Some integers related to the Hilbert scheme of points in the plane.

Question

This question is related to another question posed on this site.

Let me recall the construction: Let $A:=k[x,y]/I$ with $k$ the complex numbers (or any algebraically closed field) and $\dim_k(A)< \infty$. It follows $A$ is artinian and hence there is an isomorphism

$$\phi:A \cong A_1 \oplus \cdots \oplus A_d$$

where $A_i$ is an artinian local ring for every $i$ with maximal ideal $\mathfrak{m}_i$. Since $\dim_k(A_i)< \infty$ it follows $\mathfrak{m}_i^{l_i}=0$ for some integer $l_i \geq 1$.

Let $\mathfrak{p}_i:=A_1\oplus \cdots \oplus \mathfrak{m}_i \oplus \cdots \oplus A_d$. It follows $A/\mathfrak{p}_i \cong A_i/\mathfrak{m}_i\cong k$ and $A/\mathfrak{p}_i^{l_i} \cong A_i$. It follows

$$\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} =0.$$

By the chinese remainder theorem there is an isomorphism

$$A \cong A/(0) \cong A/\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \cong A_1 \oplus \cdots \oplus A_d$$

and since $k$ is algebraically closed it follows $\mathfrak{p}_i \cong(x-a_i,y-b_i)$ for complex numbers $(a_i,b_i)\in k^2$.

Let us lift the ideals $\mathfrak{p}_i$ to ideals $\mathfrak{q}_i \subseteq B:=k[x,y]$ with $B/\mathfrak{q}_i \cong A/\mathfrak{p}_i \cong k$. It follows the ideals $\mathfrak{q}_i \subseteq B$ are maximal ideals with

$$\mathfrak{q}_1^{l_1}\cdots \mathfrak{q}_d^{l_d} \subseteq I \subseteq \mathfrak{q}_i$$

for all $i=1,\ldots,d$.

Let us assume there is an equality of ideals

$$I:=(x-a_1,y-b_1)^{l_1}\cdots (x-a_d,y-b_d)^{l_d}$$

in $k[x,y]$, where $l_1,...,l_d$ is a set of integers $\geq 1$ satisfying a certain condition (see $C_1$ below).

Hence if our aim is to study the Hilbert scheme, we want to parametrize all length $n$ ideals, in particular we want to study the set of products of maximal ideals

$$I:=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \subseteq k[x,y]$$

with $1 \leq d \leq n$ and $l_i \geq 1$. We find the formula

$$\dim_k(k[x,y]/(x,y)^i)=\binom{i+1}{2}$$

hence if $I:=(x-a_1,y-b_1)^{l_1}\cdots (x-a_d,y-b_d)^{l_d}$

it follows

$$\dim_k(k[x,y]/I)= \sum_{j=1}^d \binom{l_j+1}{2}.$$

Note that $\dim_k(k[x,y]/(x-a,y-b)^l)=\dim_k(k[x,y]/(x,y)^l$ hence

$$\dim_k(A) = \sum_j \dim_k(k[x,y]/(x-a_j,y-b_j)^{l_j}=\sum_j \binom{l_j+1}{2}.$$

Hence when studying the Hilbert scheme we want to parametrize ideals $I=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d}$ with $1 \leq d \leq n $ and

$$(C_1)\qquad\dim_k(k[x,y]/I)= \sum_{j=1}^d \binom{l_j+1}{2}=n.$$

Question 1: Given an integer $1 \leq d \leq n$ we seek a combinatorial formula for the number $D(l_1,..,l_d,n)$ of unordered tuples of integers $(l_1,...,l_d)$ with $l_i\geq 1$ and $\sum_j \binom{l_j+1}{2}=n$:

Let $S(l_1,..,l_d,n)$ be the following set:

$$S(l_1,..,l_d,n):=\{ (l_1,..,l_d)\text{ an unordered set of integers $l_i$}. l_i \geq 1, \sum_j \binom{l_j+1}{2}=n \}$$

By definition: $D(l_1,..,l_d,n)$ is the number of elements in $S(l_1,..,l_d,n)$.

Do you know such a formula or a reference to where this type of formula is studied? I ask for an explicit reference to a study of this problem and such formulas in the litterature. If you have seen these numbers appearing in the study of the $n!$-conjecture I ask for a reference.

Note 1: I'm "imprecise" when writing $D(l_1,..,l_d,n)$ - this reflects that the numbers arise when studying the ideals

$$\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \subseteq k[x,y]$$

with $\mathfrak{p}_i:=(x-a_i,y-b_i)$.

