Quadratic ideals and Rogers–Ramanujan recursions

Abstract

We give an explicit recursive description of the Hilbert series and Gröbner bases for the family of quadratic ideals defining the jet schemes of a double point. We relate these recursions to the Rogers–Ramanujan identity and prove a conjecture of the second author, Oblomkov and Rasmussen.

1 Introduction

In this paper, we study a family of quadratic ideals defining the jet schemes for the double point $D={{\,\mathrm{Spec}\,}}\mathbf {k}[x]/x^2$. Here $\mathbf {k}$ is a field of characteristic zero. Recall that the $(n-1)$-jet scheme of X is defined as the space of formal maps ${{\,\mathrm{Spec}\,}}\mathbf {k}[t]/t^n\rightarrow X$ [11]. In the case of the double point, such a formal map is defined by a polynomial

$$\begin{aligned} x(t)=x_0+x_1t+\cdots +x_{n-1}t^{n-1}, \end{aligned}$$

such that $x(t)^2\equiv 0 \mod t^n$. By expanding this equation, we get a system of equations

$$\begin{aligned} f_1=x_0^2, f_2=2x_0x_1,\ldots , f_{n}=\sum _{i=0}^{n-1}x_{i}x_{n-1-i}. \end{aligned}$$

We denote the defining ideal of $\text {Jet}^{n-1}D\subseteq \mathbb {A}^n$ by

$$\begin{aligned} I_n:=\langle f_1,\ldots , f_n\rangle \subseteq R_n:=\mathbf {k}[x_0,\ldots ,x_{n-1}]. \end{aligned}$$

The ring $R_n$ is $\mathbb {Z}_{\ge 0}^2$-graded by assigning the grading (i, 1) to $x_i$. It is then clear that the ideal $I_n$ is bihomogeneous. Let

$$\begin{aligned} H_n(q,t)=\sum _{i,j\ge 0}\dim _k(R_n/I_n)_{i,j}q^it^j\in \mathbb {Z}[[q,t]] \end{aligned}$$

denote the bigraded Hilbert series for $R_n/I_n$. Our first main result is the following.

Theorem 1.1

The series $H_n(q,t)$ satisfies the recursion relation

$$\begin{aligned} H_n(q,t)=\frac{H_{n-2}(q,qt)+tH_{n-3}(q,q^2t)}{1-q^{n-1}t} \end{aligned}$$

with initial conditions

$$\begin{aligned} H_0(q,t)=1,\quad H_1(q,t)=1+t,\quad H_2(q,t)=\frac{1}{1 - qt} + t. \end{aligned}$$

Using this recursion relation, we obtain explicit combinatorial formulas for $H_n(q,t)$:

Theorem 1.2

The Hilbert series $H_n(q,t)$ is given by the following explicit formula:

$$\begin{aligned} H_n(q,t)=\sum _{p=0}^{\infty }\frac{\left( {\begin{array}{c}h(n,p)+1\\ p\end{array}}\right) _q\cdot q^{p(p-1)}t^p}{(1-q^{n-h(n,p)}t)\cdots (1-q^{n-1}t)}, \end{aligned}$$

where $h(n,p)=\lfloor \frac{n-p}{2}\rfloor$.

In the limit $n\rightarrow \infty$, we reprove the theorem of Bruschek et al. [4], which relates the Hilbert series of the arc space for the double point to the Rogers–Ramanujan identity. In fact, we refine their result by considering an additional grading, see Eq. (7.1) . Similar results for $n=\infty$ were obtained by Feigin–Stoyanovsky [8, 9], Lepowsky et al. [5, 6], and the second author, Oblomkov and Rasmussen in [10].

Remark 1.3

We note that in the Lie theoretic literature the variables of $R_n$ are often indexed $x_1, x_2,\ldots$ or $x_{-1}, x_{-2}, \ldots$ instead of our choice $x_0, x_1,\ldots$. We feel our normalization is more natural for purposes of commutative algebra.

Although our approach to the computation of the Hilbert series is inspired by [4], it is quite different. The key result in [4] shows that for $n=\infty$ the polynomials $f_k$ form a Gröbner basis of the ideal $I_{\infty }$. As shown below, the Gröbner basis of the ideal $I_n$ for finite n is larger and has a very subtle recursive structure. We completely describe such a basis in Theorems 4.2 and 4.6. In particular, we prove the following.

Theorem 1.4

Let $k>2$. Then the reduced Gröbner basis for $I_n$ contains $\left( {\begin{array}{c}\lfloor \frac{n-k+1}{2}\rfloor \\ k-2\end{array}}\right)$ polynomials of degree k.

Our proof of Theorem 1.1 does not use Gröbner bases at all. First, by an explicit inductive argument in Theorem 2.2 we give a complete description of the first syzygy module for $f_i$. Then, we define a “shift operator” $S: R_n\rightarrow R_{n+1}$, which sends $x_i$ to $x_{i+1}$, and identify $I_n\cap x_0R_n$ and $I_n/(I_n\cap x_0R_n)$ with the images of $I_{n-3}$ and $I_{n-2}$ under appropriate powers of S. This implies the recursion relation in Theorem 1.1.

Remark 1.5

The shift operator has a left inverse given by $x_i\mapsto x_{i-1}$ for $i\ge 1$. It can be extended to a derivation, and in the $n\rightarrow \infty$ limit this derivation has been successfully used in work of Capparelli–Lepowsky–Milas as well as Kanade [5, 12] to prove results similar to ours, using a backward induction argument. This is in contrast to our forward inductions which work for all n. We note also that in [4] a forward induction argument is used in the $n\rightarrow \infty$ limit. The representation-theoretic origin of the shift operators lies in the lattice part of the affine Weyl group of type $A_1$, but we do not pursue this connection further.

We also observe a recursive structure in the minimal free resolution of $R_n/I_n$. In particular, we prove the following:

Theorem 1.6

Let b(i, n) denote the rank of the ith term in the minimal free resolution for $R_n/I_n$, in other words the ith Betti number. Then

$$\begin{aligned} b(i,n)=b(i,n-1)+b(i-1,n-3)+b(i-2,n-3). \end{aligned}$$

As a consequence, we can compute the projective dimension of $R_n/I_n$.

Corollary 1.7

The projective dimension of $R_n/I_n$ equals $\lceil \frac{2n}{3}\rceil$.

Remark 1.8

It is easy to see that the reduced scheme $(\text {Jet}^{n-1}D)^{\text {red}}$ is a linear subspace given by the equations $x_0=\ldots =x_{\lfloor \frac{n-1}{2}\rfloor }=0$ and has dimension

$$\begin{aligned} \dim \text {Jet}^{n-1}D=n-1-\left\lfloor \frac{n-1}{2}\right\rfloor =\left\lceil \frac{n-1}{2}\right\rceil . \end{aligned}$$

A more careful analysis of the gradings in Theorem 1.6 implies another formula for the series $H_n(q,t)$ which was first conjectured in [10].

Theorem 1.9

The Hilbert series of $R_n/I_n$ has the following form:

$$\begin{aligned} H_n(q,t)= & {} \frac{1}{\prod _{i=0}^{n-1}(1-q^{i}t)}\sum _{p=0}^{\infty }(-1)^{p}\prod _{k=0}^{p-1}(1-q^{k}t) \\&\times \left( q^{\frac{5p^2-3p}{2}}t^{2p}\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) _q-q^{\frac{5p^2+5p}{2}}t^{2p+2}\left( {\begin{array}{c}n-2p-1\\ p\end{array}}\right) _q\right) . \end{aligned}$$

The paper is organized as follows. In Sect. 2 we introduce the shift operator S, describe its properties and prove Theorem 2.2 which explicitly describes all syzygies between the $f_i$. In Sect. 3, we use the shift operator to find a recursive relation for the Hilbert series and to prove Theorem 1.1. In Sect. 4, we use the recursive structure to describe a Gröbner basis for $I_n$. In Sect. 5, we give a recursive description of the minimal free resolution of $R_n/I_n$ and prove Theorem 1.6. In Sect. 6, we solve both of the above recursions explicitly (with the given initial conditions) and give two explicit combinatorial formulas for $H_n(q,t)$. Finally, in Sect. 7 we briefly discuss the limit of all these techniques at $n\rightarrow \infty$ and the connection to the Rogers–Ramanujan identity.

2 Ideals and syzygies

2.1 Ideals

Let $R_n=\mathbf {k}[x_0,\ldots ,x_{n-1}]$ and $f_k=\sum _{i=0}^{k-1}x_{i}x_{k-1-i}$. Define $I_n\subseteq R_n$ to be the ideal generated by $f_1,\ldots ,f_n$. Let $F_n$ be the free $R_n$-module with the basis $e_1,\ldots ,e_n$. Consider the map $\phi _n:F_n\rightarrow R_n$ given by the equation

$$\begin{aligned} \phi _n(\alpha _1,\ldots ,\alpha _n)=f_1\alpha _1+\ldots +f_n\alpha _n. \end{aligned}$$

The $R_n$-module ${{\,\mathrm{Ker}\,}}(\phi _n)$ is called the first syzygy module of $I_n$.

