The Eisenbud-Green-Harris Conjecture for Defect Two Quadratic Ideals

TL;DR

This paper proves the Eisenbud-Green-Harris conjecture for a specific case involving five quadrics in a polynomial ring, confirming the conjecture's validity in this scenario.

Contribution

It provides a complete proof of the EGH conjecture for ideals generated by seven quadrics in five variables with equal degrees, a previously unresolved case.

Findings

EGH conjecture holds for n=5 with all generators quadratic

Complete proof for ideals generated by n+2 quadrics containing a regular sequence

Confirms EGH conjecture in this specific quadratic case

Abstract

The Eisenbud-Green-Harris (EGH) conjecture states that a homogeneous ideal in a polynomial ring $K[x_1,\,\ldots,\,x_n]$ over a field $K$ that contains a regular sequence $f_1,\,\ldots,\, f_n$ with degrees $a_i$, $i=1,\,\ldots,\,n$ has the same Hilbert function as a lex-plus-powers ideal containing the powers $x_i^{a_i}$, $i=1,\,\ldots,\,n$. In this paper, we discuss a case of the EGH conjecture for homogeneous ideals generated by $n+2$ quadrics containing a regular sequence $f_1,\, \ldots, \, f_n$ and give a complete proof for EGH when $n=5$ and $a_1=\cdots=a_5=2$.

Equations

$$\dim_{K}\big{(}(f_{1},\,\ldots,\,f_{n})_{d+1}\cap gR_{1}\big{)}\leq d.$$

$$\dim_{K}I_{d+1}=\dim_{K}(f_{1},\,\ldots,\,f_{n})_{d+1}+\dim_{K}gR_{1}-\dim_{K}\big{(}(f_{1},\,\ldots,\,f_{n})_{d+1}\cap gR_{1}\big{)},$$

$${\dim }_K{I}_{d+1}\ge {\dim }_K(x_1^2,\dots ,x_n^2,x_1\cdots x_d{)}_{d+1}={\dim }_K(x_1^2,\dots ,x_n^2{)}_{d+1}+n-d$$

$$\dim_{K}\big{(}(f_{1},\,\ldots,\,f_{n})_{d+1}\cap gR_{1}\big{)}\leq d.$$

$${\dim }_K{I}_d={\dim }_K{L}_d\text{and}{\dim }_K{I}_{d+1}={\dim }_K{L}_{d+1}.$$

$${\dim }_K{I}_2=n+2={\dim }_K{L}_2.$$

$${\dim }_K{I}_3\ge n^2+2n-5={\dim }_K{L}_3?$$

$${\dim }_K{J}_3\ge n^2+n-2$$

$$\dim_{K}I_{3}=n^{2}+2n-\dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}-\dim_{K}\big{(}J_{3}\cap hR_{1}\big{)}.$$

$${\dim }_K{I}_3$$

$$\displaystyle=\dim_{K}\mathfrak{f}_{3}+\dim_{K}(gR_{1})-\dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}+\dim_{K}(hR_{1})-\dim_{K}\big{(}J_{3}\cap hR_{1}\big{)}$$

$$\displaystyle=n^{2}+2n-\dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}-\dim_{K}\big{(}J_{3}\cap hR_{1}\big{)},$$

$$(0,0,1,n-a,e_4,e_5,\dots ,e_5,e_4,n-a,1),$$

$$(1,n,(\genfrac{}{}{0ex}{}{n}{2})-1,(\genfrac{}{}{0ex}{}{n}{3})-n+a,(\genfrac{}{}{0ex}{}{n}{4})-e_4,\dots ,(\genfrac{}{}{0ex}{}{n}{3})-e_5,(\genfrac{}{}{0ex}{}{n}{2})-e_4,a,0).$$

$$(0,a,(\genfrac{}{}{0ex}{}{n}{2})-e_4,\dots ,(\genfrac{}{}{0ex}{}{n}{4})-e_4,(\genfrac{}{}{0ex}{}{n}{3})-n+a,(\genfrac{}{}{0ex}{}{n}{2})-1,n,1).$$

$$(0,0,2,\underset{\text{ the same as for }gA}{\underbrace{n-a,e_4,\dots ,e_4,n-a,1}}).$$

$$(1,n,(\genfrac{}{}{0ex}{}{n}{2})-2,(\genfrac{}{}{0ex}{}{n}{3})-n+a,(\genfrac{}{}{0ex}{}{n}{4})-e_4,\dots ,(\genfrac{}{}{0ex}{}{n}{3})-e_5,(\genfrac{}{}{0ex}{}{n}{2})-e_4,a,0),$$

$$(0,a,(\genfrac{}{}{0ex}{}{n}{2})-e_4,\dots ,(\genfrac{}{}{0ex}{}{n}{4})-e_4,(\genfrac{}{}{0ex}{}{n}{3})-n+a,(\genfrac{}{}{0ex}{}{n}{2})-2,n,1).$$

$${\dim }_K{\mathrm{Ann}}_{A_i}(g^{\prime })\ge {\dim }_{K}V+1.$$

$${\varphi }_g:A_i/V\to g{A}_i\text{ and }{\varphi }_h:A_i/V\to h{A}_i$$

$$T:A_i/V\to A_i/V$$

$$\dim_{K}\big{(}(f_{1},\,\ldots,\,f_{n},g_{1},\,\ldots,\,g_{\delta-1})_{3}\cap g_{\delta}R_{1}\big{)}\leq 3$$

$$x_1{g}_{\delta }$$

$$x_2{g}_{\delta }$$

$$x_3{g}_{\delta }$$

$$x_4{g}_{\delta }$$

$$L\vec{\mathbf{g}}=\vec{\mathbf{f}}\cdot (q_{i,j}).$$

$$L{g}_{\delta }\left(\begin{array}{cccc}{x}_1& x_2& x_3& x_4\end{array}\right)=\vec{\mathbf{f}}\cdot \left(\begin{array}{cccc}\overrightarrow{Q_1}& \overrightarrow{Q_2}& \overrightarrow{Q_2}& \overrightarrow{Q_4}\end{array}\right).$$

$$\vec{\mathbf{f}}\left(\begin{array}{cccccc}{x}_2\overrightarrow{Q_1}-x_1\overrightarrow{Q_2}& x_3\overrightarrow{Q_1}-x_1\overrightarrow{Q_3}& x_4\overrightarrow{Q_1}-x_1\overrightarrow{Q_4}& x_3\overrightarrow{Q_2}-x_2\overrightarrow{Q_3}& x_4\overrightarrow{Q_2}-x_2\overrightarrow{Q_4}& x_4\overrightarrow{Q_3}-x_3\overrightarrow{Q_4}\end{array}\right)$$

$$\displaystyle\Big{(}\underbrace{x_{2}\vec{Q_{1}}-x_{1}\vec{Q_{2}}}_{\tiny\begin{matrix}\text{a column vector}\\ \text{ of cubic forms}\end{matrix}}\,\,\cdots\,\,x_{4}\vec{Q_{3}}-x_{3}\vec{Q_{4}}\Big{)}=\begin{pmatrix}B_{12}\vec{\bf f}^{T}&\cdots&B_{34}\vec{\bf f}^{T}\end{pmatrix}$$

$$\Big{(}\underbrace{(x_{3}B_{12}-x_{2}B_{13}+x_{1}B_{23})}_{\tiny\begin{matrix}n\times n\,\text{matrix of}\\ \text{quadratic forms}\end{matrix}}\vec{\bf f}^{T}\,\,\cdots\,\,(x_{4}B_{23}-x_{3}B_{24}+x_{2}B_{34})\vec{\bf f}^{T}\Big{)}={\bf 0}.$$

$$C_1^{123},\dots ,C_n^{123},C_1^{124},\dots ,C_n^{124},\dots ,C_1^{234},\dots ,C_n^{234}$$

$$\begin{array}{rl}{x}_3{B}_{12}-x_2{B}_{13}+x_1{B}_{23}=& \left(\begin{array}{c}\vec{\mathbf{f}}{C}_1^{123}\\ ⋮\\ \vec{\mathbf{f}}{C}_n^{123}\end{array}\right)\\ x_4{B}_{12}-x_2{B}_{14}+x_1{B}_{24}=& \left(\begin{array}{c}\vec{\mathbf{f}}{C}_1^{124}\\ ⋮\\ \vec{\mathbf{f}}{C}_n^{124}\end{array}\right)\\ x_4{B}_{13}-x_3{B}_{14}+x_1{B}_{34}=& \left(\begin{array}{c}\vec{\mathbf{f}}{C}_1^{134}\\ ⋮\\ \vec{\mathbf{f}}{C}_n^{134}\end{array}\right)\\ x_4{B}_{23}-x_3{B}_{24}+x_2{B}_{34}=& \left(\begin{array}{c}\vec{\mathbf{f}}{C}_1^{234}\\ ⋮\\ \vec{\mathbf{f}}{C}_n^{234}\end{array}\right)\end{array}$$

Full text

1. Introduction

Let $R=K[x_{1},\,\ldots,\,x_{n}]$ be the polynomial ring in $n$ variables over a field $K$ with the homogeneous lexicographic order in which $x_{1}>\cdots>x_{n}$ and with the standard grading $R=\bigoplus\limits_{i\geq 0}R_{i}$. We denote the Hilbert function of a $\mathbb{Z}$-graded $R$-module $M$ by $\operatorname{Hilb}_{M}(i):=\dim_{K}M_{i}$, where $M_{i}$ is the homogeneous component of $M$ in degree $i$. When $I$ is a homogeneous ideal of $R$ and $M$ is $R$, or $I$, or $R/I$, the Hilbert function has value 0 when $i<0$. When the Hilbert function of $M$ is 0 in negative degree, we may discuss the Hilbert function of $M$ by giving the sequence of its values, and we refer to this sequence of integers as the O-sequence of $M$.