Note 2: We may generalize these numbers as follows: If $A:=k[x_1,..,x_n]$ is a polynomial ring in $n$ variables over $k$, we may want to "parametrize" the set of ideals $I \subseteq A$ with $dim_k(A/I)=k$ for some integer $k\geq 1$. Let $\mathfrak{p}_i:=(x_1-a(i)_1,\ldots ,x_n-a(i)_n)$ with $a(i)_j \in k$ It follows similarly (if we choose $d$ coprime maximal ideals $\mathfrak{p}_1,..,\mathfrak{p}_d$) there is an equality

$$\dim_k(A/\mathfrak{p}_i^{l_i+1})=\binom{l_i+n}{n},$$

and if $I:=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d+1}$

it follows

$$\dim_k(A/I)= \sum_j \binom{l_j+n}{n}.$$

Let $S(l_1,..,l_d,n,d,k)$ denote the set of unordered tuples $(l_1,..,l_d)$ with $l_i \geq 1$ and with

$$ \sum_j \binom{l_j+n}{n}=k.$$

Let $D(l_1,..,l_d,n,d,k)$ denote the set of elements in $S(l_1,..,l_d,n,d,k)$. I'm asking a similar question for the numbers $D(l_1,..,l_d,n,d,k)$.

Again I'm using an "imprecise" notation to indicate that these numbers arise when parametrizing ideals.

Question 2: 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 ones introduced above?

Answer 1

Question: Given an integer $1 \leq d \leq n$ we seek a combinatorial formula for the number $D(l_1,..,l_d)$ of unordered tuples of integers $(l_1,...,l_d)$ with $l_i\geq 1$ and $\sum_j \binom{l_j+1}{2}=n$. Do you know such a formula or a reference to where this type of formula is studied? I ask for an explicit reference to a study of this problem and such formulas in the litterature.

The notation $D(l_1, \ldots, l_d)$ makes little sense. Presumably you meant $D(n, d)$ or similar. (Edit: as I was writing this, you edited the question again. I won't try keeping up with a moving target.)

If you replaced $\binom{l_j+1}{2}$ with $l_j$, you would be asking for the number of integer partitions of $n$ of length $d$. These are extremely classical objects with a large variety of enumerative and other formulas, which the Wikipedia page summarizes nicely. However, in the comments you ask for an "explicit formula" for $D(n, d)$. While this is very vague, there is no such formula for integer partitions, nothing you can write in some nice closed form anyway. There's no chance in my mind that your problem will have a more explicit solution.

Nonetheless, you can likely use standard generating function approaches to get useful information here. What exactly is useful to you is not clear at present, but here's one observation. Let $m_i$ denote the number of $l_j$ equal to $i$. The corresponding "exponential notation" is $\{l_1, \ldots, l_d\} = 1^{m_1} 2^{m_2} \cdots n^{m_n}$. In general you want $m_1 + \cdots + m_n = d$ and $m_1 \binom{1+1}{2} + m_2 \binom{2+1}{2} + \cdots + m_n \binom{n+1}{2} = n$. (Of course $m_n=0$ here, well for $n>1$; this was just convenient.) The key thing is that the $m_i$ are arbitrary non-negative integers. Using the geometric series and standard manipulations with generating functions gives

\begin{align*} \sum_{n,d=0}^\infty D(n, d) x^n y^d = \prod_{i=1}^\infty \frac{1}{1 - yx^{\binom{i+1}{2}}}. \end{align*}

Answer 2

Answer: The following construction gives a relation with Question 2 and the Chinese Remainder Theorem (CRT) via Noether nomalization lemma (NNL). 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 euqlity $J_1\cdots J_d=(0)$. 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 ain inclusion $J_i \subseteq \mathfrak{p}_i$ it follow $I_i \subseteq \mathfrak{q}_i$. There is an integer $l_i$ with

$$J_i=\mathfrak{p}_{l_i+1} \subsetneq \mathfrak{p}_i^{l_i}$$

and it follows there are inclusions

$$\mathfrak{q}_i^{l_i+1} \subseteq I_i \subseteq \mathfrak{q}_i^{l_i}$$

Lemma. Given any cofinite ideal $I \subseteq A$, it follows there are maximal ideals $\mathfrak{q}_1,..,\mathfrak{q}_d$ and integers $l_1,..,l_d\geq 1$ 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.

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. We say such an ideal $I$ is an "$(\mathfrak{m},l)$-squeezed 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 space $\mathfrak{m}^l/\mathfrak{m}^{l+1}$ 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 all cofinite ideals $I \subseteq A$ of length $n$ as a subset

$$"Hilb^n(k[x,y])" \subseteq H(n).$$

Note: $"Hilb^n(k[x,y])"$ is a set, not a scheme.