# Symbolic powers of generalized star configurations of hypersurfaces

Kuei-Nuan Lin and Yi-Huang Shen  Department of Mathematics, The Penn State
University, McKeesport, PA, 15132, USA<EMAIL_ADDRESS>CAS Wu Wen-Tsun Key
Laboratory of Mathematics, School of Mathematical Sciences, University of
Science and Technology of China, Hefei, Anhui, 230026, P.R. China
<EMAIL_ADDRESS>
###### Abstract.

We introduce the class of sparse symmetric shifted monomial ideals. These
ideals have linear quotients and their Betti numbers are computed. Using this,
we prove that the symbolic powers of the generalized star configuration ideal
are sequentially Cohen–Macaulay under some mild genericness assumption. With
respect to these symbolic powers, we also consider the Harbourne–Huneke
containment problem and establish the Demailly-like bound.

2020 Mathematics Subject Classification. 13A15, 13A50, 13D02, 14N20, 52C35.

Keyword: Betti numbers, linear quotients, star configuration, symbolic power,
containment problem, stable Harbourne–Huneke conjecture.

##  1\. Introduction

In the simplest case, a _star configuration_ of codimension $c$ in
${\mathbb{P}}^{n}$ is a certain union of linear subspaces $V_{1},\dots,V_{p}$
each of codimension $c$. The study of various aspects of generalizations of
star configurations has attracted considerable attention recently; see for
example, [MR2492444], [MR3197149], [MR3003727], [MR3683102],
[arXiv:1912.04448], [MR4177276] and [arXiv:1906.08346], to name just a few. As
a sequel to our previous work [arXiv:1912.04448] along this line, we continue
to investigate the symbolic powers of the generalized star configuration of
hypersurfaces.

Let $R={\mathbb{K}}[x_{0},\dots,x_{n}]$ be a standard graded polynomial ring
over an infinite ${\mathbb{K}}$ and denote its graded maximal ideal by
$\mathfrak{m}$. Suppose that $s\geq n+1$ and let
$\mathcal{F}=\\{f_{1},\dots,f_{s}\\}$ be a set of forms in $R$ satisfying some
genericness condition. For a fixed choice of positive integers $a$ and $b$
such that $a\leq bs$, the _generalized star configuration ideal_ that we shall
study here is the _uniform $a$-fold product ideal_
$I_{a}(\mathcal{F}^{b})\coloneqq I_{a}(f_{1}^{b}\cdots f_{s}^{b})$.

In general, the _$a$ -fold product ideal_ $I_{a}(f_{1}^{b_{1}}\cdots
f_{s}^{b_{s}})$ is generated by the $a$-fold products of the forms
$f_{1},\dots,f_{s}$ with multiplicities $b_{1},\dots,b_{s}$ respectively:

$\Braket{f_{1}^{n_{1}}\cdots f_{s}^{n_{s}}:0\leq n_{i}\leq b_{i}\text{ for
each $i$ such that $\sum\nolimits_{i}n_{i}=a$}}.$

It has also been studied extensively recently; see for example
[zbMATH06759435] and [MR3864202]. This notion was originally introduced as a
nice tool for determining the minimum distance of linear codes in the coding
theory. And it also emerges naturally when dealing with the higher order
Orlik–Terao algebra of hyperplane arrangements.

The $a$-fold product ideal $I_{a}(f_{1}^{b_{1}}\cdots f_{s}^{b_{s}})$ is
_uniform_ when the multiplicities $b_{1},\dots,b_{s}$ are all identical, say,
to some positive integer $b$. Whence, we will write it as
$I_{a}(\mathcal{F}^{b})$ for simplicity. It is treated as the _generalized
star configuration ideal_ , since when $b=1$, we will get back the star
configuration ideal of hypersurfaces

$I_{c,\mathcal{F}}\coloneqq\bigcap_{1\leq i_{1}<\cdots<i_{c}\leq
s}\braket{f_{i_{1}},\dots,f_{i_{c}}}$

for $c=s-a+1$, which were studied for instance in [MR3683102] and [MR4177276].
In their work, the set $\mathcal{F}$ is often assumed to be _$c$ -generic_,
namely any subset of size at most $c+1$ will form a regular sequence.

Our current work will focus on the study of the symbolic powers of the
generalized star configuration ideal $I_{a}(\mathcal{F}^{b})$. Recall that
give an ideal $I$ in a ring $R$, the _$m$ -th symbolic power_ of $I$ is
defined by

$I^{(m)}\coloneqq\bigcap_{\mathfrak{p}\in\operatorname{Ass}(R/I)}I^{m}R_{\mathfrak{p}}\cap
R.$

For instance, when $I$ is the defining ideal of a reduced affine scheme over
an algebraically closed field of characteristic zero, Zariski and Nagata
showed that $I^{(m)}$ is generated by the polynomials whose partial
derivatives of orders up to $m-1$ vanish on this scheme. And in the previous
work [arXiv:1912.04448], we have already studied the resurgence and symbolic
defect of $I_{a}(\mathcal{F}^{b})$, as well as the Betti table and the primary
decomposition of its symbolic powers.

In [MR3003727] and [MR3683102], Geramita, Harbourne, Migliore, and Nagel
showed that the symbolic powers of the ordinary star configuration ideal
define arithmetically Cohen–Macaulay subschemes. It is then natural to ask if
the same property still holds in the generalized case. Notice that the ideal
$I_{a}(\mathcal{F}^{b})$ is no longer unmixed for $b\geq 2$. Thus, the correct
question can be put as: whether the symbolic powers of this ideal are all
sequentially Cohen–Macaulay? We provide a positive answer under some mild
genericness assumption.

In this work, we also study the Harbourne–Huneke containment problem, which
was originally raised in [MR3115195]. After [arXiv:2009.05022], a generalized
version can be stated as: given a homogeneous ideal $I$ of big height $h$ in a
standard graded ring with the maximal homogeneous ideal $\mathfrak{m}$, does
the inequality

$I^{(\ell(h+m-1))}\subseteq\mathfrak{m}^{\ell(h-1)}(I^{(m)})^{\ell}$

hold for all $m,\ell\geq 1$? The comparison between ordinary powers of an
ideal and its symbolic ideals is wildly open in general; see, for example,
[MR2629595] and [MR1881923]. We provide a positive answer for Harbourne–Huneke
containment problem in the case $I=I_{a}(\mathcal{F}^{b})$ under some mild
genericness assumption. As a by-product, we establish the Demailly-like bound
in this case as well.

As usual, for a homogeneous ideal $J$, we will write $\alpha(J)$ for the least
degree of nonzero forms in this ideal. Then, our main results can be
summarized in the following.

######  Theorem 1.1.

Let $a$ and $b$ be positive integers such that $(b-1)s+1<a\leq bs$. We assume
further that $n\geq bs-a+1$ and $\mathcal{F}=\\{f_{1},\dots,f_{s}\\}$ is a set
of $(bs-a+1)$-generic $d$-forms in $R={\mathbb{K}}[x_{0},\dots,x_{n}]$. Then
the following properties hold for the generalized star configuration ideal
$I=I_{a}(\mathcal{F}^{b})$ whose big height is known to be $h=bs-a+1$.

  1. a

The symbolic powers of $I$ are all sequentially Cohen–Macaulay.

  2. b

The ideal $I$ satisfies

$I^{(\ell(h+m-1)-h+k)}\subseteq\mathfrak{m}^{d((\ell-1)(h-1)+k-1)(bh-(bs-a))}(I^{(m)})^{\ell}$

for all positive integers $k,\ell$, and $m$.

  3. c

The inequality

$\frac{\alpha(I^{(\ell)})}{\ell}\geq\frac{\alpha(I^{(m)})+h-1}{m+h-1}$

holds for all $\ell,m\geq 1$.

##  2\. Sparse symmetric shifted ideals

To show the sequentially Cohen–Macaulayness of the symbolic powers of the
generalized star configuration ideal $I_{a}(\mathcal{F}^{b})$, we have to
introduce first the sparse symmetric shifted ideals. These squarefree monomial
ideals can be viewed as variations of the symmetric shifted ideals introduced
in [MR4108337]. Likewise, the new ideals have linear quotients (2.6) and the
Betti tables can be obtained combinatorially with ease (2.18).

###  2.1. Definition and the linear quotient property

The ground ring that we shall consider is the polynomial ring
$Z={\mathbb{K}}[z_{i,j}:1\leq i\leq s,1\leq j\leq b]$ over an arbitrary field
${\mathbb{K}}$. For each tuple
${\bm{\mu}}=(\mu_{1},\dots,\mu_{s})\in\\{0,1,2,\dots,b\\}^{s}$, a monomial of
the form ${\bm{z}}^{{\bm{\mu}}}\coloneqq\prod_{i=1}^{s}z_{i,\mu_{i}}$ will be
called a _sparse squarefree monomial_. Here, when $\mu_{i}=0$, $z_{i,\mu_{i}}$
is understood to be the identity element $1\in Z$.

Recall that a tuple
${\bm{\lambda}}=(\lambda_{1},\dots,\lambda_{s})\in{\mathbb{Z}}_{\geq 0}^{s}$
is called a _partition_ of length $s$, if
$\lambda_{1}\leq\cdots\leq\lambda_{s}$. For the sparse squarefree monomial
$f={\bm{z}}^{{\bm{\mu}}}$ above, we may suppose that the multi-set

$\\{\mu_{1},\dots,\mu_{s}\\}=\\{\lambda_{1}\leq\cdots\leq\lambda_{s}\\}.$

Now, we write

${\bm{\lambda}}(f)\coloneqq(\lambda_{1},\dots,\lambda_{s})\in{\mathbb{Z}}_{\geq
0}^{s},$

and call it the _partition_ associated to $f$. If
${\bm{\lambda}}(f)=(0,\dots,0,e_{1},\dots,e_{d})$ where $e_{1}\geq 1$, then
the _support degree_ of both $f$ and ${\bm{\lambda}}={\bm{\lambda}}(f)$ are

$\operatorname{suppdeg}(f)=\operatorname{suppdeg}({\bm{\lambda}})\coloneqq d,$

and the _weight_ of both $f$ and ${\bm{\lambda}}$ are

$w(f)=|{\bm{\lambda}}|\coloneqq
e_{1}+\cdots+e_{d}=\lambda_{1}+\cdots+\lambda_{s}=\mu_{1}+\cdots+\mu_{s}.$

Related, a sparse squarefree monomial in $Z$ whose support degree is $s$, is
called a _rainbow monomial_ in [arXiv.1912.03898, Subsection 1.3].

Let
${\bm{\mu}}^{\prime}=(\mu_{1}^{\prime},\dots,\mu_{s}^{\prime})\in\\{0,1,2,\dots,b\\}^{s}$
be another tuple. We will write ${\bm{\mu}}^{\prime}\leq_{0}{\bm{\mu}}$ if

  * •

$\mu_{i}^{\prime}\leq\mu_{i}$ for each $i$, and

  * •

$\mu_{i}^{\prime}=0$ if and only if $\mu_{i}=0$.

Notice that the permutation group $\mathfrak{S}_{s}$ has an action on $Z$,
such that for each permutation $\sigma\in\mathfrak{S}_{s}$ and a variable
$z_{i,j}$ in $Z$, we have $\sigma(z_{i,j})=z_{\sigma(i),j}$.

######  Definition 2.1.

A squarefree monomial proper ideal $J$ of $Z$ will be called _sparse symmetric
shifted_ if it satisfies the following requirements.

  1. (S-1)

The ideal $J$ is generated by some sparse squarefree monomials.