In 1927, Macaulay [13] showed that the Hilbert function of any homogeneous ideal of $R$ is attained by a lexicographic ideal in $R$. Later, in Kruskal-Katona’s theorem [11, 12], it is shown that the polynomial ring $R$ in Macaulay’s result can be replaced with the quotient $R/(x_{1}^{2},\,\ldots,\,x_{n}^{2})$. After this result, Clement and Lindström, in [5], generalized the result to $R/(x_{1}^{a_{1}},\,\ldots,\,x_{n}^{a_{n}})$ if $a_{1}\leq\cdots\leq a_{n}<\infty$.

In [7] Eisenbud, Green and Harris conjectured a generalization of the Clement-Lindström result. Let ${\operatorname{\underline{\mathbf{a}}}}=(a_{1},\,\ldots,\,a_{n})\in\mathbb{N}^{n}$, where $2\leq a_{1}\leq\ldots\leq a_{n}$.

Conjecture 1.1** (Eisenbud-Green-Harris ($\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}$) Conjecture [7]).**

If $I$ is a homogeneous ideal in $R=K[x_{1},\,\ldots,\,x_{n}]$ containing a regular sequence $f_{1},\,f_{2},\,\ldots,\,f_{n}$ with degrees $\deg f_{i}=a_{i}$, then there is a monomial ideal $\operatorname{\mathcal{L}}=(x_{1}^{a_{1}},\,\ldots,\,x_{n}^{a_{n}})+J$, where $J$ is a lexicographic ideal in $R$, such that $R/\operatorname{\mathcal{L}}$ and $R/I$ have the same Hilbert function.

Although there has been some progress on the conjecture, it remains open. The conjecture is shown to be true for $n=2$ by Richert in [14]. Francisco [8] shows the conjecture for almost complete intersections. Caviglia and Maclagan in [2] prove the result if $a_{i}>\sum\limits_{j=1}^{i-1}(a_{j}-1)$ for $2\leq i\leq n$. The rapid growth required for the degrees does not yield much insight into cases like the one in which the regular sequence consists of quadratic forms. When $n=3$, Cooper in [6] proves the EGH conjecture for the cases where $(a_{1},a_{2},a_{3})=(2,a_{2},a_{3})$ and $(a_{1},a_{2},a_{3})=(3,a_{2},a_{3})$ with $a_{2}\leq a_{3}\leq a_{2}+1$.

One of the most intriguing cases is when $a_{1}=\cdots=a_{n}=2$ for any $n\geq 2$, which is the case for which Eisenbud, Green and Harris originally stated their conjecture. It is known that the conjecture holds for homogeneous ideals minimally generated by generic quadrics: the case where $\operatorname{char}K=0$ was proved by Herzog and Popescu [10] and the case of arbitrary characteristic was proved by Gasharov [9] around the same time. There have been several other results on the EGH conjecture. More recently, the case when every $f_{i}$, $i=1,\,\ldots,\,n,$ in the regular sequence is a product of linear forms is settled by Abedelfatah in [1], and results on the EGH conjecture using linkage theory are given by Chong [4].

In this paper we focus on the case when the degrees of the elements of the regular sequence are $a_{1}=\cdots=a_{n}=2$. In [14], Richert claimed that the conjecture for quadratic regular sequences is true for $2\leq n\leq 5$, but this work has not been published, and other researchers have been unable to verify this for $n=5$ thus far. Chen, in [3], has given a proof for the case where $n\leq 4$ when $a_{1}=\cdots=a_{n}=2$.

In $\S 2$ we recall some definitions and results from the papers of Francisco [8], Caviglia-Maclagan [2] and Chen [3]. In $\S 3$ we study homogeneous ideals $I$ generated by $n+2$ quadratic forms in $n$ variables containing a regular sequence of length $n$, and Theorem 3.17 shows that there is a monomial ideal $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{n}^{2})+J$, where $J$ is a lexicographic ideal in $R$, such that $R/I$ and $R/\operatorname{\mathcal{L}}$ have the same Hilbert function in degree $2$ and $3$ (i.e., $\operatorname{EGH}_{(2,\,\ldots,\,2),n}(2)$ holds: see Definition 2.5). In $\S 4$ we give a proof to the claim of Richert for the quadratic regular sequence case when $n=5$.

2. Background and Preliminaries

In this section we recall some definitions and state some known results that are used throughout the paper.

Definition 2.1**.**

Let $u=x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$ and $v=x_{1}^{b_{1}}\cdots x_{n}^{b_{n}}$ be monomials in $R$ of the same degree. We say that $u$ is greater than $v$ with respect to the lexicographic (or lex) order if there exists an $i$ such that $a_{i}>b_{i}$ and $a_{j}=b_{j}$ for all $j<i$.

A monomial ideal $J\subseteq R$ is called a lexicographic ideal (or lex ideal) if, for all degrees $d$, the $d$-th degree component of $J$, denoted by $J_{d}$, is spanned over the base field $K$ by an initial segment of the degree $d$ monomials in the lexicographic order.

Definition 2.2**.**

Given $2\leq a_{1}\leq\cdots\leq a_{n}$, a lex-plus-powers ideal (LPP ideal) $\operatorname{\mathcal{L}}$ is a monomial ideal in $R$ that can be written as $\operatorname{\mathcal{L}}=(x_{1}^{a_{1}},\,\ldots,\,x_{n}^{a_{n}})+J$ where $J$ is a lex ideal in $R$.

This definition agrees with the one in [2]. Some authors require that the $x_{i}^{a_{i}}$ be minimal generators of $\operatorname{\mathcal{L}}$, which we do not. However, since we consider only nondegenerate homogeneous ideals in this paper, i.e., ideals contained in $(x_{1},\,\ldots,\,x_{n})^{2}$, in the case where $a_{1}=\cdots=a_{n}=2$ it is automatic that the $x_{i}^{2}$ are minimal generators of the ideal under consideration.

In [8] Francisco showed the following for almost complete intersections.

Theorem 2.3** (Francisco [8]).**

Let integers $2\leq a_{1}\leq a_{2}\leq\cdots\leq a_{n}$ and $d\geq a_{1}$ be given. Let the ideal $I$ have minimal generators $f_{1},\,\ldots,\,f_{n},\,g$ where $f_{1},\,\ldots,\,f_{n}$ form a regular sequence with $\deg f_{i}=a_{i}$ and $g$ has degree $d$. Let $\operatorname{\mathcal{L}}=(x_{1}^{a_{1}},\,\ldots,\,x_{n}^{a_{n}},m)$ be the lex-plus-powers ideal where $m$ is the greatest monomial in lex order in degree $d$ that is not in $(x_{1}^{a_{1}},\,\ldots,\,x_{n}^{a_{n}})$. Then $\operatorname{Hilb}_{R/I}(d+1)\leq\operatorname{Hilb}_{R/\operatorname{\mathcal{L}}}(d+1)$.

Note that, necessarily, $d\leq\sum_{i=1}^{n}(a_{i}-1)$, since $(f_{1},\,\ldots,\,f_{n})$ contains all forms of degree larger than that. If $a_{1}=\cdots=a_{n}=2$, then $d\leq n$.

The following corollary is an immediate consequence of Theorem 2.3 above. If $g\in R$ is a nonzero form of degree $i$ we write $gR_{j}$ for the vector space $\{gh:h\in R_{j}\}\subseteq R_{i+j}$.

Corollary 2.4**.**

Let $I=(f_{1},\,\ldots,\,f_{n},g)$ be an almost complete intersection as in Theorem 2.3 above such that $a_{1}=\cdots=a_{n}=2$. Then

[TABLE]

Proof.

We can write

[TABLE]

where $\dim_{K}gR_{1}=n$. Then by Theorem 2.3, we have

[TABLE]

Since $\operatorname{Hilb}_{R/(f_{1},\,\ldots,\,f_{n})}(i)=\operatorname{Hilb}_{R/(x_{1}^{2},\,\ldots,\,x_{n}^{2})}(i)$ for all $i\geq 0$, we can conclude that

[TABLE]

∎

The next statement is a weaker version of the $\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}$ conjecture. It focuses on the Hilbert function of the given homogeneous ideal only at the two consecutive degrees $d$ and $d+1$ for some non-negative integer $d$.

