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Grad. student in mathematics

가환대수학

Homogeneous localization of graded rings and graded modules

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We examine how to localize an arbitrary graded ring, and more generally an arbitrary graded module. In this post, unless stated otherwise, we assume all graded rings are \(\mathbb{N}_{\geq 0}\)-graded, and fix \(A=\bigoplus A_i\) and \(M=\bigoplus M_i\). Then for any \(n\),

\[M(n)_k=M_{n+k}\qquad\text{for all $k$}\]

naturally endows \(M(n)\) with the structure of a graded \(A\)-module.

Ideal Quotient

First, recall the definition of the ideal quotient for an arbitrary ring \(A\) and two ideals \(\mathfrak{a}, \mathfrak{b}\) of \(A\). (§Basic Notions)

Definition 1 For a ring \(A\) and two ideals \(\mathfrak{a}, \mathfrak{b}\) of \(A\), we define the ideal quotient by the formula

\[(\mathfrak{a}:\mathfrak{b})=\{a\in A\mid a \mathfrak{b}\subseteq \mathfrak{a}\}.\]

Then \((\mathfrak{a}:\mathfrak{b})\) is trivially closed under addition, and for any \(x\in A\) and \(a\in (\mathfrak{a}:\mathfrak{b})\),

\[xa \mathfrak{b}\subseteq x \mathfrak{a}\subseteq \mathfrak{a}\]

holds, so \(xa\in (\mathfrak{a}:\mathfrak{b})\). Thus \((\mathfrak{a}:\mathfrak{b})\) is indeed an ideal.

Properties of Homogeneous Ideals

We showed in [Algebraic Structures] §Graded Rings, ⁋Proposition 6 that any homogeneous ideal is always generated by homogeneous elements; using this, we can prove the following Lemma 2.

Lemma 2 For a graded ring \(A\) and homogeneous ideals \(\mathfrak{a},\mathfrak{b}\) of \(A\), the following hold.

  1. \(\sqrt{\mathfrak{a}}\) is a homogeneous ideal.
  2. \((\mathfrak{a}:\mathfrak{b})\) is a homogeneous ideal.
  3. Suppose that whenever homogeneous elements \(a,b\in A\) satisfy \(ab\in \mathfrak{a}\), either \(a\in \mathfrak{a}\) or \(b\in \mathfrak{a}\) holds. Then \(\mathfrak{a}\) is a prime ideal.
Proof

  1. First, we show that \(\sqrt{\mathfrak{a}}\) is a homogeneous ideal. That is, for any \(x\in \sqrt{\mathfrak{a}}\), when we write \(x\) as a sum of homogeneous elements

    \[x=x_{d_1}+\cdots+x_{d_l},\quad d_1 < \cdots < d_l\tag{$\ast$}\]

    we must show that each \(x_i\) belongs to \(\sqrt{\mathfrak{a}}\). From \(x\in \sqrt{\mathfrak{a}}\), there exists a suitable \(k\) such that \(x^k\in \mathfrak{a}\). Without loss of generality, assuming that \(x_l\) has the largest degree in the above expression (\(\ast\)), when we write \(x^k\) as a sum of homogeneous elements, \(x_l^k\) is the unique element in degree \(k\deg x_l\). Since \(x^k\in \mathfrak{a}\) and \(\mathfrak{a}\) is a homogeneous ideal, we have \(x_l^k\in \mathfrak{a}\), i.e., \(x_l\in \sqrt{\mathfrak{a}}\). Then since \(x-x_l\in\sqrt{\mathfrak{a}}\), we repeat the same argument.

  2. Let \(x\in (\mathfrak{a}:\mathfrak{b})\). As above, when we write \(x\) as a sum of homogeneous elements (\(\ast\)), we must show that each \(x_i\) belongs to \((\mathfrak{a}:\mathfrak{b})\). Let an arbitrary homogeneous element \(b\) generating \(\mathfrak{b}\) be given. Then \(x_ib\) is the homogeneous component of \(xb\in \mathfrak{a}\) in degree \(\deg x_i+\deg b\), and since \(\mathfrak{a}\) is a homogeneous ideal, \(x_ib\in \mathfrak{a}\).

