# Applications of the Chinese remainder Theorem to the study of the Hilbert scheme of points and $(\mathfrak{m},l)$-squeezed ideals.

The following construction gives a relation between the Chinese Remainder Theorem (CRT), the Noether nomalization lemma (NNL) and cofinite ideals in finitely generated $$k$$-algebras.

Let $$k$$ be any field and let $$A$$ be any finitely generated $$k$$-algebra and $$I \subseteq A$$ a cofinite ideal. By Atiyah-Macdonald Thm 8.7 (AM) and the NNL, since $$B:=A/I$$ is Artinian there is a decomposition

$$B \cong B_1\oplus \cdots \oplus B_d$$

with $$(B_i, \mathfrak{m}_i)$$ an Artinian local ring for all $$i$$. Since $$dim_k(B_i)< \infty$$ it follows there is an integer $$l_i \geq 2$$ with $$\mathfrak{m}_i^{l_i}=0$$. Let

$$\mathfrak{p}_i:=B_1\oplus \cdots \oplus \mathfrak{m}_i \oplus \cdots \oplus B_d \subseteq B,$$

and let

$J_i:=B_1 \oplus \cdots \oplus (0)\oplus \cdots \oplus B_d.$$It follows $$J_i \subseteq \mathfrak{p}_i$$ and that $$\mathfrak{p}_i$$ is a maximal ideal. There is an equality $$J_1\cdots J_d=(0)$$. The ideals $$\mathfrak{p}_i, \mathfrak{p}_j$$ are coprime for $$i \neq j$$. Similar for $$J_i,J_j$$. Hence there is an equality $$(0)=J_1\cdots J_d = J_1 \cap \cdots \cap J_d.$$ Let $$p: A \rightarrow A/I$$ and let $$I_i:=p^{-1}(J_i)$$. It follows $$I:=p^{-1}((0))=p^{-1}(J_1\cdots J_d)=p^{-1}(J_1 \cap \cdots \cap J_d)=$$ $$p^{-1}(J_1) \cap \cdots \cap p^{-1}(J_d)=I_1\cap \cdots \cap I_d=I_1\cdots I_d.$$ This is because the ideals $$I_i,I_j$$ are coprime when $$i\neq j$$. We may lift the maximal ideals $$\mathfrak{p}_i$$ to maximal ideals $$\mathfrak{q}_i:=p^{-1}(\mathfrak{p}_i) \subseteq A$$ and it follows the ideals $$\mathfrak{q}_i, \mathfrak{q}_j$$ are coprime when $$i \neq j$$. Since there is an inclusion $$J_i \subseteq \mathfrak{p}_i$$, it follow $$I_i \subseteq \mathfrak{q}_i$$. There is an integer $$l_i$$ with $$J_i=\mathfrak{p}_i^{l_i+1}$$ and it follows there are inclusions $$\mathfrak{q}_i^{l_i+1} \subseteq I_i \subseteq \mathfrak{q}_i^{l'_i} \subseteq \mathfrak{q}_i.$$ Lemma. Given any cofinite ideal $$I \subseteq A$$, it follows there are maximal ideals $$\mathfrak{q}_1,..,\mathfrak{q}_d$$ and integers $$l_1,..,l_d, l_1',..,l_d' \geq 1$$ and cofinite ideals $$\mathfrak{q}_i^{l_i+1} \subseteq I_i \subseteq \mathfrak{q}_i^{l'_i}$$ with $$I=I_1\cdots I_d.$$ Proof: The construction is given above QED. Example: If $$A:=k[x,y], I:=(x^m,y^n)$$ with $$m \leq n$$ it follows we may choose $$i=1, \mathfrak{p}_1:=(x,y)$$, $$l_1:=m+n+1, l_1':=m$$. We get inclusions $$(x,y)^{m+n+1} \subseteq (x^m,y^n) \subseteq (x,y)^m.$$ Note: A product of powers of maximal ideals is cofinite, and by the Lemma we may study ideals "squeezed" between powers of maximal ideals $$\mathfrak{m}^{l+1} \subseteq I \subseteq \mathfrak{m}^{l'}$$ to obtain all cofininite ideals. Definition: We say such an ideal $$I$$ is an "$$\mathfrak{m}$$-squeezed ideal". We use the word "squeeze" here because the ideal $$I$$ is "squeezed" between two powers of a maximal ideal. Note that if $$A$$ is a regular ring of dimension $$d$$ it follows $$\mathfrak{m}^l/\mathfrak{m}^{l+1} \cong Sym^l(\mathfrak{m}/\mathfrak{m}^2)$$ hence we have good control on the vector spaces $$\mathfrak{m}^l/\mathfrak{m}^{l+1}$$ and $$A/\mathfrak{m}^{l+1}$$ when $$A$$ is regular. Example: If $$A:=k[x,y]$$ with $$k$$ an algebraically closed field we get the following: Consider the set $$\{(a_1,b_1,l_1),\ldots ,(a_d,b_d,l_d) \}\in Sym^d(k^2 \times \mathbb{Z})$$ with $$(a_i,b_i,l_i)\in k^2 \times \mathbb{Z}, l_i\geq 1$$ and $$\sum_j \binom{l_j+1}{2}=n.