  2. (S-2)

For each minimal monomial generator $f={\bm{z}}^{{\bm{\mu}}}\in G(J)$ with
${\bm{\mu}}=(\mu_{1},\dots,\mu_{s})$, if
$f^{\prime}={\bm{z}}^{{\bm{\mu}}^{\prime}}$ with
${\bm{\mu}}^{\prime}=(\mu_{1}^{\prime},\dots,\mu_{s}^{\prime})$ and
${\bm{\mu}}^{\prime}\leq_{0}{\bm{\mu}}$, then $f^{\prime}\in J$.

  3. (S-3)

The ideal $J$ is invariant under the action of $\mathfrak{S}_{s}$.

######  Remark 2.2.

We may imagine the variables in $Z$ displayed in the canonical matrix
${\bm{Z}}=(z_{i,j})_{s\times b}$. Then, roughly speaking, the sparse symmetry
shifted ideal above has sparseness, vertical (upward) shiftiness, and
horizontal symmetry. And these properties correspond to the (S-1), (S-2), and
(S-3) respectively. Meanwhile, the symmetric shifted ideal, introduced in
[MR4108337], has both horizontal shiftiness and horizontal symmetry, since it
lacks the sparse structure here.

######  Lemma 2.3.

Let $J$ be a sparse symmetric shifted ideal in $Z$. Suppose that there is some
sparse squarefree monomial $f=\prod_{i}z_{\sigma(i),\mu_{i}}\in J$ for some
$\sigma\in\mathfrak{S}_{s}$. If
$f^{\prime}=\prod_{i}z_{\sigma^{\prime}(i),\mu_{i}^{\prime}}$ for some
$\sigma^{\prime}\in\mathfrak{S}_{s}$ with
$(\mu_{1}^{\prime},\dots,\mu_{s}^{\prime})\leq_{0}(\mu_{1},\dots,\mu_{s})$,
then $f^{\prime}\in J$ as well.

###### Proof.

By the symmetry condition (S-3), it suffices to consider the case when
$\sigma=\sigma^{\prime}=\operatorname{id}\in\mathfrak{S}_{s}$. Since $f\in J$,
there exists some minimal monomial generator $g=\prod_{i}z_{i,\tau_{i}}\in
G(J)$ such that $\tau_{i}=\mu_{i}$ or $0$ for each $i$. Define
$\tau_{i}^{\prime}=\min(\tau_{i},\mu_{i}^{\prime})$ for each $i$. Since
$(\tau_{1}^{\prime},\dots,\tau_{s}^{\prime})\leq_{0}(\tau_{1},\dots,\tau_{s})$,
the monomial $g^{\prime}\coloneqq\prod_{i}z_{i,\tau_{i}^{\prime}}\in J$ by the
shiftiness condition (S-2). As $g^{\prime}$ divides $f^{\prime}$, this implies
that $f^{\prime}\in J$ as well. ∎

######  Corollary 2.4.

Let $J$ be a sparse symmetric shifted ideal in $Z$. Suppose that
$f=\prod_{i}z_{\sigma(i),\lambda_{i}}\in J$ where
${\bm{\lambda}}(f)=(\lambda_{1},\dots,\lambda_{s})$ and
$\sigma\in\mathfrak{S}_{s}$. If $\lambda_{i_{1}}=0<k\leq\lambda_{i_{2}}$ for
some $i_{1}$ and $i_{2}$, then $f^{\prime}\coloneqq
fz_{\sigma(i_{1}),k}/z_{\sigma(i_{2}),\lambda_{i_{2}}}\in J$.

######  Remark 2.5.

Suppose that the ideal $J\subset Z$ is generated by some sparse squarefree
monomials. If $f$ is a monomial in $Z$, then the colon ideal $J:_{Z}f$ is also
generated by some sparse squarefree monomials that are coprime to $f$. To see
this, it suffices to notice that when $g$ is a sparse squarefree monomial,
then $\braket{g}:f$ is generated by the sparse squarefree monomial
$g/\gcd(f,g)$.

Here is the first main result of this section.

######  Theorem 2.6.

If $J$ is a sparse symmetric shifted ideal in $Z$, then it has linear
quotients.

###### Proof.

Inspired by the treatment in [MR4108337], we first consider the lexicographic
ordering $>_{{\operatorname{lex}}}$ on $Z$ where the variables
$z_{i,j}>z_{i^{\prime},j^{\prime}}$ if and only if

  * •

$j<j^{\prime}$, or

  * •

$j=j^{\prime}$ and $i<i^{\prime}$.

Thus, among those squarefree monomials of degree $s$, $z_{1,1}z_{2,1}\cdots
z_{s,1}$ is the initial (biggest) one with respect to
$>_{{\operatorname{lex}}}$.

Next, we take a total order $\prec$ on ${\mathbb{Z}}_{\geq 0}^{s}$ defined by

${\bm{a}}=(a_{1},\dots,a_{s})\prec{\bm{b}}=(b_{1},\dots,b_{s})$

if and only if

  * •

$\operatorname{suppdeg}({\bm{a}})<\operatorname{suppdeg}({\bm{b}})$, or

  * •

$\operatorname{suppdeg}({\bm{a}})=\operatorname{suppdeg}({\bm{b}})$ and
$|{\bm{a}}|<|{\bm{b}}|$, or

  * •

$\operatorname{suppdeg}({\bm{a}})=\operatorname{suppdeg}({\bm{b}})$ and
$|{\bm{a}}|=|{\bm{b}}|$, while the leftmost non-zero entry of
$(a_{1}-b_{1},\dots,a_{s}-b_{s})$ is negative.

Now, we are ready to introduce a total order on the minimal generating set
$G(J)$, which, by abuse of notation, will still be denoted by $\prec$. For
$f_{1}$ and $f_{2}$ in $G(J)$, we define $f_{1}\prec f_{2}$ if and only if

  * •

${\bm{\lambda}}(f_{1})\prec{\bm{\lambda}}(f_{2})$, or

  * •

${\bm{\lambda}}(f_{1})={\bm{\lambda}}(f_{2})$ and
$f_{1}>_{{\operatorname{lex}}}f_{2}$.

Then, we sort the squarefree monomials in $G(J)$ with respect to $\prec$. To
establish the linear quotient property, we take arbitrary minimal monomial
generator $f\in G(J)$ and let

$J_{f}\coloneqq\braket{g\in G(J):g\prec f}.$

The remaining task is to show that the colon ideal $J_{f}:_{Z}f$ is linear for
each such $f$. We can accomplish this task with the following two steps.

  1. i

First of all, we describe explicitly the set $\mathcal{G}_{f}$ of variables
that belong to this colon ideal.

  2. ii

Next, we prove that the set $\mathcal{G}_{f}$ actually generates the colon
ideal $J_{f}:f$ in general. Thus, $J_{f}:f$ is linear, as expected.

They are established in the coming 2.7 and 2.8 respectively. And consequently,
the ideal $J$ has linear quotients. ∎

In the remaining of this subsection, we fix a monomial $f$ in $G(J)$. The
first task is to describe the set $\mathcal{G}_{f}$ which collects the
variables in the colon ideal $J_{f}:f$. Suppose that the partition
${\bm{\lambda}}(f)=(\lambda_{1},\dots,\lambda_{s})$. Now, there is some
permutation $\sigma\in\mathfrak{S}_{s}$ such that
$f=\prod_{i=1}^{s}z_{\sigma(i),\lambda_{i}}$. Recall that for $i$ with
$\lambda_{i}=0$, $z_{\sigma(i),\lambda_{i}}$ is understood to be the identity
element $1\in Z$. And by abuse of notation, we define
$\max(f)\coloneqq\max\Set{\sigma(k):\lambda_{k}=\lambda_{s}}$.

######  Lemma 2.7.

The set $\mathcal{G}_{f}$ is the disjoint union of the following three types
of sets:

  * •

$\\{z_{\sigma(i),j}:1\leq j<\lambda_{i}\text{ and }\lambda_{i}>0\\}$,

  * •

$\\{z_{\sigma(i),j}:1\leq j<\lambda_{s}\text{ and }\lambda_{i}=0\\}$, and

  * •

$\\{z_{\sigma(i),\lambda_{s}}:\sigma(i)<\max(f)\text{ and }\lambda_{i}=0\\}$.

A quick example illustrating the linear generating set $\mathcal{G}_{f}$ above
is given in 3.4.

###### Proof.

To confirm the assertion, roughly speaking, we have two cases.

  1. A

Firstly, we consider the case when $\lambda_{i}>0$. We will show that
$z_{\sigma(i),j}\in\mathcal{G}_{f}$ if and only if $1\leq j<\lambda_{i}$.

    1. i

When $1\leq j<\lambda_{i}$, we look at the squarefree monomial
$f^{\prime}\coloneqq fz_{\sigma(i),j}/z_{\sigma(i),\lambda_{i}}$. By (S-2), we
have $f^{\prime}\in J$. Thus, we can find suitable $f^{\prime\prime}\in G(J)$
that divides $f^{\prime}$. Obviously, we have
$\operatorname{suppdeg}(f^{\prime\prime})\leq\operatorname{suppdeg}(f^{\prime})=\operatorname{suppdeg}(f)$
and $w(f^{\prime\prime})\leq w(f^{\prime})<w(f)$. It follows that
$f^{\prime\prime}\prec f$. Now,
$z_{\sigma(i),j}f\in\braket{f^{\prime}}\subseteq\braket{f^{\prime\prime}}\subseteq
J_{f}$. Consequently, $z_{\sigma(i),j}\in J_{f}:f$, as expected.

    2. ii

Suppose that $j=\lambda_{i}$. It follows from 2.5 that
$z_{\sigma(i),\lambda_{i}}\notin\mathcal{G}_{f}$.

    3. iii

When $j>\lambda_{i}$, we suppose for contradiction that
$z_{\sigma(i),j}\in\mathcal{G}_{f}$. Since now $fz_{\sigma(i),j}\in J_{f}$,
there exists some $f^{\prime\prime}\in G(J_{f})$ such that $f^{\prime\prime}$
divides $fz_{\sigma(i),j}$. If $z_{\sigma(i),j}$ does not divide this
$f^{\prime\prime}$, then $f^{\prime\prime}$ divides $f$ directly,
contradicting the minimality of $f$. Thus, $z_{\sigma(i),j}$ divides
$f^{\prime\prime}$. Since $f^{\prime\prime}$ is a sparse monomial,
$z_{\sigma(i),\lambda_{i}}$ does not divide $f^{\prime\prime}$. Consequently,
$f^{\prime\prime}$ divides the squarefree monomial $f^{\prime}\coloneqq
fz_{\sigma(i),j}/z_{\sigma(i),\lambda_{i}}$. Now, $f^{\prime}\in J$ such that
$\operatorname{suppdeg}(f^{\prime})=\operatorname{suppdeg}(f)$ and
$|{\bm{\lambda}}(f^{\prime})|>|{\bm{\lambda}}(f)|$. Hence, we don’t have
$f^{\prime}\prec f$. Consequently, $f^{\prime\prime}\neq f^{\prime}$ and
$\operatorname{suppdeg}(f^{\prime\prime})<\operatorname{suppdeg}(f^{\prime})=\operatorname{suppdeg}(f)$.
Next, we look at $f^{\prime\prime\prime}\coloneqq
f^{\prime\prime}z_{\sigma(i),\lambda_{i}}/z_{\sigma(i),j}$, which belongs to
$J$ by (S-2). Notice that $f^{\prime\prime\prime}$ properly divides $f$, as
$f^{\prime\prime}$ properly divides $f^{\prime}$. This contradicts the
minimality of $f$. Therefore, $z_{\sigma(i),j}\notin J_{f}:f$, as expected.

  2. B

Next, we consider the case when $\lambda_{i}=0$. We will show that
$z_{\sigma(i),j}\in\mathcal{G}_{f}$ if and only if $1\leq j<\lambda_{s}$, or
$j=\lambda_{s}$ and $\sigma(i)<\max(f)$.