Definition 2.5** ($\mathbf{\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}(d)}$).**

Following Caviglia-Maclagan [2], we say that “$\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}(d)$ holds” if for any homogeneous ideal $I\in K[x_{1},\,\ldots,\,x_{n}]$ containing a regular sequence of degrees $\operatorname{\underline{\mathbf{a}}}=(a_{1},\,\ldots,\,a_{n})$, where $2\leq a_{1}\leq\cdots\leq a_{n}$, there exists a lex-plus-powers ideal $\operatorname{\mathcal{L}}$ containing $\{x_{i}^{a_{i}}:1\leq i\leq n\}$ such that

[TABLE]

Lemma 2.6**.**

The condition $\operatorname{EGH}_{(d,\,\ldots,\,d),n}(d)$ on a polynomial ring $R=K[x_{1},\,\ldots,\,x_{n}]$ is equivalent to the statement that for the ideal $I$ generated by $n+\delta$ $K$-linearly independent forms of degree $d$ containing a regular sequence of quadrics, one has that $\dim_{K}I_{d+1}\geq\dim_{K}{\operatorname{\mathcal{L}}}_{d+1}$, where $\operatorname{\mathcal{L}}=(x_{1}^{d},\,\ldots,\,x_{n}^{d})+J^{\prime}$ and $J^{\prime}$ is minimally generated by the greatest in lex order $\delta$ forms of degree $d$ not already in $(x_{1}^{d},\,\ldots,\,x_{n}^{d})$.

Proof.

If there is an LPP ideal $(x_{1}^{d},\,\ldots,\,x_{n}^{d})+J$, where $J$ is a lex ideal, with the same Hilbert function as $I$ in degrees $d$ and $d+1$, it is clear that $J_{d}$ must be spanned over $K$ by the specified generators of $J^{\prime}$, so that $(x_{1}^{d},\,\ldots,\,x_{n}^{d})+J^{\prime}\subseteq(x_{1}^{d},\,\ldots,\,x_{n}^{d})+J$, which implies the specified inequality on the Hilbert functions. Moreover, when that inequality holds we may increase $\operatorname{\mathcal{L}}:=(x_{1}^{d},\,\ldots,\,x_{n}^{d})+J^{\prime}$ to an LPP ideal with the same Hilbert function as $I$ in degrees $d$ and $d+1$: if $\Delta=\operatorname{Hilb}_{I}(d+1)-\operatorname{Hilb}_{\operatorname{\mathcal{L}}}(d+1)$, we may simply include the greatest (in lex order) $\Delta$ forms of degree $d+1$ not already in $\operatorname{\mathcal{L}}$. ∎

Remark 2.7**.**

We shall eventually be focused on $\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}(d)$ in the case where $a_{1}=\cdots=a_{n}=d=2$, simply referred as $\operatorname{EGH}_{(2,\,\ldots,\,2),n}(2)$ or $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$. We shall routinely make use of this lemma in this case of quadratic regular sequence and $d=2$.

Lemma 2.8** (Caviglia-Maclagan [2]).**

*Fix $\operatorname{\underline{\mathbf{a}}}=(a_{1},\,\ldots,\,a_{n})\in\mathbb{N}^{n}$ where $2\leq a_{1}\leq a_{2}\leq\cdots\leq a_{n}$ and set $s=\sum\limits_{i=1}^{n}(a_{i}-1).$ Then for any $0\leq d\leq s-1$, $\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}(d)$ holds if and only if $\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}$(s-1-d) holds.

Furthermore, the $\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}$ conjecture holds if and only if $\operatorname{EGH}_{\operatorname{\underline{\mathbf{a}}},n}(d)$ holds for all degrees $d\geq 0$.*

From now on, we always assume $\operatorname{\underline{\mathbf{a}}}=\operatorname{\underline{\mathbf{2}}}2=(2,\,\ldots,\,2)$ for $n\geq 2$, unless it is stated otherwise.

Remark 2.9**.**

For any $n\geq 2$, $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(0)$ holds trivially. In [3, Proposition 2.1], Chen showed that $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(1)$ is true for any $n\geq 2$.

Chen proved the following.

Theorem 2.10** (Chen [3]).**

The $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}$ conjecture holds when $2\leq n\leq 4$.

Chen’s proof of this uses Lemma 2.8 above, and the observation that, when $n=4$, to demonstrate that the $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,4}$ conjecture is true, it suffices to show that $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,4}(0)$ and $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,4}(1)$ are true.

3. $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ for defect two ideals

In this section, we focus on the homogeneous ideals in $K[x_{1},\,\ldots,\,x_{n}]$ for $n\geq 5$ that are generated by $n+2$ quadratic forms containing a regular sequence. In particular, we study their Hilbert functions in degree $3$.

Definition 3.1**.**

If $I$ is a homogeneous ideal minimally generated by $n+\delta$ forms that contain a regular sequence of length $n$, then $I$ is said to be a defect $\delta$ ideal.

Clearly, when $\delta=0$ then $I$ is generated by a regular sequence, it is a complete intersection, and we understand the Hilbert function completely. If $\delta=1$, then $I$ is an almost complete intersection.

Definition 3.2**.**

We call a homogeneous ideal a quadratic ideal if it is generated by quadratic forms.

Let $I=(f_{1},\,\ldots,\,f_{n},g,h)$ be a homogeneous ideal minimally generated by $n+2$ quadrics where $f_{1},\,\ldots,\,f_{n}$ form a regular sequence. We call such an ideal a defect two ideal generated by quadrics or simply a defect two quadratic ideal. More generally, if a quadratic ideal is a defect $\delta$ ideal, then we call it defect $\delta$ quadratic ideal.

Example 3.3**.**

The lex-plus-powers ideal $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{n}^{2},x_{1}x_{2},x_{1}x_{3})$ in $R$ is also a defect two quadratic ideal.

Further, for any homogeneous defect two quadratic ideal $I$, we have the equality

[TABLE]

Main Question 3.4** ($\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ for defect two quadratic ideals).**

For any $n\geq 5$, is it true that

[TABLE]

An affirmative answer for this question is proved completely in Theorem 3.17 below.

Notation 3.5**.**

Throughout the rest of this paper we write $\mathfrak{f}$ for the ideal $(f_{1},\,\ldots,\,f_{n})R$ when $f_{1},\,\ldots,\,f_{n}$ is a regular sequence of quadratic forms, and in the defect $\delta$ quadratic ideal case we write $\mathfrak{g}$ for the additional generators $g_{1},\,\ldots,\,g_{\delta}$ of the quadratic ideal. Here, $f_{1},\,\ldots,\,f_{n},g_{1},\,\ldots,\,g_{\delta}$ are assumed to be linearly independent over $K$. Moreover, henceforth, we write $J$ for the ideal $\mathfrak{f}+(g_{1},\,\ldots,\,g_{\delta-1})$. However, when $\delta=1$ or $2$ we may write $g,\,h$ for $g_{1},g_{2}$, so that whenever $\delta=2$ we henceforth write $J$ for the ideal $\mathfrak{f}+(g_{1})=\mathfrak{f}+(g)$. We denote the graded Gorenstein Artin $K$-algebra $R/\mathfrak{f}$ by $A$.

We know that, if $a_{1}=\cdots=a_{n}=\deg g=2$, Theorem 2.3 shows that

[TABLE]

and then Corollary 2.4 gives \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}\leq 2.

Remark 3.6**.**

In [3, Proposition 3.7] Chen gave a positive answer to the Question 3.4 for defect two quadratic ideals $I=\mathfrak{f}+(g,h)$ if \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=2. We shall make repeated use of this fact in the sequel.

In this section we show $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ for a defect two quadratic ideal $I=\mathfrak{f}+(g,h)$ under the condition that \dim_{K}\big{(}\mathfrak{f}_{3}\cap g^{\prime}R_{1}\big{)}\leq 1 for all $g^{\prime}\in Kg+Kh-\{0\}$: this covers all the cases for which Chen’s result in Proposition 3.6 is not applicable.

Lemma 3.7**.**

As in Notation 3.5, $J$ is the defect 1 quadratic ideal $\mathfrak{f}+gR$. Then:

[TABLE]

*Consequently, for the cases that are not covered by the Proposition 3.6 we have:

(i)* If \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=1 then \dim_{K}I_{3}=n^{2}+2n-1-\dim_{K}\big{(}J_{3}\cap hR_{1}\big{)}, and $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ holds for a defect two quadratic ideal $I$ if and only if \dim_{K}\big{(}J_{3}\cap hR_{1}\big{)}\leq 4.

(ii)* If \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=0 then \dim_{K}I_{3}=n^{2}+2n-\dim_{K}\big{(}J_{3}\cap hR_{1}\big{)}, and $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ holds for $I$ if and only if \dim_{K}\big{(}J_{3}\cap hR_{1}\big{)}\leq 5.*

Proof.