  3. Finally, assuming the given condition, let us express arbitrary two elements \(x,y\in A\) as sums of homogeneous elements

    \[x=x_{d_1}+\cdots+x_{d_m},\quad y=y_{e_1}+\cdots+y_{e_n},\qquad d_1<\cdots< d_m,\quad e_1<\cdots< e_n.\]

    For contradiction, assume that \(xy\in \mathfrak{a}\) but \(x\not\in \mathfrak{a}\) and \(y\not\in \mathfrak{a}\). Then by assumption, at least one of the \(x_{d_i}\) must satisfy \(x_{d_i}\not\in \mathfrak{a}\). Among such \(i\), let \(k\) be the largest, and similarly define \(y_{e_l}\). Now consider the homogeneous component of the element \(xy\) of \(A\) in degree \(d_k+e_l\). This element can be written in the form

    \[(xy)_{d_k+e_l}=\sum_{d_i+e_j=d_k+e_l}x_{d_i}y_{e_j}\]

    and in this case, every term on the right-hand side other than \(x_{d_k}y_{e_l}\) must satisfy either \(d_i>d_k\) or \(e_j>e_l\). By the definition of \(d_k\) and \(e_l\), all such terms are elements of \(\mathfrak{a}\). But since \(xy\in \mathfrak{a}\) and \(\mathfrak{a}\) is a homogeneous ideal, \((xy)_{d_k+e_l}\) is also an element of \(\mathfrak{a}\), from which we obtain a contradiction.

In particular, just as localization at a prime ideal of an arbitrary ring was an important example, localization at a homogeneous prime ideal \(\mathfrak{p}\) is also an important example when \(A\) is a graded ring. Therefore, the third result of the above lemma is especially worth remembering.

Anyway, we start with the general case first. The following proposition can be checked by simply observing how the degrees of elements behave, and its proof is also trivial.

Proposition 3 Suppose all elements of a multiplicative subset \(S\) of \(A\) are homogeneous. Then for any homogeneous element \(x\in M_n\) and \(s\in S\), defining \(x/s\in S^{-1}M\) to lie in degree \(n-\deg s\), the module \(S^{-1}M\) acquires the structure of a \(\mathbb{Z}\)-graded \(A\)-module. If \(M=A\), this grading is also compatible with the multiplication defined on \(S^{-1}A\), making \(S^{-1}A\) a \(\mathbb{Z}\)-graded ring.

Via the inclusion \((S^{-1}A)_0 \rightarrow S^{-1}A\), we can regard \(S^{-1}A\) as an \((S^{-1}A)_0\)-module. Then since the degree \(0\) part \((S^{-1}A)_0\) of \(S^{-1}A\) is closed under multiplication, \((S^{-1}A)_0\) is an \((S^{-1}A)_0\)-algebra. In general, multiplication is not defined on \(S^{-1}M\), but similarly, considering the degree \(0\) part \((S^{-1}M)_0\) of \(S^{-1}M\), this has an \((S^{-1}A)_0\)-module structure.

One of the particularly important examples is the case where \(S=\{1,f,f^2,\cdots\}\) for a homogeneous element \(f\in A_1\) of degree \(1\); in this case, \(S^{-1}A\) can be recovered from \((S^{-1}A)_0\).

Proposition 4 In the above situation, the following isomorphism

\[S^{-1}A\cong (S^{-1}A)_0[T, T^{-1}]\]

holds. Here \(T\) is a formal variable of degree \(1\), and the right-hand side \((S^{-1}A)_0[T, T^{-1}]\) is defined by

\[(S^{-1}A)_0[T_1, T_2]/(T_1T_2-1).\]
Proof

Defining the function \(\{T_1,T_2\} \rightarrow S^{-1}A\) by \(T_1\mapsto f, T_2\mapsto f^{-1}\), by [Algebraic Structures] §Algebras, ⁋Proposition 8 we obtain an \((S^{-1}A)_0\)-algebra homomorphism

\[(S^{-1}A)_0[T_1,T_2] \rightarrow S^{-1}A.\]