    1. i

When $1\leq j<\lambda_{s}$, we look at the squarefree monomial
$f^{\prime}\coloneqq fz_{\sigma(i),j}/z_{\sigma(s),\lambda_{s}}$, which
belongs to $J$ by 2.4. Then, we can argue as in the item Ai of A to see that
$z_{\sigma(i),j}\in J_{f}:f$.

    2. ii

If $j>\lambda_{s}$, we suppose for contradiction that
$z_{\sigma(i),j}\in\mathcal{G}_{f}$. As in the item Aiii of A, there exists
some $f^{\prime\prime}\in G(J_{f})$ such that $f^{\prime\prime}$ divides
$fz_{\sigma(i),j}$, and $z_{\sigma(i),j}$ divides $f^{\prime\prime}$ by the
minimality of $f$. Furthermore, since $f^{\prime\prime}\prec f$, we have
$\operatorname{suppdeg}(f^{\prime\prime})\leq\operatorname{suppdeg}(f)$ and
$f^{\prime\prime}$ properly divides $fz_{\sigma(i),j}$. We have two subcases
to consider here.

      1. a

If $\operatorname{suppdeg}(f^{\prime\prime})=\operatorname{suppdeg}(f)$, there
is some $i^{\prime}\neq i$ such that $\lambda_{i^{\prime}}>0$ and
$f^{\prime\prime}=fz_{\sigma(i),j}/z_{\sigma(i^{\prime}),\lambda_{i^{\prime}}}$.
Since $\operatorname{suppdeg}(f^{\prime\prime})=\operatorname{suppdeg}(f)$
while $w(f^{\prime\prime})>w(f)$, this contradicts the condition
$f^{\prime\prime}\prec f$.

      2. b

If $\operatorname{suppdeg}(f^{\prime\prime})<\operatorname{suppdeg}(f)$, there
exists some subset

$\mathcal{I}\subseteq\Set{k:1\leq k\leq s,\lambda_{k}>0},$

such that $|\mathcal{I}|\geq 2$, $i\notin\mathcal{I}$ and
$f^{\prime\prime}=fz_{\sigma(i),j}\big{/}\prod_{k\in\mathcal{I}}z_{\sigma(k),\lambda_{k}}$.
We can fix an arbitrary $k_{0}\in\mathcal{I}$ and look at

$f^{\prime\prime\prime}\coloneqq
f^{\prime\prime}z_{\sigma(k_{0}),\lambda_{k_{0}}}/z_{\sigma(i),j}=f\bigg{/}\prod_{k\in\mathcal{I}\setminus\\{k_{0}\\}}z_{\sigma(k),\lambda_{k}}.$

Since $j>\lambda_{k_{0}}$, it follows from 2.3 that $f^{\prime\prime\prime}\in
J$ by considering the pair $(f^{\prime\prime},f^{\prime\prime\prime})$.
However, $f^{\prime\prime\prime}$ divides $f$ properly. This contradicts the
minimality of $f$.

Since we get contradiction in both subcases,
$z_{\sigma(i),j}\notin\mathcal{G}_{f}$, as expected.

    3. iii

If $\sigma(i)<\max(f)$, we look at the squarefree monomial
$f^{\prime}\coloneqq fz_{\sigma(i),\lambda_{s}}/z_{\max(f),\lambda_{s}}$,
which belongs to $J$ by (S-3). Notice that although
${\bm{\lambda}}(f^{\prime})={\bm{\lambda}}(f)$, we have
$f^{\prime}>_{{\operatorname{lex}}}f$. Thus, $f^{\prime}\prec f$, which in
turn implies $z_{\sigma(i),\lambda_{s}}\in\mathcal{G}_{f}$.

    4. iv

It remains to consider the situation when $\sigma(i)\geq\max(f)$. Since
$\sigma(i)=0<\lambda_{s}$, this amounts to saying that $\sigma(i)>\max(f)$.
Now, we suppose for contradiction that
$z_{\sigma(i),\lambda_{s}}\in\mathcal{G}_{f}$. Like in the case Bii, we can
find some $f^{\prime\prime}\in G(J_{f})$ such that $z_{\sigma(i),\lambda_{s}}$
divides $f^{\prime\prime}$ and $f^{\prime\prime}$ properly divides
$fz_{\sigma(i),\lambda_{s}}$.

      1. a

If $\operatorname{suppdeg}(f^{\prime\prime})=\operatorname{suppdeg}(f)$, there
is some $i^{\prime}\neq i$ such that $\lambda_{i^{\prime}}>0$ and
$f^{\prime\prime}=fz_{\sigma(i),\lambda_{s}}/z_{\sigma(i^{\prime}),\lambda_{i^{\prime}}}$.
We always have $\lambda_{i^{\prime}}\leq\lambda_{s}$. If
$\lambda_{i^{\prime}}<\lambda_{s}$, then $w(f^{\prime\prime})>w(f)$,
contradicting the assumption that $f^{\prime\prime}\prec f$. On the other
hand, if $\lambda_{i^{\prime}}=\lambda_{s}$, then
${\bm{\lambda}}(f^{\prime\prime})={\bm{\lambda}}(f)$. As
$\sigma(i)>\max(f)\geq\sigma(i^{\prime})$, we have
$f>_{{\operatorname{lex}}}f^{\prime\prime}$. And this again contradicts the
assumption that $f^{\prime\prime}\prec f$.

      2. b

If $\operatorname{suppdeg}(f^{\prime\prime})<\operatorname{suppdeg}(f)$, the
proof will be similar to that in item B(ii)b of Bii, with $j$ there replaced
by $\lambda_{s}$ here. Note that we can still apply 2.3.

Since we get contradiction in both subcases,
$z_{\sigma(i),\lambda_{s}}\notin\mathcal{G}_{f}$, as expected.

Thus, we have proved the 2.7. ∎

By the definition of $\mathcal{G}_{f}$, we have
$\braket{\mathcal{G}_{f}}\subseteq J_{f}:f$. The remaining task is to show
this is actually equality.

######  Lemma 2.8.

We have $\braket{\mathcal{G}_{f}}=J_{f}:f$ in general.

###### Proof.

Suppose for contradiction that we only have
$\braket{\mathcal{G}_{f}}\subsetneq J_{f}:f$. Then, we can find suitable $g\in
G(J_{f}:f)$ such that $g\notin\braket{\mathcal{G}_{f}}$. We have mentioned in
2.5 that this $g$ is a sparse squarefree monomial which is coprime to $f$.
Thus, there exist some nonempty
$\mathcal{J}\subseteq[s]\coloneqq\\{1,2,\dots,s\\}$ and positive integers
$\lambda_{j}^{\prime}$ for each $j\in\mathcal{J}$, such that
$g=\prod_{j\in\mathcal{J}}z_{\sigma(j),\lambda_{j}^{\prime}}$. Thus, as
$g\notin\mathcal{G}_{f}$ and $g$ is coprime to $f$, for each
$j\in\mathcal{J}$, we have:

  1. i

if $\lambda_{j}>0$, then $\lambda_{j}^{\prime}>\lambda_{j}$;

  2. ii

if $\lambda_{j}=0$, then either $\lambda_{j}^{\prime}>\lambda_{s}$, or
$\lambda_{j}^{\prime}=\lambda_{s}$ and $\sigma(j)>\max(f)$.

As $gf\in J_{f}$, we can find some $f^{\prime}\in G(J_{f})$ dividing $gf$.
Since $g$ is a minimal generator of $J_{f}:f$ while $f^{\prime}$ is a sparse
squarefree monomial, we must have
$f^{\prime}=g\prod_{j\in\mathcal{J}^{\prime}}z_{\sigma(j),\lambda_{j}}$ for
some
$\mathcal{J}^{\prime}\subset\Set{j\in[s]\setminus\mathcal{J}:\lambda_{j}>0}$.
Furthermore, as $f^{\prime}\prec f$, we have
$\operatorname{suppdeg}(f^{\prime})\leq\operatorname{suppdeg}(f)$. Now, we
have two cases to discern.

  1. a

The first case is when
$\operatorname{suppdeg}(f^{\prime})<\operatorname{suppdeg}(f)$, which amounts
to saying that

$\left|\Set{j\in\mathcal{J}:\lambda_{j}=0}\right|<\left|\Set{j\notin\mathcal{J}\sqcup\mathcal{J}^{\prime}:\lambda_{j}>0}\right|.$

Here, $\sqcup$ indicates that the union is disjoint. We may take an arbitrary
subset
$\mathcal{J}^{\prime\prime}\subset\Set{j\notin\mathcal{J}\sqcup\mathcal{J}^{\prime}:\lambda_{j}>0}$
with a bijection $\varphi$ from $\mathcal{J}^{\prime\prime}$ to
$\Set{j\in\mathcal{J}:\lambda_{j}=0}$. Since $f^{\prime}\in J$, the product
element

$\prod_{j\in\mathcal{J}:\lambda_{j}>0}z_{\sigma(j),\lambda_{j}^{\prime}}\cdot\prod_{j\in\mathcal{J}^{\prime}}z_{\sigma(j),\lambda_{j}}\cdot\prod_{j\in\mathcal{J}^{\prime\prime}}z_{\sigma(j),\lambda_{\varphi(j)}^{\prime}}\in
J$

by the symmetry condition (S-3). In the first part of this product, each
$\lambda_{j}^{\prime}>\lambda_{j}$. Meanwhile, in the third part, each
$\lambda_{\varphi(j)}^{\prime}\geq\lambda_{s}\geq\lambda_{j}$. Thus, by the
shiftiness condition (S-2), we have

$f^{\prime\prime}\coloneqq\prod_{j\in\mathcal{J}:\lambda_{j}>0}z_{\sigma(j),\lambda_{j}}\cdot\prod_{j\in\mathcal{J}^{\prime}}z_{\sigma(j),\lambda_{j}}\cdot\prod_{j\in\mathcal{J}^{\prime\prime}}z_{\sigma(j),\lambda_{j}}\in
J.$

As
$\operatorname{suppdeg}(f^{\prime\prime})=\operatorname{suppdeg}(f^{\prime})<\operatorname{suppdeg}(f)$,
$f^{\prime\prime}$ properly divides $f$, contradicting the minimality of $f$.

  2. b

The second case is when
$\operatorname{suppdeg}(f^{\prime})=\operatorname{suppdeg}(f)$. If we apply
the notation in the previous case, then this simply means
$\Set{j\in[s]:\lambda_{j}>0}$ is a subset of the disjoint union
$\mathcal{J}\sqcup\mathcal{J}^{\prime}\sqcup\mathcal{J}^{\prime\prime}$. It
then follows from $f^{\prime}\prec f$ that $w(f^{\prime})\leq w(f)$, which is
just

$\sum_{\begin{subarray}{c}j\in\mathcal{J}\\\
\lambda_{j}>0\end{subarray}}\lambda_{j}^{\prime}+\sum_{j\in\mathcal{J}^{\prime}}\lambda_{j}+\sum_{j\in\mathcal{J}^{\prime\prime}}\lambda_{\varphi(j)}^{\prime}\leq\sum_{\begin{subarray}{c}j\in\mathcal{J}\\\
\lambda_{j}>0\end{subarray}}\lambda_{j}+\sum_{j\in\mathcal{J}^{\prime}}\lambda_{j}+\sum_{j\in\mathcal{J}^{\prime\prime}}\lambda_{j}.$

By the comparisons stated in the previous case, this is possible only when all
of the following hold:

    1. 1

there is no $j\in\mathcal{J}$ such that $\lambda_{j}>0$;

    2. 2

for each $j\in\mathcal{J}^{\prime\prime}$,
$\lambda_{j}=\lambda_{s}=\lambda_{\varphi(j)}^{\prime}$.

