On invariants of certain symmetric algebras

Abstract

Let ${\mathrm{Syz}}_1(\mathfrak {m})$ be the first syzygy of the graded maximal ideal $\mathfrak {m}$ of a polynomial ring $K[x_1,\ldots ,x_n]$ over a field K. The multiplicity and (Castelnuovo–Mumford) regularity of the symmetric algebra ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ are estimated by using the theory of s-sequences. It is proved that the multiplicity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ is 1 when $n\ge 5$, and $n-2$ is an upper bound for its regularity. In virtue of Gröbner bases, this bound is shown to be reached provided $n\le 5$.

1 Introduction

Let R be a Noetherian ring and $M=(f_1,\ldots ,f_n)$ be a finitely generated R-module. The symmetric algebra ${\mathrm{Sym}}(M)$ of M is a quotient ring of the polynomial ring $R[y_1,\ldots ,y_n]$ over R. Considering this presentation, s-sequences were introduced to study the properties of symmetric algebras in [5] (cf. [7, 12]). If M is generated by an s-sequence, one obtains exact values for the dimension ${\mathrm{dim}}({\mathrm{Sym}} (M))$ and the multiplicity $\mathrm{e}({\mathrm{Sym}} (M))$, and bounds for the depth ${\mathrm{depth}}({\mathrm{Sym}} (M))$ and the (Castelnuovo–Mumford) regularity ${{\text {reg}}} ({\mathrm{Sym}} (M))$ by the same invariants of some special quotients of R by the annihilator ideals.

Let K be a field, $K[x_1,\ldots ,x_n]$ be a polynomial ring over K and $\mathfrak {m}=(x_1,\ldots ,x_n)$. The first syzygy of $\mathfrak {m}$ is denoted by ${\mathrm{Syz}}_1(\mathfrak {m})$. Our topic is the symmetric algebra ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$. In [10], the authors obtained the dimension and depth of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$. In this paper, we will continue to study the multiplicity and regularity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$.

We calculate the multiplicity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ in Sect. 3. In order to get the regularity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$, we need to estimate the regularity of the initial ideals of certain annihilator ideals in Sect. 4, where some new results in [3] and [8] are applied. In Sect. 5, an upper bound for the regularity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ is given. When $n\le 5$, using Buchberger’s algorithm, we find a set of minimal generators for the second syzygy module of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$. The degrees of these generators give a lower bound for the regularity. Then we obtain an equality for the regularity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ provided $n\le 5$.

2 Preliminaries

Let R be a Noetherian ring and $M=(f_1,\ldots ,f_n)$ be a finitely generated R-module. Then M has a presentation

$$\begin{aligned} R^m \longrightarrow R^n \longrightarrow M \longrightarrow 0 \end{aligned}$$

with a relation matrix $A = (a_{ij})_{m\times n}$. The symmetric algebra ${\mathrm{Sym}}(M)$ has the presentation

$$\begin{aligned} R[y_1, \ldots , y_n]/J, \end{aligned}$$

where $J = (g_1, \ldots , g_m)$ and $g_i = \sum _{j=1}^n a_{ij}y_j$, $i = 1, \ldots , m$.

Let $P = R[y_1, \ldots ,y_n]$ which is a graded R-algebra. Then J is a graded ideal, and ${\mathrm{Sym}}(M)$ is a graded R-algebra. Assign degree one to each variable $y_i$ and degree zero to the elements of R. Let < be a monomial order induced by $y_1< \cdots < y_n$. For any $f \in P$, $f= \sum _\alpha a_\alpha y^\alpha$, we put in$(f) = a_\alpha y^\alpha$ where $y^\alpha$ is the largest monomial with respect to the given order such that $a_\alpha \ne 0$. We call in(f) the initial term of f and define the ideal

$$\begin{aligned} \text {in}(J) = (\text {in}(f) : f \in J), \end{aligned}$$

which is generated by monomials in $y_1, \ldots , y_n$ with coefficients in R and is finitely generated since P is Noetherian.

For $i = 1, \ldots , n$, we set $M_i = \sum _{j=1}^i Rf_j$ and let $I_i = M_{i-1}:_R f_i = \{a \in R : af_i \in M_{i-1}\}$. We also set $I_0 = 0$. Then $I_i$ is the annihilator ideal of the cyclic module $M_i/M_{i-1}$. It is clear that

$$\begin{aligned} (I_1y_1, \ldots , I_ny_n) \subseteq \text {in}(J), \end{aligned}$$

and the two ideals coincide in degree one.

Definition 2.1

The generators $f_1, \ldots , f_n$ of M are called an s-sequence (with respect to <), if

$$\begin{aligned} (I_1y_1, \ldots , I_ny_n) = \text {in}(J). \end{aligned}$$

If, in addition, $I_1 \subseteq I_2 \subseteq \cdots \subseteq I_n$, then $f_1, \ldots , f_n$ is called a strong s-sequence.

Let $S=K[x_1,\ldots ,x_d]$ be a polynomial ring over a field K and M a finitely generated graded S-module. Let

$$\begin{aligned} \cdots \rightarrow F_j{\mathop {\rightarrow }\limits ^{\phi _j}}\cdots \rightarrow F_1{\mathop {\rightarrow }\limits ^{\phi _1}} F_0\rightarrow M\rightarrow 0 \end{aligned}$$

be a graded minimal free resolution of M, where $F_j=\oplus _iS(-a_{ji})$. ${\mathrm{Im}}(\phi _j)$ is called the j-th syzygy module of M. One says that M is m-regular if $a_{ji}-j\le m$ for all i, j and defines the Castelnuovo–Mumford regularity (or regularity) of M by

$$\begin{aligned} {\mathrm{reg}}(M)={\mathrm{min}}\{m:\, M \text{ is } m\text{-regular }\}. \end{aligned}$$

Let J be a graded ideal of S. Notice that the i-th syzygy of J is just the $(i+1)$-th syzygy of S / J. It follows that ${\mathrm{reg}}(J)={\mathrm{reg}}(S/J)+1$. On the other hand, if g is a minimal generator of J, then ${\mathrm{reg}}(J)\ge {\mathrm{deg}}(g)$, and if h is a minimal generator of the first syzygy module of J, which is the second syzygy module of S / J, then ${\mathrm{reg}}(J)\ge {\mathrm{deg}}(h)-1$, and so on. For the properties of the regularity, we refer to [1].

