Blowup Algebras

In this post, for a ring \(A\), we fix an ideal \(\mathfrak{a}\) and define two graded \(A\)-algebras built 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 on \(\gr_\mathfrak{a}A\) is given 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 taking, for \(a,b\) respectively, representatives \(\tilde{a}\in \mathfrak{a}^k, \tilde{b}\in \mathfrak{a}^l\), computing their product \(\tilde{a}\tilde{b}\), and then restricting it 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 chose different representatives \(\tilde{a}',\tilde{b}'\), and for some \(x\in \mathfrak{a}^{k+1}\) and \(y\in \mathfrak{a}^{l+1}\), let \(\tilde{a}'=\tilde{a}+x,\tilde{b}'=\tilde{b}+y\). 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 define the following.

Definition 3 For a ring \(A\), an arbitrary ideal of \(A\), \(\mathfrak{a}\), and an \(A\)-module \(M\), a filtration

\[M=M_0\supseteq M_1\supseteq\cdots\]

is called an \(\mathfrak{a}\)-filtration if for every \(k\), \(\mathfrak{a}M_k\subseteq M_{k+1}\) holds. Furthermore, if there exists some \(n\) such that whenever \(k>n\), \(\mathfrak{a}M_k=M_{k+1}\) holds, the filtration is said to be \(\mathfrak{a}\)-stable.

Now, for an arbitrary \(\mathfrak{a}\)-filtration

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

the associated graded module of \(M\) with respect to \(\mathcal{J}\) is defined by

\[\gr_\mathcal{J}M=M/M_1\oplus M_1/M_2\oplus\cdots\]

.

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 \(x\in M_l/M_{l+1}\), one takes representatives \(\tilde{a}\in \mathfrak{a}^k\), \(\tilde{x}\in M_l\) and restricts \(\tilde{a}\tilde{x}\) to \(M_{k+l}/M_{k+l+1}\), and a computation similar to that of Lemma 2 shows that this is well defined. In particular, when \(M=A\) and the \(M_i\) satisfy, for all \(i,j\), \(M_iM_j\subseteq M_{i+j}\) as ideals of \(A\), then just as in Definition 1, \(\gr_\mathcal{J}A\) also carries a ring structure, and this is likewise called the associated graded ring with respect to the filtration \(\mathcal{J}\).

The following now holds.

Proposition 4 Let \(M\) be a finitely generated module equipped with an \(\mathfrak{a}\)-stable filtration \(\mathcal{J}\), and suppose that for \(\mathcal{J}\), every term \(M_k\) 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 some \(n\) 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, among the components of \(\gr_\mathcal{J}M\), if we collect elements generating

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

these 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 \(A\) refers to 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 \(\mathfrak{a}\Bl_\mathfrak{a}A=\bigoplus_{n\geq 0}\mathfrak{a}^{n+1}\), so \(\Bl_\mathfrak{a}A/\mathfrak{a}\Bl_\mathfrak{a}A=\gr_\mathfrak{a}A\). On the other hand, more generally, for any \(A\)-module \(M\) and \(\mathfrak{a}\)-filtration \(\mathcal{J}: M_0\supseteq M_1\supseteq\cdots\), with the following expression

\[\Bl_\mathcal{J}M =M\oplus M_1\oplus\cdots\]

defining it, \(\Bl_\mathcal{J}M\) becomes, from \(\mathfrak{a}^kM_l\subseteq M_{k+l}\), a graded \(\Bl_\mathfrak{a}A\)-module. Now the following holds.

Proposition 6 For a Noetherian ring \(A\) and a finitely generated \(A\)-module \(M\), \(M\)’s \(\mathfrak{a}\)-filtration \(\mathcal{J}\) 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, there exists some \(n\) such that these generators can be chosen to be contained in \(\Bl_\mathcal{J}M\)’s first \(n\) terms. Now, rewriting each of them as a sum of homogeneous elements, \(\Bl_\mathcal{J}M\) is generated by these homogeneous elements. From this, we see 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 its submodule \(M'\). If

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

is an \(\mathfrak{a}\)-stable filtration, then the following filtration induced from this

\[\mathcal{J}':\quad M'\supseteq M'\cap M_1\supseteq M'\cap M_2\supseteq\cdots\]

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, as a submodule of \(\Bl_\mathcal{J}M\), \(\Bl_{\mathcal{J}'}M'\) 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 \(\mathfrak{a}\) of \(A\), and a finitely generated \(A\)-module \(M\). Then the following hold.

