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.
In this post we fix an ideal \(\mathfrak{a}\) of a ring \(A\) and define two graded \(A\)-algebras constructed from it.
Associated graded module
Definition 1 The associated graded ring of a ring \(A\) with respect to \(\mathfrak{a}\) is defined by
\[\gr_\mathfrak{a}A= A/\mathfrak{a}\oplus \mathfrak{a}/\mathfrak{a}^2\oplus\cdots\]In the above definition, the multiplication in \(\gr_\mathfrak{a}A\) is given as follows: for any \(a\in \mathfrak{a}^k/\mathfrak{a}^{k+1}\) and \(b\in \mathfrak{a}^l/\mathfrak{a}^{l+1}\), their product \(ab\) is obtained by first choosing representatives \(\tilde{a}\in \mathfrak{a}^k\) and \(\tilde{b}\in \mathfrak{a}^l\), computing their product \(\tilde{a}\tilde{b}\), and then restricting this to \(\mathfrak{a}^{k+l}/\mathfrak{a}^{k+l+1}\).
Lemma 2 The multiplication in \(\gr_\mathfrak{a}A\) defined above is well defined.
Proof
Suppose we choose different representatives \(\tilde{a}',\tilde{b}'\), and write \(\tilde{a}'=\tilde{a}+x\) and \(\tilde{b}'=\tilde{b}+y\) for suitable \(x\in \mathfrak{a}^{k+1}\) and \(y\in \mathfrak{a}^{l+1}\). Then
\[\tilde{a}'\tilde{b}'=\tilde{a}\tilde{b}+y\tilde{a}+x\tilde{b}+xy\]and since \(x\tilde{b},y\tilde{a}\in \mathfrak{a}^{k+l+1}\) and \(xy\in \mathfrak{a}^{k+l+2}\subseteq \mathfrak{a}^{k+l+1}\), the proof is complete.
To generalize this to \(A\)-modules, we make the following definition.
Definition 3 For a ring \(A\), any ideal \(\mathfrak{a}\) of \(A\), and an \(A\)-module \(M\), a filtration
\[M=M_0\supseteq M_1\supseteq\cdots\]is called an \(\mathfrak{a}\)-filtration if \(\mathfrak{a}M_k\subseteq M_{k+1}\) holds for all \(k\). Furthermore, if there exists some \(n\) such that \(\mathfrak{a}M_k=M_{k+1}\) whenever \(k>n\), then this filtration is called \(\mathfrak{a}\)-stable.
Now, for any \(\mathfrak{a}\)-filtration
\[\mathcal{J}:\quad M=M_0\supseteq M_1\supseteq\cdots\]we define the associated graded module of \(M\) with respect to \(\mathcal{J}\) as
\[\gr_\mathcal{J}M=M/M_1\oplus M_1/M_2\oplus\cdots\]In the above definition, \(\gr_\mathcal{J}M\) carries a \(\gr_\mathfrak{a}A\)-module structure: for any \(a\in \mathfrak{a}^k/\mathfrak{a}^{k+1}\) and \(x\in M_l/M_{l+1}\), we choose representatives \(\tilde{a}\in \mathfrak{a}^k\) and \(\tilde{x}\in M_l\) and restrict \(\tilde{a}\tilde{x}\) to \(M_{k+l}/M_{k+l+1}\); one checks that this is well defined by a calculation similar to Lemma 2. In particular, when \(M=A\) and the \(M_i\) are ideals of \(A\), then just as in Definition 1, \(\gr_\mathcal{J}A\) also carries a ring structure, and this too is called the associated graded ring with respect to the filtration \(\mathcal{J}\).
We now have the following.
Proposition 4 Suppose we are given an \(\mathfrak{a}\)-stable filtration \(\mathcal{J}\) of a finitely generated module \(M\), and assume that every term \(M_k\) of \(\mathcal{J}\) is a finitely generated submodule of \(M\). Then \(\gr_\mathcal{J}A\) is a finitely generated \(\gr_\mathfrak{a}A\)-module.