Lemma 2.1

One has

$$\begin{aligned} \sum _{i=0}^{n}(n-3i)x_if_{n+1-i}=0. \end{aligned}$$

(2.1)

Proof

Indeed,

$$\begin{aligned} \sum _{i=0}^{n}(n-3i)x_if_{n+1-i}=\sum _{i+k+l=n}(n-3i)x_ix_kx_l. \end{aligned}$$

The coefficient at each monomial $x_ix_kx_l$ equals

$$\begin{aligned}(n-3i)+(n-3k)+(n-3l)=3n-3(i+k+l)=3n-3n=0.\end{aligned}$$

$\square$

For $0<k<n$, define

$$\begin{aligned} \mu _k:=(-2kx_k,(-2k+3)x_{k-1},\ldots ,kx_0,0,\ldots ,0)\in F_n. \end{aligned}$$

By (2.1), we have $\phi _n(\mu _k)=0$. Denote also $\nu _{ij}=f_ie_j-f_je_i$ (for $i\ne j$). It is clear that $\phi _n(\nu _{ij})=0$. The main result of this section is the following.

Theorem 2.2

The first syzygy module ${{\,\mathrm{Ker}\,}}(\phi _n)$ is generated by $\mu _k$ and $\nu _{i,j}$ over $R_n$.

We prove Theorem 2.2 in Sect. 2.4.

Remark 2.3

For $n=\infty$ a similar result was obtained by Kanade [12], whose constructions are motivated by the affine Lie algebra $\widehat{\mathfrak {sl}_2}$. In particular, the ideals $I_n$ in Kanade’s picture, as well as in, e.g., Capparelli–Lepowsky–Milas [6], correspond to the principal subspaces of level 1 standard modules for $\widehat{\mathfrak {sl}_2}.$ See also [8, 10].

2.2 The shift operator

We define a ring homomorphism $S:R_n\rightarrow R_{n+1}$ by the equation $S(x_i)=x_{i+1}$. Note that S is injective and we can uniquely write any polynomial in $R_n$ in the form

$$\begin{aligned} f=x_0f'+S(f''),\ f'\in R_n,f''\in R_{n-1}. \end{aligned}$$

The following equation is clear from the definition and will be very useful:

$$\begin{aligned} f_n=2x_0x_{n-1}+S(f_{n-2}). \end{aligned}$$

(2.2)

By abuse of notation, denote also $S:F_n\rightarrow F_{n+2}$ the map which is given by

$$\begin{aligned} S(\alpha _1,\ldots ,\alpha _n)=(0,0,S(\alpha _1),\ldots ,S(\alpha _n)). \end{aligned}$$

(2.3)

Lemma 2.4

Let $\alpha \in F_n$. Then $\phi _{n+2}(S(\alpha ))$ is divisible by $x_0$ if and only if $\phi _n(\alpha )=0$.

Proof

By (2.2) we have

$$\begin{aligned} \phi _{n+2}(S(\alpha ))= \sum _{i=1}^{n} S(\alpha _i)f_{i+2} \equiv S\left( \sum _{i=1}^{n} \alpha _if_{i}\right) \mod x_0. \end{aligned}$$

Therefore $\phi _{n+2}(S(\alpha ))$ is divisible by $x_0$ if and only if $S(\sum \alpha _if_{i})$ is divisible by $x_0$. But since no shift contains $x_0$, this happens if and only if

$$\begin{aligned} S\left( \sum \alpha _if_{i}\right) =0\Leftrightarrow \sum \alpha _if_{i}=\phi _n(\alpha )=0. \end{aligned}$$

$\square$

Since $\phi _n(\mu _k)=\phi _{n}(\nu _{ij})=0$, by Lemma 2.4 the images of $S(\mu _k)$ and $S(\nu _{ij})$ under $\phi _{n+2}$ are divisible by $x_0$. The following lemma describes these images explicitly.

Lemma 2.5

One has $\phi _{n+2}(S(\mu _k))=(2k+6)x_{k+3}f_1+(2k+3)x_{k+2}f_2-(k+3)x_0f_{k+4}$, $\phi _{n+2}(S(\nu _{ij}))=2x_0x_{j+1}f_{i+2}-2x_0x_{i+1}f_{j+2}$.

Proof

By definition,

$$\begin{aligned} S(\mu _k)= & {} (0,0,-2kx_{k+1},(-2k+3)x_{k},\ldots ,kx_1,0,\ldots ,0) \\= & {} \mu _{k+3}+(2k+6)x_{k+3}e_1+(2k+3)x_{k+2}e_2-(k+3)x_0e_{k+4}, \end{aligned}$$

so

$$\begin{aligned} \phi _{n+2}(S(\mu _k))=(2k+6)x_{k+3}f_1+(2k+3)x_{k+2}f_2-(k+3)x_0f_{k+4}. \end{aligned}$$

Also, $S(\nu _{ij})=S(f_i)e_{j+2}-S(f_j)e_{i+2}$, so

$$\begin{aligned} \phi _{n+2}(S(\nu _{ij}))= & {} S(f_i)f_{j+2}-S(f_j)f_{i+2} \\= & {} (f_{i+2}-2x_0x_{i+1})f_{j+2}-(f_{j+2}-2x_0x_{j+1})f_{i+2}\\= & {} 2x_0x_{j+1}f_{i+2}-2x_0x_{i+1}f_{j+2}. \end{aligned}$$

$\square$

Corollary 2.6

One has

$$\begin{aligned} \phi _{n+2}(S(\mu _k))= & {} (2k+3)x_{k+2}f_2-(k+3)x_0S(f_{k+2}) \\= & {} kx_{k+2}f_{2}-(k+3)x_0S^2(f_{k}). \end{aligned}$$

Proof

$$\begin{aligned} \phi _{n+2}(S(\mu _k))= & {} (2k+6)x_{k+3}f_1+(2k+3)x_{k+2}f_2-(k+3)x_0f_{k+4} \\= & {} (2k+6)x_{k+3}f_1+(2k+3)x_{k+2}f_2 \\&\quad -\,(k+3)(2x_0^2x_{k+3}+2x_0x_1x_{k+2}+x_0S^2(f_{k}))\\= & {} (2k+3)x_{k+2}f_2-(k+3)x_0S(f_{k+2}) \\= & {} kx_{k+2}f_{2}-(k+3)x_0S^2(f_{k}). \end{aligned}$$

$\square$

Example 2.7

$\mu _1=(-2x_1,x_0)$, so $S(\mu _1)=(0,0,-2x_2,x_1)$, and

$$\begin{aligned} \phi _4(S(\mu _1))= & {} -\,2x_2(2x_0x_2+x_1^2)+x_1(2x_0x_3+2x_1x_2)\\= & {} 2x_3x_0x_1-4x_0x_2^2=x_3f_2-4x_0S^2(x_0^2). \end{aligned}$$

Lemma 2.8

The polynomial $x_1S(f_{n-2})$ can be expressed via $f_1,\ldots ,f_{n-1}$ modulo $x_0$.

Proof

We have $(n-3)x_0f_{n-2}+(n-6)x_1f_{n-3}+\cdots -2(n-3)x_{n-2}f_0=0$, so

$$\begin{aligned} (n-3)x_1S(f_{n-2})+(n-6)x_2S(f_{n-3})+\cdots -2(n-3)x_{n-1}S(f_0)=0. \end{aligned}$$

It remains to notice that $S(f_i)\equiv f_{i+2}\mod x_0$. $\square$

Lemma 2.9

Assume that ${{\,\mathrm{Ker}\,}}(\phi _{n-2})$ is generated by $\mu _k$ and $\nu _{i,j}$ and suppose that $\phi _n(\alpha )$ is divisible by $x_0$. Then $\alpha _n=Ax_0+Bx_1+\sum _{i=3}^{n-1} \gamma _i f_i$ for some A, B, and $\gamma _i$.