Consequently, ${\bm{\lambda}}(f^{\prime})={\bm{\lambda}}(f)$ and
$\varphi(\mathcal{J}^{\prime\prime})=\Set{j\in\mathcal{J}:\lambda_{j}=0}=\mathcal{J}$.
Furthermore, since

$\mathcal{J}^{\prime}\sqcup\mathcal{J}^{\prime\prime}\subseteq\Set{j:\lambda_{j}>0}\subseteq\mathcal{J}\sqcup\mathcal{J}^{\prime}\sqcup\mathcal{J}^{\prime\prime},$

it follows again from the above item b1 that

$\Set{j\in[s]:\lambda_{j}>0}=\mathcal{J}^{\prime}\sqcup\mathcal{J}^{\prime\prime}.$

And as $\mathcal{J}=\varphi(\mathcal{J}^{\prime\prime})$, we have
$\lambda_{j}^{\prime}=\lambda_{s}$ for each $j\in\mathcal{J}$, by the above
item b2. Hence, we can observe that

     * •

for each $j\in[s]\setminus\mathcal{J}^{\prime}$ with
$\lambda_{j}\neq\lambda_{j}^{\prime}$, $j$ belongs to
$\mathcal{J}^{\prime\prime}$ or $\mathcal{J}$, which means that
$(\lambda_{j},\lambda_{j}^{\prime})$ equals either $(\lambda_{s},0)$ or
$(0,\lambda_{s})$ correspondingly;

     * •

for each $j\in[s]\setminus\mathcal{J}^{\prime}$ with
$(\lambda_{j},\lambda_{j}^{\prime})=(0,\lambda_{s})$, one has
$j\in\mathcal{J}$ and $\sigma(j)>\max(f)$ by the property of $g$.

Since $\mathcal{J}\neq\varnothing$, it follows that
$f>_{{\operatorname{lex}}}f^{\prime}$, contradicting to the fact
$f^{\prime}\in G(J_{f})$.

Therefore, we have shown the claimed equality
$\braket{\mathcal{G}_{f}}=J_{f}:f$. ∎

Now, $J_{f}:f$ is generated by the variables in $\mathcal{G}_{f}$. In
particular, it is linear. Since this holds for each $f\in G(J)$, $J$ has
linear quotients, as expected.

###  2.2. Betti Numbers

In the following, $J$ is a sparse symmetric shifted ideal. Like [MR4108337,
Corollary 5.7], we want to calculate the graded Betti number of $J$. The main
arguments for that purpose follow closely from those in [MR4108337, Section
5]. Some necessary details are included here, mainly to keep the exposition
self-contained.

Let $\Lambda(J)\coloneqq\Set{{\bm{\lambda}}(f):f\in G(J)}$ be the set of
partitions of the minimal monomial generators of $J$. Without loss of
generality, we may assume that
$\Lambda(J)=\\{{\bm{\lambda}}^{(1)},\dots,{\bm{\lambda}}^{(t)}\\}$ with
${\bm{\lambda}}^{(1)}\prec\cdots\prec{\bm{\lambda}}^{(t)}$. Here, $\prec$ is
the total order we introduced in the proof of 2.6. For each $k$ with $1\leq
k\leq t$, let $J_{\leq k}\subseteq J$ be the sparse symmetric shifted ideal
with $\Lambda(J_{\leq
k})=\\{{\bm{\lambda}}^{(1)},\dots,{\bm{\lambda}}^{(k)}\\}$. Thus, we have a
short exact sequence

$0\to J_{\leq k-1}\to J_{\leq k}\to J_{\leq k}/J_{\leq k-1}\to 0.$

Notice that the quotient in the above sequence can be better recognized as

$J_{\leq k}/J_{\leq
k-1}\cong\braket{\sigma({\bm{z}}^{{\bm{\lambda}}^{(k)}}):\sigma\in\mathfrak{S}_{s}}\left/\left(\braket{\sigma({\bm{z}}^{{\bm{\lambda}}^{(k)}}):\sigma\in\mathfrak{S}_{s}}\cap
J_{\leq k-1}\right).\right.$

######  Lemma 2.9.

Assume that ${\bm{\lambda}}^{(k)}=(\lambda_{1},\dots,\lambda_{s})$. Then, the
intersection ideal
$K\coloneqq\braket{\mathfrak{S}_{s}{\bm{z}}^{{\bm{\lambda}}^{(k)}}}\cap
J_{\leq k-1}$ is the sum of the ideal

$K_{1}\coloneqq\braket{\sigma({\bm{z}}^{{\bm{\lambda}}^{(k)}}z_{i,j}):\sigma\in\mathfrak{S}_{s},\lambda_{i}>0\text{
and }1\leq j<\lambda_{i}}$

with the ideal

$K_{2}\coloneqq\braket{\sigma({\bm{z}}^{{\bm{\lambda}}^{(k)}}z_{i,j}):\sigma\in\mathfrak{S}_{s},\lambda_{i}=0\text{
and }1\leq j<\lambda_{s}}.$

###### Proof.

We first show the containment “$K\supseteq K_{1}+K_{2}$”. Let
$f={\bm{z}}^{{\bm{\mu}}}$ be the first monomial in $G(J_{\leq k})\setminus
G(J_{\leq k-1})$ with respect to the order $\prec$ on $G(J)$. Then, for some
$\tau\in\mathfrak{S}_{s}$, we have
$f=\tau({\bm{z}}^{{\bm{\lambda}}^{(k)}})=\prod_{i}z_{\tau(i),\lambda_{i}}$,
and $J_{f}:f=J_{\leq k-1}:f=\braket{\mathcal{G}_{f}}$. Notice that in this
case,

$\mathcal{G}_{f}=\Set{z_{\tau(i),j}:\lambda_{i}>0\text{ and }1\leq
j<\lambda_{i}}\cup\Set{z_{\tau(i),j}:\lambda_{i}=0\text{ and }1\leq
j<\lambda_{s}}.$ (1)

Clearly, the lexicographic order is used here to make the third type of sets
in 2.7 empty, via the item BBiii in its proof. Now, for each
$z_{\tau(i),\lambda_{j}}\in\mathcal{G}_{f}$, one has
$fz_{\tau(i),\lambda_{j}}=\tau({\bm{z}}^{{\bm{\lambda}}^{(k)}}z_{i,j})\in
J_{\leq k-1}$. And the containment “$\supseteq$” follows from the
$\mathfrak{S}_{s}$ actions.

Next, we show the opposite containment “$K\subseteq K_{1}+K_{2}$”. For this
purpose, we take arbitrary squarefree monomial $g$ in
$\braket{\mathfrak{S}_{s}{\bm{z}}^{{\bm{\lambda}}^{(k)}}}\cap J_{\leq k-1}$.
We will also utilize the generator $f=\tau({\bm{z}}^{{\bm{\lambda}}^{(k)}})$
in the previous paragraph. Now, $\tau^{\prime}(g)\in\braket{f}$ for some
$\tau^{\prime}\in\mathfrak{S}_{s}$. Thus, $\tau^{\prime}(g)=g_{1}f$ for some
squarefree monomial $g_{1}$. Since $\tau^{\prime}(g)\in J_{\leq k-1}$,
$g_{1}\in J_{\leq k-1}:f=\braket{\mathcal{G}_{f}}$, which was described in the
above equation (1). Now the containment “$\subseteq$” follows from the
$\mathfrak{S}_{s}$ actions. ∎

In the following, for each partition
${\bm{\lambda}}=(\lambda_{1},\dots,\lambda_{s})$, we will write
$N^{{\bm{\lambda}}}$ for the quotient module

$\displaystyle\frac{\braket{\sigma({\bm{z}}^{{\bm{\lambda}}}):\sigma\in\mathfrak{S}_{s}}}{\braket{\sigma({\bm{z}}^{{\bm{\lambda}}}z_{i,j}):\lambda_{i}>0,1\leq
j<\lambda_{i},\sigma\in\mathfrak{S}_{s}}+\braket{\sigma({\bm{z}}^{{\bm{\lambda}}}z_{i,j}):\lambda_{i}=0,1\leq
j<\lambda_{s},\sigma\in\mathfrak{S}_{s}}}.$

######  Corollary 2.10.

One has $J_{\leq k}/J_{\leq k-1}\cong N^{{\bm{\lambda}}^{(k)}}$.

Next, we want to describe the ${\mathbb{K}}$-linear structure of
$N^{{\bm{\lambda}}}$. Let $N_{num}^{{\bm{\lambda}}}$ and
$N_{denom}^{{\bm{\lambda}}}$ be the squarefree monomial ideals in the
numerator and denominator of $N^{{\bm{\lambda}}}$ respectively. Since
$N^{{\bm{\lambda}}}$ is a squarefree module, it suffices to take an arbitrary
squarefree monomial $g\in N_{num}^{{\bm{\lambda}}}$ and ask when $g\notin
N_{denom}^{{\bm{\lambda}}}$. For each $i\in[s]$, if some $z_{i,j}$ divides
$g$, let $\mu_{i}=\min\Set{j:\text{$z_{i,j}$ divides $g$}}$. If no such
$z_{i,j}$ exists for this chosen $i$, let $\mu_{i}=0$. Thus, we have a tuple
${\bm{\mu}}=(\mu_{1},\dots,\mu_{s})\in{\mathbb{Z}}_{\geq 0}^{s}$ for this
squarefree monomial $g$. Additionally, we may assume that the partition
associated with ${\bm{z}}^{{\bm{\mu}}}$ is
${\bm{\lambda}}^{\prime}=(\lambda_{1}^{\prime},\dots,\lambda_{s}^{\prime})$.
Suppose that $\theta=\operatorname{suppdeg}({\bm{\lambda}})$ and
$\theta^{\prime}=\operatorname{suppdeg}({\bm{\lambda}}^{\prime})$. Thus,
$\lambda_{s-\theta}=0<\lambda_{s-\theta+1}$ and similarly
$\lambda_{s-\theta^{\prime}}^{\prime}=0<\lambda_{s-\theta^{\prime}+1}^{\prime}$.

######  Lemma 2.11.

With the notations as above, the squarefree monomial $g$ belongs to
$N_{num}^{{\bm{\lambda}}}\setminus N_{denom}^{{\bm{\lambda}}}$ if and only if

$\theta\leq\theta^{\prime}\quad$ and
$\quad\lambda_{s-\theta+k}=\lambda_{s-\theta^{\prime}+k}^{\prime}$ for
$k=1,2,\dots,\theta$. (2)

###### Proof.

Suppose that the squarefree monomial $g$ belongs to
$N_{num}^{{\bm{\lambda}}}\setminus N_{denom}^{{\bm{\lambda}}}$. By symmetry,
we may assume that ${\bm{z}}^{{\bm{\lambda}}}$ divides $g$. We observe the
following two facts.

  1. a

Firstly, whenever $\lambda_{i}>0$, $z_{i,\lambda_{i}}$ divides $g$. Therefore,
$\mu_{i}\leq\lambda_{i}$. However, when $\mu_{i}<\lambda_{i}$ for one such
$i$, then $g\in N_{denom}^{{\bm{\lambda}}}$ as well. Thus, we have precisely
$\mu_{i}=\lambda_{i}$ whenever $\lambda_{i}>0$.

  2. b

Secondly, for each $i$ with $\lambda_{i}=0$, if $1\leq\mu_{i}<\lambda_{s}$,
then $g\in N_{denom}^{{\bm{\lambda}}}$ as well. Thus either $\mu_{i}=0$ or
$\mu_{i}\geq\lambda_{s}$.

From these observations, we see immediately that condition (2) holds.

