Our goal here is to prove that the quotient of a graded ring by a graded ideal is graded. We will approach this from a slightly more general perspective.

Let $R$ be a commutative ring with 1 and let $S$ be a semigroup. Suppose we have a family $\mathcal{A} = \{A_i\}_S$ of $(R,R)$-bimodules such that $ra = ar$ for all $a$ and $r$. Suppose further that we have a family $\Psi = \{\psi_{i,j} : A_i \times A_j \rightarrow A_{ij}\}_{i,j \in S}$ of $R$-bilinear mappings such that $\psi_{i,jk}(a, \psi_{j,k}(b,c)) = \psi_{ij,k}(\psi_{i,j}(a,b),c)$ for all $i,j,k \in S$ and $a \in A_i$, $b \in A_j$, and $c \in A_k$. Finally, define an operator $\cdot$ on the $R$-module $\bigoplus_S A_i$ by $(a_i) \cdot (b_j) = (\sum_{ij=k} \psi_{i,j}(a_i,b_j))$.

We claim that $\bigoplus_S A_i$ is a ring under this multiplication. It suffices to show that $\cdot$ is associative and distributes over addition. To that end, let $(a_i), (b_j), (c_k) \in \bigoplus_S A_s$.

1. Note the following.
 $\left[ (a_i) \cdot (b_j) \right] \cdot (c_k)$ = $(\sum_{ij = t} \psi_{i,j}(a_i,b_j)) \cdot (c_k)$ = $(\sum_{tk=u} \psi_{t,k}(\sum_{ij=t}(a_i,b_j),c_k))$ = $(\sum_{tk = u} \sum_{ij=t} \psi_{ij,k}(\psi_{i,j}(a_i,b_j),c_k)$ = $(\sum_{it = u} \sum_{jk=t} \psi_{i,jk}(a_i, \psi_{j,k}(b_j,c_k)))$ = $(\sum_{it=u} \psi_{i,t}(a_i, \sum_{jk=t}(b_j,c_k)))$ = $(a_i) \cdot (\sum_{jk=t} \psi_{j,k}(b_j,c_k)$ = $(a_i) \cdot \left[ (b_j) \cdot (c_k) \right]$.

So $\cdot$ is associative.

2. Note the following.
 $(a_i) \cdot \left[ (b_j) + (c_j) \right]$ = $(a_i) \cdot (b_j+c_j)$ = $(\sum_{ij=k} \psi_{i,j}(a_i, b_j+c_j))$ = $(\sum_{ij=k} \psi_{i,j}(a_i,b_j) + \psi_{i,j}(a_i,c_j))$ = $(\sum_{ij=k} \psi_{i,j}(a_i,b_j)) + (\sum_{ij = k} \psi_{i,j}(a_i,c_j))$ = $\left[ (a_i) \cdot (b_j) \right] + \left[ (a_i) \cdot (c_j) \right]$.

So $\cdot$ distributes over $+$ from the left; distributivity from the right is proved similarly.

So $\bigoplus_S A_s$ is a ring under this multiplication; we will call this the $S$-graded sum of the $A_i$, and denote it by $\mathsf{Gr}^S_R(\mathcal{A},\Psi)$.

Now we claim that if $S$ is a monoid with identity $1$, and if $A_1 = R$, and finally if $\psi_{1,k}(r,a) = ra$ and $\psi_{k,1}(a,r) = ar$ for all $k \in S$, $a \in A_k$, and $r \in R$, then in fact $\mathsf{GR}^S_R(\mathcal{A},\Psi)$ is a unital ring and moreover an $R$-algebra via the injection map $R \rightarrow \mathsf{Gr}^S_R(\mathcal{A},\Psi)$. To see this, consider the element $e \in \mathsf{Gr}^S_R(\mathcal{A},\Psi)$ with $e_1 = 1$ and $e_s = 0$ otherwise. Certainly then $e \cdot (a_i) = (a_i) \cdot e = (a_i)$ for all $a$. So $\mathsf{Gr}^S_M(\mathcal{A},\Psi)$ is a unital ring. Moreover, note that since $ra = ar$ for all $r \in R$, we have $R \subseteq Z(\mathsf{Gr}^S_R(\mathcal{A},\Psi)$. In particular, this ring is an $R$-algebra.