We have:

[TABLE]

and then (i) and (ii) are immediate. ∎

Remark 3.8**.**

Let $n=5$, so that $\mathfrak{f}=(f_{1},\,\ldots,\,f_{5})$. For a defect two quadratic ideal $I=(\mathfrak{f},g,h)\subseteq K[x_{1},\,\ldots,\,x_{5}]$, if \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=0 then clearly \dim_{K}\big{(}(\mathfrak{f},\,g)_{3}\cap hR_{1}\big{)}\leq\dim_{K}(hR_{1})\leq 5, therefore $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ holds for such an ideal $I$. However, we must give an argument to cover all possible cases, that is, when \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=1, to be able to confirm $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for every defect two quadratic ideal. In the last section, we discuss the EGH conjecture for $n=5$ and $a_{1}=\cdots=a_{5}=2$ in detail.

Next, we proceed with two useful lemmas.

Lemma 3.9**.**

Let $A$ be the graded Gorenstein Artin $K$-algebra $R/\mathfrak{f}$ with $\dim_{K}A_{1}=n$. Let $g,h$ be two quadratic forms such that $gA_{1}=hA_{1}$. Then $\operatorname{Ann}_{A_{1}}g=\operatorname{Ann}_{A_{1}}h$.

Moreover, $\operatorname{Ann}_{A_{i}}(g)=\operatorname{Ann}_{A_{i}}(h)$ if $i\neq n-2$.

Proof.

Suppose that the linear annihilator space of $g$, $\operatorname{Ann}_{A_{1}}g$, has dimension $a$ and $gA_{1}=hA_{1}$. Thus $gA_{1}$ has dimension $n-a$ and clearly $hA_{1}$ and $\operatorname{Ann}_{A_{1}}h$ have dimensions $n-a$ and $a$, respectively.

Notice that $gA(-2)\cong A/\operatorname{Ann}_{A}(g)$, hence it is Gorenstein and it has a symmetric O-sequence

[TABLE]

where $e_{i}$ denotes the dimension of $[gA]_{i}$ and $e_{i}=e_{n-i+2}$ for $2\leq i\leq n$. Then the Hilbert function of $A/gA$ is

[TABLE]

Since $\operatorname{Ann}_{A}(g)\cong\operatorname{Hom}_{K}(A/gA,A)\cong(A/gA)^{\vee}$, the Hilbert function of $\operatorname{Ann}_{A}(g)$ is

[TABLE]

Recall that $gA_{1}=hA_{1}$, $gA_{i}=hA_{i}$ for all $i\geq 2$, so $(g,h)A$ has the Hilbert function

[TABLE]

Then the O-sequence of $A/(g,h)$ becomes

[TABLE]

and it follows that $\operatorname{Ann}_{A}(g,h)$ has the Hilbert function

[TABLE]

We know that $\operatorname{Ann}_{A}(g,h)=\operatorname{Ann}_{A}(g)\cap\operatorname{Ann}_{A}(h)$, and in degree $1$, $\operatorname{Ann}_{A}(g,h)$ has dimension $a$, so $\operatorname{Ann}_{A}(g,h)=\operatorname{Ann}_{A_{1}}(g)=\operatorname{Ann}_{A_{1}}(h)$. Further, $\operatorname{Ann}_{A}(g)$ and $\operatorname{Ann}_{A}(h)$ are the same in every degrees except in degree $n-2$. ∎

Lemma 3.10**.**

Let $g,h$ be two quadratic forms in a graded Gorenstein Artin $K$-algebra $A$ such that $gA_{i}=hA_{i}$ and $g,h$ have the same annihilator space $V$ in $A_{i}$ for some $i\geq 1$. Then there exists $g^{\prime}\in Kg+Kh-\{0\}$ such that

[TABLE]

Proof.

Consider the multiplication maps by $g$ and $h$,

[TABLE]

whose images $gA_{i}$, $hA_{i}$ are subspaces in $A_{i+2}$ and $gA_{i}=hA_{i}$ by assumption. Then there is a automorphism

[TABLE]

such that $g\ell=hT(\ell)$ for any $\ell\in A_{i}/V$. However, $T$ has at least one nonzero eigenvector $u$ with $T(u)=cu$ for some $c\in K$. Say $\ell_{u}$ be a form in degree $i$ represented by this eigenvector $u$ in $A_{i}$ and not in the annihilator space $V$, thus $g\ell_{u}=hc\ell_{u}$. Then there is a quadratic form $g^{\prime}:=g-ch\in Kg+Kh-\{0\}$ such that $g^{\prime}$ annihilated by the space $V$ and also by $\ell_{u}\in A_{i}\setminus V$. Hence $\dim_{K}\operatorname{Ann}_{A_{i}}(g^{\prime})\geq\dim_{K}V+1$. ∎

From now on, $I=(f_{1},\,\ldots,\,f_{n},g,h)=\mathfrak{f}+(g,h)$ is a homogeneous ideal where \dim_{K}\big{(}\mathfrak{f}_{3}\cap g^{\prime}R_{1}\big{)}\neq 2 for a quadratic form $g^{\prime}\in Kg+Kh-\{0\}$, which means that $\dim_{K}g^{\prime}A_{1}\neq n-2$. Therefore $\dim_{K}g^{\prime}A_{1}$ is either $n$ or $n-1$.

Proposition 3.11**.**

For the graded Gorenstein Artin $K$-algebra $A$, if $gA_{1}=hA_{1}$ with $\dim_{K}gA_{1}=n-1=\dim_{K}hA_{1}$, that is $\dim_{K}(\mathfrak{f}_{3}\cap gR_{1})=\dim_{K}(\mathfrak{f}_{3}\cap hR_{1})=1$, then $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ holds for the homogeneous defect two quadratic ideal $I=\mathfrak{f}+(g,h)$.

Proof.

Since $\dim_{K}\operatorname{Ann}_{A_{1}}(g)=\dim_{K}\operatorname{Ann}_{A_{1}}(h)=1$ there is some $g^{\prime}\in Kg+Kh-\{0\}$ with $\dim_{K}\operatorname{Ann}_{A_{i}}(g^{\prime})=2$ by Lemma 3.10. In consequence, \dim_{K}\big{(}\mathfrak{f}_{3}\cap g^{\prime}R_{1}\big{)}=2, and so we are done by Proposition 3.6. ∎

Proposition 3.12**.**

For the graded Gorenstein Artin $K$-algebra $A$, if $\dim_{K}gA_{1}=\dim_{K}hA_{1}=n$, then there exists a quadratic form $g^{\prime}$ in $Kg+Kh$ with a nonzero linear annihilator in $A$.

Proof.

By assumption $\dim_{K}A_{1}=\dim_{K}gA_{1}=\dim_{K}hA_{1}=n$, and so we may consider again the multiplication maps $\phi_{g}:A_{1}\to gA_{1}$ and $\phi_{h}:A_{1}\to hA_{1}$. Then we obtain a automorphism $T:A_{1}\to A_{1}$ and there exists an nonzero linear form $\ell\in A_{1}$ such that $T(\ell)=c\ell$ for some $c\in K$, that is $g\ell=ch\ell$. Consider $g^{\prime}=g-ch\in Kg+Kh$. Clearly, $\ell\in\operatorname{Ann}_{A_{1}}(g^{\prime})$. ∎

Next we assume that there is a linear annihilator $L\in A_{1}$ of $g$ where $Lh\neq 0$ over the Gorenstein ring $A=R/\mathfrak{f}$. This case may come up either when $\dim_{K}gA_{1}=\dim_{K}hA_{1}=n-1$ and the linear annihilator spaces $\operatorname{Ann}_{A_{1}}(g)$ and $\operatorname{Ann}_{A_{1}}(h)$ are distinct, or when $\dim_{K}gA_{1}=n-1$ and $\dim_{K}hA_{1}=n$.

We shall make repeated use of the following result, which is Lemma 3.3 of Chen’s paper [3].

Lemma 3.13** (Chen [3]).**

If $f_{1},\,\ldots,\,f_{n}$ is a regular sequence of 2-forms in $R$ and we have a relation $u_{1}f_{1}+u_{2}f_{2}+\cdots+u_{n}f_{n}=0$ for some $t$-forms $u_{1},\,\ldots,\,u_{n}$, then $u_{1},\,\ldots,\,u_{n}\in(f_{1},\,\ldots,\,f_{n})_{t}$. More precisely, we have that $t\geq 2$ and there exists a skew-symmetric $n\times n$ matrix $B$ of $(t-2)$-forms such that $(u_{1}\,u_{2}\,\cdots\,u_{n})=(f_{1}\,f_{2}\,\cdots\,f_{n})B$. ∎

Proposition 3.14**.**

Let $I=\mathfrak{f}+\mathfrak{g}$ be a defect $\delta$, where $2\leq\delta\leq n-1$, quadratic ideal of $R$ as in Notation 3.5. If there is a linear form $L$ in $\operatorname{Ann}_{A}(g_{1},\,\ldots,\,g_{\delta-1})$ such that $Lg_{\delta}\neq 0$ in $A$, then

[TABLE]

Chen [3] used an argument involving the Koszul relations on $(x_{1},\,\ldots,\,x_{r})$ for $r\leq n$ while introducing another proof for Theorem 2.3. In the proof of this proposition we use a very similar argument.