Conversely, suppose that condition (2) is satisfied. It is clear that
${\bm{z}}^{{\bm{\lambda}}}$ divides $\tau({\bm{z}}^{{\bm{\lambda}}^{\prime}})$
for some permutation $\tau\in\mathfrak{S}_{s}$. Meanwhile,
$\tau^{\prime}({\bm{z}}^{{\bm{\lambda}}^{\prime}})={\bm{z}}^{{\bm{\mu}}}$
divides $g$ for some permutation $\tau^{\prime}\in\mathfrak{S}_{s}$. Thus,
$g\in\braket{\sigma({\bm{z}}^{{\bm{\lambda}}}):\sigma\in\mathfrak{S}_{s}}=N_{num}^{{\bm{\lambda}}}$.
As for testing whether $g\in N_{denom}^{{\bm{\lambda}}}$, we are reduced to
the following two cases by symmetry.

  1. i

Suppose that $g$ is divisible by ${\bm{z}}^{{\bm{\lambda}}}z_{i_{0},j}$ for
some $1\leq j<\lambda_{i_{0}}$ with $\lambda_{i_{0}}>0$. Then, whenever
$\lambda_{i}>0$, we have $\mu_{i}\leq\lambda_{i}$. Furthermore,
$\mu_{i_{0}}<\lambda_{i_{0}}$. This will break the assumptions in (2).

  2. ii

Suppose that $g$ is divisible by ${\bm{z}}^{{\bm{\lambda}}}z_{i_{0},j}$ for
some $1\leq j<\lambda_{s}$ with $\lambda_{i_{0}}=0$. Then, whenever
$\lambda_{i}>0$, we have $\mu_{i}\leq\lambda_{i}$. Furthermore,
$1\leq\mu_{i_{0}}<\lambda_{s}$. This still breaks the assumptions in (2).

Thus, $g\in N_{num}^{{\bm{\lambda}}}\setminus N_{denom}^{{\bm{\lambda}}}$. ∎

It is time to introduce a series of new notations.

######  Notation 2.12.

Given any partition ${\bm{\lambda}}=(\lambda_{1},\dots,\lambda_{s})$, we may
assume that

${\bm{\lambda}}=(0,\dots,0,\lambda_{k_{1}},\lambda_{k_{1}+1},\dots,\lambda_{k_{2}},\lambda_{s},\dots,\lambda_{s})$

where $\lambda_{k_{1}}\neq 0$ and $\lambda_{k_{2}}<\lambda_{s}$.

  * •

Let
$\pi_{{\bm{\lambda}}}\coloneqq(\lambda_{k_{1}},\lambda_{k_{1}+1},\dots,\lambda_{k_{2}})$
be the non-zero and non-maximal piece of ${\bm{\lambda}}$. Write
$p({\bm{\lambda}})\coloneqq
k_{2}-k_{1}+1=\left|\Set{j:1\leq\lambda_{j}<\lambda_{s}}\right|$ for the
length of this piece and
$r({\bm{\lambda}})\coloneqq\left|\Set{j:\lambda_{j}=\lambda_{s}}\right|$ for
the length of the remaining maximal piece.

  * •

For any $p({\bm{\lambda}})$-subset
$A=\\{i_{1},i_{2},\dots,i_{p({\bm{\lambda}})}\\}\subseteq[s]$ with
$i_{1}<i_{2}<\cdots<i_{p({\bm{\lambda}})}$, and any positive integer $a$, let
$Z_{A,a}\coloneqq{\mathbb{K}}[z_{i,j}\in Z:i\in A\text{ and }j\geq a]$. When
$a=1$, we may write this subring simply as $Z_{A}$ and write
${\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}}\coloneqq\prod_{j=1}^{p({\bm{\lambda}})}z_{i_{j},\lambda_{k_{1}+j-1}}\in
Z_{A}$.

  * •

With respect to the subset $A$ above, we write
${A}^{\complement}\coloneqq[s]\setminus A$ for the complement set, and
similarly introduce $Z_{{A}^{\complement},a}\coloneqq{\mathbb{K}}[z_{i,j}\in
Z:i\in{A}^{\complement}\text{ and }j\geq a]$ for the positive integer $a$.

  * •

For the complement set $A^{\complement}$, we will write
$V_{A^{\complement},{\bm{\lambda}}}$ for the monomial ideal in
$Z_{A^{\complement},\lambda_{s}}$ generated by the sparse squarefree monomials
of the form $z_{t_{1},\lambda_{s}}z_{t_{2},\lambda_{s}}\cdots
z_{t_{r({\bm{\lambda}})},\lambda_{s}}$ with
$t_{1},t_{2},\dots,t_{r({\bm{\lambda}})}$ distinct in $A^{\complement}$.

Notice that we have a natural group action of $\mathfrak{S}_{A}$ on $Z_{A,a}$,
which can be derived from the action of $\mathfrak{S}_{s}$ on $Z_{A,a}$.

######  Remark 2.13.

Fix a sparse squarefree monomial
$m=\prod_{j=1}^{p({\bm{\lambda}})}z_{i_{j},\mu_{j}}\in\mathfrak{S}_{A}{\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}}$
with ${\bm{\mu}}=(\mu_{1},\dots,\mu_{p({\bm{\lambda}})})$. Surely we have
$mm^{\prime}\in\mathfrak{S}_{s}{\bm{z}}^{{\bm{\lambda}}}$ for each sparse
monomial $m^{\prime}\in G(V_{A^{\complement},{\bm{\lambda}}})$. It is not
difficult to deduce from 2.11 that the annihilating ideal of the image of
$mm^{\prime}$ is given by

$\braket{z_{i,j}:\text{either $i\in A$ and $1\leq j<\mu_{i}$, or $i\notin A$
and $1\leq j<\lambda_{s}$}},$ (3)

which is independent of the choice of $m^{\prime}$.

######  Notation 2.14.

Inspired by the description of (3), we introduce the following subrings,
submodule, and ideals.

  * •

For any given
$m=\prod_{j=1}^{p({\bm{\lambda}})}z_{i_{j},\mu_{j}}\in\mathfrak{S}_{A}{\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}}$
with ${\bm{\mu}}=(\mu_{1},\dots,\mu_{p({\bm{\lambda}})})$, let

$Z_{m}^{\vartriangle}\coloneqq{\mathbb{K}}[z_{i,j}:\text{either $i\in A$ and
$1\leq j<\mu_{i}$, or $i\notin A$ and $1\leq j<\lambda_{s}$}],$

and write $\mathfrak{m}_{Z_{m}^{\vartriangle}}$ for its graded maximal ideal.
Similarly, let

$Z_{m}^{\triangledown}\coloneqq{\mathbb{K}}[z_{i,j}:\mu_{i}\leq j\leq b\text{
and }i\in A]\subset Z_{A}.$

Roughly speaking, the variables in $Z_{m}^{\vartriangle}$ are in the upper
half of the canonical matrix ${\bm{Z}}$ consider in 2.2, while the variables
in $Z_{m}^{\triangledown}$ are in the lower half of the matrix.

  * •

For this given $m$, we write $N_{A,m}^{{\bm{\lambda}}}$ for the submodule
generated by the image of $mV_{A^{\complement},{\bm{\lambda}}}$ in
$N^{{\bm{\lambda}}}$.

  * •

Additionally, we write $V_{A^{\complement},{\bm{\lambda}}}^{*}$ for the
squarefree Veronese ideal of degree $r({\bm{\lambda}})$ in
${\mathbb{K}}[z_{i,\lambda_{s}}:i\in A^{\complement}]$, which is generated by
the squarefree monomials of degree $r({\bm{\lambda}})$ in this ring.

######  Remark 2.15.

It is clear that

$V_{A^{\complement},{\bm{\lambda}}}=V_{A^{\complement},{\bm{\lambda}}}^{*}Z_{A^{\complement},\lambda_{s}}\cong
V_{A^{\complement},{\bm{\lambda}}}^{*}\otimes_{{\mathbb{K}}}Z_{A^{\complement},\lambda_{s}+1}$
(4)

is an extended ideal from $V_{A^{\complement},{\bm{\lambda}}}^{*}$. Meanwhile,
we also have that multigraded isomorphism

$N_{A,m}^{{\bm{\lambda}}}\cong
Z_{m}^{\vartriangle}/\mathfrak{m}_{Z_{m}^{\vartriangle}}\otimes_{\mathbb{K}}mZ_{m}^{\triangledown}\otimes_{{\mathbb{K}}}V_{A^{\complement},{\bm{\lambda}}}^{*}\otimes_{\mathbb{K}}Z_{A^{\complement},\lambda_{s}+1}.$
(5)

Thus, we can deduce from 2.11 and the multidegree information in (5) the
following conclusion.

######  Corollary 2.16.

We have the following decomposition as multigraded $Z$-modules:

$N^{{\bm{\lambda}}}=\bigoplus_{\begin{subarray}{c}A\subset[s]\\\
|A|=p({\bm{\lambda}})\end{subarray}}\bigoplus_{m\in\mathfrak{S}_{A}{\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}}}N_{A,m}^{{\bm{\lambda}}}.$

Notice that the monomial $m$ sits only in $Z_{A}$ and has degree
$\operatorname{deg}({\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}})=p({\bm{\lambda}})$.
Since $V_{A^{\complement},{\bm{\lambda}}}^{*}$ is the squarefree Veronese
ideal of degree $r({\bm{\lambda}})$ in a suitable ring. It follows from (4)
that $V_{A^{\complement},{\bm{\lambda}}}$ in $Z_{A^{\complement},\lambda_{s}}$
has a linear resolution with regularity $r({\bm{\lambda}})$. Consequently,
both $N_{A,m}^{{\bm{\lambda}}}$ and $N^{{\bm{\lambda}}}$ have linear
resolutions with regularity
$p({\bm{\lambda}})+r({\bm{\lambda}})=\operatorname{suppdeg}({\bm{\lambda}})$.

Now, we can state the second main result of this section.

######  Theorem 2.17.

If $J$ is a sparse symmetric shifted ideal, then as
${\mathbb{K}}[\mathfrak{S}_{s}]$-modules, we have

$\operatorname{Tor}_{i}(J,{\mathbb{K}})_{i+d}\cong\bigoplus_{\begin{subarray}{c}{\bm{\lambda}}\in\Lambda(J)\\\
\operatorname{suppdeg}({\bm{\lambda}})=d\end{subarray}}\operatorname{Tor}_{i}(N^{{\bm{\lambda}}},{\mathbb{K}}).$

Its proof is literally the same as the argument for [MR4108337, Theorem 5.5].
There is no need repeating it here.

It remains to give a closed formula of the graded Betti numbers of $J$ using
the information from $\Lambda(J)$. To achieve this, consider the case when
$\operatorname{suppdeg}({\bm{\lambda}})=d$, $A\subset[s]$ with
$|A|=p({\bm{\lambda}})$, and
$m=\prod_{j=1}^{p({\bm{\lambda}})}z_{i_{j},\mu_{j}}\in\mathfrak{S}_{A}{\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}}$.
Notice that the dimension of $Z_{m}^{\vartriangle}$ is precisely

$\displaystyle p^{\vartriangle}({\bm{\lambda}})$
$\displaystyle\coloneqq|\pi_{{\bm{\lambda}}}|-p({\bm{\lambda}})+(\lambda_{s}-1)(s-p({\bm{\lambda}}))$
$\displaystyle=|{\bm{\lambda}}|+s(\lambda_{s}-1)-\lambda_{s}\operatorname{suppdeg}({\bm{\lambda}}).$

Thus, we can deduce from (5) that

$\displaystyle\beta_{i,i+d}(N_{A,m}^{{\bm{\lambda}}})$
$\displaystyle=\beta_{i}(Z_{m}^{\vartriangle}/\mathfrak{m}_{Z_{m}^{\vartriangle}}\otimes_{\mathbb{K}}Z_{m}^{\triangledown}\otimes_{{\mathbb{K}}}V_{A^{\complement},{\bm{\lambda}}}^{*}\otimes_{\mathbb{K}}Z_{A^{\complement},\lambda_{s}+1})$
$\displaystyle=\sum_{k+\ell=i}\beta_{\ell}(Z_{m}^{\vartriangle}/\mathfrak{m}_{Z_{m}^{\vartriangle}})\beta_{k}(V_{A^{\complement},{\bm{\lambda}}}^{*})$
$\displaystyle=\sum_{k+\ell=i}\binom{p^{\vartriangle}({\bm{\lambda}})}{\ell}\binom{s-p({\bm{\lambda}})}{r({\bm{\lambda}})+k}\binom{r({\bm{\lambda}})+k-1}{k};$

see also [MR4105544, Theorem 2.1].