Lemma 2.2

([5, Propositions 2.4 and 2.6]) Suppose that $f_1,\ldots ,f_n$ form a strong s-sequence and have the same degree. Let $d={\mathrm{dim}}({\mathrm{Sym}}(M))$. Then

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}(M))=\sum _{r\ge 0, {\mathrm{dim}}(R/I_r)=d-r}\mathrm{e}(R/I_r), \end{aligned}$$

and

$$\begin{aligned} {\mathrm{reg}}({\mathrm{Sym}}(M))\le {\mathrm{max}}\{{\mathrm{reg}}(I_r): r=1,\ldots ,n\}. \end{aligned}$$

Assume from now on that $n\ge 3$. Let K be a field, $S = K[x_1, \ldots , x_n]$ be a polynomial ring and $\mathfrak {m}= (x_1, \ldots , x_n)$ be the graded maximal ideal of S. Denote the first syzygy of $\mathfrak {m}$ by ${\mathrm{Syz}}_1(\mathfrak {m})$. From the Koszul complex of S with respect to $x_1,\ldots ,x_n$, one has a presentation of ${\mathrm{Syz}}_1(\mathfrak {m})$ as an S-module

$$\begin{aligned} S^{\left( {\begin{array}{c}n\\ 3\end{array}}\right) } \longrightarrow S^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) } \longrightarrow {\mathrm{Syz}}_1(\mathfrak {m}) \longrightarrow 0. \end{aligned}$$

It follows that the symmetric algebra of ${\mathrm{Syz}}_1(\mathfrak {m})$ has the presentation

$$\begin{aligned} {\mathrm{Sym}}_S({\mathrm{Syz}}_1(\mathfrak {m})) = S[y_{ij}: 1\le i<j\le n]/J, \end{aligned}$$

where J is the ideal of $S[y_{ij}: 1\le i<j\le n]$ generated by the set

$$\begin{aligned} \left\{ x_i y_{jk} - x_j y_{ik} + x_k y_{ij}: 1 \le i< j < k \le n\right\} . \end{aligned}$$

Since the generators of ${\mathrm{Syz}}_1(\mathfrak {m})$ do not form an s-sequence in general with respect to the term order $x_i<y_{12}<y_{13}<\cdots <y_{n-1,n}$, we cannot apply the theory of s-sequences in this form. However, the Jacobian dual of ${\mathrm{Syz}}_1(\mathfrak {m})$ can help us.

Let $Q=K[y_{ij}:1\le i<j\le n]$. Then the Jacobian dual ${\mathrm{Syz}}_1(\mathfrak {m})^{\vee }$ of ${\mathrm{Syz}}_1(\mathfrak {m})$ is a Q-module with a presentation

$$\begin{aligned} Q^{\left( {\begin{array}{c}n\\ 3\end{array}}\right) } \longrightarrow Q^n \longrightarrow {\mathrm{Syz}}_1(\mathfrak {m})^{\vee } \longrightarrow 0, \end{aligned}$$

and

$$\begin{aligned} {\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))\cong {\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})^{\vee })\cong Q[x_1,\ldots ,x_n]/J, \end{aligned}$$

cf. [11]. We will use this new presentation of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ to estimate its multiplicity and regularity.

Lemma 2.3

([7, Lemma 3.1 and Proposition 3.3]) Let $Y=(y_{ij})_{n\times n}$ be the skew-symmetric matrix where $y_{ij}=-y_{ji}$ and $y_{ii}=0$. Then the set

$$\begin{aligned} \{x_iy_{jk}-x_jy_{ik}+x_ky_{ij}:1\le i<j<k\le n\}\cup \{x_rP_2(Y):1\le r\le n\} \end{aligned}$$

is a Gröbner basis of J with respect to the term order

$$\begin{aligned} x_n>x_{n-1}>\cdots>x_1>y_{1n}>y_{1,n-1}>\cdots>y_{12}>y_{2n}>y_{2,n-1}>\cdots >y_{n-1,n}, \end{aligned}$$

where $P_2(Y)$ is the set of all 4-Pfaffians of Y.

Let $x_i^*$ be the image of $x_i$ in ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$, $i = 1, \ldots , n$. Then $x_1^*,\ldots , x_n^*$ is a strong s-sequence with the annihilator ideal

$$\begin{aligned} I_r= & {} (\{y_{ij}:1\le i< j< r\}\cup P_2(Y))\\= & {} (\{y_{ij}: 1\le i< j< r\}\cup \{ y_{il}y_{jk}-y_{ik}y_{jl}+y_{ij}y_{kl}:1\le i< j< k <\ell \le n\}),\\&r=1,\ldots ,n, \end{aligned}$$

which are ideals of Q.

Notice that $I_1=I_2$, and when $n=3$, $I_1=I_2=0$.

Then, by Lemma 2.2,

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=\sum _{r\ge 0, {\mathrm{dim}}(Q/I_r)=d-r}\mathrm{e}(Q/I_r), \end{aligned}$$

where $d={\mathrm{dim}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))$, and

$$\begin{aligned} {\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))\le {\mathrm{max}}\{{\mathrm{reg}}(I_r): r=1,\ldots ,n\}. \end{aligned}$$

3 Multiplicity of the symmetric algebra

For the multiplicity of the symmetric algebra ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$, we have the following equalities.