Proof.

As in Notation 3.5, let $J=\mathfrak{f}+(g_{1},\,\ldots,\,g_{\delta-1})$, and denote the row vector of the regular sequence $f_{1},\,\ldots,\,f_{n}$ by $\vec{\bf f}$ and the row vector of quadratic forms $g_{1},\,\ldots,\,g_{\delta-1}$ by $\vec{\bf g}$.

Suppose $\dim_{K}(J_{3}\cap g_{\delta}R_{1})\geq 4$, and without loss of generality we may assume that

[TABLE]

where $\vec{\ell_{i}}$ and $\vec{p_{i}}$ are column vectors of linear forms of lengths $\delta-1$ and $n$, respectively.

We assume that there is a linear form $L$ such that $Lg_{i}=0$ for each $i=1,\,\ldots,\,\delta-1$ but $Lg_{\delta}\neq 0$ in $A$. Then we get am $n\times(\delta-1)$ matrix $(q_{i,j})=\begin{pmatrix}\vec{q_{1}}&\vec{q_{2}}&\cdots&\vec{q}_{\delta-1}\end{pmatrix}$ of linear forms such that

[TABLE]

We observe that each $x_{i}Lg_{\delta}$ is in $\mathfrak{f}$, and write $x_{i}Lg_{\delta}=\vec{\bf f}\cdot\vec{Q_{i}}$ where $\vec{Q_{i}}$ is a column of quadratic forms for $i=1,2,3,4$. Therefore:

[TABLE]

Let $M_{1}=\begin{pmatrix}x_{2}&x_{3}&x_{4}&0&0&0\\ -x_{1}&0&0&x_{3}&x_{4}&0\\ 0&-x_{1}&0&-x_{2}&0&x_{4}\\ 0&0&-x_{1}&0&-x_{2}&-x_{3}\end{pmatrix}$. Note that $\begin{pmatrix}x_{1}&x_{2}&\cdots&x_{4}\end{pmatrix}\cdot M_{1}=0$. Multiplying the equation (1) by $M_{1}$ from right gives that $\vec{\bf f}\cdot(\vec{Q}_{1}\ \vec{Q}_{2}\ vecQ_{3}\ \vec{Q}_{4})=0$, and so all entries are 0 in

[TABLE]

By Lemma 3.13, there are alternating $n\times n$ matrices $B_{12},B_{13},B_{14},B_{23},B_{24},B_{34}$ of linear forms such that

[TABLE]

Similarly, consider the matrix $M_{2}=\begin{pmatrix}x_{3}&x_{4}&0&0\\ -x_{2}&0&x_{4}&0\\ 0&-x_{2}&-x_{3}&0\\ x_{1}&0&0&x_{4}\\ 0&x_{1}&0&-x_{3}\\ 0&0&x_{1}&x_{2}\end{pmatrix}$ such that $M_{1}\cdot M_{2}={\bf 0}$ and multiply equation (2) by $M_{2}$ from right to obtain:

[TABLE]

Then again by Lemma 3.13, there are alternating $n\times n$ matrices

[TABLE]

of scalars such that

[TABLE]

Repeating the previous steps with $M_{3}=\begin{pmatrix}x_{4}\\ -x_{3}\\ x_{2}\\ -x_{1}\end{pmatrix}$, so that $M_{2}\cdot M_{3}={\bf 0}$, we get

[TABLE]

and then for all $i=1,2,\,\ldots,\,n$ we obtain

[TABLE]

Then, finally, $x_{4}C^{123}_{i}-x_{3}C^{124}_{i}+x_{2}C^{134}_{i}-x_{1}C^{234}_{i}=0$ for all $i=1,2,...,n$. Hence,

[TABLE]

Thus, in (3) we get $x_{3}B_{12}-x_{2}B_{13}+x_{1}B_{23}=0$. This shows that $x_{3}$ divides every entry in $x_{2}B_{13}-x_{1}B_{23}$. Therefore we may rewrite $B_{13}=x_{3}\widetilde{B_{13}}+D_{13}$ and $B_{23}=x_{3}\widetilde{B_{23}}+D_{23}$, where $\widetilde{B_{13}}$ and $\widetilde{B_{23}}$ are alternating matrices of scalars, $D_{13}$ and $D_{23}$ are alternating matrices of linear forms that do not contain $x_{3}$, and $x_{2}D_{13}-x_{1}D_{23}=0$. We obtain the following

[TABLE]

Returning to equation (2), we obtain $x_{2}\vec{Q_{1}}-x_{1}\vec{Q_{2}}=B_{12}\vec{\bf f}^{T}=(x_{2}\widetilde{B_{13}}-x_{1}\widetilde{B_{23}})\vec{\bf f}^{T}$. Consequently,

[TABLE]

which tells us that $x_{1}$ divides every entry of $\vec{Q_{1}}-\widetilde{B_{13}}\vec{\bf f}^{T}$. It follows that

[TABLE]

This shows that Lg_{\delta}=\vec{\bf f}\frac{1}{x_{1}}\Big{(}\vec{Q_{1}}-\widetilde{B_{13}}\vec{\bf f}^{T}\Big{)}\in(f_{1},\,\ldots,\,f_{n})_{3}, which contradicts our assumption $L\notin\operatorname{Ann}_{A}(g_{\delta})$. ∎

Corollary 3.15**.**

Let $I=\mathfrak{f}+\mathfrak{g}\subseteq R$ be a defect $\delta$ quadratic ideal with $2\leq\delta\leq n-1$. Suppose that

[TABLE]

Then

[TABLE]

*where $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{n}^{2})+(x_{1}x_{2},x_{1}x_{3},\,\ldots,\,x_{1}x_{\delta+1})$ is the defect $\delta$ lex-plus-powers ideal of $R$.

That is, $\operatorname{EGH}_{{\bf 2},n}(2)$ holds for any defect $\delta$ quadratic ideal with property $(\dagger)$.*

Proof.

Notice that $\dim_{K}\mathcal{L}_{3}=n^{2}+n\delta-\frac{\delta(\delta+3)}{2}$. We use induction on $\delta$. Let $J=\mathfrak{f}+(g_{1},\,\ldots,\,g_{\delta-1})$ be the defect $\delta-1$ quadratic ideal.

[TABLE]

∎

We notice that a special case of Corollary 3.15 when $\delta=2$ shows that the inequality is strict.

Corollary 3.16**.**

Let $I=\mathfrak{f}+(g,h)$ be a defect two ideal generated by quadrics in $R$. If $\operatorname{Ann}_{A_{1}}(g)=\operatorname{Span}\{L\}$ for some $L\in R_{1}$ and $L$ does not annihilate $h$ in $A=R/\mathfrak{f}$, then

[TABLE]

Proof.

The result follows from Proposition 3.14 as

[TABLE]

which is $\geq n^{2}+2n-4$.

∎

Finally, we give an affirmative answer to the Main Question 3.4.

Theorem 3.17**.**

Let $I=\mathfrak{f}+(g,h)\subseteq R=K[x_{1},\,\ldots,\,x_{n}]$ for $n\geq 5$ be a defect two ideal quadratic ideal. Then

[TABLE]

More precisely, $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$ holds for homogeneous defect two quadratic ideals in $R$ for any $n\geq 5$.

Proof.

If the given defect two ideal satisfies Proposition 3.6 , then, by Chen’s result, the theorem is proved.

Assume that \dim_{K}\big{(}\mathfrak{f}_{3}\cap g^{\prime}R_{1}\big{)}\neq 2 for any $g^{\prime}\in Kg+Kh\setminus\{0\}$. If \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=\dim_{K}\big{(}\mathfrak{f}_{3}\cap hR_{1}\big{)}=0, by Proposition 3.12, we can always find another quadratic form $g^{\prime}\in Kg+Kh\setminus\{0\}$ so that $g^{\prime}$ has a linear annihilator in $A$. Then we can apply Corollary 3.16. If \dim_{K}\big{(}\mathfrak{f}_{3}\cap gR_{1}\big{)}=\dim_{K}\big{(}\mathfrak{f}_{3}\cap hR_{1}\big{)}=1 and the same linear form annihilates both $g$ and $h$ in $A$, by Proposition 3.11. we have a situation contradicts our assumption. ∎

Corollary 3.18**.**

$\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}(2)$* holds for every defect two ideal containing a regular sequence of quadratic forms.*

Proof.