Meanwhile, we will define the term
$\operatorname{type}(\pi_{{\bm{\lambda}}})\coloneqq(t_{1},\dots,t_{\lambda_{s-1}})$
where $t_{k}=\left|\\{j:\lambda_{j}=k\\}\right|$ for $1\leq
k\leq\lambda_{s-1}$. And it is clear that
$\left|\mathfrak{S}_{A}{\bm{z}}_{A}^{\pi_{{\bm{\lambda}}}}\right|=\frac{p({\bm{\lambda}})!}{\operatorname{type}(\pi_{{\bm{\lambda}}})!}$,
where $\operatorname{type}(\pi_{{\bm{\lambda}}})!\coloneqq t_{1}!t_{2}!\cdots
t_{\lambda_{s-1}}!$. Consequently, we have the following closed formula for
Betti numbers.

######  Corollary 2.18.

If $J$ is a sparse symmetric shifted ideal, then $\beta_{i,i+d}(J)$ is given
by

$\sum_{\begin{subarray}{c}{\bm{\lambda}}\in\Lambda(J)\\\
\operatorname{suppdeg}({\bm{\lambda}})=d\end{subarray}}\left(\sum_{k+\ell=i}\frac{p({\bm{\lambda}})!}{\operatorname{type}(\pi_{{\bm{\lambda}}})!}\binom{p^{\vartriangle}({\bm{\lambda}})}{\ell}\binom{s}{p({\bm{\lambda}})}\binom{s-p({\bm{\lambda}})}{r({\bm{\lambda}})+k}\binom{r({\bm{\lambda}})+k-1}{k}\right).$

A quick example using this formula is given in 3.4.

##  3\. Sequentially Cohen–Macaulay property of uniform $a$-fold product
ideals

In this section, we will show that the symbolic powers of the uniform $a$-fold
product ideal are sequentially Cohen–Macaulay under some generic conditions.
From now on, $\mathcal{F}_{{\bm{z}}}$ will be the list of variables
$\\{z_{1},\dots,z_{s}\\}$ in the polynomial ring
$T={\mathbb{K}}[z_{1},\dots,z_{s}]$. And the field ${\mathbb{K}}$ is infinite.

###  3.1. Standard monomial case

######  Proposition 3.1.

Suppose that
$I=\bigcap_{k=1}^{t}I_{c_{k},\mathcal{F}_{\bm{z}}}^{(m_{k})}\subset T$ is an
intersection of symbolic powers of monomial star configuration ideals. Then,
$I$ is sequentially Cohen–Macaulay.

###### Proof.

To show that this ideal is sequentially Cohen–Macaulay, we will apply the
polarization technique. And here is a list of facts that we will utilize. They
are in Propositions 2.3, 2.5 and 4.11 of [MR2184792] respectively.

  * •

We have $\operatorname{Pol}(I_{1}\cap
I_{2})=\operatorname{Pol}(I_{1})\cap\operatorname{Pol}(I_{2})$.

  * •

The polarization of $\braket{z_{i_{1}},\dots,z_{i_{r}}}^{m}$ has the following
irredundant irreducible primary decomposition:

$\operatorname{Pol}(\braket{z_{i_{1}},\dots,z_{i_{r}}}^{m})=\bigcap_{\begin{subarray}{c}1\leq
e_{j}\leq m,\\\ \sum_{j}\\!e_{j}\leq
m+r-1\end{subarray}}\braket{z_{i_{1},e_{1}},\dots,z_{i_{r},e_{r}}}.$

  * •

The ideal $I$ is sequentially Cohen–Macaulay if and only if
$\operatorname{Pol}(I)$ is so.

Now, we will apply the polarization to the ideals in the intersection

$I=\bigcap_{k=1}^{t}I_{c_{k},\mathcal{F}_{\bm{z}}}^{(m_{k})}=\bigcap_{k=1}^{t}\bigcap_{1\leq
i_{1}<\cdots<i_{c_{k}}\leq s}\braket{z_{i_{1}},\dots,z_{i_{c_{k}}}}^{m_{k}}.$

Notice that the squarefree monomial ideal $\operatorname{Pol}(I)$ is
sequentially Cohen–Macaulay, if and only if the Alexander dual ideal
$\operatorname{Pol}(I)^{\vee}$ is componentwise linear, by [MR2724673, Theorem
8.2.20]. And now it suffices to show that $\operatorname{Pol}(I)^{\vee}$ has
linear quotients, by [MR2724673, Theorem 8.2.15].

Observe that $\operatorname{Pol}(I)^{\vee}$ is generated by monomials of the
form $\prod_{j=1}^{c_{k}}z_{i_{j},e_{j}}$, where $1\leq k\leq t$, $1\leq
i_{1}<\cdots<i_{c_{k}}\leq s$, $1\leq e_{j}\leq m_{k}$ and $\sum_{j}e_{j}\leq
m_{k}+c_{k}-1$. It is clear that $\operatorname{Pol}(I)^{\vee}$ is sparse
symmetric shifted. Thus, by 2.6, this ideal has linear quotients, as expected.
And this completes our proof for 3.3. ∎

In the remaining of this paper, we will stay with the following setup.

######  Setting 3.2.

Let $a,b$ be positive integers such that $(b-1)s+1<a\leq bs$. And we introduce
the positive integers

$h\coloneqq bs-a+1,\quad c_{0}\coloneqq
s-\left\lfloor\frac{a-1}{b}\right\rfloor\quad\text{and}\quad\mu_{a}^{0}\coloneqq
a-b(s-c_{0}).$

Notice that if $a\leq(b-1)s+1$, then $I_{a}(\mathcal{F}_{{\bm{z}}}^{b})$ is
not saturated by [arXiv:1912.04448, Proposition 4.3]. And under 3.2, the big
height of $I_{a}(\mathcal{F}_{\bm{z}}^{b})$ is precisely $h$ by
[arXiv:1912.04448, Remark 4.5]. The main result of this subsection is the
following.

######  Proposition 3.3.

Assume the notations in 3.2. Then, the symbolic powers of
$I_{a}(\mathcal{F}_{\bm{z}}^{b})=I_{a}(z_{1}^{b}\cdots z_{s}^{b})$ are all
sequentially Cohen–Macaulay.

###### Proof.

It follows from [arXiv:1912.04448, Theorem 4.6] that the symbolic power

$I_{a}(\mathcal{F}_{\bm{z}}^{b})^{(m)}=\bigcap_{c=c_{0}}^{h}I_{c,\mathcal{F}_{\bm{z}}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}$
(6)

for each positive integer $m$. Now, it remains to apply 3.1. ∎

######  Example 3.4.

Here, we consider a small example where
$I=I_{6}(z_{1}^{2}z_{2}^{2}z_{3}^{2}z_{4}^{2})$ in
${\mathbb{K}}[z_{1},z_{2},z_{3},z_{4}]$. Now, the squarefree monomial ideal
$J=\operatorname{Pol}(I)^{\vee}$ in
${\mathbb{K}}[z_{1,1},z_{2,1},z_{3,1},z_{4,1},z_{1,2},z_{2,2},z_{3,2},z_{4,2}]$
is sparse symmetric shifted, and has $22$ minimal monomial generators. We can
arrange these monomials using the total order $\prec$ in the proof of 2.6, and
denote them by $u_{1},\dots,u_{22}$. Table 1 shows all the linear quotients
$J_{u_{i}}:u_{i}$, as described by 2.7. After applying either the well-known
formula [MR2724673, Corollary 8.2.2] or 2.18, we can obtain the following
Betti diagram for $J$:

$\begin{array}[]{rrrrrrr}&0&1&2&3&4&5\\\ \text{total}:&22&75&115&94&40&7\\\
1:&18&56&79&60&24&4\\\ 2:&4&19&36&34&16&3\end{array}$

Both the linear quotients and the Betti diagram agree with the outputs of
Macaulay2 [M2].

Table 1. Linear quotients of a sparse symmetric shifted ideal

$i$ | $u_{i}$ | $J_{u_{i}}:u_{i}$ | $\max(u_{i})$
---|---|---|---
1 | $z_{1,1}z_{2,1}$ | - | $2$
2 | $z_{1,1}z_{3,1}$ | $\braket{z_{2,1}}$ | $3$
3 | $z_{1,1}z_{4,1}$ | $\braket{z_{3,1},z_{2,1}}$ | $4$
4 | $z_{2,1}z_{3,1}$ | $\braket{z_{1,1}}$ | $3$
5 | $z_{2,1}z_{4,1}$ | $\braket{z_{3,1},z_{1,1}}$ | $4$
6 | $z_{3,1}z_{4,1}$ | $\braket{z_{2,1},z_{1,1}}$ | $4$
7 | $z_{1,1}z_{2,2}$ | $\braket{z_{4,1},z_{3,1},z_{2,1}}$ | $2$
8 | $z_{1,1}z_{3,2}$ | $\braket{z_{2,2},z_{4,1},z_{3,1},z_{2,1}}$ | $3$
9 | $z_{1,1}z_{4,2}$ | $\braket{z_{3,2},z_{2,2},z_{4,1},z_{3,1},z_{2,1}}$ | $4$
10 | $z_{2,1}z_{1,2}$ | $\braket{z_{4,1},z_{3,1},z_{1,1}}$ | $1$
11 | $z_{2,1}z_{3,2}$ | $\braket{z_{1,2},z_{4,1},z_{3,1},z_{1,1}}$ | $3$

$i$ | $u_{i}$ | $J_{u_{i}}:u_{i}$ | $\max(u_{i})$
---|---|---|---
12 | $z_{2,1}z_{4,2}$ | $\braket{z_{3,2},z_{1,2},z_{4,1},z_{3,1},z_{1,1}}$ | $4$
13 | $z_{3,1}z_{1,2}$ | $\braket{z_{4,1},z_{2,1},z_{1,1}}$ | $1$
14 | $z_{3,1}z_{2,2}$ | $\braket{z_{1,2},z_{4,1},z_{2,1},z_{1,1}}$ | $2$
15 | $z_{3,1}z_{4,2}$ | $\braket{z_{2,2},z_{1,2},z_{4,1},z_{2,1},z_{1,1}}$ | $4$
16 | $z_{4,1}z_{1,2}$ | $\braket{z_{3,1},z_{2,1},z_{1,1}}$ | $1$
17 | $z_{4,1}z_{2,2}$ | $\braket{z_{1,2},z_{3,1},z_{2,1},z_{1,1}}$ | $2$
18 | $z_{4,1}z_{3,2}$ | $\braket{z_{2,2},z_{1,2},z_{3,1},z_{2,1},z_{1,1}}$ | $3$
19 | $z_{1,2}z_{2,2}z_{3,2}$ | $\braket{z_{4,1},z_{3,1},z_{2,1},z_{1,1}}$ | $3$
20 | $z_{1,2}z_{2,2}z_{4,2}$ | $\braket{z_{3,2},z_{4,1},z_{3,1},z_{2,1},z_{1,1}}$ | $4$
21 | $z_{1,2}z_{3,2}z_{4,2}$ | $\braket{z_{2,2},z_{4,1},z_{3,1},z_{2,1},z_{1,1}}$ | $4$
22 | $z_{2,2}z_{3,2}z_{4,2}$ | $\braket{z_{1,2},z_{4,1},z_{3,1},z_{2,1},z_{1,1}}$ | $4$

###  3.2. General case

Notice that in the standard monomial case, the big height of
$I_{a}(\mathcal{F}_{\bm{z}}^{b})$ is $h=bs-a+1$ by [arXiv:1912.04448, Remark
4.5]. Thus, for the general case when $I=I_{a}(\mathcal{F}^{b})$, we need the
following genericness assumption.