Proof
Since \(\mathcal{J}\) is an \(\mathfrak{a}\)-stable filtration, there exists an \(n\) such that \(\mathfrak{a}M_k=M_{k+1}\) holds for all \(k>n\). Hence, for such \(k\) we have \((\mathfrak{a}/\mathfrak{a}^2)(M_k/M_{k+1})=M_{k+1}/M_{k+2}\). Therefore, if we collect elements that generate the components
\[M_0/M_1, M_1/M_2,\ldots, M_{n+1}/M_{n+2}\]of \(\gr_\mathcal{J}M\), then they generate all of \(\gr_\mathcal{J}M\). The desired claim now follows from the assumption that each \(M_i\) is finitely generated.
Blowup Algebras
Definition 5 For a ring \(A\) and an ideal \(\mathfrak{a}\), the blowup algebra of \(\mathfrak{a}\) in \(A\) is the graded \(A\)-algebra
\[\Bl_\mathfrak{a}A=A\oplus \mathfrak{a}\oplus \mathfrak{a}^2\oplus\cdots\cong A[t \mathfrak{a}]\subseteq A[t]\]Then it is obvious that \(\Bl_\mathfrak{a}A/\mathfrak{a}\Bl_\mathfrak{a}A=\gr_\mathfrak{a}A\). More generally, for any \(A\)-module \(M\) and \(\mathfrak{a}\)-filtration \(\mathcal{J}: M_0\supseteq M_1\supseteq\cdots\), one easily checks that \(\Bl_\mathcal{J}M\) defined by the formula
\[\Bl_\mathcal{J}M =M\oplus M_1\oplus\cdots\]is a graded \(\Bl_\mathfrak{a}A\)-module. We now have the following.
Proposition 6 An \(\mathfrak{a}\)-filtration \(\mathcal{J}\) of \(M\) is \(\mathfrak{a}\)-stable if and only if \(\Bl_\mathcal{J}M\) is finitely generated as a \(\Bl_\mathfrak{a}A\)-module.
Proof
First, if \(\Bl_\mathcal{J}M\) is finitely generated, then there exists an \(n\) such that all of its generators are contained in the first \(n\) terms of \(\Bl_\mathcal{J}M\). Replacing each of them by a sum of homogeneous elements, we see that \(\Bl_\mathcal{J}M\) is generated by these homogeneous elements. From this we conclude that \(\mathcal{J}\) is \(\mathfrak{a}\)-stable. This argument also works in the opposite direction.
Artin–Rees Lemma
We now prove the following useful Artin–Rees lemma.
Lemma 7 (Artin–Rees) Fix a Noetherian ring \(A\) and an ideal \(\mathfrak{a}\subseteq A\), and fix a finitely generated \(A\)-module \(M\) and a submodule \(M'\) of it. If
\[\mathcal{J}:\quad M=M_0\supseteq M_1\supseteq\cdots\]is an \(\mathcal{a}\)-stable filtration, then the induced filtration
\[\mathcal{J}':\quad M'\supseteq M'\cap M_1\supseteq M'\cap M_2\supseteq\cdots\]is also \(\mathfrak{a}\)-stable.
Proof
Since \(\mathcal{J}\) is \(\mathfrak{a}\)-stable, \(\Bl_\mathcal{J}M\) is finitely generated as a \(\Bl_\mathfrak{a}A\)-module. On the other hand, \(\Bl_\mathfrak{a}A\) is a finitely generated \(A\)-algebra and \(A\) is Noetherian, so by §Basic Notions, §§Finiteness Condition, \(\Bl_\mathfrak{a}A\) is also Noetherian. Therefore, the submodule \(\Bl_{\mathcal{J}'}M'\) of \(\Bl_\mathcal{J}M\) is also finitely generated, and applying Proposition 6 again gives the desired result.
Corollary 8 (Krull Intersection Theorem) Fix a Noetherian ring \(A\), an ideal \(\mathfrak{a}\) of it, and a finitely generated \(A\)-module \(M\). Then the following hold.
- There exists \(a\in \mathfrak{a}\) such that \((1-a)\left(\bigcap_1^\infty \mathfrak{a}^i M\right)=0\).
- If \(A\) is a domain or a local ring and \(\mathfrak{a}\) is a proper ideal, then \(\bigcap \mathfrak{a}^i=0\).