Theorem 3.1

If $n\ne 4$, then

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=1, \end{aligned}$$

and, if $n=4$ then

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=5. \end{aligned}$$

Proof

Let $d={\mathrm{dim}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))$. Then, by [10, Theorem 4.1], $d={\mathrm{max}}\{\frac{n(n-1)}{2},2n-1\}$. Let us calculate the multiplicity $\mathrm{e}(Q/I_r)$ with ${\mathrm{dim}}(Q/I_r)=d-r$. Notice that $I_1=I_2$ and, by [10, Proposition 3.4], ${\mathrm{dim}}(Q/I_r)=2n-1-r$ for $r\ge 2$.

Firstly, suppose that $n=3$. In this case, $d=5$, $I_1=I_2=0$, $I_3=(y_{12})$ and ${\mathrm{dim}}(Q/I_3)=2=d-3$. Then

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1({\mathfrak {m}})))=\mathrm{e}(Q/I_3)=1. \end{aligned}$$

Now assume that $n=4$. Then $d=7$ and ${\mathrm{dim}}(Q/I_r)+r=d$ for $r=2,3,4$. We have

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=\mathrm{e}(Q/I_2)+\mathrm{e}(Q/I_3)+\mathrm{e}(Q/I_4). \end{aligned}$$

Notice that

$$\begin{aligned} Q/I_2= & {} Q/(P_2(Y)),\\ Q/I_3= & {} K[y_{13},y_{14},y_{23},y_{24},y_{34}]/(y_{14}y_{23}-y_{13}y_{24}),\\ Q/I_4= & {} K[y_{14},y_{24},y_{34}]. \end{aligned}$$

For the multiplicity of a Pfaffian ideal, by [6, Theorem 5.6], we have the following result: If Y is a $2r\times 2r$ generic skew matrix of indeterminates and $R=K[Y]$, then

$$\begin{aligned} \mathrm{e}(R/(P_r(Y)))={\mathrm{det}}\left[ {2\atopwithdelims ()-i+j+1}-{2\atopwithdelims ()-i-j+1}\right] _{i,j=1,\ldots ,r-1}, \end{aligned}$$

from which we have

$$\begin{aligned} \mathrm{e}(Q/I_2)={2\atopwithdelims ()1}-{2\atopwithdelims ()-1}=2. \end{aligned}$$

For the multiplicity of a determinantal ideal, there is a well-known result (cf. [6]): If X is an $n\times n$ generic matrix of indeterminates and $R=K[X]$, then

$$\begin{aligned} \mathrm{e}(R/({\mathrm{det}}(X)))={\mathrm{det}}\left[ {2n-i-j\atopwithdelims ()n-i}\right] _{i,j=1,\ldots ,n-1}, \end{aligned}$$

which implies that

$$\begin{aligned} \mathrm{e}(Q/I_3)=\mathrm{e}(K[y_{13},y_{14},y_{23},y_{24}]/(y_{14}y_{23}-y_{13}y_{24}))={2\atopwithdelims ()1}=2. \end{aligned}$$

It is clear that $\mathrm{e}(Q/I_4)=1$. Hence, in this case, $\mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=5$.

Finally, suppose that $n\ge 5$. Then $d=\frac{n(n-1)}{2}\ne r+{\mathrm{dim}}(Q/I_r)$ for $r=1,\ldots ,n$. Hence

$$\begin{aligned} \mathrm{e}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=\mathrm{e}(Q)=1. \end{aligned}$$

The proof is complete. $\square$

4 Regularity of annihilators

Let us estimate the regularity ${\mathrm{reg}}(I_r)$ of an annihilator ideal $I_r$. Notice that $Q/I_r=Q_r/I'_r$ where $Q_r=K[y_{ij}:1\le i<j\le n,j\ge r]$ and

$$\begin{aligned} I'_r= & {} (\{y_{il}y_{jk}-y_{ik}y_{jl}:1\le i<j<r\le k<l\le n\}\\&\cup \{y_{il}y_{jk}-y_{ik}y_{jl}+y_{ij}y_{kl}:1\le i<j;r\le j<k<l\le n\}). \end{aligned}$$

Then ${\mathrm{reg}}(I_r)={\mathrm{reg}}(I'_r)$.

In [10], in order to calculate the dimension and depth of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$, the following result was proved.

Lemma 4.1

([10, Lemma 3.3 and Proposition 3.4]) $Q_r/I'_r$ is Cohen–Macaulay of dimension $2n-1-r$ for $r\ge 2$ and

$$\begin{aligned} {\mathrm{in}}(I'_r)=({\mathrm{in}}(A)\,:\, A \text{ is } \text{ a } 2\text{-minor } \text{ of } Y'_r) \end{aligned}$$

with the term order

$$\begin{aligned} y_{1n}>y_{1,n-1}>\cdots>y_{12}>y_{2n}>y_{2,n-1}>\cdots>y_{23}>\cdots \cdots >y_{n-1,n}, \end{aligned}$$

where

$$\begin{aligned} Y'_r=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} y_{1r}&{}y_{1,r+1}&{}\cdots &{}y_{1n}\\ &{}\cdots &{}\cdots &{}\\ y_{r-1,r}&{}y_{r-1,r+1}&{}\cdots &{}y_{r-1,n}\\ &{}y_{r,r+1}&{}\cdots &{}y_{rn}\\ &{}&{}\cdots &{}\\ &{}&{}&{}y_{n-1,n} \end{array}\right) . \end{aligned}$$