######  Setting 3.5.

With the assumptions in 3.2, we assume further that $n\geq bs-a+1$ and
$\mathcal{F}=\\{f_{1},\dots,f_{s}\\}$ is a set of $(bs-a+1)$-generic $d$-forms
in $R={\mathbb{K}}[x_{0},\dots,x_{n}]$.

From now on, we will fix a positive integer $m$ and investigate the symbolic
power $I_{a}(\mathcal{F}^{b})^{(m)}$. It follows from [arXiv:1912.04448,
Theoorem 4.6] that

$I_{a}(\mathcal{F}^{b})^{(m)}=\bigcap_{c=c_{0}}^{h}I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}.$

Inspired by the discussion in [MR2184792, Subsection 4.2], we simultaneously
look at the ideals

$K_{j}\coloneqq\bigcap_{c=c_{0}}^{h-j}I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}\qquad\text{and}\qquad
K_{j}({\bm{z}})\coloneqq\bigcap_{c=c_{0}}^{h-j}I_{c,\mathcal{F}_{{\bm{z}}}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}$

for $j=0,1,\dots,h-c_{0}$. We may also write $K_{h-c_{0}+1}=R$ and
$K_{h-c_{0}+1}({\bm{z}})=T$. Thus, we have a filtration of submodules

$0=\overline{I_{a}(\mathcal{F}^{b})^{(m)}}=\overline{K_{0}}\subseteq\overline{K_{1}}\subseteq\cdots\subseteq\overline{K_{h-c_{0}}}\subseteq
R/I_{a}(\mathcal{F}^{b})^{(m)},$

where $\overline{\phantom{m}}$ means the corresponding image in
$R/I_{a}(\mathcal{F}^{b})^{(m)}$. And we have a similar filtration of
submodules for $T/I_{a}(\mathcal{F}_{{\bm{z}}}^{b})^{(m)}$. Now,
$\overline{K_{j+1}({\bm{z}})}/\overline{K_{j}({\bm{z}})}\cong
K_{j+1}({\bm{z}})/K_{j}({\bm{z}})$ is either $0$ or Cohen–Macaulay of expected
dimension for each $j$, by [MR2184792, Theorem A.4] and 3.3. It remains to
check if each $\overline{K_{j+1}}/\overline{K_{j}}$ still has the similar
property.

######  Lemma 3.6.

  1. a

If $\overline{K_{j+1}}\neq\overline{K_{j}}$, then
$\operatorname{dim}(R)-\operatorname{dim}(\overline{K_{j+1}}/\overline{K_{j}})=h-j$.

  2. b

Similarly, if $\overline{K_{j+1}({\bm{z}})}\neq\overline{K_{j}({\bm{z}})}$,
then
$\operatorname{dim}(T)-\operatorname{dim}(\overline{K_{j+1}({\bm{z}})}/\overline{K_{j}({\bm{z}})})=h-j$
as well.

###### Proof.

In the first case,

$\displaystyle\overline{K_{j+1}}/\overline{K_{j}}$
$\displaystyle\cong{K_{j+1}}/{K_{j}}=K_{j+1}/(K_{j+1}\cap
I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))})$
$\displaystyle\cong(K_{j+1}+I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))})/I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}\subset
R/I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}$

for $c=h-j$. Thus,
$\operatorname{Ass}(\overline{K_{j+1}}/\overline{K_{j}})\subseteq\operatorname{Ass}(R/I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))})$.
But $R/I_{c,\mathcal{F}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}$ is Cohen–Macaulay of
dimension $\operatorname{dim}(R)-c$ by [MR3683102, Corollary 3.7]. This shows
that
$\operatorname{dim}(\overline{K_{j+1}}/\overline{K_{j}})=\operatorname{dim}(R)-c$
as well. The second case is similar. ∎

######  Lemma 3.7.

Let $\varphi:T={\mathbb{K}}[z_{1},\dots,z_{s}]\to R$ be the homomorphism
induced by $z_{i}\mapsto f_{i}$ for each $i$. Suppose that
${\bm{F}}_{\bullet}$ is a graded minimal free resolution of
$T/K_{j}({\bm{z}})$ for
$\mathcal{F}_{{\bm{z}}}=\\{z_{1},\dots,z_{s}\\}\subseteq T$. Then,
$K_{j}=\varphi(K_{j}({\bm{z}}))R$ and ${\bm{F}}_{\bullet}\otimes_{T}R$ is a
graded minimal free resolution of $R/K_{j}$.

###### Proof.

The argument will be similar to the proof of [MR3683102, Theorem 3.3]. We
prove by induction on $s-\hat{c}$, where
$\hat{c}\coloneqq\max\\{c:\text{$\mathcal{F}$ is $c$-generic}\\}$. Obviously,
$h\leq\hat{c}\leq s-1$. If $s-\hat{c}=1$, $\mathcal{F}$ forms a regular
sequence. Whence, $\varphi$ is flat and the claim is clear by
[arXiv:1912.04448, Remark 2.1] or [MR3683102, Lemma 3.1] and its proof.

When $s-\hat{c}>1$, let $y$ be a new variable over $R$. For each $i$, let
$f_{i}^{\prime}$ be a general $d$-form in the ideal $(f_{i},y)\subset R[y]$.
Now, consider the new homomorphism $\gamma:T\to R[y]$ induced by $z_{i}\mapsto
f_{i}^{\prime}$. Notice that
$\mathcal{F}^{\prime}\coloneqq\\{f_{1}^{\prime},\dots,f_{s}^{\prime}\\}$ is a
set of $(\hat{c}+1)$-generic forms in $R[y]$. Whence, by induction,
${\bm{F}}_{\bullet}\otimes_{T}R[y]$ is a graded minimal free resolution of
$R[y]/K_{j}^{\prime}$ where
$K_{j}^{\prime}=\bigcap_{c=c_{0}}^{h-j}I_{c,\mathcal{F}^{\prime}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}$.
Meanwhile, we have the graded isomorphism

$R[y]/(K_{j}^{\prime},y)\cong R/K_{j}.$

Thus, the last piece of the proof is to show that $y$ is a non-zero-divisor of
$R[y]/K_{j}^{\prime}$.

Suppose for contradiction that $y$ is not regular modulo the ideal
$K_{j}^{\prime}$, then $y\in\mathfrak{p}$ for some associated prime
$\mathfrak{p}$. Notice that this is equivalent to saying that $y$ is not
regular when modulo $I_{c,\mathcal{F}^{\prime}}^{(m(\mu_{a}^{0}+b(c-c_{0})))}$
for some $c$ with $c_{0}\leq c\leq h-j$. But this is impossible by the proof
of [MR3683102, Theorem 3.3]. ∎

Here is the main result of this section.

######  Theorem 3.8.

Assuming 3.5, the symbolic powers of
$I_{a}(\mathcal{F}^{b})=I_{a}(f_{1}^{b}\cdots f_{s}^{b})$ are all sequentially
Cohen–Macaulay.

###### Proof.

We will fix a positive integer $m$ and look at the symbolic power
$I_{a}(\mathcal{F}^{b})^{(m)}$. After applying the notations $K_{j}$ and
$K_{j}({\bm{z}})$ introduced before 3.6, we have a short exact sequence

$0\to K_{j+1}({\bm{z}})/K_{j}({\bm{z}})\to T/K_{j}({\bm{z}})\to
T/K_{j+1}({\bm{z}})\to 0.$

We may assume that $K_{j}\neq K_{j+1}$. Hence, by 3.7, $K_{j}({\bm{z}})\neq
K_{j+1}({\bm{z}})$. Furthermore,

$\operatorname{Tor}_{k}^{T}(T/K_{j}({\bm{z}}),R)=0=\operatorname{Tor}_{k}^{T}(T/K_{j+1}({\bm{z}}),R)$

for $k\geq 1$. If we take the well-known induced long exact sequence, then we
obtain $\operatorname{Tor}_{k}^{T}(K_{j+1}({\bm{z}})/K_{j}({\bm{z}}),R)=0$ as
well for all $k\geq 1$. Furthermore, since $(T/K_{j}({\bm{z}}))\otimes R\cong
R/K_{j}$, $(T/K_{j+1}({\bm{z}}))\otimes R\cong R/K_{j+1}$ and
$\operatorname{Tor}_{1}(T/K_{j+1}({\bm{z}}),R)=0$, we see immediately that
$(K_{j+1}({\bm{z}})/K_{j}({\bm{z}}))\otimes R\cong K_{j+1}/K_{j}$. Therefore,
if ${\bm{G}}_{\bullet}$ is a graded minimal free resolution of
$K_{j+1}({\bm{z}})/K_{j}({\bm{z}})$, then ${\bm{G}}_{\bullet}\otimes R$ will
be a graded minimal free resolution of $K_{j+1}/K_{j}$. In particular,
$\operatorname{pd}(K_{j+1}/K_{j})=\operatorname{pd}(K_{j+1}({\bm{z}})/K_{j}({\bm{z}}))$.

Since $K_{j+1}({\bm{z}})/K_{j}({\bm{z}})$ is Cohen–Macaulay, it follows from
3.6 and the Auslander–Buchsbaum formula that $K_{j+1}/K_{j}$ is also
Cohen–Macaulay of the expected dimension. Therefore,
$I_{a}(\mathcal{F}^{b})^{(m)}$ is sequentially Cohen–Macaulay. ∎

##  4\. Harbourne–Huneke containment problem for the uniform $a$-fold product
ideal

Recall that for any homogeneous ideal $I$ in a standard graded ring,
$\alpha(I)$ stands for the least degree of a homogeneous polynomial in this
ideal. This final section is mainly concerned with the following two questions
involving this invariant, which were raised for _radical_ homogeneous ideals
in [arXiv:2009.05022, Section 3].

######  Question 4.1.

[Demailly-like bound] Let $R$ be a polynomial ring over a field ${\mathbb{K}}$
and let $I$ be a homogeneous ideal of big height $h$ in $R$. Does the
inequality

$\frac{\alpha(I^{(n)})}{n}\geq\frac{\alpha(I^{(m)})+h-1}{m+h-1}$

hold for all $n,m\geq 1$?

Notice that the _Waldschmidt constant_

$\widehat{\alpha}(I)\coloneqq\lim_{n}\frac{\alpha(I^{(n)})}{n}$

actually agrees with $\inf_{n}\frac{\alpha(I^{n})}{n}$. Thus, the above
inequality is simply

$\widehat{\alpha}(I)\geq\frac{\alpha(I^{(m)})+h-1}{m+h-1}.$

######  Question 4.2.

[General version of the Harbourne–Huneke containment for radical homogeneous
ideal] Let $R$ be a polynomial ring over a field ${\mathbb{K}}$ with the
maximal homogeneous ideal $\mathfrak{m}$. If $I$ is a homogeneous ideal of big
height $h$ in $R$, does the inequality

$I^{(\ell(h+m-1))}\subseteq\mathfrak{m}^{\ell(h-1)}(I^{(m)})^{\ell}$

holds for all $\ell,m\geq 1$?

