Facts about graded rings and graded ideals

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.

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