This result follows from Lemma 2.6 and Theorem 3.17. ∎

4. The $\operatorname{EGH}$ conjecture when $n=5$ and $a_{1}=\cdots=a_{5}=2$

In this section $R=K[x_{1},\ldots,\,x_{5}]$ and $I=(f_{1},\,\ldots,\,f_{5})+(g_{1},\,\ldots,\,g_{\delta})=\mathfrak{f}+\mathfrak{g}$ is a homogeneous defect $\delta$ ideal in $R$, where $f_{1},\,\ldots,\,f_{5}$ is a regular sequence of quadrics and $\deg g_{j}\geq 2$ for $j=1,\,\ldots,\,\delta$. Throughout, we shall write $A:=R/\mathfrak{f}$, which is a graded Gorenstein local Artin ring. We will show the existence of a lex-plus-powers ideal $\mathcal{L}\subseteq R$ containing $x_{i}^{2}$ for $i=1,\,\ldots,\,5$ with the same Hilbert function as $I$ by proving the following main theorem.

Theorem 4.1**.**

The EGH conjecture holds for all homogeneous ideals containing a regular sequence of quadrics in $K[x_{1},\,\ldots,\,x_{5}]$.

Lemma 2.8 of Caviglia-Maclagan tells us that $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(d)$ holds if and only if $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(5-d-1)$ holds. Thus it will be enough to show $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(d)$ when $d=0,1,2$. By Remark 2.9 we know that $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(d)$ is true when $d=0,1$, therefore $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(3)$ and $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(4)$ both hold as well.

Our goal in this section is to prove $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for any homogeneous ideal containing a regular sequence of quadrics: this will complete the proof of $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}$. To achieve this, it suffices to understand $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for quadratic ideals with arbitrary defect $\delta$ (but, of course, $\delta\leq 10$, since $\dim_{K}R_{2}=15$), by Lemma 2.6.

Remark 4.2**.**

As a result of Corollary 3.18, we see that $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,n}$ holds for any defect $\delta=2$ quadratic ideal in $K[x_{1},\,\ldots,\,x_{n}]$ for $n=5$.

To accomplish our goal we will prove $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for defect $\delta\geq 3$ quadratic ideals. In the next subsection, we prove that if one knows the case where $\delta=3$, on obtains all the cases for $\delta\geq 4$. In the final subsection we finish the proof by establishing $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for $\delta=3$.

Quadratic ideals with defect $\delta\geq 4$

Lemma 4.3**.**

If $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ holds for all defect three quadratic ideals, then it holds for all quadratic ideals with defect $\delta\geq 4$.

Proof.

Let $I=(f_{1},\,\ldots,\,f_{5},g_{1},g_{2},g_{3},g_{4})=\mathfrak{f}+\mathfrak{g}\subseteq R$ be a defect $4$ homogeneous ideal generated by quadrics, where $f_{1},\,\ldots,\,f_{5}$ form a regular sequence. By assumption the defect three quadratic ideal $J=\mathfrak{f}+(g_{1},g_{2},g_{3})\subseteq I$ satisfies $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$, that is, $\dim_{K}J_{3}\geq 31$.

Let $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{5}^{2},x_{1}x_{2},x_{1}x_{3},x_{1}x_{4},x_{1}x_{5})$ be the LPP ideal with $\dim_{K}\operatorname{\mathcal{L}}_{2}=\dim_{K}I_{2}=9$. Then we get $\dim_{K}I_{3}\geq\dim_{K}J_{3}\geq 31=\dim_{K}\operatorname{\mathcal{L}}_{3}$, as we need for the case of defect $\delta=4$.

Now assume $5\leq\delta\leq 10$. Let ${}^{\delta}\!I$ denote an arbitrary defect $\delta$ quadratic ideal, and let ${}^{\delta}\!{{\mathcal{L}}}$ denote the lex-plus-power ideal with defect $\delta\geq 5$. More precisely, ${}^{\delta}\!{{\mathcal{L}}}:=(x_{1}^{2},\,\ldots,\,x_{5}^{2})+(m_{1},\,\ldots,\,m_{\delta})$ where $m_{i}$ are the next greatest quadratic square-free monomials with respect to lexicographic order. We need to show that $\operatorname{Hilb}_{R/{}^{\delta}\!I}(3)\leq\operatorname{Hilb}_{R/{}^{\delta}\!{{\mathcal{L}}}}(3)$.

We assume that $\operatorname{Hilb}_{R/{}^{\delta}\!I}(3)\geq\operatorname{Hilb}_{R/{}^{\delta}\!{{\mathcal{L}}}}(3)+1$, and we shall obtain a contradiction.

Using duality for Gorenstein rings, we know that for $0\leq d\leq 5$ we have that

[TABLE]

Then, for $d=3$, using the assumption we get

[TABLE]

We next show that $\dim_{K}(\mathfrak{f}:{}^{\delta}\!I)_{1}=0$. If there is a nonzero linear form $\ell\in\mathfrak{f}:{}^{\delta}\!I$ then $\dim_{K}\operatorname{Ann}_{A_{2}}\ell A\geq\delta\geq 5$, so we get that $\dim_{K}A_{3}/\ell A_{2}\geq 5$. On the other hand, we see that $A_{3}/\ell A_{2}\cong[R/(\bar{f_{1}},\,\ldots,\,\bar{f_{4}},\bar{f_{5}},l)]_{3}$ where the $\bar{f_{i}}$ are the images of the $f_{i}$, and the dimension of $[R/(\bar{f_{1}},\,\ldots,\,\bar{f_{4}},\bar{f_{5}},l)]_{3}$ as a $K$-vector space is at most $4$.

Then we can find a defect $\gamma$ quadratic ideal ${}^{\gamma}\!J\subseteq\mathfrak{f}:{}^{\delta}\!I$ for $\gamma=3,2,1$ if the defect of ${}^{\delta}\!I$ is $\delta=5$ or $\delta=6,7$ or $\delta=8,9,10$, respectively. We then have the inequalities shown below, where the first is obvious and the second follows by comparison with Hilbert functions of quotients by LPP ideals in degree 3 and the fact that, by assumption, EGH${}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ holds for quadratic ideals with defect less than or equal to three.

[TABLE]

However, each of the cases above contradicts the following equality:

[TABLE]

Thus, we get $\operatorname{Hilb}_{R/{}^{\delta}\!I}(3)\leq\operatorname{Hilb}_{R/{}^{\delta}\!{{\mathcal{L}}}}(3)$ for any defect $\delta\geq 5$ quadratic ideal ${}^{\delta}\!I$ in $R$. ∎

Defect three quadratic ideals

Lemma 4.4**.**

Let $I=\mathfrak{f}+(g_{1},g_{2},g_{3})$ be a defect three quadratic ideal in the polynomial ring $R$. Then, for any $1\leq i_{1}<i_{2}\leq 3$,

[TABLE]

and, furthermore, $\dim_{K}(\mathfrak{f}:(g_{1},g_{2},g_{3}))_{1}\leq 1$.

Proof.

Suppose that $\dim_{K}(\mathfrak{f}:(g_{1},g_{2}))_{1}\geq 2$, and assume there are $\ell_{1},\ell_{2}\in R_{1}$ such that $\ell_{i}g_{1},\ell_{i}g_{2}\in\mathfrak{f}$ for both $i=1,2$. Without loss of generality we assume that $\ell_{1}=x_{1}$ and $\ell_{2}=x_{2}$.

Therefore, we can write $(x_{1},x_{2},f_{1},\,\ldots,\,f_{5})\subseteq\mathfrak{f}:I$. Then

[TABLE]

which is a contradiction. ∎

Hence, working in the graded Gorenstein Artin $K$-algebra $A=R/\mathfrak{f}$, we have from the lemma just above that $\operatorname{Ann}_{A_{1}}(g_{1},g_{2})$ is a $K$-vector space of dimension at most one, and, therefore

[TABLE]

since $\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})\subseteq\operatorname{Ann}_{A_{1}}(g_{1},g_{2})$.

Remark 4.5**.**

By Remark 4.2 we know that for any defect two quadratic ideal $J$ in $R$, $\dim_{K}J_{3}$ is at least $30$. Then $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ holds for the defect three quadratic ideals $I$ containing a defect two quadratic ideal $J$ with $\dim_{K}J_{3}\geq 31$, as $\operatorname{Hilb}_{R/I}(3)\leq\operatorname{Hilb}_{R/J}\leq 4$.

We henceforth focus on defect three quadratic ideals $I=\mathfrak{f}+(g_{1},g_{2},g_{3})$ in $R$ such that every defect two quadratic ideal $J\subseteq I$ containing $\mathfrak{f}$ has $\dim_{K}J_{3}=30$.

For such defect three quadratic ideals, we observe the following.

Remark 4.6**.**

Consider the ideal $\mathcal{I}=(g_{1},g_{2},g_{3})A$ in the Gorenstein ring $A$ such that any ideal $(g_{i_{1}},g_{i_{2}})A$ contained in $\mathcal{I}$ has degree three component of dimension $\dim_{K}(g_{i_{1}},g_{i_{2}})A_{1}=5$. Assuming that $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1})=1$, we have that $\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})=\operatorname{Ann}_{A_{1}}(g_{1})$.