Let $Z_{nr}'$ be the mirror symmetry of $Y_r'$:

$$\begin{aligned} Z'_{nr}=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} z_{11}&{}z_{12}&{}\cdots &{}z_{1,n-r}&{}z_{1,n-r+1}\\ &{}&{}\cdots &{}\cdots &{}\\ z_{r-1,1}&{}z_{r-1,2}&{}\cdots &{}z_{r-1,n-r}&{}z_{r-1,n-r+1}\\ z_{r1}&{}z_{r2}&{}\cdots &{}z_{r,n-r}&{}\\ &{}&{}\cdots &{}&{}\\ \ z_{n-1,1}&{}&{}&{}&{} \end{array}\right) , \end{aligned}$$

i.e., $z_{ij}=y_{i,n-j+1}$, and let the term order $<'$ be as the following

$$\begin{aligned} z_{11}>'z_{12}>'\cdots>'z_{1,n-r+1}>'z_{21}>'z_{22}>'\cdots>'z_{2,n-r+1}>'\cdots \cdots >'z_{n-1,1}. \end{aligned}$$

Notice that, by changing the variables from $y_{ij}$ to $z_{ij}$ and the term order from < to $<'$, $Q_r$ and ${\mathrm{in}}(I'_r)$ remain the same. Then

$$\begin{aligned} {\mathrm{in}}(I'_r)=({\mathrm{in}}_{<'}(B)\,:\, B \text{ is } \text{ a } 2\text{-minor } \text{ of } Z'_{nr}). \end{aligned}$$

Let $I_2(Z'_{nr})$ be the ideal of $Q_r$ generated by all the 2-minors of $Z'_{nr}$. Since $Z'_{nr}$ is a ladder, by [9, Corollary 3.4], the set of 2-minors of $Z'_{nr}$ forms a Gröbner basis. It follows that ${\mathrm{in}}(I'_r)={\mathrm{in}}_{<'}(I_2(Z'_{nr}))$.

Set $[n]=\{1,\ldots ,n\}$. Let us identify $z_{ij}$ with its index (i, j). Then $Z'_{nr}$ is an ideal poset of $[n]\times [n]$ where $(i,j)\le (k,l)$ if and only if $i\le k$ and $j\le l$ (cf. [4, §9.1.2]). For any set S, denote the set of indeterminates $x_s, s\in S$, by $x_S$. Let $L(2,Z'_{nr})$ be the monomial ideal of $K[x_{[2]\times Z'_{nr}}]$ generated by the following monomials:

$$\begin{aligned} x_{1p}x_{2q},\,\, p,q\in Z'_{nr},p\le q. \end{aligned}$$

By [2, Theorem 2.4], $L(2,Z'_{nr})$ is a Cohen–Macaulay ideal, hence, unmixed ($L(n,Z'_{nr})$ is just $I_{2,2}(Z'_{nr})$ with the notation of [2]). Then Theorem 4.4 of [8] claims that the regularity of $K[x_{[2]\times Z'_{nr}}]/L(2,Z'_{nr})$ is just the maximal cardinality of an antichain in $Z'_{nr}$, where an antichain in $Z'_{nr}$ is a sequence of points in $Z'_{nr}$ with the property that any two points are incomparable. From the shape of $Z'_{nr}$, it is easy to see that this maximal cardinality is $n-r+1$. Then, we have proved the following

Lemma 4.2

${\mathrm{reg}}(K[x_{[2]\times Z'_{nr}}]/L(2,Z'_{nr}))=n-r+1.$

In [3], the authors developed one method by which one gets the same regularity by cutting down a regular sequence from $K[x_{[2]\times Z'_{nr}}]/L(2,Z'_{nr})$.

Let

$$\begin{aligned} \phi \,:\, [2]\times Z'_{nr}\rightarrow & {} [n+1]\times [n+1]\\ (1,i,j)\mapsto & {} (i,j)\\ (2,i,j)\mapsto & {} (i+1,j+1), \end{aligned}$$

and $L^{\phi }(2,Z'_{nr})$ be the monomial ideal of $K[x_{[n+1]\times [n+1]}]$ generated by the following monomials:

$$\begin{aligned} x_{ij}x_{i'+1,j'+1},\,\, (i,j),(i',j')\in Z'_{nr}, (i,j)\le (i',j'). \end{aligned}$$

Then, by [3, Corollary 2.3], the following result holds.

Lemma 4.3

([3, Corollary 2.3]) The quotient rings $K[x_{[2]\times Z'_{nr}}]/L(2,Z'_{nr})$ and

$K[x_{[n+1]\times [n+1]}]/L^{\phi }(2,Z'_{nr})$ have the same regularity.

Notice that the generators of $L^{\phi }(2,Z'_{nr})$ are just the initial terms of the 2-minors of $Z'_{n+2,r+1}$ (identifying $x_{ij}$ with $z_{ij}$) with respect to the term order $<'$, which form a Gröbner basis as we have noted before. It follows that $L^{\phi }(2,Z'_{nr})={\mathrm{in}}_{<'}(I_2(Z'_{n+2,r+1}))$.

Then we have the following crucial lemma.

Lemma 4.4

For $r=3,\ldots ,n$,

$$\begin{aligned} {\mathrm{reg}}({\mathrm{in}}_{<'}(I_2(Z'_{nr})))=n-r+1. \end{aligned}$$

Proof

It is because

$$\begin{aligned} {\mathrm{reg}}({\mathrm{in}}_{<'}(I_2(Z'_{nr})))= & {} {\mathrm{reg}}(K[x_{[n+1]\times [n+1]}]/L^{\phi }(2,Z'_{n-2,r-1}))+1\\= & {} {\mathrm{reg}}(K[x_{[2]\times Z'_{n-2,r-1}}]/L(2,Z'_{n-2,r-1}))+1\\= & {} n-r+1, \end{aligned}$$

where the last equality follows from Lemma 4.2. $\square$

5 Regularity of the symmetric algebra

Now, we can estimate the regularity of the symmetric algebra ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$.

