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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

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    $\begingroup$ In your first lemma, if one takes $A=k[x,y]$ and $I=(x,y^n)$, can you tell me what are $\mathfrak{q}_i, I_i,l_i$? $\endgroup$
    – Mohan
    Jun 13 at 17:01
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    $\begingroup$ 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. $\endgroup$
    – KReiser
    Jun 15 at 9:34
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    $\begingroup$ 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. $\endgroup$
    – KReiser
    Jun 15 at 9:48
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    $\begingroup$ 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. $\endgroup$ Jun 17 at 21:46
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    $\begingroup$ @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. $\endgroup$ Jun 18 at 12:45

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