Proof

As stated above, we can write $\alpha _i=x_0\alpha '_i+S(\alpha ''_{i-2})$ for $i\ge 3$. Since $f_1$ and $f_2$ are divisible by $x_0$, we get

$$\begin{aligned} \phi _n(S(\alpha ''))=\sum _{i=3}^nS(\alpha ''_{i-2})f_i\equiv \sum _{i=1}^{n} \alpha _if_i\equiv 0\mod x_0. \end{aligned}$$

By Lemma 2.4 we get $\phi _{n-2}(\alpha '')=0$. By the assumption, we can write

$$\begin{aligned} \alpha ''=\sum _{k<n-2} \beta _k \mu _k+\sum _{i<j\le n-2}\gamma _{i,j}\nu _{ij}. \end{aligned}$$

Therefore

$$\begin{aligned} \alpha ''_{n-2}=\beta _{n-1}x_0+\sum _{j \le n-3}\gamma _{j,n-2}f_j, \end{aligned}$$

and

$$\begin{aligned} \alpha _n= & {} x_0\alpha '_n+S(\alpha ''_{n-2})=x_0\alpha '_n+S(\beta _{n-1})x_1 \\&+\sum _{j \le n-3}S(\gamma _{j,n-2})(f_{j+2}-2x_0x_{j+1}). \end{aligned}$$

$\square$

2.3 Examples

Before proving Theorem 2.2, we would like to present the proof for $n\le 4$.

Example 2.10

For $n=2$, we have $f_1=x_0^2$ and $f_2=2x_0x_1$, so the module of syzygies is clearly generated by $(-2x_1,x_0)=\mu _1$.

Example 2.11

Let $n=3$, suppose that $\alpha _1f_1+\alpha _2f_2+\alpha _3f_3=0$. We can write $\alpha _3=\alpha '_3x_0+\alpha ''_3$, where $\alpha ''_3$ does not contain $x_0$. Since $f_1$ and $f_2$ are divisible by $x_0$ and $f_3=2x_0x_2+x_1^2$, we get $x_1^2\alpha ''_3=0$, so $\alpha ''_3=0$. Now $\alpha =\frac{1}{2}\alpha '_3\mu _2+\gamma$, where $\gamma$ is a syzygy between $f_i$ with $\gamma _3=0$. By the previous example, $\gamma$ is a multiple of $\mu _1$, so the module of syzygies is actually generated by $\mu _1$ and $\mu _2$.

Example 2.12

Let $n=4$, suppose that $\alpha$ is a syzygy. We can write $\alpha _3=\alpha '_3x_0+\alpha ''_3$ and $\alpha _4=\alpha '_4x_0+\alpha ''_4$ where $\alpha ''_i$ do not contain $x_0$. Similarly to the previous case, we obtain

$$\begin{aligned} \alpha ''_3x_1^2+\alpha ''_4\cdot 2x_1x_2=0. \end{aligned}$$

(2.4)

This means that there exists some $\beta$ such that $\alpha ''_3=-2x_2\beta$ and $\alpha ''_4=x_1\beta$. Now

$$\begin{aligned}&\alpha _1x_0^2+\alpha _2\cdot 2x_0x_1+(\alpha '_3x_0-2x_2\beta )(2x_0x_2+x_1^2) \\&\quad +\,(\alpha '_4x_0+x_1\beta )(2x_0x_3+2x_1x_2)=0. \end{aligned}$$

The terms without $x_0$ cancel, and the linear terms in $x_0$ are the following:

$$\begin{aligned} x_0(2\alpha _2x_1+\alpha '_3x_1^2-4x_2^2\beta +2\alpha '4x_1x_2+2\beta x_1x_3)=0. \end{aligned}$$

Note that all terms but $-4x_2^2\beta$ are divisible by $x_1$, so $\beta$ is divisible by $x_1$, $\beta =m x_1$. Then

$$\begin{aligned} \alpha _4=\alpha '_4x_0+m x_1^2=(\alpha '_4-2x_2m)x_0+mf_3. \end{aligned}$$

By subtracting $m\nu _{3,4}+\frac{1}{3}(\alpha '_4-2x_2m)\mu _3$ from $\alpha$, we obtain a syzygy between $f_1,f_2,f_3$, and reduce to the previous case.

2.4 Syzygies

In this section, we prove Theorem 2.2 by induction on n. The base cases were covered in Sect. 2.3. Suppose that $\alpha =(\alpha _1,\ldots ,\alpha _n)\in {{\,\mathrm{Ker}\,}}(\phi _n)$, i.e., is a linear relation between $f_1,\ldots , f_n$. As stated above, write $\alpha _i=\alpha '_ix_0+S(\alpha ''_{i-2})$ for $i\ge 3$. Without loss of generality, we can assume that $\alpha '_i$ do not contain $x_0$ (otherwise we can subtract a multiple of $\nu _{1,i}$). Since

$$\begin{aligned} f_i=2x_0x_{i-1}+S(f_{i-2}), \end{aligned}$$

by collecting terms without $x_0$ we get $\sum _{i=3}^{n}S(\alpha ''_{i-2})S(f_{i-2})=0.$ This means that $\phi _{n-2}(\alpha '')=0$ and by the induction assumption we may then write

$$\begin{aligned} \alpha ^{\prime \prime }=\sum _{i=3}^{n-1} \beta _{i+1} \mu _{i-2}+\sum _{3\le j<k\le n,j\ne k} \beta _{j,k}\nu _{j-2,k-2}. \end{aligned}$$

Because

$$\begin{aligned} S(\nu _{j-2,k-2})=-S(f_{k-2})e_j+S(f_{j-2})e_k=\nu _{j,k}+2x_0x_{k}e_j-2x_0x_je_k, \end{aligned}$$

without loss of generality we can assume $\alpha ^{\prime \prime }=S(\sum _{i=3}^{n-1} \beta _{i+1} \mu _{i-2}).$ By Corollary 2.6 we get

$$\begin{aligned}\phi _n(S(\mu _{i-2}))=-(i+1)x_0S(f_{i})+(2i-1)x_{i-1}f_2,\end{aligned}$$

hence

$$\begin{aligned} \phi _n(\alpha )= & {} \alpha _1f_1+(\alpha _2+\sum _{i=3}^{n-1}(2i-1)S(\beta _{i+1})x_{i-1})f_2 \\&\quad +\sum _{i=3}^nx_0\alpha ^\prime _if_i-\sum _{i=3}^{n-1} (i+1)S(\beta _{i+1}) x_0S(f_i)=0. \end{aligned}$$

By collecting the terms linear in $x_0$, we get

$$\begin{aligned}&\bigg (\alpha _2+\sum _{i=3}^{n-1}(2i-1)S(\beta _{i+1})x_{i-1}\bigg )2x_1+\sum _{i=3}^n\alpha ^\prime _iS(f_{i-2}) \\&\qquad \quad -\sum _{i=3}^{n-1} (i+1)S(\beta _{i+1})S(f_i)=0, \end{aligned}$$

so

$$\begin{aligned} \sum _{i=3}^n\alpha ^\prime _iS(f_{i-2})-\sum _{i=3}^{n-1} (i+1)S(\beta _{i+1})S(f_i) \end{aligned}$$

is divisible by $x_1$, and

$$\begin{aligned} \sum _{i=3}^n\alpha '''_{i}f_{i-2}-\sum _{i=3}^{n-1} (i+1)\beta _{i+1}f_i \end{aligned}$$

is divisible by $x_0$, where $\alpha ^\prime _i=S(\alpha '''_i)$. By Lemma 2.9, this implies

$$\begin{aligned} \beta _n=Bx_0+Cx_1+\sum _{i=3}^{n-2} \gamma _i f_i \end{aligned}$$

for some constants B, C. Now we can rewrite

$$\begin{aligned} \alpha _n= & {} \alpha '_nx_0+S(\beta _nx_0)= \alpha '_nx_0+Bx_1^2+Cx_1x_2 \\&\quad +\sum _{i=3}^{n-3} \gamma _i x_1(f_{i+2}-2x_0x_{n-1})+\gamma _{n-2}x_1S(f_{n-2}). \end{aligned}$$

Observe that $x_1^2=f_3-2x_0x_2,x_1x_2=\frac{1}{2}(f_3-2x_0x_3)$ and by Lemma 2.8$x_1S(f_{n-2})$ can be expressed via $f_1,\ldots ,f_{n-1}$ modulo $x_0$. In other words,

$$\begin{aligned} \alpha _n=\delta x_0+\sum _{i=3}^{n-1}\delta _if_i \end{aligned}$$

for some coefficients $\delta _i$. Then $\alpha -\frac{1}{n-1}\delta \mu _{n-1}-\sum _{i=3}^{n-1}\delta _i \nu _{i,j}$ is a syzygy between $f_1,\ldots ,f_{n-1}$, so by the induction assumption it can be expressed as an $R_{n-1}$-linear combination of the $\mu _i$ and $\nu _{i,j}$.

Remark 2.13

The above proof shows that the syzygies $\nu _{1,k}$ and $\nu _{2,k}$ are not necessary, and can be expressed as linear combinations of other syzygies. Indeed, since the coefficients at $e_k$ are divisible by $x_0$, one can subtract an appropriate multiple of $\mu _{k-1}$ and get a syzygy involving $e_1,\ldots ,e_{k-1}$ only.

3 Hilbert series

In this section, we prove Theorem 3.5 by studying the relation between the ideals $I_n$ and $x_0R_n$.