Furthermore, if $g_{1}A_{1}$ is $5$-dimensional, that is, there is no linear form that annihilates $g_{1}$ in $A$, then for any quadric $g$ in $Kg_{1}+Kg_{2}+Kg_{3}$ the vector space $gA_{1}\subseteq A_{3}$ is either $3$ or $5$ dimensional.

Proof.

Let $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1})=1$, and let the linear form $L$ annihilate $g_{1}$ but not some form $g^{\prime}\in Kg_{2}+Kg_{3}$ in $A$. We define a defect two quadratic ideal

[TABLE]

in $R$. Hence, by Corollary 3.16, we know already that $\dim_{K}\,J_{3}\geq 31$, which means that $\dim_{K}(g_{1},g^{\prime})A_{1}=6$. This contradicts our assumption. Thus, $L$ must be in $\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})$. ∎

Recall that the following holds, by Proposition 3.14, when $\delta=3$.

Proposition 4.7**.**

Let $I=\mathfrak{f}+(g_{1},g_{2},g_{3})\subseteq K[x_{1},...,x_{5}]$ be a defect $3$ quadratic ideal.

As usual, let $A=R/\mathfrak{f}$. If there is a linear form $L\in\operatorname{Ann}_{A}(g_{1},g_{2})$ such that $L\notin\operatorname{Ann}_{A}(g_{3})$, then

[TABLE]

∎

When a defect three quadratic ideal $I$ satisfies the condition of the above proposition, we notice a sharp bound for $\operatorname{Hilb}_{R/I}(3)$.

Corollary 4.8**.**

Given a defect three quadratic ideal $I=\mathfrak{f}+(g_{1},g_{2},g_{3})$ in $R=K[x_{1},\,\ldots,\,x_{5}]$, and, as usual, let $A=R/\mathfrak{f}$, which is a graded Gorenstein Artin ring. If $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2})=1$ and $\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})=0$ then

[TABLE]

where $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{5}^{2},x_{1}x_{2},x_{1}x_{3},x_{1}x_{4})$.

Proof.

By assumption there is a linear form in $\operatorname{Ann}_{A}(g_{1},g_{2})$, say $L$, such that $L$ does not annihilate $g_{3}$. Hence, Proposition 4.7 gives us \dim_{K}\big{(}(\mathfrak{f}+(g_{1},g_{2}))_{3}\cap g_{3}R_{1}\big{)}\leq 3. Then we get

[TABLE]

∎

Proposition 4.9**.**

Suppose that for all quadratic forms $g$ in $Kg_{1}+Kg_{2}$, the subspace $gA_{1}$ of $A_{3}$ is a $3$-dimensional. If $\dim_{K}(g_{1},g_{2})A_{1}=5$, then $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2})=1$.

We first state the following observation in a linear algebra setting, which will be useful for the proof Proposition 4.9.

Lemma 4.10**.**

Let $S$, $T$ be linear transformations from $V$ to $W$, both $n$-dimensional vector spaces over $K$, such that $\operatorname{rank}(S)=\operatorname{rank}(T)=\operatorname{rank}(S-T)=r$, and the kernels of $S$, $T$ are disjoint. Then the images of $S$ and $T$ are contained in the same $(3r-n)$-dimensional subspace of $W$.

Proof.

$V_{0}=\ker(S-T)$ is $(n-r)$-dimensional. $S$ and $T$ are injective on $V_{0}$, since for $v\in V_{0}$, $S(v)=0$ iff $T(v)=0$, and $\operatorname{Ker}(S)\cap\operatorname{Ker}(T)=0$. Thus, $S(V_{0})=T(V_{0})$ is an $(n-r)$-dimensional space in $S(V)\cap T(V)$. Since $S(V),T(V)$ are $r$-dimensional and overlap in a space of dimension at least $n-r$, $S(V)+T(V)$ has dimension at most $r+r-(n-r)=3r-n$. ∎

Proof of Proposition 4.9.

Assume that $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2})=0$. Since all quadratic forms $g$ in $Kg_{1}+Kg_{2}$ are such that $gA_{1}\subseteq A_{3}$ has vector space dimension 3, we have from Lemma 4.10 with $n=5,r=3$, that $(Kg_{1}+Kg_{2})A_{1}\subseteq A_{3}$ is at most $4$-dimensional. Consequently,

[TABLE]

contradicting $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for defect 2 quadratic ideals. Hence, $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2})=1$. ∎

Proposition 4.11**.**

Let $I=\mathfrak{f}+(g_{1},g_{2},g_{3})$ be a defect three quadratic ideal in $R=K[x_{1},\,\ldots,\,x_{5}]$. If $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})=0$ then $\operatorname{Hilb}_{R/I}(3)\leq 4$.

Proof.

First, by Remark 4.5 we note that it suffices to consider any defect two quadratic ideal $J\subseteq I$ with $\operatorname{Hilb}_{R/J}(3)=5$.

Suppose that $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})=0$. Then, clearly, no $g_{i}$, for $i=1,2,3$ has a $1$-dimensional linear annihilator space in $A$, since, otherwise, by Remark 4.6, we obtain that $\dim_{K}\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})=1$, which contradicts our assumption. Thus, for the rest of the proof we may assume that each $g_{i}A_{1}$, $i=1,2,3$, is either $3$ or $5$ dimensional.

If all forms $g$ in $Kg_{1}+Kg_{2}+Kg_{3}$ are such that $\dim_{K}gA_{1}=3$ then we can find two independent quadratic forms whose linear annihilator spaces intersect in $1$-dimensional space, and the result follows from Corollary 4.8.

Let $g_{1}A_{1}$ be a $5$-dimensional subspace of $A_{3}$ and for every $g\in Kg_{2}+Kg_{3}$, $gA_{1}$ has dimension either $3$ or $5$.

We complete the proof by obtaining a contradiction. We assume that $\operatorname{Hilb}_{R/I}(3)=5$. In other words, the space $W=(Kg_{1}+Kg_{2}+Kg_{3})A_{1}\subseteq A_{3}$ is 5-dimensional. Then we get $W=g_{1}A_{1}=(Kg_{2}+Kg_{3})A_{1}$.

Consider the multiplication maps by $g_{1},g_{2}$ and $g_{3}$ from $A_{1}$ to the subspace $W$ of $A_{3}$. By adjusting the bases of $A_{1}$ and $W$ we can assume the matrix of $g_{1}$ is the identity matrix $\mathbb{I}_{5}$ of size $5$. Denote the matrices of $g_{2}$ and $g_{3}$ by $\alpha$ and $\beta$, respectively. We can assume that $\alpha$ and $\beta$ are both singular, and so have rank $3$, by subtracting the suitable multiples of $\mathbb{I}_{5}$ from them if they are not singular.

We see that all matrices $z\mathbb{I}_{5}+x\alpha+y\beta$ must have at most two eigenvalues, otherwise we can form a linear combination whose kernel is $1$-dimensional, which corresponds to a quadratic form with $1$-dimensional linear annihilator space. Then there are two main cases: one is that every matrix in the space spanned by $\mathbb{I}_{5},\alpha$ and $\beta$ has one eigenvalue. The other is that almost all matrices in the form $z\mathbb{I}_{5}+x\alpha+y\beta$ have two eigenvalues, since the subset with at most one eigenvalue is Zariski closed.

Define $D(x,y,z)=\det(z\mathbb{I}_{5}-x\alpha-y\beta)$, a homogeneous polynomial in $x,y,z$ of degree $5$ that is monic in $z$. Note that $D$ is also the characteristic polynomial, in $z$, of $x\alpha+y\beta$. Notice that the singular matrices in the subspace of $5\times 5$ matrices spanned by $\mathbb{I},\alpha$ and $\beta$ are defined by the vanishing of $D$.

If the determinant $D$ is square-free (as the characteristic polynomial in $z$), then the ideal $(D)$ is a radical ideal and it cannot contain a nonzero polynomial of degree less than $5$, which contradicts the fact that all size $4$ minors of a singular matrix must vanish, since in our situation these singular matrices have rank $3$. Therefore the size $4$ minors, whose degrees are at most $4$, are in the radical $(D)$.

If the determinant $D$ is not square-free, then its squared factor must be linear or quadratic: in the latter case the other factor is linear, so that in either case $D$ has a linear factor, say $z-ax-by$.

Consider the independent matrices $\alpha^{\prime}=a\mathbb{I}_{5}-\alpha$, $\beta^{\prime}=b\mathbb{I}_{5}-\beta$. Then we think of any linear combination of them, say $r\alpha^{\prime}+s\beta^{\prime}=r(a\mathbb{I}_{5}-\alpha)+s(b\mathbb{I}_{5}-\beta)=(ar+bs)\mathbb{I}_{5}-r\alpha-s\beta.$ As $z-ax-by$ is a factor of $D(x,y,z)$, and hence, $D$ vanishes for $x=r,y=s,z=ar+bs.$ This means that every linear combination of $\alpha^{\prime}$ and $\beta^{\prime}$ is singular. Therefore, we can replace $\alpha,\beta$ by $\alpha^{\prime}$ and $\beta^{\prime}$ and so we can assume that we are in the case where every linear combination of the two non-identity matrices is singular, and, if not [math], of rank $3$. By Lemma 4.10, this implies that the kernels of $\alpha^{\prime}$ and $\beta^{\prime}$ cannot be disjoint, so we are done by Proposition 4.9 and Corollary 4.8.

