This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
We investigate how to localize an arbitrary graded ring, and more generally, an arbitrary graded module. In this post, unless stated otherwise, we assume that every graded ring is \(\mathbb{N}_{\geq 0}\)-graded, and fix \(A=\bigoplus A_i\) and \(M=\bigoplus M_i\). Then for any \(n\), the module \(M(n)\) defined by
\[M(n)_k=M_{n+k}\qquad\text{for all $k$}\]naturally carries a graded \(A\)-module structure.
Ideal Quotients
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)
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 obviously closed under addition, and for any \(x\in A\) and \(a\in (\mathfrak{a}:\mathfrak{b})\), the relation
\[xa \mathfrak{b}\subseteq x \mathfrak{a}\subseteq \mathfrak{a}\]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 every homogeneous ideal is always generated by homogeneous elements; using this, we can prove the following Lemma 1.
Lemma 2 Let \(A\) be a graded ring and let \(\mathfrak{a},\mathfrak{b}\) be homogeneous ideals of \(A\). Then the following hold.
- \(\sqrt{\mathfrak{a}}\) is a homogeneous ideal.
- \((\mathfrak{a}:\mathfrak{b})\) is a homogeneous ideal.
- Suppose that whenever homogeneous elements \(a,b\in A\) satisfy \(ab\in \mathfrak{a}\), we have \(a\in \mathfrak{a}\) or \(b\in \mathfrak{a}\). Then \(\mathfrak{a}\) is a prime ideal.
Proof
-
First, we show that \(\sqrt{\mathfrak{a}}\) is a homogeneous ideal. That is, for any \(x\in \sqrt{\mathfrak{a}}\), we must show that when \(x\) is written as a sum of homogeneous elements
\[x=x_{d_1}+\cdots+x_{d_l},\quad d_1 < \cdots < d_l\tag{$\ast$}\]each \(x_i\) lies in \(\sqrt{\mathfrak{a}}\). First, since \(x\in \mathfrak{a}\), there exists an integer \(k\) such that \(x^k\in \mathfrak{a}\). Without loss of generality, assume that in the above expression (\(\ast\)), \(x_l\) has the largest degree. 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, it follows that \(x_l^k\in \mathfrak{a}\), i.e., \(x_l\in \sqrt{\mathfrak{a}}\). Since \(x-x_l\in\sqrt{\mathfrak{a}}\), we can repeat the same argument.
- Let \(x\in (\mathfrak{a}:\mathfrak{b})\). As above, writing \(x\) as a sum of homogeneous elements (\(\ast\)), we must show that each \(x_i\) belongs to \((\mathfrak{a}:\mathfrak{b})\). Let \(b\) be any homogeneous generator of \(\mathfrak{b}\). Then \(x_ib\) is the homogeneous component of \(xb\in \mathfrak{a}\) of degree \(\deg x_i+\deg b\), and since \(\mathfrak{a}\) is a homogeneous ideal, we have \(x_ib\in \mathfrak{a}\).
-
Finally, assume the given condition and write any 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 \(x_{d_i}\)’s satisfies \(x_{d_i}\not\in \mathfrak{a}\). Let \(k\) be the largest such index, and define \(y_{e_l}\) similarly. Now consider the homogeneous component of \(xy\) 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 the right-hand side of the above expression, every term except \(x_{d_k}y_{e_l}\) must satisfy either \(d_i>d_k\) or \(e_j>e_l\). By the definitions of \(d_k\) and \(e_l\), all such terms belong to \(\mathfrak{a}\). Since \(xy\in \mathfrak{a}\) and \(\mathfrak{a}\) is a homogeneous ideal, \((xy)_{d_k+e_l}\) also lies in \(\mathfrak{a}\), yielding a contradiction.
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 begin with the general case first. The following proposition can be proved simply by observing how the degrees of elements behave, and its proof is obvious.
Proposition 3 Let \(S\) be a multiplicative subset of \(A\) all of whose elements are homogeneous. Then for any homogeneous element \(x\in M_n\) and \(s\in S\), defining \(x/s\in S^{-1}M\) to be in degree \(n-\deg s\), the module \(S^{-1}M\) carries a \(\mathbb{Z}\)-graded \(A\)-module structure. 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.
We may regard \(S^{-1}A\) as an \((S^{-1}A)_0\)-module via the inclusion \((S^{-1}A)_0 \rightarrow S^{-1}A\). Then the degree \(0\) part \((S^{-1}A)_0\) of \(S^{-1}A\) is closed under multiplication, so \((S^{-1}A)_0\) is an \((S^{-1}A)_0\)-algebra. In general, multiplication is not defined on \(S^{-1}M\), but similarly the degree \(0\) part \((S^{-1}M)_0\) of \(S^{-1}M\) has an \((S^{-1}A)_0\)-module structure.