Proof
Consider the \(\mathfrak{a}\)-stable filtration
\[M\supseteq \mathfrak{a}M \supseteq \mathfrak{a}^2 M\supseteq\cdots\]of \(M\). Then by Lemma 7, the filtration
\[\left(\bigcap \mathfrak{a}^iM\right) \cap M\supseteq \left(\bigcap \mathfrak{a}^iM\right)\cap \mathfrak{a}M \supseteq \left(\bigcap \mathfrak{a}^iM\right) \cap \mathfrak{a}^2 M\supseteq\cdots\]is also \(\mathfrak{a}\)-stable. That is, there exists an \(n\) such that
\[\mathfrak{a}\left(\left(\bigcap \mathfrak{a}^iM\right)\cap \mathfrak{a}^p M\right)=\left(\bigcap \mathfrak{a}^iM\right)\cap \mathfrak{a}^{n+1} M\]holds. Simplifying the left- and right-hand sides of the above equation, we obtain
\[\mathfrak{a}\left(\bigcap \mathfrak{a}^iM\right)=\left(\bigcap \mathfrak{a}^iM\right)\]and applying §Integral Extensions, ⁋Lemma 7 gives the first result.
To prove the second result, set \(M=A\). For the element \(a\) obtained from the first result, it suffices to show that \(1-a\) is not a zerodivisor. First, since \(\mathfrak{a}\) is a proper ideal of \(A\), we have \(1-a\neq 0\), and hence there is nothing more to prove when \(A\) is a domain. If \(A\) is a local ring, then \(\mathfrak{a}\) must be contained in the (unique) maximal ideal \(\mathfrak{m}\) of \(A\), so \(a\in \mathfrak{m}\), and therefore \(1-a\) must be a unit.
Finally, we define the following.
Definition 9 Suppose we are given an \(A\)-module \(M\) with an \(\mathfrak{a}\)-filtration
\[\mathcal{J}:\qquad M=M_0\supseteq M_1\supseteq\cdots\]and the associated graded module \(\gr_\mathcal{J}M\). Then for any \(x\in M\), we define the initial form \(\initial(x)\) of \(x\) by the formula
\[\initial(x)=x+M_{k+1}\quad\text{in $M_k/M_{k+1}$,}\qquad\text{where $k$ is the greatest integer satisfying $x\in M_k$}\]In the above situation, suppose an arbitrary \(A\)-submodule \(M'\subseteq M\) is given. Regarding \(\gr_\mathcal{J}M\) as a \(\gr_\mathfrak{a}A\)-module, we can define \(\initial(M')\) to be the \(\gr_\mathfrak{a}A\)-submodule of \(\gr_\mathcal{J}M\) generated by the \(\initial(x)\) for \(x\in M'\).
Example 10 Let \(A=\mathbb{K}[\x,\y]\) and \(\mathfrak{a}=(\x,\y)\). Then \(\gr_\mathfrak{a}A\) is the graded ring whose grading is determined by the degree of polynomials. Now set \(M=A\) and consider the \(A\)-submodule (that is, the ideal of \(A\)) \(\mathfrak{b}=(\x^2, \y^2)\). Then any element of \(\mathfrak{b}\) is of the form
\[f(\x,\y)\x^2+g(\x,\y)\y^2\]so \(\initial(\mathfrak{b})\) is the homogeneous ideal of \(\gr_\mathfrak{a}A\) generated by \(\x^2\) and \(\y^2\).
However, in general \(\initial(M')\) is not generated by the initial forms of generators of \(M'\).
Corollary 11 For a Noetherian local ring \(A\) and a proper ideal \(\mathfrak{a}\) of \(A\), if \(\gr_\mathfrak{a}A\) is a domain, then so is \(A\).
Proof
Assume \(ab=0\) in \(A\); it suffices to show that \(a=0\) or \(b=0\). Then we must have \(\initial(a)\initial(b)=0\) in \(\gr_\mathfrak{a}A\), and therefore either \(\initial(x)\) or \(\initial(y)\) must be \(0\). Since \(\bigcap \mathfrak{a}^n=0\) by the corollary above, we must have \(a=0\) or \(b=0\).
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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