Lemma 3.1

One has

$$\begin{aligned} R_n/(x_0R_n+I_n)\simeq S(R_{n-2}/I_{n-2})[x_{n-1}] \end{aligned}$$

as $R_{n}$-modules, the module structure on the right coming from $S: R_{n-1}\rightarrow R_n$.

Proof

We have $x_0R_n+I_n=\langle x_0,f_1,\ldots ,f_n\rangle =\langle x_0,S(f_1),\ldots ,S(f_{n-2})\rangle$, so

$$\begin{aligned} R_n/(x_0R_n+I_n)=R_n/\langle x_0,S(f_1),\ldots ,S(f_{n-2})\rangle =S(R_{n-2}/I_{n-2})[x_{n-1}]. \end{aligned}$$

$\square$

Lemma 3.2

The subspace $x_0S^2(I_{n-3})[x_{n-1}]$ does not intersect the ideal $\langle f_1,f_2\rangle$ in $R_n$. Furthermore, $x_0S^2(I_{n-3})[x_{n-1}]+\langle f_1,f_2\rangle$ is an ideal in $R_n$ which is contained in $I_n\cap x_0R_n$.

Proof

Given a non-zero polynomial $g\in I_{n-3}$, the iterated shift $S^2(g)$ does not contain $x_0$ or $x_1$, so that $x_0S^2(g)$ is not contained in $\langle f_1,f_2\rangle$. Furthermore, $I_{n-3}$ is stable under multiplication by $x_0,\ldots ,x_{n-4}$, so $S^2(I_{n-3})$ is stable under multiplication by $x_2,\ldots ,x_{n-2}$, and $x_0S^2(I_{n-3})[x_{n-1}]$ is stable under multiplication by $x_2,\ldots ,x_{n-1}$. Multiplication by $x_0$ or $x_1$ sends the latter subspace to $\langle f_1,f_2\rangle$, so $x_0S^2(I_{n-3})[x_{n-1}]+\langle f_1,f_2\rangle$ is an ideal in $R_n$.

Finally, to prove that this ideal is contained in $I_n$, it is sufficient to prove that $x_0S^2(f_k)\in I_n$ for $k\le n-3$. On the other hand, by Corollary 2.6:

$$\begin{aligned} x_0S^2(f_k)=\frac{1}{k+3}\phi _n(S(\mu _{k}))\mod \langle f_1,f_2\rangle . \end{aligned}$$

$\square$

Lemma 3.3

One has

$$\begin{aligned} I_n\cap x_0R_n=x_0S^2(I_{n-3})[x_{n-1}]+\langle f_1,f_2\rangle . \end{aligned}$$

Proof

By Lemma 3.2, the right-hand side is a submodule of the left-hand side, so it remains to prove the reverse inclusion. We have

$$\begin{aligned} f_i=2x_0x_{i-1}+S(f_{i-2})=2x_0x_{i-1}+2x_1x_{i-2}+S^2(f_{i-4}). \end{aligned}$$

Suppose that $\sum _{i=1}^{n}\alpha _if_i\in I_n\cap x_0R_n$. Then by Lemma 2.9,

$$\begin{aligned} \alpha _n=Ax_0+Bx_1+\sum _{j}\gamma _jf_j=A'x_0+B'x_1+\sum _{j}\gamma _jS^2(f_{j-4}). \end{aligned}$$

Now by (2.1) and Corollary 2.6, $x_0f_n$ and $x_1f_n$ can be expressed as $R_n$-linear combinations of $f_1,\ldots ,f_{n-1}$ and elements of $x_0S^2(I_{n-3})[x_{n-1}]+\langle f_1,f_2\rangle$, so $\sum _{i=1}^{n}\alpha _if_i$ can be expressed as such a combination as well. Induction on n finishes the proof. $\square$

Corollary 3.4

One has

$$\begin{aligned} x_0R_n/(I_n\cap x_0R_n)=x_0S^2(R_{n-3}/I_{n-3})[x_{n-1}]. \end{aligned}$$

Proof

We have

$$\begin{aligned} x_0R_n/\langle f_1,f_2\rangle =x_0R_n/(x_0^2,x_0x_1)=x_0\mathbf {k}[x_2,\ldots ,x_{n-1}]=x_0S^2(R_{n-3})[x_{n-1}]. \end{aligned}$$

Therefore

$$\begin{aligned} x_0R_n/(I_n\cap x_0R_n)= & {} x_0R_n/(x_0S^2(I_{n-3})[x_{n-1}]+\langle f_1,f_2\rangle ) \\= & {} x_0S^2(R_{n-3}/I_{n-3})[x_{n-1}]. \end{aligned}$$

$\square$

Theorem 3.5

Let $H_n(q,t)$ denote the bigraded Hilbert series of the quotient $R_n/I_n$. Then one has the following recursion relation:

$$\begin{aligned} H_n(q,t)=\frac{H_{n-2}(q,qt)+tH_{n-3}(q,q^2t)}{1-q^{n-1}t} \end{aligned}$$

(3.1)

with initial conditions

$$\begin{aligned} H_0(q,t)=1,\quad H_1(q,t)=1+t,\quad H_2(q,t)=\frac{1}{1 - qt} + t. \end{aligned}$$

Remark 3.6

This recursion is similar, but not identical to the various recursions considered by Andrews [1,2,3] in his proofs of the Rogers–Ramanujan identity. It is also similar to the recursions recently considered by Paramonov [13] in a different context. In the $n\rightarrow \infty$ limit, Capparelli–Lepowsky–Milas [5] use analogous exact sequences for principal level 1 subspaces of the standard modules of $\widehat{\mathfrak {sl}_2}$ to arrive at a similar formula.

Proof

We have an exact sequence

$$\begin{aligned} 0\rightarrow x_0R_n/(x_0R_n\cap I_n)\rightarrow R_n/I_n\rightarrow R_n/(x_0R_n+I_n)\rightarrow 0. \end{aligned}$$

By Lemma 3.1, the Hilbert series of $R_n/(x_0R_n+I_n)$ equals $\frac{H_{n-2}(q,qt)}{1-q^{n-1}t}$, and by Corollary 3.4 the Hilbert series of $x_0R_n/(x_0R_n\cap I_n)$ equals $\frac{tH_{n-3}(q,q^2t)}{1-q^{n-1}t}$. $\square$

4 Gröbner bases

We will now compute Gröbner bases for the ideals $I_n$. Recall that a Gröbner basis for an ideal I is a subset $G=\{g_1,\ldots ,g_s\}\subset I$ such that, for a chosen monomial ordering <,

$$\begin{aligned}\langle {{\,\mathrm{LT_<}\,}}(g_1),\ldots , {{\,\mathrm{LT_<}\,}}(g_s)\rangle ={{\,\mathrm{LT_<}\,}}(I),\end{aligned}$$

where ${{\,\mathrm{LT_<}\,}}$ denotes leading term.

Let us order the monomials in $R_n$ in grevlex order, that is

$$\begin{aligned} x^\alpha <x^\beta \end{aligned}$$

if $|\alpha |<|\beta |$ or $|\alpha |=|\beta |$ and the rightmost entry of $\alpha -\beta$ is negative.

Remark 4.1

In fact, any order refining the reverse lexicographic order will work, but for definiteness and its popularity in computer algebra systems we shall fix grevlex order throughout.

Theorem 4.2

Let

$$\begin{aligned} G_1=\{f_1\}\subseteq R_1, G_2=\{f_1, f_2\}\subset R_2 \end{aligned}$$

and recursively define the sets $G_n, n\ge 3$ as follows:

$$\begin{aligned} G_n=x_0S^2(G_{n-3})\sqcup \{f_1,f_2\} \sqcup \widetilde{S}(G_{n-2}), \end{aligned}$$

where $\widetilde{S}$ is a modified shift operator as explained below. Then $G_n$ is a Gröbner basis for $I_n$.

Remark 4.3

The notation requires explanation. Note that any $G_m$ is naturally a subset of $R_n$, $n\ge m$ so we can and will identify $G_m$ inside a larger polynomial ring without explicit mention. Furthermore, we denote by $x_0S^2(G_{n-3})$ the image of $G_{n-3}$ under $S^2: R_{n-2}\rightarrow R_n$ multiplied by $x_0$. The “operator” $\widetilde{S}$ is defined on elements $p\in I_{n-2}$ as follows: write $p=\sum _{i=1}^n \varphi _i f_i$, and let

$$\begin{aligned}\widetilde{S}(p)=\sum _{i=1}^n S(\varphi _i) f_{i+2}.\end{aligned}$$

Note that by (2.2), we have $\widetilde{S}(p)=S(p)+\sum _{i=1}^nx_0x_{i+2}S(\varphi _i)\in I_{n+2}$. In particular, if $p\ne 0$ and p is homogeneous then ${{\,\mathrm{LT_<}\,}}(\widetilde{S}(p))=S({{\,\mathrm{LT_<}\,}}(p))$. Therefore, the construction of $\widetilde{S}(p)$ requires a choice if $\varphi _i$, but the leading term of the result does not depend on this choice.