$$ Here $$Sym^d(k^2 \times \mathbb{Z})$$ is the "set theoretic symmetric product" of $$k^2 \times \mathbb{Z}$$. The symmetric group on d elements $$S_d$$ acts on $$(k^2 \times \mathbb{Z})^d$$ and $$Sym^d(k^2\times \mathbb{Z}):=(k^2 \times \mathbb{Z})^d/S_d$$ is the "quotient". Let $$H^i:=Sym^i(k^2 \times \mathbb{Z})$$ and consider $$H(n):=H^1\times H^2 \cdots \times H^n.$$ We get a construction of a class of cofinite ideals $$I \subseteq A$$ of length $$n$$ (all products of powers of maximal ideals) as a subset $$"Hilb^n(k[x,y])" \subseteq H(n).$$ Note: $$"Hilb^n(k[x,y])"$$ is a set, not a scheme. The Hilbert scheme $$Hilb^n(k[x,y])$$ is the unique scheme representing the Hilbert functor. The Hilbert scheme comes equipped with a "universal family". From the above construction we see there is (at least set theoretically) a close connection between the different symmetric products $$Sym^d(k^2 \times \mathbb{Z})$$ and the set $$"Hilb^n(k[x,y])"$$. Question: Given an arbitrary field $$k$$ and an arbitrary finitely generated $$k$$-algebra $$A$$ with a confinite ideal $$I \subseteq A$$ (this means $$I$$ is an ideal with $$dim_k(A/I)< \infty$$). Are you able to give an "elementary" parametrization of all such ideals I using methods similar to the one introduced above? Given a maximal ideal $$\mathfrak{m} \subseteq A$$ are you able to construct all $$\mathfrak{m}$$-squeezed ideals $$\mathfrak{m}^{l+1} \subseteq I \subseteq \mathfrak{m}^{l'}$$? The $$\mathfrak{m}$$-squeezed ideals form the building blocks of the set of all cofinite ideals, and one wants to classify them (construct a parameter space) and to relate this classification to $$Hilb^n(Spec(A))$$. The post originates in a question posed here: Some integers related to the Hilbert scheme of points in the plane. The question(s) has relations to differential operators, Taylor maps, the $$n!$$-conjecture, the Macdonald positivity conjecture and the $$qt$$-Kostka coefficients https://en.wikipedia.org/wiki/N!_conjecture https://mathoverflow.net/questions/10014/applications-of-the-chinese-remainder-theorem/394852#394852 • In your first lemma, if one takes$A=k[x,y]$and$I=(x,y^n)$, can you tell me what are$\mathfrak{q}_i, I_i,l_i$? Jun 13 at 17:01 • This is the 16th version of this question, which is itself a repost of a portion of a question which made it to 13 revisions before being closed. That's 29 revisions. Please decide what you are going to say, say it, and then leave it alone. Jun 15 at 9:34 • When you do it 29 times it is! This post alone is in the top 5 most edited posts on the entire site for the past seven days despite existing for only two of them. That's ridiculous. Jun 15 at 9:48 • There are zero-dimensional subschemes of the plane much more complicated than monomial ideals. Look at Nakajima's book Lectures on Hilbert Schemes of Points on Surfaces, Chapter 1 to understand how to parameterize the Hilbert scheme. Jun 17 at 21:46 • @hm2020 if the idea has no chance over$\mathbb C\$, then changing the base field will not help. Further, zero-dimensional Hilbert schemes of higher-dimensional schemes may be quite pathological. Jun 18 at 12:45