Theorem 5.1

If $n\ge 3$, then

$$\begin{aligned} {\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))\le n-2. \end{aligned}$$

Proof

Notice that $I_1=I_2$, ${\mathrm{in}}_{<'}(I_2(Z_{n2}'))={\mathrm{in}}_{<'}(I_2(Z'_{n3}))$ and

$$\begin{aligned} {\mathrm{reg}}(I_r)={\mathrm{reg}}(I'_r)\le {\mathrm{reg}}({\mathrm{in}}(I'_r))={\mathrm{reg}}({\mathrm{in}}_{<'}(I_2(Z_{nr}'))). \end{aligned}$$

Then, by Lemmas 2.2 and 4.4, we have that

$$\begin{aligned} {\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))\le & {} {\mathrm{max}}\{{\mathrm{reg}}(I_r):r=2,\ldots ,n\}\\\le & {} {\mathrm{max}}\{{\mathrm{reg}}({\mathrm{in}}_{<'}(I_2(Z'_{nr}))):r=2,\ldots ,n\}\\= & {} {\mathrm{max}}\{{\mathrm{reg}}({\mathrm{in}}_{<'}(I_2(Z'_{nr}))):r=3,\ldots ,n\}\\= & {} {\mathrm{max}}\{n-r+1:r=3,\ldots ,n\}\\= & {} n-2. \end{aligned}$$

$\square$

The above theorem obtains an inequality for the regularity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$. We wish that the other direction’s inequality could also hold. For this purpose, let us check a graded minimal free resolution of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$. For any $1\le i<j<k\le n$, set $g_{ijk}=y_{ij}x_k-y_{ik}x_j+y_{jk}x_i$. Then ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))=Q[x_1,\ldots ,x_n]/J$ and J is minimally generated by $g_{ijk}$, $1\le i<j<k\le n$. We can construct the first two steps of a graded minimal free resolution of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$:

$$\begin{aligned} \cdots \longrightarrow \bigoplus _{1\le i<j<k\le n}Q[x_1,\ldots ,x_n]e_{ijk}(-2){\mathop {\longrightarrow }\limits ^{\phi _1}} Q[x_1,\ldots ,x_n]\longrightarrow {\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})) \longrightarrow 0, \end{aligned}$$

where $\phi _1(e_{ijk})=g_{ijk}$. From the generators of the first syzygies of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$, i.e., $g_{ijk}$, are all of degree 2, we see immediately from the definition of regularity that ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))\ge 1$. Hence ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=n-2$ when $n=3$. However this bound is not big enough for $n\ge 4$. We have to consider the degrees of the minimal generators of the second syzygies of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$.

Now assume that $n\ge 4$. In order to get that ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))\ge n-2$, it is enough to find one minimal generator of ${\mathrm{Ker}}(\phi _1)$ which has degree n. Since we have found a Gröbner basis for J, it is possible, as pointed out in [1, page 335], to get a set of generators for the second syzygy module of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ by Buchberger’s algorithm:

Lemma 5.2

(cf. [1, Theorem 15.10]) Let $S=K[X_1,\ldots ,X_n]$ and $g_1,\ldots ,g_s$ be a set of minimal generators of a graded ideal I of S. Suppose that $g_1,\ldots ,g_s,g_{s+1},\ldots , g_t$ form a Gröbner basis for I with respect to a term order <. Then, for any $1\le i<j\le t$, the S-pair

$$\begin{aligned} S(g_i,g_j):=m_{ji}g_i-m_{ij}g_j=\sum _uf_u^{(ij)}g_u,\,\, {\mathrm{in}}(f_u^{(ij)}g_u)<{\mathrm{in}}(m_{ji}g_i), \end{aligned}$$

where $m_{ij}=\frac{{\mathrm{in}}(g_i)}{\mathrm{gcd}({\mathrm{in}}(g_i),{\mathrm{in}}(g_j))}$. Substituting $g_{s+1},\ldots ,g_t$ in the above expressions in terms of $g_1,\ldots ,g_s$, one has that

$$\begin{aligned} m_{ji}g_i-m_{ij}g_j=\sum _{u=1}^sh_u^{(ij)}g_u,\,\, {\mathrm{in}}(h_u^{(ij)}g_u)<{\mathrm{in}}(m_{ji}g_i), 1\le i<j\le s. \end{aligned}$$

Define an S-homomorphism

$$\begin{aligned} \phi \,:\, \bigoplus _{i=1}^sSe_i\rightarrow & {} I\\ e_i\mapsto & {} g_i. \end{aligned}$$

Set $\tau _{ij}=m_{ji}e_i-m_{ij}e_j-\sum _{u=1}^sh_u^{(ij)}e_u$. Then the set $\{\tau _{ij} : 1\le i<j \le s\}$ generates ${\mathrm{Ker}}(\phi )$, i.e., the first syzygy module of I.

Notice that, once a set of generators is given as above, some generator $\tau _{i_0,j_o}$ is minimal if and only if $\tau _{i_0,j_o}$ is not a linear combination of other generators in this set. We will use this idea to find a satisfied minimal generator.

It is clear that every element of ${\mathrm{Ker}}(\phi _1)$ is a linear combination:

$$\begin{aligned} \sum _{1\le i<j<k\le n}f_{ijk}e_{ijk},\,\, f_{ijk}\in Q[x_1,\ldots ,x_n]. \end{aligned}$$

We call $f_{ijk}$ the coefficient of $e_{ijk}$, which is a polynomial in variables $x_1,\ldots ,x_n$, $y_{ij}$, $1\le i<j\le n$. Sometimes, we write such a linear combination as $f_{123}e_{123}+\cdots$. Notice that the degree of $f_{ijk}e_{ijk}$ is equal to the degree of $f_{ijk}$ plus two. Therefore, in order to get that ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))\ge n-2$, it is enough to find one minimal homogeneous generator of ${\mathrm{Ker}}(\phi _1)$

$$\begin{aligned} \sum _{1\le i<j<k\le n}f_{ijk}e_{ijk},\,\, f_{ijk}\in Q[x_1,\ldots ,x_n], \end{aligned}$$

where all the nonzero coefficients $f_{ijk}$ are of degree $n-2$. We will find such generators when $n=4$ or 5 by using Lemma 5.2.

