Commutative Algebra 53

Graded Rings


A grading on a ring A is a collection of additive subgroups A_0, A_1, \ldots \subseteq A such that

A = A_0 \oplus A_1 \oplus A_2 \oplus \ldots

as abelian groups, and A_i A_j \subseteq A_{i+j} for any i, j\ge 0, i.e..

a\in A_i, b\in A_j \implies ab \in A_{i+j}.

graded ring is a ring A with a specified grading.


The notation A = \oplus_i A_i means every a\in A can be uniquely written as a finite sum a_0 + a_1 + \ldots + a_n for some a_i \in A_i (uniqueness holds up to appending or removal of 0\in A_m). Then a_d is called the degree-d component of a.

An a\in A is said to be homogeneous of degree d if a \in A_d; we write \deg a = d. Note that d is unique if a\ne 0.


The standard example of a graded ring is the ring of polynomials B = A[X_1, \ldots, X_n] over some ring A, where B_d has A-basis given by the set of monomials X_1^{m_1} \ldots X_n^{m_n} satisfying m_i \ge 0 and \sum_i m_i = d.

Lemma 1.

A_0 is a subring of A and each A_i is a module over A_0.


For the first statement, since A_0 A_0 \subseteq A_0 it suffices to show 1 \in A_0; we may assume 1\ne 0. Write 1 = a_0 + \ldots + a_n with a_i \in A_i and a_n \ne 0. Suppose n > 0. For any b_m \in A_m we have

b_m = b_m\cdot 1 = b_m(a_0 + \ldots + a_n) = b_m a_0 + \ldots + b_m a_n.

Since the LHS is homogeneous of degree m, we have b_m a_n = 0. Thus a_n A_m = 0 for any m so we have a_n A = 0. This gives a_n = 0, a contradiction.

The second statement is clear. ♦

By definition if a, b\in A are homogeneous of degrees m and n, then ab is homogeneous of degree m+n. We also have:

Lemma 2.

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


Write a = \sum_m a_m and b = \sum_n b_n as sums of their components. Let m (resp. m’) be the minimum (resp. maximum) degree for which a_m \ne 0 (resp a_{m'} \ne 0). Similarly, let n (resp. n’) be the minimum (resp. maximum) degree for which b_n \ne 0 (resp b_{n'} \ne 0). By definition a_m b_n, a_{m'}b_{n'} \ne 0. Since ab is homogeneous we have m+n = m'+n' and thus m = m', n = n'. So a and b are homogeneous. ♦

Exercise A

Find a graded ring A, with a,b \in A-\{0\} 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.


A grading on an A-module M is a collection of additive subgroups M_0, M_1, \ldots \subseteq M such that

M = M_0 \oplus M_1 \oplus M_2 \oplus \ldots

as additive groups, and A_i M_j \subseteq M_{i+j} for any i,j \ge 0.


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 m\in M as a sum m_0 + m_1 + \ldots + m_n with m_i \in M_i, then m_d is called the degree-d component of m.
  • Also m\in M is homogeneous of degree d if m \in M_d; again write \deg m = d.

The following result is quite important for grading of submodules.

Proposition 1.

Let M be a graded A-module and N\subseteq M be a submodule. The following are equivalent.

  1. There is a generating set for N comprising of homogeneous elements.
  2. If n\in N, all homogeneous components of n lie in N.
  3. We have N = (N \cap M_0) \oplus (N \cap M_1) \oplus \ldots.


(1⇒3) It suffices to show N = \sum_i (N \cap M_i). Pick a generating set S for N comprising of non-zero homogeneous elements. For each n\in N, write n = \sum_{i=1}^k a_i m_i with a_i \in A and m_i \in S with \deg m_i = d_i. Fix d\ge 0 and let n’ be the degree-d component of n. Then n’ is the sum of the degree-d components of a_i m_i. Hence

n' = \sum_{i=1}^k a_i' m_i, where a_i' is the degree-(d-d_i) component of a_i

so n'\in N \cap M_d.

(3⇒2) Let n\in N; we can write n = n_0 + n_1 + \ldots + n_d where n_i \in N \cap M_i. Since n_i \in M_i, it is homogeneous of degree i; thus n_i 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 m\in S to obtain a homogeneous generating set.


Let M be a graded A-module. A submodule N\subseteq M is said to be graded if it satisfies the conditions of proposition 1.

An ideal \mathfrak a\subseteq A is said to be graded if it is graded as a submodule.

Proposition 1 immediately gives the following.

Corollary 1.

Given a graded module M, with collection of graded submodules (N_i), graded submodule N, and graded ideal \mathfrak a\subseteq A,

\sum_i N_i, \quad \cap_i N_i, \quad \mathfrak a N

are all graded submodules of M.