Proof

We will proceed by induction. The base cases $n=1,2$ are clear because the ideals are monomial. Consider now the ideal ${{\,\mathrm{LT_<}\,}}(I_n)$ generated by all the leading terms of elements of $I_n$. It is clear by Lemma 3.1 and the fact that S respects the reverse lexicographic order that if $g\in I_n$ is not divisible by $x_0$, its leading term is the image of a leading term in $I_{n-2}$ under S. Since we assumed $G_{n-2}$ to be a Gröbner basis, we must have ${{\,\mathrm{LT_<}\,}}(g)$ divisible by some monomial in $S({{\,\mathrm{LT_<}\,}}(G_{n-2}))$.

Similarly, if g is divisible by $x_0$, we know by Lemma 3.2 and order preservation that its leading term is the image under $x_0S^2$ of a leading term in $I_{n-3}$ or divisible by $f_1, f_2$. By the induction assumption ${{\,\mathrm{LT_<}\,}}(g)$ is then divisible by an element of $x_0S^2({{\,\mathrm{LT_<}\,}}(G_{n-3}))\sqcup \{f_1, f_2\}$. In particular, ${{\,\mathrm{LT_<}\,}}(I_n)\subseteq \langle {{\,\mathrm{LT_<}\,}}(G_n)\rangle$. But the reverse inclusion is clear, so we have

$$\begin{aligned} {{\,\mathrm{LT_<}\,}}(I_n)=\langle {{\,\mathrm{LT_<}\,}}(G_n)\rangle \end{aligned}$$

as desired, and $G_n$ is a Gröbner basis for $I_n$. $\square$

Example 4.4

We have

$$\begin{aligned} G_3&=\{f_1,f_2,f_3\},\\ G_4&=\{f_1,f_2,f_3,f_4,x_0x_2^2\},\\ G_5&=\{f_1,f_2,f_3,f_4,f_5,x_0x_2x_3\},\\ G_6&=\{f_1,\ldots , f_6, x_0x_3^2+2 x_0x_2x_4, 2 x_1x_3^2+3 x_0x_3x_4-x_0x_2x_5\}. \end{aligned}$$

Note that the last polynomial in $G_6$ can be identified with $\widetilde{S}(x_0x_2^2)\in \widetilde{S}(G_4)$. Indeed,

$$\begin{aligned} 4x_0x_2^2= & {} 2x_2(2x_0x_2+x_1^2)-x_1(2x_0x_3+2x_1x_2)+x_3(2x_0x_1) \\= & {} 2x_2f_3-x_1f_4+x_3f_2, \end{aligned}$$

so

$$\begin{aligned} \widetilde{S}(4x_0x_2^2)= & {} 2x_3f_5-x_2f_6+x_4f_4 \\= & {} 2x_3(2x_0x_4+2x_1x_3+x_2^2)-x_2(2x_0x_5+2x_1x_4+2x_2x_3) \\&\quad +\,x_4(2x_0x_3+2x_1x_2) \\= & {} 4x_1x_3^2+6x_0x_3x_4-2x_0x_2x_5. \end{aligned}$$

Remark 4.5

The Gröbner basis constructed in Theorem 4.2 is far from being reduced. The following theorem describes the reduced basis implicitly.

Since all $G_n$ contain $\{f_1,\ldots , f_n\}$ and none of their leading terms divides one another, we can throw away other polynomials in $G_n$ in a controlled manner to obtain a minimal Gröbner basis. That is to say, if the leading terms of $G_n\backslash \{g\}$ still generate the leading ideal we are in business. Therefore after appropriate reduction [7, Proposition 6 on p. 92] we get a reduced Gröbner basis with the same leading terms.

Let us call a monomial $\prod x_i^{a_i}$admissible if $a_i+a_{i+1}\le 1$ for all i, that is, it is not divisible by $x_i^2$ or by $x_ix_{i+1}$.

Theorem 4.6

Fix $k>2$. The leading terms of (t-)degree k in a reduced Gröbner basis for $I_n$ have the form $m(x){{\,\mathrm{LT_<}\,}}(f_{n+k-2})$ where m(x) is an admissible monomial of degree $k-2$ in variables $x_0,\ldots ,x_{\lfloor \frac{n+k-7}{2}\rfloor }$. The number of degree k polynomials in the reduced Gröbner basis equals $\left( {\begin{array}{c}\lfloor \frac{n-k+1}{2}\rfloor \\ k-2\end{array}}\right)$.

Remark 4.7

It is easy to see that there are no linear polynomials in the Gröbner basis (or in the ideal $I_n$), and $f_1,\ldots ,f_n$ are the only quadratic polynomials in the reduced Gröbner basis.

Proof

We prove the statement by induction in n. Suppose that it is true for $G_{n-2}$ and $G_{n-3}$. By Theorem 4.2, the leading monomials in the degree k part of $G_n$ consist of shifted degree k monomials in $G_{n-2}$, and twice shifted degree $(k-1)$ monomials in $G_{n-3}$, multiplied by $x_0$.

Consider first the case $k=3$. We will prove that the leading terms in the reduced Gröbner basis have the form $x_j{{\,\mathrm{LT_<}\,}}(f_{n+1})$ for $j\le \lfloor \frac{n-4}{2}\rfloor$. Indeed, in the first case, we get $S(x_j{{\,\mathrm{LT_<}\,}}(f_{(n-2)+1}))=x_{j+1}{{\,\mathrm{LT_<}\,}}(f_{n+1}).$ In the second case, we have to consider the polynomials $x_0S^2(f_i)$ for all $i\le n-3$. Observe that for $i\le n-4$ we get ${{\,\mathrm{LT_<}\,}}(x_0S^2(f_i))=x_0{{\,\mathrm{LT_<}\,}}(f_{i+4})$ and hence divisible by the leading term of $f_{i+4}$ and can be eliminated. For $i=n-3$ we get ${{\,\mathrm{LT_<}\,}}(x_0S^2(f_{n-3}))=x_0{{\,\mathrm{LT_<}\,}}(f_{n+1})$.

Assume now that $k>3$. In the first case, we get

$$\begin{aligned} S(m(x){{\,\mathrm{LT_<}\,}}(f_{(n-2)+k-2}))=S(m(x)){{\,\mathrm{LT_<}\,}}(f_{n+k-2}). \end{aligned}$$

If m(x) is an admissible monomial in $x_j$, $0\le j\le \lfloor \frac{(n-2)+k-7}{2}\rfloor$ then S(m(x)) is an admissible monomial in $x_j$, $1\le j\le \lfloor \frac{(n-2)+k-7}{2}\rfloor +1=\lfloor \frac{n+k-7}{2}\rfloor .$

In the second case, we get

$$\begin{aligned} x_0S^2(m(x)){{\,\mathrm{LT_<}\,}}(f_{(n-3)+(k-1)-2}))=x_0S^2(m(x)){{\,\mathrm{LT_<}\,}}(f_{n+k-2}). \end{aligned}$$

Now $S^2(m(x))$ is an admissible monomial in $x_j$, $2\le j\le \lfloor \frac{(n-3)+(k-1)-7}{2}\rfloor +2=\lfloor \frac{n+k-7}{2}\rfloor$, so $x_0S^2(m(x))$ is also an admissible in a correct set of variables. In fact, all such monomials not divisible by $x_0$ appear from the first case, and the ones divisible by $x_0$ appear from the second case.

It is easy to see that none of these leading monomials are divisible by each other. Therefore after appropriate reduction [7] we get a reduced Gröbner basis with the same leading terms.

Finally, we can count monomials of given degree k. The number of admissible monomials of degree l in s variables equals $\left( {\begin{array}{c}s-l+1\\ l\end{array}}\right)$, so the number of polynomials in $G_n$ of degree k equals

$$\begin{aligned} \left( {\begin{array}{c}1+\lfloor \frac{n+k-7}{2}\rfloor -(k-2)+1\\ k-2\end{array}}\right) =\left( {\begin{array}{c}\lfloor \frac{n-k+1}{2}\rfloor \\ k-2\end{array}}\right) . \end{aligned}$$

$\square$

Example 4.8

Let $n=12$. The reduced Gröbner basis for $I_{12}$ contains quadratic polynomials $f_1,\ldots ,f_{12}$. It also contains 5 cubic polynomials with leading terms

$$\begin{aligned} x_0x_6^2,x_1x_6^2,x_2x_6^2,x_3x_6^2,x_4x_6^2, \end{aligned}$$

6 quartic polynomials with leading terms

$$\begin{aligned} x_0x_2x_6x_7,x_0x_3x_6x_7,x_0x_4x_6x_7,x_1x_3x_6x_7,x_1x_4x_6x_7,x_2x_4x_6x_7, \end{aligned}$$

and 4 quintic polynomials with leading terms

$$\begin{aligned} x_0x_2x_4x_7^2,x_0x_2x_5x_7^2,x_0x_3x_5x_7^2,x_1x_3x_5x_7^2. \end{aligned}$$

Observe that ${{\,\mathrm{LT_<}\,}}(f_{13})=x_6^2,{{\,\mathrm{LT_<}\,}}(f_{14})=x_6x_7$, and ${{\,\mathrm{LT_<}\,}}(f_{15})=x_7^2$.