Theorem 5.3

When $n=3$ or 4, or $n=5$ and ${\mathrm{char}}(K)\ne 2$,

$$\begin{aligned} {\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))= n-2. \end{aligned}$$

Proof

We may assume that $n\ge 4$. Define an order on $Q[x_1,\ldots ,x_n]$ as follows

$$\begin{aligned} x_n>x_{n-1}>\cdots>x_1>y_{12}>y_{13}>\cdots>y_{23}>\cdots >y_{n-1,n}. \end{aligned}$$

Set $P_{ijkl}^{(r)}=x_r(y_{ij}y_{kl}-y_{ik}y_{jl}+y_{il}y_{jk})$ for $1\le i<j<k<l\le n, 1\le r\le n$. We will use the same notation for $g_{ijk}$ and $P_{ijkl}^{(r)}$ when i, j, k or i, j, k, l are only different and not necessarily in the above order.

Let us follow the Buchberger’s algorithm to get a Gröbner basis from $g_{ijk},\, 1\le i<j<k\le n$ and then, try to find a minimal first syzygy of degree 4 or 5 when $n=4$ or 5. The first step is to compute S-pairs $S(g_{ijk},g_{stl})$ where we may assume that the initial terms of $g_{ijk}$ and $g_{stl}$ are not co-prime. There are two possibilities: $(i,j)=(s,t), k\ne l$ or $(i,j)\ne (s,t), k=l$. In the first case, we need to compute $S(g_{ijk},g_{ijl})$ with $k<l$. One has

$$\begin{aligned} S(g_{ijk},g_{ijl})= & {} x_lg_{ijk}-x_kg_{ijl}\\= & {} -y_{ik}x_jx_l+y_{jk}x_ix_l+y_{il}x_jx_k-y_{jl}x_ix_k\\= & {} -x_jg_{ikl}+x_ig_{jkl}, \end{aligned}$$

which induces a generator of the first syzygy module of J:

$$\begin{aligned} x_le_{ijk}-x_ke_{ijl}+x_je_{ikl}-x_ie_{jkl}. \end{aligned}$$

Notice that this generator is of degree 3 and with coefficients in variables $x_u$. We will see that this kind of generators would not appear in the second case. We discuss the second case according to $n=4$ or 5.

Assume firstly that $n=4$. Then $S(g_{ijk},g_{stk})$ with $(i,j)\ne (s,t)$ are all the following

$$\begin{aligned} S(g_{124},g_{134})= & {} y_{13}g_{124}-y_{12}g_{134}=y_{14}g_{123}-P^{(1)}_{1234},\\ S(g_{124},g_{234})= & {} y_{23}g_{124}-y_{12}g_{234}=y_{24}g_{123}-P^{(2)}_{1234},\\ S(g_{134},g_{234})= & {} y_{23}g_{134}-y_{13}g_{234}=y_{34}g_{123}-P^{(3)}_{1234}. \end{aligned}$$

It follows that the following elements form a Gröbner basis (cf. Lemma 2.3)

$$\begin{aligned} P_{1234}^{(1)}, P_{1234}^{(2)},P_{1234}^{(3)}, g_{ijk}, 1\le i<j<k\le 4. \end{aligned}$$

Substituting $P_{1234}^{(1)}$ and $P_{1234}^{(2)}$ in the S-pair $S(P_{1234}^{(1)},P_{1234}^{(2)})=x_2P_{1234}^{(1)}-x_1P_{1234}^{(2)}$, one gets that

$$\begin{aligned} (x_2y_{14}-x_1y_{24})g_{123}+(-x_2y_{13}+x_1y_{23})g_{124}+x_2y_{12}g_{134}-x_1y_{12}g_{234}=0, \end{aligned}$$

which induces to the following degree 4 syzygy:

$$\begin{aligned} (x_2y_{14}-x_1y_{24})e_{123}+(-x_2y_{13}+x_1y_{23})e_{124}+x_2y_{12}e_{134}-x_1y_{12}e_{234}. \end{aligned}$$

It is clear that this syzygy is not a multiple of the unique degree 3 syzygy $x_4e_{123}-x_3e_{124}+x_2e_{134}-x_1e_{234}$. Therefore the above syzygy is minimal. It follows that ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))= 2$ when $n=4$.

Now assume that $n=5$. The possible cases for $S(g_{ijk},g_{stk})$ with $(i,j)\ne (s,t)$ are the following

$$\begin{aligned} S(g_{125},g_{345})= & {} y_{34}g_{125}-y_{12}g_{345}=y_{35}g_{124}-y_{45}g_{123}-P_{1345}^{(2)}+P_{2345}^{(1)},\\ S(g_{135},g_{245})= & {} y_{24}g_{135}-y_{13}g_{245}=y_{25}g_{134}+y_{45}g_{123}-P_{1245}^{(3)}-P_{2345}^{(1)},\\ S(g_{145},g_{235})= & {} y_{23}g_{145}-y_{14}g_{235}=-y_{15}g_{234}+y_{45}g_{123}-P_{1245}^{(3)}+P_{1345}^{(2)}, \end{aligned}$$

and, for all $1\le i<j<k<l\le 5$,

$$\begin{aligned} S(g_{ijl},g_{ikl})= & {} y_{ik}g_{ijl}-y_{ij}g_{ikl}=y_{il}g_{ijk}-P^{(i)}_{ijkl},\\ S(g_{ijl},g_{jkl})= & {} y_{jk}g_{ijl}-y_{ij}g_{jkl}=y_{jl}g_{ijk}-P^{(j)}_{ijkl},\\ S(g_{ikl},g_{jkl})= & {} y_{jk}g_{ikl}-y_{ik}g_{jkl}=y_{kl}g_{ijk}-P^{(k)}_{ijkl}. \end{aligned}$$

