# VECTOR-SPREAD MONOMIAL IDEALS AND
ELIAHOU–KERVAIRE TYPE RESOLUTIONS

Antonino Ficarra

###### Abstract

We introduce the class of vector-spread monomial ideals. This notion
generalizes that of $t$-spread ideals introduced by Ene, Herzog and Qureshi.
In particular, we focus on vector-spread strongly stable ideals, we compute
their Koszul cycles and describe their minimal free resolution. As a
consequence the graded Betti numbers and the Poincaré series are determined.
Finally, we consider a generalization of algebraic shifting theory for such a
class of ideals. †† _Keywords:_ monomial ideals, minimal graded resolution,
vector-spread ideals, $t$-spread ideals, Koszul homology, Eliahou–Kervaire
resolution. _2020 Mathematics Subject Classification:_ 05E40, 13B25, 13D02,
16W50, 68W30.

Department of Mathematics and Computer Sciences, Physics and Earth Sciences,
University of Messina, Viale Ferdinando Stagno d’Alcontres 31, 98166 Messina,
Italy

E-mail address<EMAIL_ADDRESS>
## Introduction

Algebraic shifting is one of the most powerful techniques in Combinatorial
Commutative Algebra [23, Chapter 11]. It is based on the simple idea of
shifting and spreading the variables of the generators of a monomial ideal in
a coherent way. The origins of this theory date back to a famous paper by
Erdös, Ko and Rado, Intersection theorems for systems of finite sets [19], and
made its way into Commutative Algebra through the work of Gil Kalai [27].
Lately, algebraic shifting theory and monomial ideals arising from shifting
operators [23] have seen a resurgence. The $t$-spread monomial ideals have
been introduced in 2019 by Ene, Herzog and Qureshi [18]. The homological and
combinatorial properties of these and some related classes of ideals are the
subject of a large body of research [1, 2, 3, 4, 5, 6, 7, 8, 13, 14, 15, 16,
18, 20, 21, 24, 28, 29, 30].

In this paper, our aim is to investigate the more general possible class of
such ideals. Let $S=K[x_{1},\dots,x_{n}]$ be a polynomial ring, with $K$ a
field. Given a vector ${\bf
t}=(t_{1},t_{2},\dots,t_{d-1})\in{\mathbb{Z}}_{\geq 0}^{d-1}$, $d\geq 2$, of
non negative integers, we say that a monomial $u=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{\ell}}\in S$, with $j_{1}\leq j_{2}\leq\dots\leq j_{\ell}$ and $\ell\leq
d$, is a vector-spread monomial of type ${\bf t}$ or simply a ${\bf t}$-spread
monomial if

$j_{i+1}-j_{i}\geq t_{i},\ \ \text{for all}\ i=1,\ldots,\ell-1.$

For instance, $u=x_{1}^{3}x_{2}x_{4}$ is $(0,0,1,2)$-spread, but not
$(1,0,1,2)$-spread. A monomial ideal $I\subseteq S$ is a ${\bf t}$-spread
monomial ideal if it is generated by ${\bf t}$-spread monomials. If $t_{i}=t$,
for all $i$, a ${\bf t}$-spread monomial is called an ordinary or uniform
$t$-spread monomial, see [18]. A ${\bf 1}=(1,1,\dots,1)$-spread monomial ideal
is in particular squarefree.

Let ${\bf t}\in{\mathbb{Z}}_{\geq 0}^{d-1}$, $d\geq 2$. We say that a ${\bf
t}$-spread monomial ideal $I\subset S=K[x_{1},\dots,x_{n}]$ is a ${\bf
t}$-spread strongly stable ideal if for any ${\bf t}$-spread monomial $u\in
I$, and all $j<i$ such that $x_{i}$ divides $u$ and $x_{j}(u/x_{i})$ is ${\bf
t}$-spread, then $x_{j}(u/x_{i})\in I$. For ${\bf t}={\bf 0}=(0,0,\dots,0)$
(${\bf t}={\bf 1}=(1,1,\dots,1)$) one obtains the classical notion of strongly
stable (squarefree strongly stable) ideals [23]. On the other hand, strongly
stable ideals have a central role in Commutative Algebra. Indeed, for a field
$K$ of characteristic zero, they appear as generic initial ideals [23].
Eliahou and Kervaire constructed their minimal free resolutions [17]. Bigatti
and Hulett showed that among all homogeneous ideals with the same Hilbert
function, the lexicographic ideals (which are also strongly stable) have the
biggest Betti numbers [12, 26]. Using shifting theory [18, 23, 27], Aramova,
Herzog and Hibi extended these results to squarefree ideals [10, 11].

In [18], it was shown that ordinary $t$-spread strongly stable ideals have
linear quotients [23, 25]. For ideals with linear quotients, the graded Betti
numbers can be easily computed. Moreover, if the so–called decomposition
function of $I$ is regular [25], then the minimal free resolution can be
explicitly described. Unfortunately, even ordinary 2-spread strongly stable
ideals do not have regular decomposition functions, as noted in [18]. Hence,
the minimal free resolution of ordinary $t$-spread strongly stable ideals
could not be determined and has remained elusive ever since.

The purpose of this paper is twofold, we construct the minimal free resolution
of vector-spread strongly stable ideals generalizing the Eliahou–Kervaire
resolution [17], and extend algebraic shifting theory to vector-spread
strongly stable ideals [27].

The paper is organized as follows. After some preliminary materials in Section
1, we introduce in Section 2 the concept of vector-spread monomials and
ideals.