5 Minimal resolution

In this section, we describe the bigraded minimal free resolutions of $I_n$ and $R_n/I_n$. We write them as follows:

$$\begin{aligned} 0\leftarrow I_n\xleftarrow {} F(1,n)\xleftarrow {} F(2,n)\xleftarrow {} F(3,n)\cdots \end{aligned}$$

and

$$\begin{aligned} 0\leftarrow R_n/I_n\xleftarrow {} R_n=F(0,n)\xleftarrow {} F(1,n)\xleftarrow {} F(2,n)\xleftarrow {} F(3,n)\cdots \end{aligned}$$

Theorem 5.1

Let F(i, n) be the i-th term in the minimal free resolution for $I_n$. Then there is an injection $F(i,n-1)\hookrightarrow F(i,n)$, and

$$\begin{aligned} F(i,n)/F(i,n-1)\simeq S(F(i-1,n-3))\oplus x_0S(F(i-2,n-3)) \end{aligned}$$

as $R_n$-modules, and the shift of a free $R_n$-module is as in (2.3). Note that the gradings in the right-hand side are shifted by the bidegree of $f_n$ (which equals $q^{n-1}t^2$).

Proof

Observe that the ideal generated by $f_1,\ldots ,f_{n-1}$ in $R_n$ is isomorphic to $I_{n-1}[x_{n-1}]$, so its minimal resolution over $R_n$ is identical to the one for $I_{n-1}$ over $R_{n-1}$ tensored over $R_{n}$. Moreover, since $I_n=\langle f_1,\ldots , f_n\rangle$, the minimal free $R_n$-resolution of $I_{n-1}[x_{n-1}]$ is naturally a subcomplex of the minimal free resolution for $I_n$. In other words, $F(i,n-1)\otimes _{R_{n-1}}R_n$ can be identified with a subspace in F(i, n), which we will by abuse of notation also denote $F(i,n-1)$. We have a short exact sequence

$$\begin{aligned} 0\rightarrow F(i,n-1)\rightarrow F(i,n)\rightarrow F(i,n)/F(i,n-1)\rightarrow 0. \end{aligned}$$

From the long exact sequence in cohomology, it is easy to see that $F(i,n)/F(i,n-1)$ is acyclic in positive degrees. Now $I_n=\langle f_1,\ldots f_n\rangle$, so $F(1,n)/F(1,n-1)\cong R_n$ is generated by a single-vector $\tilde{f}_n$ corresponding to $f_n$. Furthermore, by Theorem 2.2F(2, n) has generators corresponding to $\mu _{1},\ldots ,\mu _{n-1}$ and $\nu _{i,j}$ for $3\le i<j\le n$, so $F(2,n)/F(2,n-1)\cong R_n^{n-2}$ is spanned by the basis elements corresponding to $\mu _{n-1}$ and $\nu _{i,n}$ for $3\le i\le n-1$. The differential $d:F(2,n)\rightarrow F(1,n)$ descends to $\tilde{d}: F(2,n)/F(2,n-1) \rightarrow F(1,n)/F(1,n-1)$. It sends $\mu _{n-1}$ to $x_0\tilde{f}_n$ and $\nu _{i,n}$ to $f_{i}\cdot \tilde{f}_n$.

Therefore, the quotient complex with terms $F(i,n)/F(i,n-1)$ is isomorphic to the minimal resolution of $R_n/\langle x_0,f_3,\ldots ,f_{n-1}\rangle =R_n/\langle x_0,S(f_1),\ldots ,S(f_{n-3})\rangle$. The latter is nothing but the (shifted) minimal resolution for $I_{n-3}$ tensored with the two-term complex $R_n\xleftarrow {x_0} R_n$. $\square$

Corollary 5.2

Let b(i, n) denote the rank of F(i, n). Then

$$\begin{aligned} b(i,n)=b(i,n-1)+b(i-1,n-3)+b(i-2,n-3). \end{aligned}$$

(5.1)

Corollary 5.3

Let $H_n(q,t)$ denote the Hilbert series for $R_n/I_n$, and let $\widetilde{H}_n(q,t)=H_n(q,t)\prod _{i=0}^{n-1}(1-q^{i}t).$ Then $\widetilde{H}_n(q,t)$ satisfies the following recursion relation:

$$\begin{aligned} \widetilde{H}_n(q,t)=\widetilde{H}_{n-1}(q,t)-q^{n-1}t^2(1-t^2)\widetilde{H}_{n-3}(q,qt). \end{aligned}$$

(5.2)

Corollary 5.4

The projective dimension of $I_n$ equals $\lceil \frac{2n}{3}\rceil -1.$ The projective dimension of $R_n/I_n$ equals $\lceil \frac{2n}{3}\rceil$.

Proof

By definition, the projective dimension ${{\,\mathrm{pd}\,}}(I_n)$ is equal to the length of the minimal free (or projective) resolution. By (5.1) we have ${{\,\mathrm{pd}\,}}(I_n)={{\,\mathrm{pd}\,}}(I_{n-3})+2$. The minimal free resolutions for $I_1$, $I_2$, and $I_3$ are easy to compute:

$$\begin{aligned}&I_1\xleftarrow {\begin{pmatrix}f_1\end{pmatrix}} R_1 \\&I_2\xleftarrow {\begin{pmatrix}f_1&f_2\end{pmatrix}} R_2^2\xleftarrow {\begin{pmatrix} -2x_1\\ x_0 \end{pmatrix}} R_2 \\&I_3\xleftarrow {\begin{pmatrix} f_1&f_2&f_3 \end{pmatrix}} R_3^3\xleftarrow {\begin{pmatrix} -2x_0 &{} -4x_2 \\ x_1 &{} -x_1 \\ 0 &{} 2x_0 \end{pmatrix}} R_3^2. \end{aligned}$$

The minimal resolution of $R_n/I_n$ is one step longer than the one for $I_n$. $\square$

6 Combinatorial identities

We define

$$\begin{aligned} \left( {\begin{array}{c}a\\ b\end{array}}\right) _q=\frac{(1-q)\cdots (1-q^a)}{(1-q)\cdots (1-q^b)\cdot (1-q)\cdots (1-q^{a-b})}. \end{aligned}$$

If $a<b$, we set $\left( {\begin{array}{c}a\\ b\end{array}}\right) _q=0$. The following lemma is well known.

Lemma 6.1

The following identities holds:

$$\begin{aligned} \left( {\begin{array}{c}a\\ b\end{array}}\right) _q+q^{b+1}\left( {\begin{array}{c}a\\ b+1\end{array}}\right) _q=\left( {\begin{array}{c}a+1\\ b+1\end{array}}\right) _q=q^{a-b}\left( {\begin{array}{c}a\\ b\end{array}}\right) _q+\left( {\begin{array}{c}a\\ b+1\end{array}}\right) _q. \end{aligned}$$

Proof

One has

$$\begin{aligned} \left( {\begin{array}{c}a\\ b+1\end{array}}\right) _q=\frac{(1-q^{a-b})}{(1-q^{b+1})}\left( {\begin{array}{c}a\\ b\end{array}}\right) _q, \end{aligned}$$

hence

$$\begin{aligned} \left( {\begin{array}{c}a\\ b\end{array}}\right) _q+q^{b+1}\left( {\begin{array}{c}a\\ b+1\end{array}}\right) _q= & {} \left( {\begin{array}{c}a\\ b\end{array}}\right) _q\left( 1+q^{b+1}\frac{(1-q^{a-b})}{(1-q^{b+1})}\right) \\= & {} \left( {\begin{array}{c}a\\ b\end{array}}\right) _q\frac{(1-q^{a+1})}{(1-q^{b+1})}=\left( {\begin{array}{c}a+1\\ b+1\end{array}}\right) _q. \end{aligned}$$

$\square$

Theorem 6.2

The Hilbert series $H_n(q,t)$ is given by the following explicit formula:

$$\begin{aligned} H_n(q,t)=\sum _{p=0}^{\infty }\frac{\left( {\begin{array}{c}h(n,p)+1\\ p\end{array}}\right) _q\cdot q^{p(p-1)}t^p}{(1-q^{n-h(n,p)}t)\cdots (1-q^{n-1}t)}, \end{aligned}$$

(6.1)

where $h(n,p)=\lfloor \frac{n-p}{2}\rfloor$.