∎

In order to prove $\operatorname{EGH}_{\operatorname{\underline{\mathbf{2}}}2,5}(2)$ for every defect three quadratic ideal $I=\mathfrak{f}+(g_{1},g_{2},g_{3})$ in $R=K[x_{1},\,\ldots,\,x_{5}]$ we must also discuss the cases when there is a nonzero linear form $L\in\operatorname{Ann}_{A}(g_{1},g_{2},g_{3})$.

Proposition 4.12**.**

Let $I=\mathfrak{f}+(g_{1},g_{2},g_{3})$ be a defect three quadratic ideal in $R$. If $\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})$ is a $1$-dimensional $K$-subspace of $A_{1}$, say $KL$, then

[TABLE]

Claim 4.13**.**

One of the quadratic forms $f_{i}$ in the regular sequence has the linear factor $L$.

Proof of claim.

As $g_{1},g_{2},g_{3}\in\operatorname{Ann}_{A_{2}}(L)\subseteq A_{2}$ for $L\in\operatorname{Ann}_{A_{1}}(g_{1},g_{2},g_{3})$ we know that

[TABLE]

This tells us that $\dim_{K}LA_{2}\leq 7,$ which implies

[TABLE]

as $\dim_{K}A_{3}=10$.

Assume that $L=x_{5}$ and let $\bar{f}_{i}$ be the image of $f_{i}$ modulo $x_{5}$ .

Suppose that $\bar{\mathfrak{f}}=(\bar{f}_{1},\bar{f}_{2},\bar{f}_{3},\bar{f}_{4},\bar{f}_{5})$ is an almost complete intersection in $K[x_{1},x_{2},x_{3},x_{4}]$. Thus,

[TABLE]

However, using the Francisco’s result for almost complete intersections [8], we know that

[TABLE]

This contradicts (4).

Hence the images of $f_{i}$ modulo $L$ form a regular sequence in $K[x_{1},\,\ldots,\,x_{4}]$, that is, one of them has a linear factor $x_{5}$. ∎

As a result of the claim, after a suitable change of variables, we may assume that the linear annihilator is $L=x_{5}$ and may consider $I$ in two possible forms: either $I$ is in the form of (5) in Case 1 below, where $f_{1},\,f_{2},\,f_{3},\,f_{4},\,x_{1}x_{5}$ is the regular sequence, or $I$ is as in (6) in Case 2 below, where $f_{1},\,f_{2},\,f_{3},\,f_{4},\,x_{5}^{2}$ form a quadratic regular sequence in $I$.

Case 1. Suppose that $f_{5}=x_{1}x_{5}$. Then we can assume that $g_{1}=x_{1}x_{2},\,g_{2}=x_{1}x_{3},\,g_{3}=x_{1}x_{4}$. Furthermore, after we alter the $f_{i}$ by getting rid of all the terms containing $x_{1}$ except $x_{1}^{2}$, we may assume that the defect three quadratic ideal $I$ looks like

[TABLE]

where $f_{1},f_{2},f_{3},f_{4}$ form a regular sequence in $K[x_{2},x_{3},x_{4},x_{5}]$ and $c\in K$.

Proposition 4.14**.**

Let $I=(f_{1},\,f_{2},\,f_{3},\,f_{4}+cx_{1}^{2},\,x_{1}x_{5},\,x_{1}x_{2},\,x_{1}x_{3},\,x_{1}x_{4})$ be a defect three quadratic ideal in $R$ where $f_{1},f_{2},f_{3},f_{4}$ is an $K[x_{2},x_{3},x_{4},x_{5}]$-sequence. Then

[TABLE]

where $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{5}^{2},\,x_{1}x_{2},\,x_{1}x_{3},\,x_{1}x_{4})$.

Proof.

One can easily see that $I$ contains all cubic monomials divisible by $x_{1}$ since $x_{1}x_{i}\in I$ for all $i=2,3,4,5$ and $f_{4}$ is a quadratic form in $K[x_{2},x_{3},x_{4},x_{5}]$, therefore $x_{1}f_{4}\in I$ and so is $x_{1}^{3}$. Thus, the Hilbert functions of $R/I$ and k[x_{2},x_{3},x_{4},x_{5}]\big{/}I\cap K[x_{2},x_{3},x_{4},x_{5}] agrees in degree $3$. So \operatorname{Hilb}_{R/I}(3)=\operatorname{Hilb}_{K[x_{2},x_{3},x_{4},x_{5}]\big{/}I\cap K[x_{2},x_{3},x_{4},x_{5}]}(3)=\operatorname{Hilb}_{K[x_{2},x_{3},x_{4},x_{5}]\big{/}(f_{1},\,f_{2},\,f_{3},\,f_{4})}(3)=4 ∎

Case 2. Suppose that $f_{5}=x_{5}^{2}$ by altering the variables and generators, and then we can assume that $g_{1}=x_{1}x_{5}$, $g_{2}=x_{2}x_{5}$, $g_{3}=x_{3}x_{5}$. As we did in the case above, we get rid of all the terms containing $x_{5}$ except $x_{4}x_{5}$ in the $f_{i}$, and so the defect three quadratic ideal can be written as follows:

[TABLE]

where $f_{1},\,f_{2},\,f_{3},\,f_{4}$ form a regular sequence in $K[x_{1},x_{2},x_{3},x_{4}]$ and $c\in K$.

Lemma 4.15**.**

Let $\mathfrak{a}=(f_{1},\,f_{2},\,f_{3},\,f_{4}+x_{4}x_{5},\,x_{5}^{2}):(x_{1}x_{5},\,x_{2}x_{5},\,x_{3}x_{5})$ be the colon ideal in $R$. Then we have $\operatorname{Hilb}_{R/\mathfrak{a}}(2)=6.$

Proof.

It suffices to show $\dim_{K}\mathfrak{a}_{2}=9$.

We know that $x_{1}x_{5},\,x_{2}x_{5},\,x_{3}x_{5},\,x_{4}x_{5},\,x_{5}^{2}$ are all in $\mathfrak{a}_{2}$, and $f_{1},\,f_{2},\,f_{3},\,f_{4}\in\mathfrak{a}_{2}$ as well. Thus we see that $\dim_{K}\mathfrak{a}_{2}\geq 9$.

If there is another independent quadratic form in $\mathfrak{a}$, it must be in $R[\check{x}_{5}]$, as we have all quadratic monomials containing $x_{5}$, so call it $Q$ in $R[\check{x}_{5}]$. Then we consider the cubic form $H=x_{5}Q.$ Clearly $H$ is not in the $R_{1}$-span of $f_{1},\,f_{2},\,f_{3},\,f_{4},\,x_{5}^{2}$, therefore we can define the ideal $J=(f_{1},\,f_{2},\,f_{3},\,f_{4},\,x_{5}^{2},\,H)$, which is an almost complete intersection in $R$. Then we get \dim_{K}\big{(}(f_{1},\,f_{2},\,f_{3},\,f_{4},x_{5}^{2})_{4}\,\cap\,HR_{1}\big{)}\geq 4 as $x_{1}H,x_{2}H,x_{3}H$ and $x_{5}H$ are in $(f_{1},\,f_{2},\,f_{3},\,f_{4},x_{5}^{2})_{4}$, but by Corollary 2.4 this dimension must be at most $3$. This proves that there cannot be such a quadratic form $Q$ in $\mathfrak{a}$. ∎

Proposition 4.16**.**

Let $I=(f_{1},\,f_{2},\,f_{3},\,f_{4}+x_{4}x_{5},\,x_{5}^{2},\,x_{1}x_{5},\,x_{2}x_{5},\,x_{3}x_{5})$ be a defect three quadratic ideal in $R$ where $f_{1},f_{2},f_{3},f_{4}$ is an $R[\check{x}_{5}]$-sequence. Then

[TABLE]

where $\operatorname{\mathcal{L}}=(x_{1}^{2},\,\ldots,\,x_{5}^{2},\,x_{1}x_{2},\,x_{1}x_{3},\,x_{1}x_{4})$.

Proof.

Using the duality of Gorenstein algebras, again we can obtain

[TABLE]

where $\mathfrak{a}$ is the colon ideal $(f_{1},\,f_{2},\,f_{3},\,f_{4}+x_{4}x_{5},\,x_{5}^{2}):I$.

Then proof is done, since $\operatorname{Hilb}_{R/(f_{1},\,f_{2},\,f_{3},\,f_{4}+x_{4}x_{5},\,x_{5}^{2})}(3)=10$ and $\operatorname{Hilb}_{R/\mathfrak{a}}(2)=6$ by the above lemma. ∎