Blowup Algebras

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.

In this post, we fix an ideal \(\mathfrak{a}\) of a ring \(A\) and define two graded \(A\)-algebras arising from it.

Associated graded module

Definition 1 The associated graded ring of \(A\) with respect to \(\mathfrak{a}\) is defined as

\[\gr_\mathfrak{a}A= A/\mathfrak{a}\oplus \mathfrak{a}/\mathfrak{a}^2\oplus\cdots\]

In the above definition, the multiplication in \(\gr_\mathfrak{a}A\) is defined as follows: given arbitrary \(a\in \mathfrak{a}^k/\mathfrak{a}^{k+1}\) and \(b\in \mathfrak{a}^l/\mathfrak{a}^{l+1}\), their product \(ab\) is obtained by first computing the product \(\tilde{a}\tilde{b}\) of representatives \(\tilde{a}\in \mathfrak{a}^k\) and \(\tilde{b}\in \mathfrak{a}^l\), then restricting the result to \(\mathfrak{a}^{k+l}/\mathfrak{a}^{k+l+1}\).

Lemma 2 The multiplication on \(\gr_\mathfrak{a}A\) 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 introduce the following.

Definition 3 For a ring \(A\), an arbitrary 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 \(\mathfrak{a}M_k=M_{k+1}\) for all \(k>n\), 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 arbitrary \(a\in \mathfrak{a}^k/\mathfrak{a}^{k+1}\) and \(x\in M_l/M_{l+1}\), we choose representatives \(\tilde{a}\in \mathfrak{a}^k\) and \(\tilde{x}\in M_l\), then restrict \(\tilde{a}\tilde{x}\) to \(M_{k+l}/M_{k+l+1}\); a calculation similar to Lemma 2 shows that this is well-defined. In the special case where \(M=A\) and the \(M_i\) are ideals of \(A\), \(\gr_\mathcal{J}A\) also has a ring structure, just as in Definition 1, and this too is called the associated graded ring with respect to the filtration \(\mathcal{J}\).

We now have the following.

Proposition 4 Let \(\mathcal{J}\) be an \(\mathfrak{a}\)-stable filtration of a finitely generated module \(M\), and suppose every term \(M_k\) of \(\mathcal{J}\) is a finitely generated submodule of \(M\). Then \(\gr_\mathcal{J}M\) is a finitely generated \(\gr_\mathfrak{a}A\)-module.

Proof

Since \(\mathcal{J}\) is an \(\mathfrak{a}\)-stable filtration, there exists a suitable \(n\) such that \(\mathfrak{a}M_k=M_{k+1}\) holds for all \(k>n\). Therefore, for such \(k\), we have \((\mathfrak{a}/\mathfrak{a}^2)(M_k/M_{k+1})=M_{k+1}/M_{k+2}\). Hence, collecting only the generators of the components

\[M_0/M_1, M_1/M_2,\ldots, M_{n+1}/M_{n+2}\]

of \(\gr_\mathcal{J}M\) suffices to generate all of \(\gr_\mathcal{J}M\). The desired claim now follows from the assumption that each \(M_i\) is finitely generated.

Blowup algebra

Definition 5 For a ring \(A\) and an ideal \(\mathfrak{a}\), the blowup algebra of \(\mathfrak{a}\) in \(A\) is the following 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 \(\Bl_\mathfrak{a}A/\mathfrak{a}\Bl_\mathfrak{a}A=\gr_\mathfrak{a}A\) is obvious. More generally, for any \(A\)-module \(M\) and \(\mathfrak{a}\)-filtration \(\mathcal{J}: M_0\supseteq M_1\supseteq\cdots\), one can also easily verify that \(\Bl_\mathcal{J}M =M\oplus M_1\oplus\cdots\) defined by the formula above becomes 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 a suitable \(n\) such that these generators can be arranged to lie in the first \(n\) terms of \(\Bl_\mathcal{J}M\). If we replace them by the sums of their homogeneous components, then \(\Bl_\mathcal{J}M\) is generated by these homogeneous elements. From this we see that \(\mathcal{J}\) is \(\mathfrak{a}\)-stable. This argument works in the reverse direction as well.

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 let \(M\) be a finitely generated \(A\)-module with a submodule \(M'\). If

\[\mathcal{J}:\quad M=M_0\supseteq M_1\supseteq\cdots\]

is an \(\mathfrak{a}\)-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, ⁋Corollary 13, \(\Bl_\mathfrak{a}A\) is also Noetherian. Therefore, the submodule \(\Bl_{\mathcal{J}'}M'\) of \(\Bl_\mathcal{J}M\) is also finitely generated, and applying Proposition 6 again yields the desired result.

Corollary 8 (Krull intersection theorem) Fix a Noetherian ring \(A\), an ideal \(\mathfrak{a}\), and a finitely generated \(A\)-module \(M\). Then the following hold.

1. There exists \(a\in \mathfrak{a}\) such that \((1-a)\left(\bigcap_1^\infty \mathfrak{a}^i M\right)=0\).
2. 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 of \(M\)

\[M\supseteq \mathfrak{a}M \supseteq \mathfrak{a}^2 M\supseteq\cdots\]

Then by Lemma 7 (Artin–Rees), 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 a suitable \(n\) such that

\[\mathfrak{a}\left(\left(\bigcap \mathfrak{a}^iM\right)\cap \mathfrak{a}^p M\right)=\left(\bigcap \mathfrak{a}^iM\right)\cap \mathfrak{a}^{n+1} M\]

Simplifying the left-hand 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 show 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. Since \(\mathfrak{a}\) is a proper ideal of \(A\), we have \(1-a\neq 0\), and there is nothing more to prove when \(A\) is a domain. If \(A\) is a local ring, then \(\mathfrak{a}\) must belong to the (unique) maximal ideal \(\mathfrak{m}\) of \(A\), so \(a\in \mathfrak{m}\), and hence \(1-a\) must be a unit.

Finally, we define the following.

Definition 9 Let an \(\mathfrak{a}\)-filtration

\[\mathcal{J}:\qquad M=M_0\supseteq M_1\supseteq\cdots\]

be given on an \(A\)-module \(M\), with associated graded module \(\gr_\mathcal{J}M\). Then for any \(x\in M\), the initial form \(\initial(x)\) of \(x\) is defined 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 situation above, 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 \(\initial(M')\) as 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 with grading 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 \(\mathfrak{b}\) has the form

\[f(\x,\y)\x^2+g(\x,\y)\y^2\]

so \(\initial(\mathfrak{b})\) is the homogeneous ideal of \(\gr_\mathcal{a}A\) 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 \(b=0\). In \(\gr_\mathfrak{a}A\) we must have \(\initial(a)\initial(b)=0\), so either \(\initial(a)\) or \(\initial(b)\) must be \(0\). By the corollary above, \(\bigcap \mathfrak{a}^n=0\), and hence \(a=0\) or \(b=0\).

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References

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

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