A ring $R$ is graded if it has a direct sum decomposition $R=\bigoplus_{i\in\mathbb{Z}}R_i$ where the $R_i$ are abelian groups and $R_iR_j\subseteq R_{i+j}$. An ideal $I\subseteq R$ is graded if $I=\bigoplus_{i\in \mathbb{Z}}I_i$ where $I_i=R_i\cap I$. I want to prove that $R/I$ is a graded ring. Supposedly we do this by asserting that $R/I=\bigoplus_{i\in\mathbb{Z}} R_i/I_i$, and that $(R_i/I_i)(R_j/I_j)\subseteq R_{i+j}/I_{i+j}$ because $(r_i+I_i)(r_j+I_j)=r_ir_j+I_{i+j}$.
Here's where I am confused: how are the $R_i/I_i$ abelian subgroups of $R/I$? Seems like the best we could do is say they're isomorphic to subgroups of $R/I$. Furthermore, am I correct in saying that $(r_i+I_i)(r_j+I_j)=r_ir_j+I_{i+j}$ because that's how the multiplication in $\bigoplus_{i\in\mathbb{Z}} R_i/I_i$ is defined?
I feel like this graded quotient construction makes sense intuitively, but I'm getting caught up in these formalisms. Could anyone shed some light on this?
(My English is not vert good)
Let S be a graded ring and I be its graded ideal.$S=S_0\oplus S_1\oplus\dots$
Let $S_k/I$ denote the set $\{x+I|x\in S_k \}$, Since $S_k$ is an abelian group, $S_k/I$ is alse an abelian group. Then we proof that $S/I = S_0/I\oplus S_1/I\oplus \dots$
It's easy to check that $S/I = S_0/I+ S_1/I+ \dots$ The uniqueness check as follow: if $a_0+I+\dots+a_k+I=b_0+I+\dots+b_k+I$ then $a_0-b_0+\dots +a_k-b_k\in I=\oplus_{k=0}^{\infty}I\cap S_k$. Hence, $a_0-b_0+\dots +a_k-b_k=c_1+\dots+c_m$,where $c_i\in S_i\cap I $.Finally we fill the "blank" with $0+I$ in left side and right side,we can get $a_i-b_i\in S_i\cap I$ by $S=S_0\oplus S_1\oplus\dots$.Then $a_i+I=b_i+I$,which indicates $S/I = S_0/I\oplus S_1/I\oplus\dots$
$S_k/I\cong S_k/(I\cap S_k)$ is easy to check.