# Graded Rings

Definition.A

gradingon a ring A is a collection of additive subgroups such thatas abelian groups, and for any , i.e..

A

graded ringis a ring A with a specified grading.

**Note**

The notation means every can be *uniquely* written as a *finite* sum for some (uniqueness holds up to appending or removal of ). Then is called the **degree- d component** of .

An is said to be **homogeneous of degree d** if ; we write . Note that

*d*is unique if .

**Example**

The standard example of a graded ring is the ring of polynomials over some ring *A*, where has *A*-basis given by the set of monomials satisfying and .

Lemma 1.is a subring of A and each is a module over .

**Proof**

For the first statement, since it suffices to show ; we may assume . Write with and . Suppose *n* > 0. For any we have

Since the LHS is homogeneous of degree *m*, we have . Thus for any *m* so we have . This gives , a contradiction.

The second statement is clear. ♦

By definition if are homogeneous of degrees *m* and *n*, then is homogeneous of degree *m*+*n*. We also have:

Lemma 2.Suppose A is an integral domain with grading. If are non-zero elements such that is homogeneous, then so are a and b.

**Proof**

Write and as sums of their components. Let *m* (resp. *m’*) be the minimum (resp. maximum) degree for which (resp ). Similarly, let *n* (resp. *n’*) be the minimum (resp. maximum) degree for which (resp ). By definition . Since *ab* is homogeneous we have and thus . So *a* and *b* are homogeneous. ♦

**Exercise A**

Find a graded ring *A*, with such that *ab* is homogeneous but *a* and *b* are not. [ Hint: come back to this exercise after finishing the whole article. ]

# Graded Modules

For the rest of this article, we fix a graded ring *A*.

Definition.A

gradingon an A-module M is a collection of additive subgroups such thatas additive groups, and for any .

**Note**

*We need a fixed grading on the base ring A before we can talk about grading on A-modules.*

For example, every graded ring is a graded module over itself (with the same grading).

- As before if we write as a sum with , then is called the
**degree-**component of*d**m*. - Also is
**homogeneous of degree**if ; again write .*d*

The following result is quite important for grading of submodules.

Proposition 1.Let M be a graded A-module and be a submodule. The following are equivalent.

- There is a generating set for N comprising of homogeneous elements.
- If , all homogeneous components of n lie in N.
- We have .

**Proof**

(1⇒3) It suffices to show . Pick a generating set *S* for *N* comprising of non-zero homogeneous elements. For each , write with and with . Fix and let *n’* be the degree-*d* component of *n*. Then *n’* is the sum of the degree-*d *components of . Hence

, where is the degree- component of

so .

(3⇒2) Let ; we can write where . Since , it is homogeneous of degree *i*; thus is the degree-*i* component of *n* and it lies in *N*.

(2⇒1) Pick any generating set *S* of *N*, then take the homogeneous components of all to obtain a homogeneous generating set.

Definition.Let M be a graded A-module. A submodule is said to be

gradedif it satisfies the conditions of proposition 1.An ideal is said to be

gradedif it is graded as a submodule.

Proposition 1 immediately gives the following.

Corollary 1.Given a graded module M, with collection of graded submodules , graded submodule N, and graded ideal ,

are all graded submodules of M.

**Proof**

Since each is generated by homogeneous elements, so is ; thus is graded. Suppose , if *n’* is a homogeneous component of *n*, then for each *i*, and hence . Thus is graded. Finally, if *S* (resp. *T*) is a generating set of (resp. *N*) comprising of homogeneous elements, then is a generating set of comprising of homogeneous elements. ♦

**Exercise B**

Decide if each statement is true.

- If is a graded ideal of
*A*, then is graded. - If are graded submodules, then is a graded ideal.

# Quotient

Proposition 2.Let N be a graded submodule of a graded module M. Then M/N has a canonical grading given by

.

**Proof**

Note that we have . It remains to show *M*/*N* is a direct sum of .

Let . Write with . Then is the sum of images of in . So .

Suppose can be written as and . Then

By condition 2 of proposition 1, each so . ♦

Corollary 2.If is a graded ideal, then is a graded ring under the above grading.

**Proof**

Since , multiplication gives

♦

**Example**

1. Suppose *A* is the coordinate ring of a variety , so that . If *I*(*V*) is homogeneous, then *A* is a graded ring. In particular, the group of units in *A* is , the multiplicative group . This does not work for non-homogeneous *I*(*V*), e.g. the unit group of contains (does equality hold?).

2. We will do exercise C here, i.e. show that is not a UFD. Indeed we have the equality in *A*; we claim that are all irreducible. By lemma 2, since *Z* is homogeneous of degree 1, we can only factor it as a product of a degree-0 and a degree-1 element. But all non-zero degree-0 elements of *A* are units (see example 1). Similarly are irreducible and are clearly not associates.

3. By the same reasoning, is not a UFD.

Note: however is a UFD for all , as we will see later.

# Graded Maps

Definition.Let M and N be graded A-modules, and be A-linear. We say f is

gradedif for each i.

Immediately we have:

Lemma 3.If is a graded map of graded A-modules, then is a graded submodule of M and is a graded submodule of N.

**Proof**

If and with , then and hence is the unique decomposition of *f*(*m*) into its homogeneous components. The rest is an easy exercise. ♦

Proposition 3 (First Isomorphism Theorem).For a graded map of graded modules, we have an isomorphism

in the category of graded A-modules.

**Proof**

Let us show that *g* is graded. The grading on the LHS is given by . Since *f* is graded, *g* takes the LHS into .

Since *g* is bijective, it remains to show that is also graded. For , let . Write as a sum of homogeneous components; since is the degree-*i* homogeneous component of we have so . ♦

**Exercise C**

State and prove the remaining two isomorphism theorems.