1. \((1-a)\left(\bigcap_1^\infty \mathfrak{a}^i M\right)=0\) holds for some \(a\in \mathfrak{a}\).
2. If \(\mathfrak{a}\) is a proper ideal and \(A\) is a domain or a local ring, then \(\bigcap \mathfrak{a}^i=0\) holds.

Proof

For \(M\), consider the \(\mathfrak{a}\)-stable filtration

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

Then, by Lemma 7 (Artin-Rees), the following 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 some \(n\) such that for all \(p>n\),

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

holds. Now, simplifying the left- and right-hand sides of the above equation respectively, we obtain

\[\mathfrak{a}\left(\bigcap \mathfrak{a}^iM\right)=\left(\bigcap \mathfrak{a}^iM\right)\]

Now, since \(A\) is Noetherian and \(M\) is finitely generated, \(\bigcap \mathfrak{a}^iM\) is also finitely generated, and hence applying §Integral Extensions, ⁋Lemma 7 yields the first result.

To prove the second result, set \(M=A\). For the element \(a\) obtained in 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 \(1-a\neq 0\), from which there is nothing more to prove when \(A\) is a domain. If \(A\) is a local ring, then \(\mathfrak{a}\) is contained in the (unique) maximal ideal \(\mathfrak{m}\) of \(A\), so \(a\in \mathfrak{m}\), and from this \(1-a\) must be a unit.

Finally, we define the following.

Definition 9 Suppose we are given an \(\mathfrak{a}\)-filtration

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

on an \(A\)-module \(M\), and its associated graded module \(\gr_\mathcal{J}M\). Then for any \(x\in M\), the initial form of \(x\), \(\initial(x)\), is defined by the formula

\[\initial(x)=x+M_{k+1}\quad\text{in $M_k/M_{k+1}$,}\qquad\text{where $k$ is the greatest integer satisfying $x\in M_k$}\]

Here, if \(x\in\bigcap_k M_k\), such a \(k\) does not exist, and in this case we define \(\initial(x)=0\).

In the above situation, suppose an arbitrary \(A\)-submodule \(M'\subseteq M\) is given. Then, viewing \(\gr_\mathcal{J}M\) as a \(\gr_\mathfrak{a}A\)-module, we can define \(\initial(M')\) to be the submodule generated, for \(x\in M'\), by \(\initial(x)\) in \(\gr_\mathcal{J}M\) as a \(\gr_\mathfrak{a}A\)-submodule.

Example 10 Let \(A=\mathbb{K}[\x,\y]\) and \(\mathfrak{a}=(\x,\y)\). Then \(\gr_\mathfrak{a}A\) is a graded ring whose grading is determined by the degree of polynomials. Now set \(M=A\), and consider in \(M\) the \(A\)-submodule (i.e., ideal of \(A\)) \(\mathfrak{b}=(\x^2, \y^2)\). Then any element of \(\mathfrak{b}\) is of the form

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

and therefore \(\initial(\mathfrak{b})\) is the homogeneous ideal generated by \(\x^2, \y^2\) in \(\gr_\mathfrak{a}A\).

In general, however, \(\initial(M')\) is not generated by the initial forms of the generators of \(M'\). For instance, with \(A\) and \(\mathfrak{a}\) as above, consider \(\mathfrak{c}=(\x^2-\y^3, \x\y)\); the initial forms of the two generators are \(\x^2\) and \(\x\y\), respectively. Then, since \(\x(\x\y)-\y(\x^2-\y^3)=\y^4\), we have \(\y^4\in \mathfrak{c}\), and therefore \(\initial(\y^4)=\y^4\) belongs to \(\initial(\mathfrak{c})\); but every element of \((\x^2,\x\y)\) is divisible by \(\x\), so \(\y^4\) does not belong to it.

Corollary 11 For a Noetherian local ring \(A\) and a proper ideal of \(A\), \(\mathfrak{a}\), if \(\gr_\mathfrak{a}A\) is a domain, then so is \(A\).

Proof

In \(A\), assume \(ab=0\); it suffices to show that \(a=0\) or \(b=0\). Now, in \(\gr_\mathfrak{a}A\) we must have \(\initial(a)\initial(b)=0\), and therefore \(\initial(a)\) or \(\initial(b)\) must be \(0\). Now, since \(\bigcap \mathfrak{a}^n=0\) by Corollary 8 (Krull intersection theorem), we must have \(a=0\) 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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