Assume that ${\mathrm{char}}(K)\ne 2$. Then from the above first three equations, we can solve $P_{1245}^{(3)}$, $P_{1345}^{(2)}$ and $P_{2345}^{(1)}$:

$$\begin{aligned} P_{1245}^{(3)}= & {} \frac{1}{2}y_{45}g_{123}+\cdots ,\\ P_{1345}^{(2)}= & {} -\frac{1}{2}y_{45}g_{123}+\cdots ,\\ P_{2345}^{(1)}= & {} -\frac{1}{2}y_{45}g_{123}+\cdots . \end{aligned}$$

It follows that the following elements form a Gröbner basis (cf. Lemma 2.3)

$$\begin{aligned} P_{1245}^{(3)},P_{1345}^{(2)},P_{2345}^{(1)},P_{ijkl}^{(i)},P_{ijkl}^{(j)},P_{ijkl}^{(k)},1\le i<j<k<l\le 5, g_{ijk},1\le i<j<k\le 5. \end{aligned}$$

Notice that, as a conclusion, there are no syzygies on $\{g_{ijk}\}$ of degree 3 with coefficients in variables $y_{uv}$.

From

$$\begin{aligned} S\left( P_{1234}^{(2)},P_{2345}^{(2)}\right)= & {} y_{23}y_{45}P_{1234}^{(2)}-y_{12}y_{34}P_{2345}^{(2)}\\= & {} (y_{24}y_{35}-y_{25}y_{34})P_{1234}^{(2)}-(y_{13}y_{24}-y_{14}y_{23})P_{2345}^{(2)} \end{aligned}$$

and substituting $P_{2345}^{(2)}=y_{23}g_{245}-y_{24}g_{235}+y_{25}g_{234}$ and $P_{1234}^{(2)}=y_{24}g_{123}-y_{23}g_{124}+y_{12}g_{234}$, we get a syzygy of degree 5 with coefficients in variables $y_{uv}$

$$\begin{aligned} (*)\qquad \,\,\, y_{24}(y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34})e_{123}+\cdots . \end{aligned}$$

We will prove that the above syzygy $(*)$ is minimal. Then ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))= 3$ follows.

Notice that if the syzygy $(*)$ is not minimal, then $(*)$ should be a linear combination of some degree 4 syzygies whose monomials of the coefficient of $e_{123}$ divide the monomials of $y_{24}(y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34})$. Let us identify all such possible degree 4 syzygies. We will use the exclusive method.

The possible cases appear only in the S-pairs $S(P_{ijkl}^{(r)},g_{uvr})$ with $(u,v)\ne (i,j)$ and (k, l) or $S(P_{ijkl}^{(r)},P_{stuv}^{(r)})$ with $(i,j)=(s,t)$ or $(k,l)=(u,v)$.

For the first case $S(P_{ijkl}^{(r)},g_{uvr})$, r must be 3 or 4. When $r=3$, there are only two subcases: $S(P_{1345}^{(3)},g_{123})$ and $S(P_{2345}^{(3)},g_{123})$. When $r=4$, it is just $S(P_{2345}^{(4)},g_{234})$. From

$$\begin{aligned} S\left( P_{1345}^{(3)},g_{123}\right)= & {} y_{12}P_{1345}^{(3)}-y_{13}y_{45}g_{123}\\= & {} (-y_{14}y_{45}+y_{15}y_{34})g_{123}+y_{13}P_{1345}^{(2)}-y_{23}P_{1345}^{(1)}, \end{aligned}$$

we see that this case should be excluded because $y_{1i}$ appears in the coefficients of $e_{123}$. However, one has that

$$\begin{aligned} S\left( P_{2345}^{(3)},g_{123}\right)= & {} y_{12}P_{2345}^{(3)}-y_{23}y_{45}g_{123}\\= & {} (-y_{24}y_{35}+y_{25}y_{34})g_{123}+y_{13}P_{2345}^{(2)}-y_{23}P_{2345}^{(1)},\\ S\left( P_{2345}^{(4)},g_{124}\right)= & {} y_{12}P_{2345}^{(4)}-y_{23}y_{45}g_{124}\\= & {} (-y_{24}y_{35}+y_{25}y_{34})g_{124}+y_{14}P_{2345}^{(2)}-y_{24}P_{2345}^{(1)}, \end{aligned}$$

from which we get two syzygies:

$$\begin{aligned}&\left( \frac{1}{2}y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34}\right) e_{123}+\cdots ,\\&\quad \frac{1}{2}y_{24}y_{45}e_{123}+\cdots . \end{aligned}$$

For the second case $S(P_{ijkl}^{(r)},P_{stuv}^{(r)})$ with $(i,j)=(s,t)$ or $(k,l)=(u,v)$, there are six possibilities: $S(P_{1245}^{(r)}, P_{1345}^{(r)})$ with $r\le 4$, $S(P_{1235}^{(r)},P_{1245}^{(r)})$ with $r\le 3$, $S(P_{1234}^{(r)},P_{1245}^{(r)})$ with $r\le 3$, $S(P_{1234}^{(r)},P_{1235}^{(r)})$ with $r\le 3$, $S(P_{1345}^{(r)}, P_{2345}^{(r)})$ with $r\le 4$ and $S(P_{1245}^{(r)},P_{2345}^{(r)})$ with $r\le 4$.

