we only those $(x_1,x_2,x_3,\dots)$ where finitely many of the $x_i$ are nonzero, the decomposition is true again: if we let $I$ be the (finite) set of indices $i\in I$ for which $x_i\neq0$, then $(x_1,x_2,x_3,\dots)=\sum_{i\in I}\vec x_i$, where $\vec x_i=(0,\dots,0,x_i,0,\dots)$ is nonzero only in the $i$th index.
In general, the subring of $\prod_iR_i$ consisting of those sequences wherein only finitely many terms are nonzero is precisely the direct sum, denoted $\bigoplus_iR_i$. Intuitively, this is the resulting ring where you allow elements of the different rings to be added to gether "freely" (if $r_i\in R_i$ and $r_j\in R_j$, then $r_i+r_j$ only really makes sense if $i=j$; if $i\neq j$, then we don't know what to do, so just let their sum be this new element called $r_i+r_j$). In this way, it is essentially by definition that elements of $\bigoplus_iR_i$ are uniquely given by finite sums of elements in the individual $R_i$'s.
In fact, your disjointness of direct summands from linear algebra is still valid for direct sums of rings: if you identify $R_j$ with the subring of $\bigoplus_iR_i$ consisting of those elements whose only nonzero term is at index $j$, then $R_i\cap R_j=\{0\}$ whenever $i\neq j$.
A: I'll try my best to answer this but I'm not that familiar with graded rings. But the definition seems pretty straightforward.
1.: It's closed under multiplication since $R_0 R_0 \subset R_{0+0} = R_{0}$. Now suppose $1 \in R_n$ with $n \neq 0$. Then $1 \cdot 1 = 1 \in R_{n}$ but $(1 \cdot 1) \in R_{n+n} = R_{2n}$. This implies $1 \in R_n\cap R_{2n} = \{0\}$ (because $n \neq 2n$, see shibais answer) which is a contradiction since $1 \neq 0$. Therefore $1 \in R_0$.
2.: There is a distinction between an internal and an external direct sum (see Wikipedia). You used the internal definition for vector subspaces and therefore needed the extra assumption. When using the external direct sum we already think of the objects as being disjoint, therefore we don't need an extra assumption. The uniqueness property still holds, see shibais answer for a more detailed answer regarding this.