In Sections 3 and 4 we construct the minimal free resolution of $I$ a ${\bf
t}$-spread strongly stable ideal of $S$. As pointed out before, the method of
linear quotients is unavailable to us, as the decomposition functions of
vector-spread strongly stable ideals are, in general, non regular. Thus we use
Koszul homology as developed by Aramova and Herzog in [9]. Let $H_{i}({\bf
x};S/I)$ the $i$th homology module of ${\bf x}=x_{1},\dots,x_{n}$ with respect
to $S/I$. Due to the isomorphism $\operatorname{Tor}_{i}^{S}(K,S/I)_{j}\cong
H_{i}({\bf x};S/I)_{j}$, one can calculate the graded Betti numbers of $S/I$
as $\beta_{i,j}(S/I)=\dim_{K}H_{i}({\bf x};S/I)_{j}$. Thus one has to
determine a basis of this $K$-vector space. To do so we compute the Koszul
cycles of $S/I$. As many examples indicate, Koszul cycles of arbitrary ${\bf
t}$-spread strongly stable ideals do not have a nice expression as in the
cases ${\bf t}={\bf 0},{\bf 1}$ [9, 10] (Remark 3.10). To compute them we
introduce the following notion (Definition 3.1). If
$u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}}$ is a ${\bf t}$-spread monomial, the
${\bf t}$-spread support of $u$ is the set

${\operatorname{supp}}_{\bf t}(u):=\textstyle\bigcup\limits_{i=1}^{\ell-1}\
\big{[}j_{i},j_{i}+(t_{i}-1)\big{]},$

where $[a,b]=\\{c:a\leq c\leq b\\}$, for $a,b\in{\mathbb{Z}}_{\geq 1}$.
Furthermore, we set $[n]=\\{1,2,\dots,n\\}$, for $n\geq 1$. Let $G(I)$ be the
unique minimal set of monomial generators of $I$. For a monomial $u\in S$,
$\max(u)=\max\\{j:x_{j}\ \textup{divides}\ u\\}$. The main result of Section 3
is

Theorem 3.8. Let $I\subset S$ be a ${\bf t}$-spread strongly stable ideal.
Then, for all $i\geq 1$, the $K$-vector space $H_{i}({\bf x};S/I)$ has as a
basis the homology classes of the Koszul cycles

$e(u;\sigma)\ \ \ \textit{such that}\ \ \ u\in G(I),\ \ \
\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u),\ \ \
|\sigma|=i-1.$

Firstly, we show that the elements $e(u;\sigma)$ (Definition 3.3) are Koszul
cycles. We employ an inductive argument, as a direct proof is rather tedious
(Remark 3.5). Then, we inductively determine the basis for the Koszul
homologies of $S/I$ on partial sequences of ${\bf x}$. Note that for ${\bf
t}={\bf 0}$ (${\bf t}={\bf 1}$) the conditions that $\sigma$ must satisfy in
Theorem 3.8 are the same as in [9, Proposition 2.1] ([10, Proposition 2.2]).
So, we get a formula for the graded Betti numbers (Corollary 5.2) independent
from the characteristic of the underlying field $K$, generalizing the known
results in [9, 10, 17, 18].

In Section 4, we introduce the ${\bf t}$-spread decomposition function
(Definition 4.1). As a consequence, the differentials of the minimal free
resolution of $S/I$ are explicitly described (Theorem 4.2). Examples 3.9,
3.11, 4.3 illustrate our methods.

Finally, Section 5 is devoted to a generalization of the algebraic shifting
theory. Classically, a simplicial complex $\Delta$ on the vertex set $[n]$ is
called shifted if for all $F\in\Delta$, all $i\in F$, $j\in[n]$, $j>i$, then
$(F\setminus\\{i\\})\cup\\{j\\}\in\Delta$ [23, 27]. Note that $\Delta$ is
shifted if and only if the Stanley–Reisner ideal of $\Delta$, $I_{\Delta}$, is
an ordinary squarefree (1-spread) strongly stable ideal [23]. The usefulness
of Combinatorial shifting comes from the fact that a simplicial complex shares
the same $f$-vector of its shifted simplicial complex [23], and moreover the
$f$-vector of the shifted complex is easier to compute.

From the algebraic point of view, Algebraic shifting is defined as follows.
Let $K$ be a field of characteristic zero. Let $\operatorname{Gin}(I)$ the
generic initial ideal of $I\subset S$ with respect to the reverse
lexicographic order [23], in particular $\operatorname{Gin}(I)$ is (${\bf
0}$-spread) strongly stable. One defines
$I^{s}=(\operatorname{Gin}(I))^{\sigma}$, where $\sigma$ is the squarefree
operator that assign to each monomial $u=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$
the monomial $\sigma(u):=x_{i_{1}}x_{i_{2}+1}\cdots x_{i_{d}+(d-1)}$ and to
each monomial ideal $I$, the monomial ideal $I^{\sigma}$ with minimal
generating set $G(I^{\sigma})=\\{\sigma(u):u\in G(I)\\}$ [23, 27]. Then, the
following properties hold

  1.     1. $(\textup{Shift}_{1})$

$I^{s}$ is a squarefree strongly stable monomial ideal;

    2. $(\textup{Shift}_{2})$

$I^{s}=I$ if $I$ is a squarefree strongly stable ideal;

    3. $(\textup{Shift}_{3})$

$I$ and $I^{s}$ have the same Hilbert function;

    4. $(\textup{Shift}_{4})$

If $I\subseteq J$, then $I^{s}\subseteq J^{s}$.

We are mainly interested in the algebraic side of this construction. We
introduce an analogous “${\bf t}$-spread” algebraic shifting by the assignment
$I^{s,{\bf t}}=(\operatorname{Gin}(I))^{\sigma_{{\bf 0},{\bf t}}}$, where
$\sigma_{{\bf 0},{\bf t}}$ will be a suitable shifting operator. The ${\bf
t}$-spread versions of $(\textup{Shift}_{1})$-$(\textup{Shift}_{4})$ will be
established. For ${\bf t}={\bf 1}=(1,1,\dots,1)$, our construction returns the
classical one. In particular, $(\operatorname{Gin}(I))^{\sigma_{{\bf 0},{\bf
t}}}=I$, if $I$ is a ${\bf t}$-spread strongly stable ideal (Theorem 5.5).

## 1\. Preliminaries

### 1.1. Monomial Ideals.

For the rest of the paper, $S=K[x_{1},\dots,x_{n}]$ is the standard graded
polynomial ring in $n$ indeterminates, with $K$ a field. Let
$u=x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}\in S$ be a monomial, the
integer $\deg(u)=a_{1}+a_{2}+\dots+a_{n}$ is called the degree of $u$.

By ${\operatorname{Mon}}(S)$ we denote the set of all monomials of $S$.
Whereas, ${\operatorname{Mon}}_{\ell}(S)$ denotes the set of all monomial of
$S$ having degree $\ell$. For a monomial $u\in S$, the support of $u$ is the
set ${\operatorname{supp}}(u):=\\{i:x_{i}\ \text{divides}\ u\\}$. Furthermore,
we set $\max(u):=\max{\operatorname{supp}}(u)$ and
$\min(u):=\min{\operatorname{supp}}(u)$. For convenience we let
$\min(1)=\max(1)=n$, for the monomial $1\in S$. For $I$ a monomial ideal of
$S$, $G(I)$ denotes the unique minimal set of monomial generators of $I$,
whereas $G(I)_{\ell}:=\big{\\{}u\in G(I):\deg(u)=\ell\big{\\}}$.

