A grading on a ring A is a collection of additive subgroups such that
as abelian groups, and for any , i.e..
A graded ring is a ring A with a specified grading.
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 .
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 .
is a subring of A and each is a module over .
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:
Suppose A is an integral domain with grading. If are non-zero elements such that is homogeneous, then so are a and b.
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. ♦
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. ]
For the rest of this article, we fix a graded ring A.
A grading on an A-module M is a collection of additive subgroups such that
as additive groups, and for any .
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-d component of m.
- Also is homogeneous of degree d if ; again write .
The following result is quite important for grading of submodules.
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 .
(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
(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.
Let M be a graded A-module. A submodule is said to be graded if it satisfies the conditions of proposition 1.
An ideal is said to be graded if it is graded as a submodule.
Proposition 1 immediately gives the following.
Given a graded module M, with collection of graded submodules , graded submodule N, and graded ideal ,
are all graded submodules of M.
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. ♦
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.
Let N be a graded submodule of a graded module M. Then M/N has a canonical grading given by
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 . ♦
If is a graded ideal, then is a graded ring under the above grading.
Since , multiplication gives
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.
Let M and N be graded A-modules, and be A-linear. We say f is graded if for each i.
Immediately we have:
If is a graded map of graded A-modules, then is a graded submodule of M and is a graded submodule of N.
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.
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 . ♦
State and prove the remaining two isomorphism theorems.