Explicitly, this homomorphism is given by

\[\sum_{i,j\geq 0} a_{i,j}T_1^i T_2^j\mapsto \sum_{i,j\geq 0} a_{i,j}f^{i-j}=\sum_{d\in \mathbb{Z}}\left(\sum_{j\geq \max(-d,0)} a_{j+d,j}\right)f^d\]

and it is trivial that the ideal \((T_1T_2-1)\) is contained in the kernel of the above homomorphism. On the other hand, since \(\deg a_{i,j}=0\), the kernel of the above homomorphism is the set of polynomials satisfying the equation

\[\sum_{j\geq \max(-d,0)} a_{j+d,j}=0\qquad\text{for all $d\in\mathbb{Z}$}\]

For notational convenience, we can rewrite the above sum according to the sign of \(d\) into the following three conditions

\[\sum_{j\geq 0} a_{j,j}=0,\quad \sum_{j\geq 0} a_{j+d,j}=0,\quad \sum_{j\geq 0} a_{j,j+d}=0\qquad \text{for all $d>0$}\]

and then in each case we obtain

\[a_{0,0}=-\sum_{j\geq 1} a_{j,j},\quad a_{d,0}=-\sum_{j\geq 1} a_{j+d,j},\quad a_{0,d}=-\sum_{j\geq 1} a_{j,j+d}\qquad \text{for all $d>0$}.\]

Now substituting these into the polynomial

\[\begin{aligned}\sum_{i,j\geq 0} a_{i,j} T_1^iT_2^j&=\sum_{j\geq 0} a_{j,j}T_1^jT_2^j+\sum_{d>0}\sum_{j\geq 0} a_{j+d,j}T_1^{j+d}T_2^j+\sum_{d>0}\sum_{j\geq 0} a_{j,j+d}T_1^jT_2^{j+d}\\&=\left(a_{0,0}+\sum_{j\geq 1} a_{j,j}T_1^jT_2^j\right)+\sum_{d>0} \left(a_{d,0}T_1^d+\sum_{j\geq 1}a_{j+d,j}T_1^{j+d}T_2^j\right)+\sum_{d>0} \left(a_{0,d}T_2^d+\sum_{j\geq 1}a_{j,j+d}T_1^{j}T_2^{j+d}\right)\end{aligned}\]

the right-hand side becomes

\[\left(\sum_{j\geq 1} a_{j,j}(T_1^jT_2^j-1)\right)+\sum_{d>0}\left(\sum_{j\geq 1}a_{j+d,j}T_1^d(T_1^jT_2^j-1)\right)+\sum_{d>0}\left(\sum_{j\geq 1}a_{j,j+d}T_2^d(T_1^jT_2^j-1)\right).\]

Since each \(T_1^jT_2^j-1\) is contained in \((T_1T_2-1)\), the kernel of the above expression is exactly the ideal \((T_1T_2-1)\) in \((S^{-1}A)_0[T_1,T_2]\). On the other hand, since this homomorphism is surjective by definition, we obtain the desired result.

Homogeneous Localization

Definition 5 We call the \((S^{-1}A)_0\) and \((S^{-1}M)_0\) defined above the homogeneous localization of \(A\) and \(M\) respectively, and denote them by \(A_{(S)}\) and \(M_{(S)}\).

As with general localization, for a homogeneous element \(f\in A\), we write the homogeneous localization of \(M\) obtained by taking \(S=\{1,f,f^2,\cdots\}\) as \(M_{(f)}\), and for a homogeneous prime ideal \(\mathfrak{p}\subseteq A\), we write the homogeneous localization of \(M\) obtained by taking \(S=A\setminus \mathfrak{p}\) as \(M_{(\mathfrak{p})}\).

In the remainder of this post, for an arbitrary graded \(A\)-module \(M\) we write

\[M^{(d)}=\bigoplus_{k\geq 0} M_{kd}.\]

Then the following is a generalization of Proposition 4.