Any monomial ideal $I$ of $S$ has a unique minimal graded free resolution

${\mathbb{F}}:0\rightarrow\bigoplus_{j\geq 0}S(-j)^{\beta_{r,j}}\xrightarrow{\
d_{r}\ }\cdots\xrightarrow{\ d_{1}\ }\bigoplus_{j\geq
0}S(-j)^{\beta_{0,j}}\xrightarrow{\ d_{0}\ }I\rightarrow 0,$

where $S(-j)$ is the free $S$-module obtained by shifting the degrees of $S$
by $j$. For all $i,j\geq 0$, the numbers
$\beta_{i,j}=\beta_{i,j}(I)=\dim_{K}\operatorname{Tor}^{S}_{i}(K,I)_{j}$ are
called the graded Betti numbers of $I$, and
$\beta_{i}(I)=\sum_{j}\beta_{i,j}(I)$ is the $i$th total Betti number of $I$.
Furthermore, ${\operatorname{pd}}(I)=\max\\{i:\beta_{i}(I)\neq 0\\}$ is the
projective dimension of $I$. Whereas, the (Castelnuovo–Mumford) regularity of
$I$ is ${\operatorname{reg}}(I)=\max\\{j:\beta_{i,i+j}(I)\neq 0,\ \textup{for
some}\ i\\}$.

### 1.2. The Koszul complex.

To compute the graded Betti numbers one can use the Koszul complex [23]. Let
${\bf f}=f_{1},\dots,f_{m}$ be a sequence of elements of $S$. The Koszul
complex $K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf f};S)$ attached to the
sequence ${\bf f}$ is defined as follows: let $F$ be the free $S$-module with
basis $e_{1},\dots,e_{m}$.

  1. -

We let $K_{i}({\bf f};S)={\bigwedge}^{i}F$, for all $i=0,\dots,m$. A basis of
the free $S$-module $K_{i}({\bf f};S)$ is given by the wedge products
$e_{\tau}=e_{k_{1}}\wedge e_{k_{2}}\wedge\dots\wedge e_{k_{i}}$, where
$\tau=\\{k_{1}<k_{2}<\dots<k_{i}\\}\subseteq[m]=\\{1,\dots,m\\}$, with
$\deg(u_{\tau})=|\tau|=i$.

  2. -

We define the differential $\partial_{i}:K_{i}({\bf f};S)\rightarrow
K_{i-1}({\bf f};S)$, $i=1,\dots,m-1$ by

$\partial_{i}(e_{\tau})\ =\
\sum_{\ell=1}^{i}(-1)^{\ell+1}f_{k_{\ell}}e_{\tau\setminus\\{k_{\ell}\\}}.$

We order the wedge products lexicographically, as follows:
Let
$\sigma=\\{k_{1}<k_{2}<\dots<k_{p}\\},\tau=\\{\ell_{1}<\ell_{2}<\dots<\ell_{q}\\}\subseteq[m]$,
we define $\sigma>\tau$ if $p=q$ and for some $j\in[p]=\\{1,\dots,p\\}$ one
has

$k_{1}=\ell_{1},\ \ k_{2}=\ell_{2},\ \ \ldots,\ \ k_{j-1}=\ell_{j-1}\ \
\text{and}\ \ k_{j}<\ell_{j}.$

If $\sigma>\tau$, then we set $e_{\sigma}>e_{\tau}$. For example,

$e_{1}\wedge e_{2}\ >\ e_{1}\wedge e_{3}\ >\ e_{1}\wedge e_{4}\ >\ e_{2}\wedge
e_{3}\ >\ e_{2}\wedge e_{4}\ >\ e_{3}\wedge e_{4}.$

### 1.3. Koszul Homology.

Let $I$ be a monomial ideal of $S$, and let $\varepsilon:S\rightarrow S/I$ be
the canonical epimorphism. For all $1\leq j\leq n$, we let ${\bf x}_{j}$ be
the regular sequence ${\bf x}_{j}=x_{j},x_{j+1},\dots,x_{n}$. In particular,
${\bf x}_{1}={\bf x}=x_{1},\dots,x_{n}$. One can define the complex
$K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf
x}_{j};S/I)=K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf x}_{j};S)\otimes S/I$. We
set $\varepsilon(f)e_{\sigma}>\varepsilon(g)e_{\tau}$ if
$e_{\sigma}>e_{\tau}$. We denote by $H_{i}({\bf
x}_{j};S/I)=H_{i}(K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf x}_{j};S/I))$ the
$i$th homology module of ${\bf x}_{j}$ with respect to $S/I$, such modules are
graded. If $a\in K_{i}({\bf x}_{j};S/I)$ is a Koszul cycle, _i.e._ ,
$\partial_{i}(a)=0$, the symbol $[a]$ denotes the homology class of $a$ in
$H_{i}({\bf x}_{j};S/I)$.

Koszul homology allows us to calculate the graded Betti numbers. Indeed, we
have the isomorphism $H_{i}({\bf
x};S/I)_{j}\cong\operatorname{Tor}^{S}_{i}(K,S/I)_{j}$, see [23, Corollary
A.3.5]. Thus

$\beta_{i-1,j}(I)=\beta_{i,j}(S/I)=\dim_{K}H_{i}({\bf x};S/I)_{j}\ \
\textup{for all}\ \ i\geq 1,j\geq 0.$ (1)

We recall the following rule of multiplication: $\partial(a\wedge
b)=\partial(a)\wedge b+(-1)^{\deg a}a\wedge\partial(b)$ for $a,b\in
K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf x};S/I)$, with $a$ homogeneous.

To simplify the notations, if there is no risk of confusion, we set
$K_{i}({\bf x}_{j})=K_{i}({\bf x}_{j};S/I)$ and $H_{i}({\bf x}_{j})=H_{i}({\bf
x}_{j};S/I)$, for all $i$ and all $j=1,\dots,n$. Each module $H_{i}({\bf
x}_{j})$ is a $S=K[x_{1},\dots,x_{n}]$-module, so it is in particular a
$S/({\bf x}_{j})\cong K[x_{1},\dots,x_{j-1}]$-module, and for $j=1$, a
$K$-vector space.