Proof

By Theorem 3.5 it is sufficient to prove that the right-hand side of (6.1) satisfies the recursion relation (3.1). Let us denote the p-th term in (6.1) by $H_{n,p}(q,t)$ so that $H_n(q,t)=\sum _{p}H_{n,p}(q,t)$. We have $h(n-2,p)=h(n-3,p-1)=h(n,p)-1$, so

$$\begin{aligned} H_{n-2,p}(q,qt)= & {} \frac{\left( {\begin{array}{c}h(n,p)\\ p\end{array}}\right) _q\cdot q^{p(p-1)}t^p\cdot q^p}{(1-q^{n-h(n,p)}t)\cdots (1-q^{n-2}t)}, \\ H_{n-3,p-1}(q,q^2t)= & {} \frac{\left( {\begin{array}{c}h(n,p)\\ p-1\end{array}}\right) _q\cdot q^{(p-1)(p-2)}t^{p-1}\cdot q^{2p-2}}{(1-q^{n-h(n,p)}t)\cdots (1-q^{n-2}t)}, \end{aligned}$$

therefore

$$\begin{aligned}&H_{n-2,p}(q,qt)+tH_{n-3,p-1}(q,q^2t) \nonumber \\&\qquad \quad = \frac{q^{p(p-1)}t^p}{(1-q^{n-h(n,p)}t)\cdots (1-q^{n-2}t)}\left[ q^p\left( {\begin{array}{c}h(n,p)\\ p\end{array}}\right) _q+\left( {\begin{array}{c}h(n,p)\\ p-1\end{array}}\right) _q\right] \nonumber \\&\qquad \quad = \frac{q^{p(p-1)}t^p}{(1-q^{n-h(n,p)}t)\cdots (1-q^{n-2}t)}\left( {\begin{array}{c}h(n,p)+1\\ p\end{array}}\right) _q \nonumber \\&\qquad \quad = (1-q^{n-1}t)H_{n,p}(q,t). \end{aligned}$$

(6.2)

This proves (3.1), and the initial conditions are easy to check. $\square$

The free resolution of $I_n$ gives another formula for the Hilbert series of $R_n/I_n$.

Proposition 6.3

Let b(i, n), as stated above, denote the rank of i-th module in the free resolution of $R_n/I_n$. Then

$$\begin{aligned} b(i,n)=\sum _{p}\left[ \left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p\end{array}}\right) +\left( {\begin{array}{c}n-2p-1\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p-1\end{array}}\right) \right] . \end{aligned}$$

Remark 6.4

The terms in the first sum are non-zero if $p\le (n+1)/3$ and $i/2\le p\le i.$ The terms in the second sum are non-zero if $p\le (n-1)/3$ and $(i-1)/2\le p\le (i-1).$

Proof

Let

$$\begin{aligned} A(n,p,i)=\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p\end{array}}\right) , B(n,p,i)=\left( {\begin{array}{c}n-2p-1\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p-1\end{array}}\right) . \end{aligned}$$

Then

$$\begin{aligned}&A(n-1,p,i)+A(n-3,p-1,i-1)+A(n-3,p-1,i-2) \\&\qquad \qquad = \left( {\begin{array}{c}n-2p\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p\end{array}}\right) +\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) \left( {\begin{array}{c}p-1\\ i-p\end{array}}\right) +\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) \left( {\begin{array}{c}p-1\\ i-p-1\end{array}}\right) \\&\qquad \qquad = \left( {\begin{array}{c}n-2p\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p\end{array}}\right) +\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) \left( {\begin{array}{c}p\\ i-p\end{array}}\right) \\&\qquad \qquad =\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ i-p\end{array}}\right) =A(n,p,i). \end{aligned}$$

Similarly, $B(n-1,p,i)+B(n-3,p-1,i-1)+B(n-3,p-1,i-2)=B(n,p,i)$, so the right-hand side satisfies the recursion relation (5.1). It remains to check the base cases:

$$\begin{aligned} f(0,n)= & {} 1=\left( {\begin{array}{c}n-1\\ 0\end{array}}\right) , \\ f(1,n)= & {} n=\left( {\begin{array}{c}n-1\\ 1\end{array}}\right) +\left( {\begin{array}{c}n-3\\ 0\end{array}}\right) , \\ f(2,n)= & {} (n-1)+\left( {\begin{array}{c}n-2\\ 2\end{array}}\right) =\left( {\begin{array}{c}n-1\\ 1\end{array}}\right) +\left( {\begin{array}{c}n-3\\ 1\end{array}}\right) +\left( {\begin{array}{c}n-3\\ 2\end{array}}\right) . \end{aligned}$$

By Corollary 5.4$b(i,n)=0$ for $i>2$ and $n\le 3$. $\square$

We have the following (q, t)-analog of Proposition 6.3.

Proposition 6.5

Let $\widehat{b}(i,n)$ denote the bigraded Hilbert polynomial for the generating set in F(i, n). Then

$$\begin{aligned} \widehat{b}(i,n)= & {} \sum _{p>0}q^{\frac{5p^2-3p+(i-p)(i-p-1)}{2}}t^{2p+(i-p)}\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q \nonumber \\&+\, q^{\frac{5p^2+5p+(i-p)(i-p-1)}{2}}t^{2p+2+(i-p)}\left( {\begin{array}{c}n-2p-1\\ p\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p-1\end{array}}\right) _q. \end{aligned}$$

(6.3)

Proof

The proof is completely analogous to the proof of Proposition 6.3, but we include it here for completeness. By Theorem 5.1 we have a recursion relation

$$\begin{aligned} \widehat{b}(i,n)= & {} \widehat{b}(i,n-1)+q^{n-1}t^2\widehat{b}(i-1,n-3)(q,qt) \nonumber \\&\quad +\,q^{n-1}t^3\widehat{b}(i-2,n-3)(q,qt). \end{aligned}$$

(6.4)

We need to prove that the right-hand side of (6.3) satisfies (6.4). Let

$$\begin{aligned} \widehat{A}(n,p,i)=q^{\frac{5p^2-3p+(i-p)(i-p-1)}{2}}t^{2p+(i-p)}\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q. \end{aligned}$$

Then

$$\begin{aligned} \widehat{A}(n-3,p-1,i-1)(q,qt)= & {} q^{\frac{5p^2-9p+4+(i-p)(i-p+1)}{2}}t^{2p-2+(i-p)} \\&\quad \times \,\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) _q\left( {\begin{array}{c}p-1\\ i-p\end{array}}\right) _q, \\ \widehat{A}(n-3,p-1,i-2)(q,qt)= & {} q^{\frac{5p^2-9p+4+(i-p)(i-p-1)}{2}}t^{2p-2+(i-p-1)} \\&\quad \times \,\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) _q\left( {\begin{array}{c}p-1\\ i-p-1\end{array}}\right) _q, \end{aligned}$$

so

$$\begin{aligned}&\widehat{A}(n-3,p-1,i-1)(q,qt)+t\widehat{A}(n-3,p-1,i-2)(q,qt) \\&\qquad \qquad = q^{\frac{5p^2-9p+4+(i-p)(i-p-1)}{2}}t^{2p-2+(i-p)}\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q. \end{aligned}$$

Now

$$\begin{aligned}&\widehat{A}(n-1,p,i)+q^{n-1}t^2\widehat{A}(n-3,p-1,i-1)(q,qt) \\&\qquad +\,q^{n-1}t^3\widehat{A}(n-3,p-1,i-2)(q,qt) \\&\qquad \qquad =q^{\frac{5p^2-3p+(i-p)(i-p-1)}{2}}t^{2p+(i-p)} \\&\qquad \qquad \quad \times \, \left[ \left( {\begin{array}{c}n-2p\\ p\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q+q^{n-3p+1}\left( {\begin{array}{c}n-2p\\ p-1\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q\right] \\&\qquad \qquad =q^{\frac{5p^2-3p+(i-p)(i-p-1)}{2}}t^{2p+(i-p)}\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) _q\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q=\widehat{A}(n,p,i). \end{aligned}$$

A similar recursion holds for $\widehat{B}(n,p,i)$. It remains to check the initial conditions:

$$\begin{aligned} \widehat{b}(0,n)= & {} 1,\\ \widehat{b}(1,n)= & {} (t^2+qt^2+\ldots +q^{n-1}t^2)=qt^2\left( {\begin{array}{c}n-1\\ 1\end{array}}\right) _q+t^2\left( {\begin{array}{c}n-3\\ 0\end{array}}\right) , \\ \widehat{b}(2,n)= & {} qt^3[n-1]_q+q^5t^4\left( {\begin{array}{c}n-2\\ 2\end{array}}\right) _q \\= & {} qt^3\left( {\begin{array}{c}n-1\\ 1\end{array}}\right) _q+q^5t^4\left( {\begin{array}{c}n-3\\ 1\end{array}}\right) +q^7t^4\left( {\begin{array}{c}n-3\\ 2\end{array}}\right) _q. \end{aligned}$$

$\square$

The following result was conjectured by the second author, Oblomkov and Rasmussen in [10, Conjecture 4.1].

