## Commutative Algebra 53

Definition.

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..

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

Note

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 $a\ne 0$.

Example

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 $A_0$.

Proof

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.

Proof

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. ]

Definition.

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 $i,j \ge 0$.

Note

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

• 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 $N = (N \cap M_0) \oplus (N \cap M_1) \oplus \ldots$.

Proof

(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.

Definition.

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.

Proof

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.

# Quotient

Proposition 2.

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

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

Proof

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.

Proof

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

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

Example

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.

Definition.

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.

Proof

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.

Proof

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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