Let $1\leq j\leq n-1$. For all $i$, let $\alpha_{i}:K_{i}({\bf
x}_{j+1})\rightarrow K_{i}({\bf x}_{j})$ be the inclusion. We define also a
homomorphism $\beta_{i}:K_{i}({\bf x}_{j})\rightarrow K_{i-1}({\bf x}_{j+1})$
as follows: Any element $a\in K_{i}({\bf x}_{j})$ can be written uniquely as
$a=e_{j}\wedge b+c$, for unique $b\in K_{i-1}({\bf x}_{j+1})$ and $c\in
K_{i}({\bf x}_{j+1})$, we set $\beta_{i}(a)=b$. One immediately verifies that
$\beta_{i}\circ\alpha_{i}=0$.

We can construct the following short exact sequence of complexes

$0\rightarrow K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf
x}_{j+1};S/I)\xrightarrow{\ \alpha\ }K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf
x}_{j};S/I)\xrightarrow{\ \beta\ }K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf
x}_{j+1};S/I)[-1]\rightarrow 0,$ (2)

where $[-1]$ denotes the shifting of the homological degree by $-1$; indeed
the following diagram is commutative with exact rows

$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i+2}}$$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i+2}}$$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i+1}}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K_{i+1}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i+1}}$$\scriptstyle{\alpha_{i+1}}$$\textstyle{K_{i+1}({\bf
x}_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta_{i+1}}$$\scriptstyle{\partial_{i+1}}$$\textstyle{K_{i}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i}}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K_{i}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i}}$$\scriptstyle{\alpha_{i}}$$\textstyle{K_{i}({\bf
x}_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta_{i}}$$\scriptstyle{\partial_{i}}$$\textstyle{K_{i-1}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i-1}}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K_{i-1}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i-1}}$$\scriptstyle{\alpha_{i-1}}$$\textstyle{K_{i-1}({\bf
x}_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta_{i-1}}$$\scriptstyle{\partial_{i-1}}$$\textstyle{K_{i-2}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{i-2}}$$\textstyle{0}$$\textstyle{\vdots}$$\textstyle{\vdots}$$\textstyle{\vdots}$

Therefore, we can apply the homology functor $H$ to the short exact sequence
of complexes (2). By abuse of notations, we denote the induced maps
$H_{i}(\alpha),H_{i}(\beta)$ again by $\alpha_{i},\beta_{i}$, for all $i$. So,
we have the long exact sequence

$\textstyle{\cdots\
\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta_{i+1}\
\ \ \ \ \ }$$\textstyle{H_{i+1}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\alpha_{i+1}}$$\textstyle{H_{i+1}({\bf
x}_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta_{i+1}}$$\textstyle{H_{i}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\ \ \ \ \ \delta_{i}}$ $\scriptstyle{\delta_{i}\ \ \ \ \ \
}$$\textstyle{H_{i}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\ \alpha_{i}}$$\textstyle{H_{i}({\bf
x}_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta_{i}\
}$$\textstyle{H_{i-1}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\ \ \ \ \ \delta_{i-1}}$ $\scriptstyle{\delta_{i-1}\ \ \ \ \ \
}$$\textstyle{H_{i-1}({\bf
x}_{j+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\alpha_{i-1}}$$\textstyle{\ \ \ \cdots\ \ \
\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{0}\
}$$\textstyle{H_{0}({\bf
x}_{j})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$
(3)

where the maps $\delta_{i}$ are the connecting homomorphisms. It can be
verified that $\delta_{i}$ is multiplication by $\pm x_{j}$, _i.e._ ,
$\delta_{i}([a])=\pm x_{j}[a]$, see [23, Theorem A.3.3].

## 2\. Basic concepts

From now on, ${\bf t}=(t_{1},\dots,t_{d-1})$ is a vector of non negative
integers and $d\geq 2$.

###### Definition 2.1

Let $u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}}\in S$ be a monomial of degree
$\ell\leq d$, with $1\leq j_{1}\leq j_{2}\leq\dots\leq j_{\ell}\leq n$. We say
that $u$ is a vector-spread monomial of type ${\bf t}$, or simply a _${\bf t}$
-spread monomial_, if

$j_{i+1}-j_{i}\geq t_{i},\ \ \text{for all}\ i=1,\ldots,\ell-1.$

In particular, any variable $x_{j}$ is ${\bf t}$-spread. We assume that $u=1$
is ${\bf t}$-spread too. Whereas, we say that a monomial ideal $I\subseteq S$
is a vector-spread monomial ideal of type ${\bf t}$, or simply a _${\bf t}$
-spread monomial ideal_, if all monomials $u\in G(I)$ are ${\bf t}$-spread.

For instance, the monomial $u=x_{1}^{3}x_{2}x_{4}x_{5}$ is
$(0,0,1,2,1)$-spread, but it is not $(1,0,1,2,1)$-spread. For a ${\bf
t}$-spread monomial ideal $I\subset S$, it is $G(I)_{k}=\varnothing$ for
$k>d$. Let ${\bf 0}=(0,0,\dots,0)$ be the null vector with $d-1$ components.
All monomials of degree $\ell\leq d$ are ${\bf 0}$-spread. If $t_{i}\geq 1$
for all $i$, a ${\bf t}$-spread monomial is squarefree [23].

###### Definition 2.2

In our context, if $t_{i}=t\geq 0$, for all $i=1,\dots,d-1$, we say that a
${\bf t}$-spread monomial (ideal) $u\in{\operatorname{Mon}}(S)$, ($I\subseteq
S$), is an uniform or ordinary ${\bf t}$-spread monomial (ideal). Such
definition agrees with that given in [18]. In this case, we drop the bold
character “${\bf t}$” and we simply speak of $t$-spread monomial ideals.