Since each N_i is generated by homogeneous elements, so is \sum N_i; thus \sum N_i is graded. Suppose n\in \cap_i N_i, if n’ is a homogeneous component of n, then for each i, n \in N_i \implies n' \in N_i and hence n' \in \cap N_i. Thus \cap N_i is graded. Finally, if S (resp. T) is a generating set of \mathfrak a (resp. N) comprising of homogeneous elements, then \{an : a\in S, n\in T\} is a generating set of \mathfrak aN comprising of homogeneous elements. ♦

Exercise B

Decide if each statement is true.

  • If \mathfrak a\subseteq A is a graded ideal of A, then r(\mathfrak a) is graded.
  • If N, P \subseteq M are graded submodules, then (N : P) = \{a \in A: aP \subseteq N\} is a graded ideal.



Proposition 2.

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

(M/N)_i := M_i / (N\cap M_i) \hookrightarrow M/N.


Note that we have A_i (M/N)_j \subseteq (M/N)_{i+j}. It remains to show M/N is a direct sum of (M/N)_i.

Let m\in M. Write m = \sum_i m_i with m_i \in M_i. Then m+N \in M/N is the sum of images of m_i + (N \cap M_i) \in M/(N\cap M_i) in M/N. So M/N = \sum_i (M/N)_i.

Suppose m + N can be written as \sum_i [m_i + (N\cap M_i)] and \sum_i [m_i' + (N\cap M_i)]. Then

m+N = (\sum_i m_i) + N = (\sum_i m_i') + N \implies \sum_i (m_i - m_i') \in N.

By condition 2 of proposition 1, each m_i - m_i' \in N so m_i + (N\cap M_i) = m_i' + (N\cap M_i). ♦

Corollary 2.

If \mathfrak a\subseteq A is a graded ideal, then A/\mathfrak a is a graded ring under the above grading.


Since A_i A_j \subseteq A_{i+j}, multiplication gives

A_i / (\mathfrak a \cap A_i) \times A_j / (\mathfrak a \cap A_j) \longrightarrow A_{i+j} / (\mathfrak a \cap A_{i+j}).


1. Suppose A is the coordinate ring of a variety V\subseteq \mathbb A^n, so that A \cong k[X_1, \ldots, X_n]/I(V). If I(V) is homogeneous, then A is a graded ring. In particular, the group of units in A is k^*, the multiplicative group k - \{0\}. This does not work for non-homogeneous I(V), e.g. the unit group of k[X, Y]/(XY - 1) contains \{ c X^n : n \in \mathbb Z, c \in k^*\} (does equality hold?).

2. We will do exercise C here, i.e. show that A = \mathbb C[X, Y, Z]/(Z^2 - X^2 - Y^2) is not a UFD. Indeed we have the equality Z \cdot Z = (X+iY)(X - iY) in A; we claim that Z, X+iY, X-iY \in A 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 X+iY, X-iY are irreducible and Z, X+iY, X-iY are clearly not associates.

3. By the same reasoning, \mathbb C[W, X, Y, Z]/(Z^2 - W^2 - X^2 - Y^2) is not a UFD.

Note: however \mathbb C[X_1, \ldots, X_n]/(X_1^2 + \ldots + X_n^2) is a UFD for all n\ge 5, as we will see later.


Graded Maps


Let M and N be graded A-modules, and f:M\to N be A-linear. We say f is graded if f(M_i) \subseteq N_i for each i.

Immediately we have:

Lemma 3.

If f:M\to N is a graded map of graded A-modules, then \mathrm{ker} f is a graded submodule of M and \mathrm{im} f is a graded submodule of N.


If m \in M and m = m_0 + \ldots + m_d with m_i \in M_i, then f(m_i) \in N_i and hence f(m) = f(m_0) + \ldots + f(m_d) 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 f:M\to N of graded modules, we have an isomorphism

g : M/\mathrm{ker} f \longrightarrow \mathrm{im } f, \quad m + \mathrm{ker } f \mapsto f(m).

in the category of graded A-modules.


Let us show that g is graded. The grading on the LHS is given by i\mapsto M_i / (\mathrm{ker} f \cap M_i). Since f is graded, g takes the LHS into f(M_i) = N_i.

Since g is bijective, it remains to show that g^{-1} is also graded. For n\in N_i, let m = g^{-1}(n). Write m = m_0 + \ldots + m_d as a sum of homogeneous components; since g(m_i) is the degree-i homogeneous component of g(m) we have g(m) = g(m_i) so m = m_i. ♦

Exercise C

State and prove the remaining two isomorphism theorems.


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