Proposition 6 Fix a homogeneous element \(f\in A\) of degree \(d\). Then the following isomorphism

\[M_{(f)}\cong M^{(d)}/(f-1)M^{(d)}\]

holds.

Proof

The given isomorphism is obtained by applying the first isomorphism theorem to a suitable \(u:M^{(d)} \rightarrow M_{(f)}\), where \(u\) is obtained through the formula

\[u_k:M_{kd} \rightarrow M_{(f)};\qquad x\mapsto x/f^k.\]

Then it is not difficult to show that \(u\) is surjective and that \(\ker u=(f-1)M^{(d)}\).

If \(\deg f=1\), then the above isomorphism can be written as \(M_{(f)}\cong M/(f-1)M\).

On the other hand, if \(S\) contains at least one element of degree \(1\), then applying Proposition 4 to each element, we obtain the following.

Proposition 7 If \(S\) is a homogeneous multiplicative set containing at least one element of degree \(1\), then \(S^{-1}A\cong (S^{-1}A)_0[T,T^{-1}]\) holds.

Proof

This is essentially the same proof as Proposition 4: choose an element \(f\) of degree \(1\) belonging to \(S\) and define the homomorphism \((S^{-1}A)_0[T_1,T_2] \rightarrow S^{-1}A\) in the same way as in the proof of Proposition 4. Then that the kernel of this homomorphism is \((T_1T_2-1)\) can be shown by the same proof, and that this homomorphism is surjective can be easily shown by using the fact that an arbitrary element \(a/s\) of \(S^{-1}A\) of degree \(d\) can be written in the form

\[\frac{a}{s}=\frac{af^d}{s}\frac{1}{f^d}.\]

In particular, fix a homogeneous prime ideal \(\mathfrak{p}\) and assume \(A_1\not\subset \mathfrak{p}\). Let \(S\) be the multiplicative subset consisting of homogeneous elements not belonging to \(\mathfrak{p}\); then there exists at least one nonzero \(f\in A_1\) such that \(f\in S\), so by the above proposition we obtain

\[S^{-1}A\cong A_{(\mathfrak{p})}[T,T^{-1}].\]

Proposition 8 In the above situation, let \(\mathfrak{q}\) be the image of \(\mathfrak{p}\) under the homomorphism \(p:A \rightarrow A/(f-1)\). Then \(\mathfrak{q}\) is a prime ideal, and the formula

\[A_{(\mathfrak{p})}\cong\left(A/(f-1)\right)_\mathfrak{q}\]

holds.

Proof

Consider the ring homomorphism \(q:A \rightarrow A/\mathfrak{p}\), and let \(\bar{f}\) be the image of \(f\) under \(q\). Then

\[\frac{A/(f-1)}{\mathfrak{q}}\cong \frac{A/\mathfrak{p}}{(\bar{f}-1)}\]

and by Proposition 6, the right-hand side is in turn isomorphic to \((A/\mathfrak{p})[f^{-1}]_0\). But since \(\mathfrak{p}\) is a prime ideal, \(A/\mathfrak{p}\) is an integral domain and therefore the localization \((A/\mathfrak{p})[f^{-1}]\) is also an integral domain, and hence \((A/\mathfrak{p})[f^{-1}]_0\) is also an integral domain. From this we know that \(\mathfrak{q}\) is a prime ideal of \(A/(f-1)\). For convenience, writing \(\mathfrak{a}=(f-1)\), the desired isomorphism arises from comparing the following homomorphisms

\[A \overset{a\mapsto a/1}{\longrightarrow} S^{-1}A \overset{f\mapsto T}{\longrightarrow} A_{(\mathfrak{p})}[T, T^{-1}] \overset{T\mapsto 1}{\longrightarrow} A_{(\mathfrak{p})}\]

and

\[A\overset{a\mapsto a+\mathfrak{a}}{\longrightarrow}A/\mathfrak{a}\overset{a+\mathfrak{a}\mapsto\frac{a+\mathfrak{a}}{1}}{\longrightarrow}(A/\mathfrak{a})_\mathfrak{q}.\]

References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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