One particularly important example 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 given degree \(1\), and the right-hand side \((S^{-1}A)_0[T, T^{-1}]\) is defined as
\[(S^{-1}A)_0[T_1, T_2]/(T_1T_2-1).\]Proof
Define the map \(\{T_1,T_2\} \rightarrow S^{-1}A\) by \(T_1\mapsto f\) and \(T_2\mapsto f^{-1}\). Then 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 obvious that the ideal \((T_1T_2-1)\) is contained in the kernel of this homomorphism. On the other hand, since \(\deg a_{i,j}=0\), the kernel of the above homomorphism consists of the 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 may rewrite the above sum according to the sign of \(d\) as the 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
\[a_{0,0}=-\sum_{j\geq 1} a_{j,j},\quad a_{d,0}=-\sum_{j\geq 1} a_{j+d,j},\quad a_{0,d}=-\sum_{j\geq 1} a_{j,j+d}\qquad \text{for all $d>0$}.\]Substituting these into the polynomial
\[\begin{aligned}\sum_{i,j\geq 0} a_{i,j} T_1^iT_2^j&=\sum_{j\geq 0} a_{j,j}T_1^jT_2^j+\sum_{d>0}\sum_{j\geq 0} a_{j+d,j}T_1^{j+d}T_2^j+\sum_{d>0}\sum_{j\geq 0} a_{j,j+d}T_1^jT_2^{j+d}\\&=\left(a_{0,0}+\sum_{j\geq 1} a_{j,j}T_1^jT_2^j\right)+\sum_{d>0} \left(a_{d,0}T_1^d+\sum_{j\geq 1}a_{j+d,j}T_1^{j+d}T_2^j\right)+\sum_{d>0} \left(a_{0,d}T_2^d+\sum_{j\geq 1}a_{j,j+d}T_1^{j}T_2^{j+d}\right)\end{aligned}\]yields the right-hand side
\[\left(\sum_{j\geq 1} a_{j,j}(T_1^jT_2^j-1)\right)+\sum_{d>0}\left(\sum_{j\geq 1}a_{j+d,j}T_1^d(T_1^jT_2^j-1)\right)+\sum_{d>0}\left(\sum_{j\geq 1}a_{j,j+d}T_2^d(T_1^jT_2^j-1)\right).\]Since each \(T_1^jT_2^j-1\) lies in \((T_1T_2-1)\), the kernel of the above map is exactly the ideal \((T_1T_2-1)\) in \((S^{-1}A)_0[T_1,T_2]\). Meanwhile, this homomorphism is surjective by definition, so 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 ordinary localization, for a homogeneous element \(f\in A\), we write \(M_{(f)}\) for the homogeneous localization of \(M\) obtained from \(S=\{1,f,f^2,\cdots\}\), and for a homogeneous prime ideal \(\mathfrak{p}\subseteq A\), we write \(M_{(\mathfrak{p})}\) for the homogeneous localization of \(M\) obtained from \(S=A\setminus \mathfrak{p}\).
For the remainder of this post, for any 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 map \(u:M^{(d)} \rightarrow M_{(f)}\), where \(u\) is given by 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 \(\deg f=1\), 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 \(1\), we obtain the following by applying Proposition 4 to each such element.
Proposition 7 If \(S\) is a homogeneous multiplicative set containing at least one element of degree \(1\), then \(S^{-1}A\cong (S^{-1}A)_0[T,T^{-1}]\).
Proof
This is proved in essentially the same way as Proposition 4: choose an element \(f\in S\) of degree \(1\) and define the homomorphism \((S^{-1}A)_0[T_1,T_2] \rightarrow S^{-1}A\) in the same manner as in the proof of Proposition 4. Then the fact that the kernel of this homomorphism is \((T_1T_2-1)\) can be shown by the same argument, and its surjectivity follows easily from the observation that any 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 \(A_1\not\subset \mathfrak{p}\). If \(S\) is the multiplicative subset consisting of homogeneous elements not belonging to \(\mathfrak{p}\), then there exists a nonzero \(f\in A_1\) with \(f\in S\), 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 \((A/\mathfrak{p})[f^{-1}]_0\). Since \(\mathfrak{p}\) is a prime ideal, \(A/\mathfrak{p}\) is an integral domain; hence the localization \((A/\mathfrak{p})[f^{-1}]\) is also an integral domain, and therefore \((A/\mathfrak{p})[f^{-1}]_0\) is an integral domain as well. It follows that \(\mathfrak{q}\) is a prime ideal of \(A/(f-1)\). For convenience, write \(\mathfrak{a}=(f-1)\). Then the desired isomorphism arises by comparing the 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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