We will drop the radical requirement and study these questions for
$I=I_{a}(\mathcal{F}^{b})$ under the assumptions in 3.5. It is not difficult
to see that a positive answer to 4.2 would imply an affirmative answer to 4.1.
Thus, we will focus on 4.2. Actually, like [arXiv:2009.05022, Theorem 3.6], a
stronger result can be achieved. Notice that the big height of the ideal
$I_{a}(\mathcal{F}^{b})$ is $h=bs-a+1$ while the height is the positive
integer $c_{0}$ by [arXiv:1912.04448, Remark 4.5]. Furthermore, we have seen
in the proof of [arXiv:1912.04448, Proposition 4.7] that $c_{0}=h$ precisely
when $b=1$ or $a=bs$. The $b=1$ case is just the standard star configuration
case studied in [arXiv:2009.05022]. Meanwhile, the $a=bs$ case is the trivial
one when the ideal is actually principal. Thus, in the following, we may
assume that $c_{0}<h$. Consequently, $b>1$, $a<bs$ and $h-1>0$.

We present two technical lemmas first.

######  Lemma 4.3.

We have $(h-1)(\mu_{a}^{0}+b(c-c_{0}))\geq c$ for each $c$ with $c_{0}\leq
c\leq h$.

###### Proof.

Notice that the coefficients of $c$ are $(h-1)b$ and $1$ on the two sides.
Since $(h-1)b\geq 1$, it suffices to show the inequality for $c=c_{0}$,
namely, to show

$(h-1)\mu_{a}^{0}\geq c_{0}.$ (7)

After putting back the definitions of $h$, $\mu_{a}^{0}$ and $c_{0}$, this
inequality is actually

$(bs-a)\left(a-b\left\lfloor\frac{a-1}{b}\right\rfloor\right)\geq
s-\left\lfloor\frac{a-1}{b}\right\rfloor.$

And it can be simplified into

$(bs-a)a-s\geq\left((bs-a)b-1\right)\left\lfloor\frac{a-1}{b}\right\rfloor.$

Since $bs>a$ and $b>1$, we have $(bs-a)b-1>0$. Thus, it suffices to show that

$(bs-a)a-s\geq\left((bs-a)b-1\right)\cdot\frac{a-1}{b},$

which is equivalent to saying that

$(b-1)(bs-a)\geq 1.$

However, the last inequality is clear. Thus, we have successfully shown the
inequality (7). And this completes our proof for 4.3. ∎

In the following, for each positive integer $c$ such that $c_{0}\leq c\leq h$,
we write $\widehat{\mu}_{c}$ for the positive integer
$\mu_{a}^{0}+b(c-c_{0})$.

######  Lemma 4.4.

Let $k,\ell$, and $m$ be positive integers. Then, we have

$(\ell(h+m-1)-h+k)\widehat{\mu}_{c}-c(\ell-1)\geq\ell m\widehat{\mu}_{c}$

for each $c$ with $c_{0}\leq c\leq h$.

###### Proof.

This artificially irregular inequality can be quickly modified into

$\ell\left((h-1)\widehat{\mu}_{c}-c\right)\geq(h-k)\widehat{\mu}_{c}-c.$ (8)

Since $(h-1)\widehat{\mu}_{c}\geq c$ by 4.3, we only need to show (8) for
$\ell=1$, i.e., to show that

$(h-1)\widehat{\mu}_{c}-c\geq(h-k)\widehat{\mu}_{c}-c.$

However, this holds automatically, since both $k$ and $\widehat{\mu}_{c}$ are
positive integers. ∎

Now, we are ready for the main theorem of this section.

######  Theorem 4.5.

Under the assumptions in 3.5, the ideal $I=I_{a}(\mathcal{F}^{b})$ satisfies

$I^{(\ell(h+m-1)-h+k)}\subseteq\mathfrak{m}^{d((\ell-1)(h-1)+k-1)\widehat{\mu}_{h}}(I^{(m)})^{\ell}$
(9)

for all positive integers $k,\ell$ and $m$.

Note that $\widehat{\mu}_{h}$ above is simply
$\mu_{a}^{0}+b(h-c_{0})=bh-(bs-a)$.

###### Proof.

Firstly, we consider the standard monomial case when
$I=I_{a}(\mathcal{F}_{{\bm{z}}}^{b})$ for
$\mathcal{F}_{\bm{z}}=\\{z_{1},\dots,z_{s}\\}$ in
$T={\mathbb{K}}[z_{1},\dots,z_{s}]$. Within this ring, by abuse of notation,
we will write ${\bm{z}}^{\bm{\mu}}\coloneqq z_{1}^{\mu_{1}}\cdots
z_{s}^{\mu_{s}}$ for any tuple
${\bm{\mu}}=(\mu_{1},\dots,\mu_{s})\in{\mathbb{Z}}_{\geq 0}^{s}$. As usual,
$|{\bm{\mu}}|$ is simply $\mu_{1}+\cdots+\mu_{s}$, and ${\bm{\mu}}_{\leq c}$
is short for the sub-tuple $(\mu_{1},\dots,\mu_{c})$. Since the permutation
group $\mathfrak{S}_{s}$ has a natural action on $T$, for any monomial ideal
$K$ that is invariant under this action, we will introduce

$\mathcal{P}(K)\coloneqq\Set{{\bm{\lambda}}\text{ is a partition of length
$s$}:{\bm{z}}^{{\bm{\lambda}}}\in K}$

and, by abuse of notation,

$\Lambda(K)\coloneqq\Set{{\bm{\lambda}}:{\bm{z}}^{{\bm{\lambda}}}\text{ is a
minimal monomial generator of $K$}}.$

Take an arbitrary partition
${\bm{\lambda}}=(\lambda_{1},\dots,\lambda_{s})\in\mathcal{P}(I_{a}(\mathcal{F}_{{\bm{z}}}^{b})^{(m^{\prime})})$
where $m^{\prime}\coloneqq\ell(h+m-1)-h+k$. Due to the decomposition in (6),
this amounts to saying that

${\bm{\lambda}}\in\mathcal{P}(I_{c,\mathcal{F}_{{\bm{z}}}}^{(m^{\prime}\widehat{\mu}_{c})})\qquad\text{for
each $c$ with $c_{0}\leq c\leq h$,}$

namely,

$|{\bm{\lambda}}_{\leq c}|\geq m^{\prime}\widehat{\mu}_{c}\qquad\text{for each
$c$ with $c_{0}\leq c\leq h$}$

by [MR4108337, Proposition 4.1].

In the following, for each $j\leq h$, we write
$\lambda_{j}^{\prime}\coloneqq\left\lfloor\frac{\lambda_{j}}{\ell}\right\rfloor$
and $\widehat{\lambda}_{j}\coloneqq\lambda_{j}-\ell\lambda_{j}^{\prime}$.
Since $\widehat{\lambda}_{j}\leq\ell-1$ for such $j$, we have $\sum_{j\leq
c}\widehat{\lambda}_{j}\leq c(\ell-1)$. Thus,

$\sum_{j\leq c}\ell\lambda_{j}^{\prime}=\sum_{j\leq
c}(\lambda_{j}-\widehat{\lambda}_{j})=\sum_{j\leq c}\lambda_{j}-\sum_{j\leq
c}\widehat{\lambda}_{j}\\\ \geq m^{\prime}\widehat{\mu}_{c}-c(\ell-1).$

As a quick consequence of 4.4, we have

$\sum_{j\leq c}\lambda_{j}^{\prime}\geq m\widehat{\mu}_{c}$

for each such $c$. It follows again from [MR4108337, Proposition 4.1] and the
decomposition in (6) that

${\bm{\lambda}}^{\prime}\coloneqq(\lambda_{1}^{\prime},\lambda_{2}^{\prime},\dots,\lambda_{h}^{\prime},\lambda_{h}^{\prime},\dots,\lambda_{h}^{\prime})\in\mathcal{P}(I_{a}(\mathcal{F}_{{\bm{z}}}^{b})^{(m)}).$

Therefore, we can find suitable partition
${\bm{\lambda}}^{\prime\prime}=(\lambda_{1}^{\prime\prime},\dots,\lambda_{s}^{\prime\prime})\in\Lambda(I_{a}(\mathcal{F}_{{\bm{z}}}^{b})^{(m)})$
such that ${\bm{\lambda}}^{\prime\prime}\leq{\bm{\lambda}}^{\prime}$
componentwise. In the part (b) of the proof of [arXiv:1912.04448, Proposition
4.7], we have shown that $|{\bm{\lambda}}_{\leq
h}^{\prime\prime}|=m\widehat{\mu}_{h}$. Furthermore,
$\lambda_{j}^{\prime\prime}=\lambda_{h}^{\prime\prime}$ whenever $h\leq j\leq
s$. Now, $\ell{\bm{\lambda}}^{\prime\prime}\leq{\bm{\lambda}}$ componentwise.
And it remains to estimate
$|{\bm{\lambda}}|-\ell|{\bm{\lambda}}^{\prime\prime}|$. It is clear that

$\displaystyle|{\bm{\lambda}}|-\ell|{\bm{\lambda}}^{\prime\prime}|$
$\displaystyle=\sum_{j=1}^{s}({\bm{\lambda}}_{j}-\ell{\bm{\lambda}}_{j}^{\prime\prime})\geq\sum_{j=1}^{h}({\bm{\lambda}}_{j}-\ell{\bm{\lambda}}_{j}^{\prime\prime})=|{\bm{\lambda}}_{\leq
h}|-\ell|{\bm{\lambda}}_{\leq h}^{\prime\prime}|$ $\displaystyle\geq
m^{\prime}\widehat{\mu}_{h}-\ell m\widehat{\mu}_{h}=(\ell(h+m-1)-h+k-\ell
m)\widehat{\mu}_{h}$
$\displaystyle=\left((\ell-1)(h-1)+k-1\right)\widehat{\mu}_{h}.$

Consequently, for $I=I_{a}(\mathcal{F}_{{\bm{z}}}^{b})$, we have

$I^{(\ell(h+m-1)-h+k)}\subseteq\mathfrak{m}^{\left((\ell-1)(h-1)+k-1\right)\widehat{\mu}_{h}}(I^{(m)})^{\ell},$
(10)

establishing the containment (9) in this standard monomial case.

In the final step, we consider the general uniform star configuration case,
namely when $I=I_{a}(\mathcal{F}^{b})$. Notice that both
$I^{(\ell(h+m-1)-h+k)}$ and $I^{(m)}$ are generated by monomials in the forms
in $\mathcal{F}$; cf. [arXiv:1912.04448, Proposition 4.11]. Thus, it follows
directly from (10) that

$I^{(\ell(h+m-1)-h+k)}\subseteq\mathfrak{m}^{d\left((\ell-1)(h-1)+k-1\right)\widehat{\mu}_{h}}(I^{(m)})^{\ell},$

where $d$ is the common degree of the forms in $\mathcal{F}$. ∎

The following result generalizes [arXiv:2009.05022, Corollary 3.7].

######  Corollary 4.6.

The Harbourne–Huneke type containment (4.2) and the Demailly-like bound (4.1)
hold for the ideal $I=I_{a}(\mathcal{F}^{b})$ under the assumptions in 3.5.

###### Proof.

The Harbourne–Huneke type containment follows from (9), since both $d$ and
$\widehat{\mu}_{h}$ are positive integers, while $k$ can be specially taken to
be $h$. By applying the identical arguments in the proof of [arXiv:2009.05022,
Corollary 3.7], we see that the Demailly-like bound also holds. ∎

######  Acknowledgment.

The authors are grateful to the software system Macaulay2 [M2], for serving as
an excellent source of inspiration. The second author is partially supported
by the “Anhui Initiative in Quantum Information Technologies” (No. AHY150200).

## References