Let $T=K[x_{1},x_{2},\dots,x_{n},\dots]$ be the polynomial ring in infinitely
many variables. We let ${\operatorname{Mon}}(T;{\bf t})$ to be the set of all
${\bf t}$-spread monomials of $T$. Analogously, ${\operatorname{Mon}}(S;{\bf
t})$ denotes the set of all ${\bf t}$-spread monomials of $S$. Furthermore,
for all $0\leq\ell\leq d$, we define the following sets

$\displaystyle{\operatorname{Mon}}_{\ell}(T;{\bf t})$
$\displaystyle:=\big{\\{}u\in{\operatorname{Mon}}(T;{\bf
t}):\deg(u)=\ell\big{\\}},$ $\displaystyle{\operatorname{Mon}}_{\ell}(S;{\bf
t})$ $\displaystyle:=\big{\\{}u\in{\operatorname{Mon}}(S;{\bf
t}):\deg(u)=\ell\big{\\}}.$

Note that ${\operatorname{Mon}}_{\ell}(S;{\bf
t})={\operatorname{Mon}}_{\ell}(T;{\bf t})\cap S$,
${\operatorname{Mon}}_{\ell}(S;{\bf t})=\varnothing$ for $\ell>d$, and
${\operatorname{Mon}}(S;{\bf t})$ is the disjoint union of the sets
${\operatorname{Mon}}_{\ell}(S;{\bf t})$, $\ell=0,\dots,d$.

Sometimes, we may use the abbreviation $M_{n,\ell,{\bf t}}$ for
${\operatorname{Mon}}_{\ell}(S;{\bf t})$. For instance,

$M_{5,4,(1,0,2)}=\big{\\{}x_{1}x_{2}^{2}x_{4},x_{1}x_{2}^{2}x_{5},x_{1}x_{2}x_{3}x_{5},x_{1}x_{3}^{2}x_{5},x_{2}x_{3}^{2}x_{5}\big{\\}}.$

In order to compute the cardinality of the sets $M_{n,\ell,{\bf t}}$, we
introduce a new shifting operator, see [11]. Let ${\bf 0}=(0,0,,\dots,0)$ be
the null vector with $d-1$ components, we define the map $\sigma_{{\bf 0},{\bf
t}}:{\operatorname{Mon}}(T;{\bf 0})\rightarrow{\operatorname{Mon}}(T;{\bf
t})$, by setting $\sigma_{{\bf 0},{\bf t}}(1)=1$, $\sigma_{{\bf 0},{\bf
t}}(x_{i})=x_{i}$ and for all monomials $u=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{\ell}}\in{\operatorname{Mon}}(T;{\bf 0})$ with $j_{1}\leq
j_{2}\leq\dots\leq j_{\ell}$, $2\leq\ell\leq d$,

$\displaystyle\sigma_{{\bf 0},{\bf t}}(x_{j_{1}}x_{j_{2}}\cdots
x_{j_{\ell}}):=\prod_{k=1}^{\ell}x_{j_{k}+\sum_{s=1}^{k-1}t_{s}}.$

Whereas, $\sigma_{{\bf t},{\bf t}}:{\operatorname{Mon}}(T;{\bf
t})\rightarrow{\operatorname{Mon}}(T;{\bf t})$ denotes the identity function
of ${\operatorname{Mon}}(T;{\bf t})$.

###### Lemma 2.3

The map $\sigma_{{\bf 0},{\bf t}}$ is a bijection.

Proof. We define the map $\sigma_{{\bf t},{\bf 0}}:{\operatorname{Mon}}(T;{\bf
t})\rightarrow{\operatorname{Mon}}(T;{\bf 0})$, by setting $\sigma_{{\bf
t},{\bf 0}}(1)=1$, $\sigma_{{\bf t},{\bf 0}}(x_{i})=x_{i},\ \text{for all}\
i\in{\mathbb{N}}$, and for all monomials $u=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{\ell}}\in{\operatorname{Mon}}(T;{\bf t})$ with $1\leq j_{1}\leq
j_{2}\leq\dots\leq j_{\ell}$, and $2\leq\ell\leq d$,

$\displaystyle\sigma_{{\bf t},{\bf 0}}(x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}})$
$\displaystyle:=\textstyle\prod\limits_{k=1}^{\ell}x_{j_{k}-\sum_{s=1}^{k-1}t_{s}}.$

One immediately verifies that $\sigma_{{\bf 0},{\bf t}}\circ\sigma_{{\bf
t},{\bf 0}}=\sigma_{{\bf t},{\bf t}}$ and $\sigma_{{\bf t},{\bf
0}}\circ\sigma_{{\bf 0},{\bf t}}=\sigma_{{\bf 0},{\bf 0}}$. $\Box$

In particular, the restriction $\sigma_{{\bf t},{\bf 0}}|_{M_{n,\ell,{\bf
t}}}$ is a injective map whose image is the set
$M_{n-(t_{1}+t_{2}+\ldots+t_{\ell-1}),\ell,{\bf
0}}={\operatorname{Mon}}_{\ell}(K[x_{1},\dots,x_{n-(t_{1}+t_{2}+\ldots+t_{\ell-1})}])$.
Thus,

###### Corollary 2.4

For all $0\leq\ell\leq d$,

$|M_{n,\ell,{\bf t}}|=\binom{n+(\ell-1)-\sum_{j=1}^{\ell-1}t_{j}}{\ell}.$

Now, we introduce three fundamental classes of ${\bf t}$-spread ideals.

###### Definition 2.5

Let $U$ be a non empty subset of $M_{n,\ell,{\bf t}}$, $\ell\leq d$. We say
that

  1. -

$U$ is a ${\bf t}$-spread stable set, if for all $u\in U$, and all $j<\max(u)$
such that $x_{j}(u/x_{\max(u)})$ is ${\bf t}$-spread, then
$x_{j}(u/x_{\max(u)})\in U$;

  2. -

$U$ is a ${\bf t}$-spread strongly stable set, if for all $u\in U$, and all
$j<i$ such that $x_{i}$ divides $u$ and $x_{j}(u/x_{i})$ is ${\bf t}$-spread,
then $x_{j}(u/x_{i})\in U$;

  3. -

$U$ is a ${\bf t}$-spread lexicographic set, if for all $u\in U$, $v\in
M_{n,\ell,{\bf t}}$ such that $v\geq_{{\operatorname{lex}}}u$, then $v\in U$,
where $\geq_{{\operatorname{lex}}}$ is the lexicographic order with
$x_{1}>x_{2}>\dots>x_{n}$ [23].

We assume the empty set $\varnothing$ to be a ${\bf t}$-spread stable,
strongly stable and lexicographic set. Whereas, for $I$ a ${\bf t}$-spread
ideal of $S$, we say that $I$ is a ${\bf t}$-spread stable, strongly stable,
lexicographic ideal, if $U_{\ell}=I\cap M_{n,\ell,{\bf t}}$ is a ${\bf
t}$-spread stable, strongly stable, lexicographic set, respectively, for all
$\ell=0,\dots,d$.

