Graded Rings

Graded Rings

When the index set \(I\) is a commutative monoid, we agreed to call the direct sum \(\bigoplus_{i\in I} A_i\) of a family \((A_i)_{i\in I}\) of abelian groups a graded abelian group. At that time, since there were no conditions on \(A_i\), this was not a particularly interesting definition, but now that \(A_i\) is equipped with a multiplication structure, this definition becomes more meaningful.

Definition 1 Let a commutative monoid \(I\) and an \(I\)-indexed family of abelian groups \((A_i)_{i\in I}\) be given. If the multiplication structure defined on \(\bigoplus_{i\in I} A_i\) makes it a ring, and additionally satisfies the condition

\[A_i A_j\subseteq A_{i+j}\qquad\text{for all $i,j\in I$}\]

then \(A\) is called an \(I\)-graded ring. Elements belonging to \(A_i\) are called homogeneous elements.

By definition, any element of \(A\) can be uniquely expressed as a finite sum of homogeneous elements.

Proposition 2 Suppose every element of \(I\) is cancellable and \(\bigoplus_{i\in I} A_i\) is a graded ring. Then \(A_0\) is a subring of \(A\).

Proof

From \(A_0A_0\subseteq A_0\), it is clear that \(A_0\) is closed under multiplication. Therefore, it suffices to show that the multiplicative identity \(1\) of \(A=\bigoplus A_i\) belongs to \(A_0\). Let \(1=\sum_{i\in I} \epsilon_i\). Then for any \(\alpha\in A_j\),

\[\alpha=1\alpha=\sum_{i\in I} \epsilon_i\alpha\in A_j\]

and therefore for all \(i\neq 0\), we have \(\epsilon_i\alpha=0\), and only for \(i=0\) does \(\epsilon_0\alpha=\alpha\) hold. Now since any element of \(A\) can be expressed as a sum of homogeneous elements, the proof is complete.

In most cases, we are interested in the case \(I=\mathbb{Z}\) or \(I= \mathbb{N}\). Thus the precondition of Proposition 2 is satisfied.

Example 3 For any abelian group \(G\), the free ring \(F(G)\) generated by \(A\) is an \(\mathbb{N}\)-graded ring.

Graded Ring Homomorphisms

Definition 4 For a commutative monoid \(I\) and two \(I\)-graded rings \(A,A'\), a ring homomorphism \(\phi:A \rightarrow A'\) is called a graded homomorphism if \(\phi(A_i)\subseteq A_i'\) holds for all \(i\in I\).

It is not difficult to see that \(I\)-graded rings and \(I\)-graded homomorphisms form a category \(\bgr_I\Ring\).

Homogeneous Ideals and Quotients of Graded Rings

For a graded ring \(A=\bigoplus_{i\in I} A_i\), the quotient ring \(A/\mathfrak{a}\) by an ideal \(\mathfrak{a}\) of \(A\) is not always a graded ring.

Example 5 Fix a ring \(A\) and consider the polynomial ring with coefficients in \(A\):

\[A[\x]=\{\alpha_n\x^n+\cdots+\alpha_1\x+\alpha_0\mid \alpha_i\in A\}\]

This carries a graded ring structure by the decomposition

\[A[\x]=\bigoplus_{n\geq 0} A\x^n\]

Now consider the ideal \((\x-1)\) generated by \(\x-1\). As rings,

\[A[\x]/(\x-1)\cong A\]

Explicitly, this isomorphism is obtained by applying the first isomorphism theorem to the evaluation map

\[\alpha_n\x^n +\cdots+\alpha_1\x+\alpha_0\quad \mapsto\quad \alpha_n+\cdots+\alpha_1+\alpha_0\]

However, the above homomorphism is not a graded homomorphism.

To avoid this, we introduce the notion of homogeneous ideals.

Proposition 6 For an \(I\)-graded ring \(A=\bigoplus_{i\in I} A_i\) and an ideal \(\mathfrak{a}\) of \(A\), the following are all equivalent:

1. \(\mathfrak{a}\) is the sum of the \(\mathfrak{a}\cap A_i\).
2. When any element of \(\mathfrak{a}\) is decomposed into homogeneous elements, each of these elements also belongs to \(\mathfrak{a}\).
3. \(\mathfrak{a}\) is generated by homogeneous elements.

Proof

As elements of \(A\), all elements of \(\mathfrak{a}\) are uniquely expressed as sums of homogeneous elements. Thus, the equivalence of the first two conditions is clear, and that condition 1 implies condition 3 is also clear. Now assume the third condition and prove the second. Suppose \(\mathfrak{a}\) is generated by homogeneous elements \((x_j)_{j\in J}\). Then any \(x\in \mathfrak{a}\) can be written as

\[x=\sum_{j\in J} \alpha_j x_j,\qquad\text{$(\alpha_j)_{j\in J}$ finitely supported}\]

Now each \(\alpha_j\) can again be expressed as a sum of homogeneous elements

\[\alpha_j=\sum_{k\in K_j} \alpha_{jk},\qquad \text{$(\alpha_{jk})_{k\in K_j}$ finitely supported}\]

Thus

\[x=\sum_{j\in J}\sum_{k\in K_j}\alpha_{jk}x_j,\qquad \text{$(\alpha_{jk})_{j\in J,k\in K_j}$ finitely supported}\]

and the \(\alpha_{jk}x_j\) are all homogeneous elements and all belong to \(\mathfrak{a}\). From this, we can show condition 2.

An ideal satisfying the above equivalent conditions is called a homogeneous ideal. Then the following holds.

Proposition 7 For a homogeneous ideal \(\mathfrak{a}\), \(A/\mathfrak{a}\) is a graded ideal, and its decomposition is given by

\[A/\mathfrak{a}=\bigoplus_{i\in I}A_i/(\mathfrak{a}\cap A_i)\]

The proof is clear and is omitted.