Localization of Graded Rings

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 \(n\), \(M(n)\) defined by

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

naturally has 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, §§Basic Definitions)

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 of \(A\), \(\mathfrak{a},\mathfrak{b}\), the following hold.

1. \(\sqrt{\mathfrak{a}}\) is a homogeneous ideal.
2. \((\mathfrak{a}:\mathfrak{b})\) is a homogeneous ideal.
3. Suppose that \(\mathfrak{a}\neq A\), and 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, let us show that \(\sqrt{\mathfrak{a}}\) is a homogeneous ideal. That is, for any \(x\in \sqrt{\mathfrak{a}}\), when \(x\) is expressed 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 \(x_{d_i}\) belongs to \(\sqrt{\mathfrak{a}}\). First, since \(x\in \sqrt{\mathfrak{a}}\), there exists some \(k\) such that \(x^k\in \mathfrak{a}\). Meanwhile, without loss of generality, assuming that \(x_{d_l}\) has the largest degree in the above expression (\(\ast\)), when \(x^k\) is expressed as a sum of homogeneous elements, \(x_{d_l}^k\) is the unique element in degree \(k\deg x_{d_l}\). Now, since \(x^k\in \mathfrak{a}\) and \(\mathfrak{a}\) is a homogeneous ideal, we know that \(x_{d_l}^k\in \mathfrak{a}\), that is, \(x_{d_l}\in \sqrt{\mathfrak{a}}\). After this, since \(x-x_{d_l}\in\sqrt{\mathfrak{a}}\), we can repeat the same argument.

2. Let \(x\in (\mathfrak{a}:\mathfrak{b})\). As above, when \(x\) is expressed as the sum of homogeneous elements (\(\ast\)), we must show that each \(x_{d_i}\) belongs to \((\mathfrak{a}:\mathfrak{b})\). Now suppose an arbitrary homogeneous element generating \(\mathfrak{b}\), \(b\), is given. Then \(x_{d_i}b\) is the homogeneous component of the element \(xb\in \mathfrak{a}\) in degree \(\deg x_{d_i}+\deg b\), and since \(\mathfrak{a}\) is a homogeneous ideal, \(x_{d_i}b\in \mathfrak{a}\). Since \(\mathfrak{b}\) is generated by homogeneous elements \(b_j\), for any \(\sum_j a_jb_j\in \mathfrak{b}\) we have \(x_{d_i}\sum_j a_jb_j=\sum_j a_j(x_{d_i}b_j)\in \mathfrak{a}\), and therefore \(x_{d_i}\in (\mathfrak{a}:\mathfrak{b})\).

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

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

   Contrary to the conclusion, suppose 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 in \(A\), consider the homogeneous component of the element \(xy\) corresponding to 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 here, every term on the right-hand side of the above equation 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 \(\mathfrak{a}\). However, since \(xy\in \mathfrak{a}\) and \(\mathfrak{a}\) is a homogeneous ideal, \((xy)_{d_k+e_l}\) is also an element of \mathfrak{a}\(, and by assumption\)x_{d_k}\in \mathfrak{a}\(or\)y_{e_l}\in \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, even when \(A\) is a graded ring, localization at a homogeneous prime ideal \(\mathfrak{p}\) is an important example. Therefore, the third result of the above lemma is especially worth remembering.

In any case, we first begin with the general case. The following proposition can be proved by simply observing how the degrees of elements behave, and its proof is also trivial.

Proposition 3 Suppose that in \(A\), all elements of a multiplicative subset \(S\) are homogeneous. Then for any homogeneous element \(x\in M_n\) and \(s\in S\), if we define \(x/s\in S^{-1}M\) to lie in degree \(n-\deg s\), \(S^{-1}M\) has 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 in \(S^{-1}A\) the degree \(0\) part \((S^{-1}A)_0\) is closed under multiplication, \(S^{-1}A\) is an \((S^{-1}A)_0\)-algebra. In general, multiplication is not defined on \(S^{-1}M\), but similarly, considering for \(S^{-1}M\) the degree \(0\) part \((S^{-1}M)_0\), this has an \((S^{-1}A)_0\)-module structure.

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

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 giving 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

If we define a 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 here it is clear 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 collection 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 convenience of notation, 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

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

Here, since each \(T_1^jT_2^j-1\) is contained in \((T_1T_2-1)\), the kernel of the above expression is exactly the \((S^{-1}A)_0[T_1,T_2]\)-ideal \((T_1T_2-1)\). On the other hand, for any \(a\in A_n\) and \(k\geq 0\), since \(a/f^k=(a/f^n)f^{n-k}\) and \(a/f^n\in (S^{-1}A)_0\), this homomorphism is surjective, and hence 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 \(f\in A\), taking \(S=\{1,f,f^2,\cdots\}\), we write the homogeneous localization of \(M\) obtained from it as \(M_{(f)}\), and for a homogeneous prime ideal \(\mathfrak{p}\subseteq A\), taking the set of homogeneous elements not belonging to \(\mathfrak{p}\) to be \(S\), we write the homogeneous localization of \(M\) obtained from it 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 For \(d\geq 1\), fix a degree \(d\) homogeneous element \(f\in A\). 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: it suffices to choose from \(S\) a degree \(1\) element \(f\) and, in the same manner as in the proof of Proposition 4, define the homomorphism \((S^{-1}A)_0[T_1,T_2] \rightarrow S^{-1}A\). 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 degree \(d\) element of \(S^{-1}A\), \(a/s\), can be written in the form

\[\frac{a}{s}=\left(\frac{a}{s}f^{-d}\right)f^d\]

so that the first factor can be taken as an element of \((S^{-1}A)_0\).

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

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

Proposition 8 In the above situation, under the homomorphism \(p:A \rightarrow A/(f-1)\), let the image of \(\mathfrak{p}\) be \(\mathfrak{q}\). 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 under \(q\), let the image of \(f\) be \(\bar{f}\). 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 \(A/(f-1)\). If for convenience we write \(\mathfrak{a}=(f-1)\), 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

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

yields the desired isomorphism.

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References

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

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