Since

$$\begin{aligned} S\left( P_{1245}^{(r)},P_{1345}^{(r)}\right) =y_{13}P_{1245}^{(r)}-y_{12}P_{1345}^{(r)}= y_{14}P_{1235}^{(r)}-y_{15}P_{1234}^{(r)}, \end{aligned}$$

and its coefficients are all with some $y_{1s}$, we see immediately that this possibility is excluded. Similarly for

$$\begin{aligned} S\left( P_{1235}^{(r)},P_{1245}^{(r)}\right) =y_{45}P_{1235}^{(r)}-y_{35}P_{1245}^{(r)}=y_{15}P_{2345}^{(r)}-y_{25}P_{1345}^{(r)}, \end{aligned}$$

the coefficients of $g_{123}$ in the explains of $y_{45}P_{1235}^{(r)}$, $y_{35}P_{1245}^{(r)}$ and $y_{25}P_{1345}^{(r)}$ do not divide any monomials of $y_{24}(y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34})$ in any cases; this possibility should also be excluded. Now consider $S(P_{1234}^{(r)},P_{1245}^{(r)})$ with $r\le 3$. In the equality

$$\begin{aligned} S\left( P_{1234}^{(r)},P_{1245}^{(r)}\right) =y_{45}P_{1234}^{(r)}-y_{34}P_{1245}^{(r)}= -y_{24}P_{1345}^{(r)}+y_{14}P_{2345}^{(r)}, \end{aligned}$$

only when $r=2$, $y_{45}P_{1234}^{(r)}$ contains $y_{24}y_{45}g_{123}$ and $y_{24}P_{1345}^{(r)}$ contains $-\frac{1}{2}y_{24}y_{45}g_{123}$. Thus, in this possibility, one gets only one required syzygy: $\frac{1}{2}y_{24}y_{45}e_{123}+\cdots$.

For the two possibilities that $S(P_{1234}^{(r)},P_{1235}^{(r)})$ with $r\le 3$ and $S(P_{1345}^{(r)}, P_{2345}^{(r)})$ with $r\le 4$, in the equalities

$$\begin{aligned} S\left( P_{1234}^{(r)},P_{1235}^{(r)}\right)= & {} y_{35}P_{1234}^{(r)}-y_{34}P_{1235}^{(r)}= -y_{23}P_{1345}^{(r)}+y_{13}P_{2345}^{(r)},\\ S\left( P_{1345}^{(r)},P_{2345}^{(r)}\right)= & {} y_{23}P_{1345}^{(r)}-y_{13}P_{2345}^{(r)}= -y_{35}P_{1234}^{(r)}+y_{34}P_{1235}^{(r)}, \end{aligned}$$

only when $r=2$, $y_{23}P_{1345}^{(r)}$ has $-\frac{1}{2}y_{23}y_{45}g_{123}$, $y_{35}P_{1234}^{(r)}$ has $y_{24}y_{35}g_{123}$, and $y_{34}P_{1235}^{(r)}$ has $y_{25}y_{34}g_{123}$. It turns out, in these two possibilities, there is only one required syzygy: $(-\frac{1}{2}y_{23}y_{45}+y_{24}y_{35}-y_{25}y_{34})e_{123}+\cdots$.

Finally, for $S(P_{1245}^{(r)},P_{2345}^{(r)})$ with $r\le 4$, in the equality

$$\begin{aligned} S\left( P_{1245}^{(r)},P_{2345}^{(r)}\right) =y_{23}P_{1245}^{(r)}-y_{12}P_{2345}^{(r)}=-y_{25}P_{1234}^{(r)}+y_{24}P_{1235}^{(r)}, \end{aligned}$$

only when $r=3$, $y_{23}P_{1245}^{(r)}$ has $\frac{1}{2}y_{23}y_{45}g_{123}$; when $r=2$, $y_{25}P_{1234}^{(r)}$ has $y_{24}y_{25}g_{123}$, and when $r=3$, $y_{25}P_{1234}^{(r)}$ has $y_{25}y_{34}g_{123}$; when $r=2$, $y_{24}P_{1235}^{(r)}$ has $y_{24}y_{25}g_{123}$, and when $r=3$, $y_{24}P_{1235}^{(r)}$ has $y_{24}y_{35}g_{123}$. Therefore, only when $r=3$, there is a syzygy $(\frac{1}{2}y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34})e_{123}+\cdots$.

To summarize the results obtained, there are only two degree 4 syzygies whose monomials of the coefficient of $e_{123}$ divide the monomials of $y_{24}(y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34})$:

$$\begin{aligned} \left( \frac{1}{2}y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34}\right) e_{123}+\cdots \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2}y_{24}y_{45}e_{123}+\cdots . \end{aligned}$$

It is clear that they cannot generate $y_{24}(y_{23}y_{45}-y_{24}y_{35}+y_{25}y_{34})e_{123}+\cdots$. Therefore the syzygy $(*)$ is minimal, as required. $\square$

Remark 5.4

When $n=3$, the equality ${\mathrm{reg}}({\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})))=1$ can be easily seen because ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m})) =Q[x_1,x_2,x_3]/(x_1y_{23}-x_2y_{13}+x_3y_{12})$ and $x_1y_{23}-x_2y_{13}+x_3y_{12}$ is homogeneous of degree two.

On the other hand, when $n=6$, one might hope naturally to find one degree 6 second syzygy by using the same method as above. Unfortunately it is impossible because, in this case, we can find a set of generators of degree at most 5 for the first syzygy of elements $\{g_{ijk}\}$ by using the computer algebra system CoCoA. Therefore, to get an equality for the regularity of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$ in the case $n\ge 6$, one might need to find a satisfied third syzygy of ${\mathrm{Sym}}({\mathrm{Syz}}_1(\mathfrak {m}))$, which is challenging.