For ${\bf t}={\bf 0}=(0,0,\dots,0)$ we obtain the classical notions of stable,
strongly stable and lexicographic sets and ideals [23]. For ${\bf t}={\bf
1}=(1,1,\dots,1)$, the squarefree analogues [10]. Finally, if ${\bf
t}=(t,t,\dots,t)$ we have the ordinary $t$-spread stable, strongly stable and
lexicographic sets and ideals, as in [18].

The following hierarchy of ${\bf t}$-spread monomial ideals of $S$ holds

${\bf t}$-spread lexicographic ideals $\implies$ ${\bf t}$-spread strongly
stable ideals

$\implies$ ${\bf t}$-spread stable ideals.

The next lemma provides the existence of a standard decomposition for all
${\bf t}$-spread monomials belonging to a ${\bf t}$-spread strongly stable
ideal.

###### Lemma 2.6

Let $I$ be a ${\bf t}$-spread strongly stable ideal of $S$, and $w\in I$ a
${\bf t}$-spread monomial. Then, there exist unique monomials $u\in G(I)$ and
$v\in{\operatorname{Mon}}(S)$ such that $w=uv$ and $\max(u)\leq\min(v)$.

Proof. The statement holds when $w\in G(I)$. In such a case, $w=w\cdot 1$,
with $w\in G(I)$, $1\in{\operatorname{Mon}}(S)$ and $\max(w)\leq n=\min(1)$.
Otherwise, there exists a ${\bf t}$-spread monomial $u\in G(I)$ such that $u$
divides $w$ and $\deg(u)<\deg(w)$. We choose $u$ to be of minimal degree. Then
$w=uv$, for a suitable monomial $v\in{\operatorname{Mon}}(S)$. Write
$w=x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}}$, then
$u=x_{j_{k_{1}}}x_{j_{k_{2}}}\cdots x_{j_{k_{s}}}$ for $1\leq
k_{1}<k_{2}<\dots<k_{s}<\ell$. Now, $u_{1}=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{s}}\in I$, as $u_{1}$ is ${\bf t}$-spread and $I$ is ${\bf t}$-spread
strongly stable. Moreover, $u_{1}$ divides $w$ and it is a minimal generator.
Otherwise, if there exists $u_{2}\in G(I)$ such that $u_{2}$ divides $u_{1}$
and $\deg(u_{2})<\deg(u_{1})$, then $\deg(u_{2})<\deg(u)$ and $u_{2}$ divides
$w$, an absurd for the choice of $u$. Hence $w=u_{1}v_{1}$ with
$u_{1}=x_{j_{1}}\cdots x_{j_{s}}\in G(I)$ and $v_{1}=x_{j_{s+1}}\cdots
x_{j_{\ell}}\in{\operatorname{Mon}}(S)$. Clearly, the monomials $u_{1}$ and
$v_{1}$ satisfying the statement are unique. $\Box$

As a consequence we have the following

###### Corollary 2.7

Let $I$ be a ${\bf t}$-spread monomial ideal of $S$. Then, the following
conditions are equivalent:

  1. (i)

$I$ is a ${\bf t}$-spread strongly stable ideal;

  2. (ii)

for all $u\in G(I)$, $i\in{\operatorname{supp}}(u)$, $j<i$ such that
$x_{j}(u/x_{i})$ is a ${\bf t}$-spread monomial, then $x_{j}(u/x_{i})\in I$.

Proof. (i)$\implies$(ii) is obvious. For the converse, let $w\in I$ be a ${\bf
t}$-spread monomial, $i\in{\operatorname{supp}}(w)$ and $j<i$ such that
$w_{1}=x_{j}(w/x_{i})$ is ${\bf t}$-spread, we need to prove that $w_{1}\in
I$. Write $w=uv$, with $u$ and $v$ as in Lemma 2.6. If
$i\notin{\operatorname{supp}}(u)$, then $u$ divides $w_{1}$, and so $w_{1}\in
I$. Otherwise, if $i\in{\operatorname{supp}}(u)$, then $j<i\leq\max(u)$ and so
$j\notin{\operatorname{supp}}(v)$. Thus,
$w_{1}=x_{j}(w/x_{i})=x_{j}(u/x_{i})v=u_{1}v$ with $u_{1}=x_{j}(u/x_{i})$, and
$u_{1}$ is ${\bf t}$-spread, as $w_{1}$ is. By (ii), $u_{1}\in I$. Hence
$u_{1}$ divides $w_{1}$ and so $w_{1}\in I$. $\Box$

## 3\. Koszul cycles of vector-spread strongly stable ideals

The main computational tool in this paper is Theorem 3.8. It allows to
calculate a basis of the homology modules of the Koszul complex
$K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf x};S/I)$, where ${\bf
x}=x_{1},\dots,x_{n}$ and $I$ is a ${\bf t}$-spread strongly stable ideal of
$S$.

The symbol $[n]$ denotes the set $\\{1,2,\dots,n\\}$, where
$n\in{\mathbb{Z}}_{\geq 1}$. If $j,k\geq 1$ are integers, we set
$[j,k]:=\\{\ell\in\mathbb{N}:j\leq\ell\leq k\\}$, and $[j,k]\neq\varnothing$
if and only if $j\leq k$. If $j=k=0$, we set $[0,0]=[0]:=\varnothing$. For a
monomial $u\in S$, $u\neq 1$, we set $u^{\prime}=u/x_{\max(u)}$.

The next combinatorial tool will be fundamental for our aim.

###### Definition 3.1

Let $u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}}\in{\operatorname{Mon}}(S;{\bf
t})$ a ${\bf t}$-spread monomial of $S$, with $1\leq j_{1}\leq\dots\leq
j_{\ell}\leq n$. The ${\bf t}$-spread support of $u$ is the following subset
of $[n]$:

${\operatorname{supp}}_{\bf t}(u):=\textstyle\bigcup\limits_{i=1}^{\ell-1}\
\big{[}j_{i},j_{i}+(t_{i}-1)\big{]}.$

Note that ${\operatorname{supp}}_{\bf 0}(u)=\varnothing$, and if $u$ is
squarefree, ${\operatorname{supp}}_{\bf
1}(u)={\operatorname{supp}}(u/x_{\max(u)})=\big{\\{}j_{1},j_{2},\dots,j_{\ell-1}\big{\\}}$,
where ${\bf 1}=(1,1,\dots,1)\in{\mathbb{Z}}_{\geq 0}^{d-1}$.