Next we prove that this algebra has an appropriate universal property. Suppose $B$ is an $R$-algebra and that we have a family $\varphi_i : A_i \rightarrow B$ of $R$-module homomorphisms such that $\varphi_{ij}(\psi_{i,j}(a,b)) = \varphi_i(a)\varphi_j(b)$ for all $i \in S$ and all $a$ and $b$. Then there exists a unique $R$-algebra homomorphism $\theta : \mathsf{Gr}^S_R(\mathcal{A},\Psi) \rightarrow B$ such that $\theta \circ \iota_i = \varphi_i$ for each $i \in S$. It suffices to show that the unique induced module homomorphism (via the universal property of coproducts of modules) is an $R$-algebra homomorphism. To that end, let $(a_i), (b_j) \in \mathsf{Gr}^S_R(\mathcal{A},\Psi)$. We have $\theta((a_i)(b_j)) = \theta((\sum_{ij=k} \psi_{i,j}(a_i,b_j)))$ $= (\theta \circ \iota_k)(\sum_{ij=k} \psi_{i,j}(a_i,b_j))$ for each $k$. Then $\varphi_k(\sum_{ij=k} \psi_{i,j}(a_i,b_j))$ $= \sum_{ij=k} \varphi_{ij}(\psi_{i,j}(a_i,b_j))$ $= \sum_{ij=k} \varphi_i(a_i)\varphi_j(b_j)$ $= \theta((a_i)) \theta((b_i))$. (Recall that $\theta((a_i)) = \sum \varphi(a_i)$.)

Now suppose we have a family of submodules $\{B_s\}_S$, with $I_s \subseteq B_s$, such that $\psi_{i,j}[B_i,A_j], \psi_{i,j}[A_i,B_j] \subseteq B_{ij}$ for all $i,j \in S$. Certainly $\bigoplus_S B_s \subseteq \mathsf{Gr}^S_R(\mathcal{A},\Psi)$ is an ideal; we will call ideals of this form $S$-graded.

Let $\mathcal{A}$, $\Psi$, and $\{B_i\}$ be as above. For each pair $i,j \in S$, define $\overline{\psi} : A_i/B_i \times A_j/B_j \rightarrow A_{ij}/B_{i,j}$ by $\overline{\psi}_{i,j}(a_i+B_i, a_j+B_j) = \psi_{i,j}(a_i,b_j) + B_{ij}$. We claim that $\overline{\psi}$ is well-defined and $R$-bilinear. To see well-definedness, note that if $a_i - a_i^\prime \in B_i$ and $a_j - a_j^\prime \in B_j$, then by bilinearity we have $\psi_{i,j}(a_i,a_j) - \psi_{i,j}(a_i^\prime,a_j) \in B_{i,j}$ and $\psi_{i,j}(a_i^\prime,a_j) - \psi_{i,j}(a_i^\prime,a_j^\prime) \in B_{i,j}$. Thus $\psi_{i,j}(a_i,a_j) - \psi_{i,j}(a_i^\prime,a_j^\prime) \in B_{i,j}$, and so $\overline{\psi}_{i,j}$ is well-defined. Bilinearity is clear. Moreover, it is straightforward to show that $\overline{\psi}_{ij,k}(\overline{\psi}_{i,j}(\overline{a}_i,\overline{b}_j),\overline{c}_k) = \overline{\psi}_{i,jk}(\overline{a}_i, \overline{\psi}_{j,k}(\overline{b}_j,\overline{c}_k))$. Thus we may construct the $S$-graded sum $\mathsf{Gr}^S_R(\overline{A},\overline{\Psi})$.

Let $\mathcal{B} = \{B_i\}_S$, and for each pair $i,j \in S$, let $\psi|_{i,j}$ denote the restriction of $\psi_{i,j}$ to $B_i \times B_j$. Let $\Psi|_B$ denote the set of these restrictions. Then the ideal $\bigoplus_S B$ is in fact $\mathsf{Gr}^S_R(\mathcal{B},\Psi|_B)$.

We claim that $\mathsf{Gr}^S_R(\mathcal{A},\Psi)/\mathsf{Gr}^S_R(\mathcal{B},\Psi|_B) \cong_R \mathsf{Gr}^S_R(\overline{\mathcal{A}},\overline{\Psi})$. To see this, it suffices to show that the usual module isomorphism $(\bigoplus A_i)/(\bigoplus B_i) \rightarrow \bigoplus A_i/B_i$ given by $\overline{(a_i)} \mapsto (\overline{a_i})$ preserves multiplication. To that end, note that $\overline{(a_i)(b_j)} = \overline{(\sum_{ij=k} \psi_{i,j}(a_i,b_j))}$ $= (\sum_{ij=k} \overline{\psi}_{i,j}(\overline{a}_i,\overline{b}_j))$ $= \overline{(a_i)} \overline{(b_j)}$, as desired.