Hilbert function of a monomial ideal generated by degree two square free monomials

Question

Let $R=K[x_1,...,x_n]$ be a polynomial ring over a field $K$ (one can assume $K$ is the field of complex numbers).
Let $I=\langle m_1,...,m_l\rangle= \oplus I_j$ (where $I_j$ is $j^{th}$ graded piece of $I$ and $m_i$ are degree two square free monomials).
What is $K$-dimension of $R_j/I_j ?$

Answer 1

I doubt you will get a satisfying answer. The reason is that it (e.g., the Hilbert-Poincaré series) depends deeply on the divisions between the (factors of the) $m_i$. The following works for explicit and not too long sequences $m_1,\dots m_l$, but it's not good enough for a general answer. $\newcommand{\HP}{\mathrm{HP}}$ We can and should make use of the following.

Let $A = K[x_1,\dots x_n]$, $\mathfrak{a}\subset A$ an ideal and $f\in A$ a homogenous element of degree $d$. Then $$\HP_{A/(\mathfrak{a}+(f))}(t) = \HP_{A/\mathfrak{a}}(t)-t^d\HP_{A/(\mathfrak{a}:(f))}(t).$$

Define $\mathfrak{a}_i := (m_i,m_{i+1},\dots m_l)$ and $\mathfrak{b}_i := \mathfrak{a}_i : (m_{i-1})$. Then we inductively get $$\HP_{A/\mathfrak{a}_1} = \HP_{A/\mathfrak{a}_2}(t)-t^2\HP_{A/(\mathfrak{a}_2:(m_1))}(t) = ...\\=\HP_{A}(t)-t^2\sum_{i=1}^l\HP_{A/\mathfrak{b}_{i+1}}(t).$$

Now if we try to compute $\HP_{A/\mathfrak{b}_{i}}(t)$, we get a similar formula, but the ideals involved become less predictable.

We can decompose $\mathfrak{b}_i = \mathfrak{a}_{i+1}:(m_i)$ into a degree one and a degree two part. The degree one part is generated by certain $x_k$, $k\in I_i$, and the degree two part is generated by $\mathfrak{a}_{i+1}$, again. However, if we only take the $m_j$, $j\geq i+1$, which are not divided by any $x_k$, $k\in I_i$, i.e., $\mathfrak{c}_{i} := (m_j\mid j\geq i+1;\,x_k\nmid m_j\forall k\in I_i)$, then $\mathfrak{c}_{i}:(x_k) = \mathfrak{c}_{i}$ and so $$\HP_{A/\mathfrak{b}_{i}}(t) = (1+|I_i|t)\HP_{A/\mathfrak{c}_{i}}.$$

We reduced the problem to ($l$ times) the same problem with a shorter ideal, generated by a subsequence of $m_1,\dots m_l$. This makes the above an algorithm, but it's not good enough to give a clean representation of the Hilbert-Poincaré series. It's the best I can do unless you tell me that the $m_i$ are a regular sequence or give me an explicit one.

For example, if $m_1,\dots m_l$ is a regular sequence, then $$\HP_{A/(m_1,\dots m_l)}(t) = \frac{(1-t^2)^l}{(1-t)^n}.$$ The $i$-th coefficient of $(1-t)^{-n}$ is $\binom{n+i-1}{n-1}$ and $(1-t^2)^l = \sum_{i=0}^l\binom{l}{i}(-1)^it^{2i}$. Therefore $$\dim_K((A/(m_1,\dots m_l))_k) = \sum_{i+2j=k}(-1)^{j}\binom{l}{j}\binom{n+i-1}{n-1}.$$

Answer 2

Like Ben, I do not think you will get a satisfactory answer. More precisely, I will show that computing the degree of the Hilbert polynomial is equivalent to computing the size of the maximal clique of a graph. Since the latter problem is NP-HARD, if you had a description of the Hilbert polynomial which was explicit and short enough to easily read off the degree, you would have solved MAX-CLIQUE and shown P=NP, which is not expected to be true.

Define a graph $\Gamma$ with vertices $\{ 1,2,\ldots, n \}$ and an edge $(i,j)$ if and only if $x_i x_j$ is NOT in the ideal $I$. Then a basis for $R/I$ is the monomials which are supported on cliques of $\Gamma$. Given a clique $C$, the number of degree $d$ monomials with support precisely $C$ is $\binom{d-1}{|C|-1}$. (For example, if $C = \{ i, j \}$, then we are talking about the $d-1$ monomials $x_1^{d-1} x_2$, $x_1^{d-2} x_2^2$, ..., $x_1 x_2^{d-1}$.)

Letting $\mathcal{C}$ be the set of cliques of $\Gamma$, we see that the hilbert Function is $$h(d) = \sum_{C \in \mathcal{C}} \binom{d-1}{|C|-1}$$ So a clique of size $C$ contributes a polynomial of degree $|C|-1$, and we see that the degree of the Hilbert polynomial is the size of the largest clique.

Answer 3

Let $R/I$ be a complete intersection where $I=(f_1,...,f_g)$. Then $$\text{HS}(R/I,t)= \frac{\displaystyle \prod_{i=1}^{g}(1-t^{e_i})}{(1-t)^{n+1}}$$ where $e_i := \deg(f_i)$.

And so, according to the above, the Hilbert function is really just the coefficient of $t^j$ in the given generating function (expand the product etc).