Let us explain now the combinatorial meaning of the vector-spread support. Let
$u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}}\in{\operatorname{Mon}}(S;{\bf t})$,
$u\neq 1$. For any $k\in[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$,
we define $k^{(u)}:=\min\\{j\in{\operatorname{supp}}(u):j>k\\}$. We note that
$k^{(u)}$ always exists as $k<\max(u)$ and
$\max(u)\in{\operatorname{supp}}(u)$. An easy calculation shows that
$w=x_{k}(u/x_{k^{(u)}})$ is again a ${\bf t}$-spread monomial, and moreover,
if $I$ is a ${\bf t}$-spread strongly stable ideal and $u\in I$, then $w\in I$
also, by definition. This property will be crucial in order to construct our
Koszul cycles. For instance,

###### Example 3.2

Let $u=x_{1}^{2}x_{2}x_{4}x_{6}x_{8}\in{\operatorname{Mon}}(S;(0,0,1,2,1))$,
$S=K[x_{1},\dots,x_{8}]$. We have
${\operatorname{supp}}_{(0,0,1,2,1)}(x_{1}x_{1}x_{2}x_{4}x_{6}x_{8})=\\{2,4,5,6\\}$,
and $[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)=\\{1,3,7\\}$. For
$k=1$, $k^{(u)}=2$, for $k=3$, $k^{(u)}=4$ and for $k=7$, $k^{(u)}=\max(u)=8$.
Note that $x_{1}(u/x_{2})=x_{1}^{3}x_{4}x_{6}x_{8}$,
$x_{3}(u/x_{4})=x_{1}^{2}x_{2}x_{3}x_{6}x_{8}$ and
$x_{7}(u/x_{8})=x_{1}^{2}x_{2}x_{4}x_{6}x_{7}$ are all $(0,0,1,2,1)$-spread
monomials.

Let $I$ be a ${\bf t}$-spread strongly stable ideal of $S$. We are going to
construct suitable cycles of $K_{i}({\bf x};S/I)=K_{i}({\bf x})$.

We shall make the following conventions. For a non empty subset
$A\subseteq[n]$, we set ${\bf x}_{A}=\prod_{i\in A}x_{i}$ and
$e_{A}=\bigwedge_{i\in A}e_{i}$, whereas for $A=\varnothing$, ${\bf
x}_{\varnothing}=1$ and $e_{\tau}\wedge e_{\varnothing}=e_{\varnothing}\wedge
e_{\tau}=e_{\tau}$ for any non empty subset $\tau\subseteq[n]$. We take
account of repetitions. For example, if $A=\\{1,1,2,3\\}$, then ${\bf
x}_{A}=x_{1}^{2}x_{2}x_{3}$ and $e_{A}=e_{1}\wedge e_{1}\wedge e_{2}\wedge
e_{3}=0$.

Let $u\in S$ be a ${\bf t}$-spread monomial and
$\sigma=\\{k_{1}<k_{2}<\dots<k_{i-1}\\}\subseteq[\max(u)-1]$ with
$|\sigma|=i-1$, $i\geq 1$. For each $\ell=1,\dots,i-1$, we define

$k_{\ell}^{(u)}=j_{\ell}:=\min\big{\\{}j\in{\operatorname{supp}}(u):j>k_{\ell}\big{\\}}.$

Clearly, $j_{1}\leq j_{2}\leq\dots\leq j_{i-1}\leq\max(u)$. If
$F=\\{k_{s_{1}},\dots,k_{s_{m}}\\}\subseteq\sigma$, we set

$F^{(u)}:=\\{j_{s_{1}}\leq j_{s_{2}}\leq\dots\leq
j_{s_{m}}\\}=\big{\\{}\min\\{j\in{\operatorname{supp}}(u):j>k_{s_{\ell}}\\}:\ell=1,\dots,m\big{\\}}.$

###### Definition 3.3

Let $u\in S$ be a ${\bf t}$-spread monomial. We set
$u^{\prime}=u/x_{\max(u)}$. Let
$\sigma=\\{k_{1}<k_{2}<\dots<k_{i-1}\\}\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)$ with $|\sigma|=i-1$, $i\geq 1$. Let $I$ be any ${\bf t}$-spread
strongly stable ideal of $S$ such that $u\in I$ and let
$\varepsilon:S\rightarrow S/I$ be the canonical map. We define the following
element of $K_{i}({\bf x};S/I)$:

$\displaystyle e(u;\sigma)\ $ $\displaystyle:=\
\sum_{F\subseteq\sigma}(-1)^{u(\sigma;F)}\varepsilon({\bf
x}_{F}(u^{\prime}/{\bf x}_{F^{(u)}}))e_{\sigma\setminus F}\wedge
e_{F^{(u)}}\wedge e_{\max(u)}$ (4) $\displaystyle=\
\varepsilon(u^{\prime})e_{\sigma}\wedge e_{\max(u)}+\sum_{\varnothing\neq
F\subseteq\sigma}(-1)^{u(\sigma;F)}\varepsilon({\bf x}_{F}(u^{\prime}/{\bf
x}_{F^{(u)}}))e_{\sigma\setminus F}\wedge e_{F^{(u)}}\wedge e_{\max(u)},$

with $u(\sigma;\varnothing):=0$ and for $F\neq\varnothing$,
$F\subseteq\sigma$, $u(\sigma;F)$ is defined recursively as follows:

  1. -

if $\max(\sigma)=k_{i-1}\notin F$, then
$u(\sigma;F):=u(\sigma\setminus\\{k_{i-1}\\};F)+|F|$;

  2. -

if $\max(\sigma)=k_{i-1}\in F$, then

$\displaystyle u(\sigma;F)\ :=$ $\displaystyle\
u(\sigma\setminus\\{k_{r}\in\sigma:j_{r}=j_{i-1}\\};F\setminus\\{k_{r}\in
F:j_{r}=j_{i-1}\\})$
$\displaystyle\phantom{:}+(|\\{k_{r}\in\sigma:j_{r}=j_{i-1}\\}|-1)(|F|+1)+1.$

The definition of the $u(\sigma;F)$’s will became clear in the proof of
Proposition 3.4.