Theorem 6.6

The Hilbert series of $R_n/I_n$ has the following form:

$$\begin{aligned} H_n(q,t)= & {} \frac{1}{\prod _{i=0}^{n-1}(1-q^{i}t)}\sum _{p=0}^{\infty }(-1)^{p}\prod _{k=0}^{p-1}(1-q^{k}t) \nonumber \\&\times \left( q^{\frac{5p^2-3p}{2}}t^{2p}\left( {\begin{array}{c}n-2p+1\\ p\end{array}}\right) _q-q^{\frac{5p^2+5p}{2}}t^{2p+2}\left( {\begin{array}{c}n-2p-1\\ p\end{array}}\right) _q\right) .\nonumber \\ \end{aligned}$$

(6.5)

Proof

It is clear that $H_n(q,t)=\frac{1}{\prod _{i=0}^{n-1}(1-q^{i}t)}\sum _{i=0}^{\infty }(-1)^{i}\widehat{b}(i,n).$ The latter can be computed by (6.3), and it remains to use the identity

$$\begin{aligned} \prod _{k=0}^{p-1}(1-q^{k}t)=\sum _{j=0}^{p}(-1)^{j}q^{j(j-1)/2}t^{j}\left( {\begin{array}{c}p\\ j\end{array}}\right) . \end{aligned}$$

$\square$

7 Limit at $n\rightarrow \infty$

In the limit $n\rightarrow \infty$ both formulas for the Hilbert series simplify. Indeed, for fixed p we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\begin{array}{c}n\\ p\end{array}}\right) _q=\frac{1}{(1-q)\cdots (1-q^{p})}, \end{aligned}$$

so we can take the limit of all the above results.

Proposition 7.1

The limit of the Hilbert series $H_n(q,t)$ has the following form:

$$\begin{aligned} H_{\infty }(q,t)=\sum _{p=0}^{\infty }\frac{q^{p(p-1)}t^p}{(1-q)(1-q^2)\cdots (1-q^{p})}. \end{aligned}$$

(7.1)

Proposition 7.2

The limit of the bigraded rank of the ith syzygy module F(i, n) equals

$$\begin{aligned} \widehat{b}(i,\infty )= & {} \sum _{p>0}\left( q^{\frac{5p^2-3p+(i-p)(i-p-1)}{2}}t^{2p+(i-p)}\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q\frac{1}{(1-q)\cdots (1-q^{p})} \right. \nonumber \\&+\, q^{\frac{5p^2+5p+(i-p)(i-p-1)}{2}}t^{2p+2+(i-p)}\left( {\begin{array}{c}p\\ i-p-1\end{array}}\right) _q \nonumber \\&\left. \times \,\frac{1}{(1-q)\cdots (1-q^{p})}\phantom {\left( {\begin{array}{c}p\\ i-p\end{array}}\right) _q}\right) . \end{aligned}$$

(7.2)

Proposition 7.3

The limit of the Hilbert series $H_n(q,t)$ has the following form:

$$\begin{aligned} H_n(q,t)= & {} \frac{1}{\prod _{i=0}^{\infty }(1-q^{i}t)}\sum _{p=0}^{\infty }(-1)^{p}\prod _{k=0}^{p-1}\frac{1-q^{k}t}{1-q^{k+1}} \nonumber \\&\quad \times \,\left( q^{\frac{5p^2-3p}{2}}t^{2p}-q^{\frac{5p^2+5p}{2}}t^{2p+2}\right) . \end{aligned}$$

(7.3)

The equality between the right-hand sides of (7.3) and (7.1) was proved in [9, Theorem 3.3.2(b)]. At $t=1$ and $t=q$, one recovers more familiar Rogers–Ramanujan identities.

The following proposition concerning Gröbner bases in the limit was proved first in [4], but we give an alternative proof here. In fact, [4] use a slightly different basis of Bell polynomials. In [14, Section 17], a vertex-algebraic proof of essentially the same fact was also obtained. Yet another proof can be obtained by taking the limit in Theorem 4.6, as follows.

Proposition 7.4

For $n\rightarrow \infty$, the polynomials $f_i$ form a Gröbner basis for the ideal $I_{\infty }$.

Before embarking on the proof, we record the following lemmas concerning Gröbner bases here for the convenience of the reader.

Lemma 7.5

([7] Proposition 8 on p. 106). Given $(g_1,\ldots , g_s)\in F_s$, the S-pairs

$$\begin{aligned} S_{ij}:=\frac{{{\,\mathrm{lcm}\,}}({{\,\mathrm{LT_<}\,}}(g_i),{{\,\mathrm{LT_<}\,}}(g_j))}{{{\,\mathrm{LT_<}\,}}(g_i)}e_i-\frac{{{\,\mathrm{lcm}\,}}({{\,\mathrm{LT_<}\,}}(g_i),{{\,\mathrm{LT_<}\,}}(g_j))}{{{\,\mathrm{LT_<}\,}}(g_j)}e_j \end{aligned}$$

(7.4)

form a homogeneous basis for the syzygies on $\{{{\,\mathrm{LT_<}\,}}(g_1),\ldots ,{{\,\mathrm{LT_<}\,}}(g_s)\}$.

Lemma 7.6

([7] Proposition 9 on p. 107) Let $I=\langle g_1,\ldots , g_s\rangle$. Then $G=\{g_1,\ldots , g_s\}$ is a Gröbner basis for I if and only if every element of a homogeneous basis for the syzygies on ${{\,\mathrm{LT_<}\,}}(G)$ reduces to zero modulo G.

Lemma 7.7

([7] Proposition 4 on p.103) $G=\{g_1,\ldots , g_s\}\subset R_n$, and suppose $g_i, g_j\in G$ have relatively prime leading monomials. Then the S-polynomial

$$\begin{aligned} S(g_i,g_j):= & {} \phi _n(S_{ij})=\frac{{{\,\mathrm{lcm}\,}}({{\,\mathrm{LT_<}\,}}(g_i),{{\,\mathrm{LT_<}\,}}(g_j))}{{{\,\mathrm{LT_<}\,}}(g_i)}g_j \nonumber \\&\quad -\,\frac{{{\,\mathrm{lcm}\,}}({{\,\mathrm{LT_<}\,}}(g_i),{{\,\mathrm{LT_<}\,}}(g_j))}{{{\,\mathrm{LT_<}\,}}(g_j)}g_j \end{aligned}$$

(7.5)

reduces to zero modulo G.

Proof of Proposition 7.4

Consider $S(f_i,f_j)$. By Lemma 7.7, $\gcd ({{\,\mathrm{LT_<}\,}}(f_i), {{\,\mathrm{LT_<}\,}}(f_j))=1$ implies that $S(f_i,f_j)$ reduces to zero modulo $\{f_k\}_{k=1}^\infty$. Write $i=2q+r$, where $r=0,1$. Then ${{\,\mathrm{LT_<}\,}}(f_i)=x_q^2$ if i is even and ${{\,\mathrm{LT_<}\,}}(f_i)=2x_qx_{q+1}$ if i is odd. So the only case we need to consider is $j=i+1$. In this case, we have

$$\begin{aligned} {{\,\mathrm{lcm}\,}}({{\,\mathrm{LT_<}\,}}(f_i),{{\,\mathrm{LT_<}\,}}(f_{i+1}))= {\left\{ \begin{array}{ll} 2x_q^2x_{q+1}, &{} i \text { even}\\ 2x_qx_{q+1}^2, &{} i \text { odd}. \end{array}\right. } \end{aligned}$$

Additionally

$$\begin{aligned} S(f_i,f_{i+1})={\left\{ \begin{array}{ll} 2x_{q+1}f_i-x_qf_{i+1}, &{} i \text { even}\\ x_q f_i-2x_{q+1}f_{i+1}, &{} i \text { odd}. \end{array}\right. } \end{aligned}$$

But from (2.1) it follows that these S-pairs appear in the relations $\phi _n(\mu _{n-1})=0$ for $n\gg 0$. Since $n=\infty$, we always have these relations in $I_\infty$. Additionally, moving the S-pair to the right-hand side we reduce $S(f_i, f_{i+1})\equiv 0$ modulo $\{f_k\}_{k=1}^\infty$. In particular, Lemma 7.6 implies that $\{f_k\}_{k=1}^\infty$ is a Gröbner basis for $I_\infty$. $\square$