The element $e(u;\sigma)\in K_{i}({\bf x})$ is well defined. Indeed, if for
some $k_{p}<k_{q}$, $k_{p},k_{q}\in\sigma$ we have $j_{p}=j_{q}=j$, then
$x_{j}^{2}$ may not divide $u$, however, in such case the wedge product
$e_{\sigma\setminus F}\wedge e_{F^{(u)}}\wedge e_{\max(u)}$ is zero, as
$j_{p},j_{q}\in F^{(u)}$ and $e_{j_{p}}\wedge e_{j_{q}}=e_{j}\wedge e_{j}=0$.
The same reasoning applies if $j_{p}=\max(u)$, for some $p$. Moreover,
$\varepsilon(u^{\prime})e_{\sigma}\wedge e_{\max(u)}$ is the biggest summand
of $e(u;\sigma)$ with respect to the order on the wedge products we have
defined in Section 1.

As noted before, for $u\in I$ a ${\bf t}$-spread monomial and
$k_{\ell}\in[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$, then
$x_{k_{\ell}}(u/x_{j_{\ell}})\in I$, _i.e._ ,
$\varepsilon(x_{k_{\ell}}(u/x_{j_{\ell}}))=0$, as $I$ is a ${\bf t}$-spread
strongly stable ideal.

###### Proposition 3.4

Let $I\subseteq S$ be a ${\bf t}$-spread strongly stable ideal. For all $i\geq
1$, the elements

$e(u;\sigma)\ \ \text{such that}\ \ u\in G(I),\ \
\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u),\ \
|\sigma|=i-1,$

are cycles of $K_{i}({\bf x};S/I)$.

###### Remark 3.5

Proving the above proposition in the cases $i=1,2,3$ is straightforward. For
the general case, $e(u;\sigma)$ is a sum of $2^{i-1}$ terms, if
$|\sigma|=i-1$. A direct verification of the equation
$\partial_{i}(e(u;\sigma))=0$ is nasty. Therefore, we employ an inductive
argument. After verifying two base cases $(i=1,2)$, we assume that
$e(u;\vartheta)$ is a cycle for all proper subsets $\vartheta\subset\sigma$.
Depending on some cases, we suitably write $e(u;\sigma)$ in terms of the
$e(u;\vartheta)$’s, equations (5) and (6). It is from these equations that we
obtained our coefficients $u(\sigma;F)$, $F\subseteq\sigma$, by observing that
we must change sign each time we exchange two consecutive basis elements
$e_{k},e_{\ell}$ in a non zero wedge product involving them. Finally, using
the rule of multiplication, $\partial(a\wedge b)=\partial(a)\wedge
b+(-1)^{\deg a}a\wedge\partial(b)$ for a $a,b\in
K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf x};S/I)$, with $a$ homogeneous, we
complete our proof.

We begin by giving the decomposition for the $e(u;\sigma)$’s mentioned above.

###### Lemma 3.6

Let $u\in S$ be a ${\bf t}$-spread monomial, and let
$\sigma=\\{k_{1}<\dots<k_{i-1}\\}\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)$, $|\sigma|=i-1$, $i\geq 2$. Then the following hold:

  1. (a)

If $j_{i-1}=\max(u)$, setting $\tau=\sigma\setminus\\{k_{i-1}\\}$, then

$e(u;\sigma)=-e(u;\tau)\wedge e_{k_{i-1}}.$ (5)

  2. (b)

If $j_{i-1}\neq\max(u)$, setting
$\ell=\min\\{\ell\in[i-1]:j_{\ell}=j_{i-1}\\}$, $v=x_{k_{i-1}}u/x_{j_{i-1}}$
and $\rho=\sigma\setminus\\{k_{\ell},k_{\ell+1},\dots,k_{i-2},k_{i-1}\\}$,
then

$e(u;\sigma)=-e(u;\tau)\wedge e_{k_{i-1}}+(-1)^{i-1-\ell}e(v;\rho)\wedge
e_{k_{\ell}}\wedge e_{k_{\ell+1}}\wedge\dots\wedge e_{k_{i-2}}\wedge
e_{j_{i-1}}.$ (6)

Proof. (a) Suppose that $j_{i-1}=\max(u)$. In such a case, for all
$F\subseteq\sigma$ with $k_{i-1}\in F$, the corresponding term of
$e(u;\sigma)$ is zero, as $e_{j_{i-1}}\wedge e_{\max(u)}=e_{\max(u)}\wedge
e_{\max(u)}=0$. Moreover, for all $F\subseteq\sigma$ such that
$k_{i-1}=\max(\sigma)\notin F$, that is $F\subseteq\tau$, we have
$u(\sigma;F)=u(\tau;F)+|F|$, hence
$(-1)^{u(\sigma;F)}(-1)^{|F|+1}=-(-1)^{u(\tau;F)+2|F|}=-(-1)^{u(\tau;F)}$. So,
we obtain the desired formula (5),

$\displaystyle e(u;\sigma)$
$\displaystyle=\phantom{-\Big{(}}\sum_{F\subseteq\tau}(-1)^{u(\sigma;F)}\varepsilon({\bf
x}_{F}(u^{\prime}/{\bf x}_{F^{(u)}}))e_{\sigma\setminus F}\wedge
e_{F^{(u)}}\wedge e_{\max(u)}$
$\displaystyle=\phantom{-\Big{(}}\sum_{F\subseteq\tau}(-1)^{u(\sigma;F)}(-1)^{|F|+1}\varepsilon({\bf
x}_{F}(u^{\prime}/{\bf x}_{F^{(u)}}))e_{\tau\setminus F}\wedge
e_{F^{(u)}}\wedge e_{\max(u)}\wedge e_{k_{i-1}}$
$\displaystyle=-\big{(}\sum_{F\subseteq\tau}(-1)^{u(\tau;F)}\varepsilon({\bf
x}_{F}(u^{\prime}/{\bf x}_{F^{(u)}}))e_{\tau\setminus F}\wedge
e_{F^{(u)}}\wedge e_{\max(u)}\big{)}\wedge e_{k_{i-1}}$
$\displaystyle=-\phantom{\Big{(}}e(u;\tau)\wedge e_{k_{i-1}}.$

(b) Suppose $j_{i-1}\neq\max(u)$. Note that $v=x_{k_{i-1}}(u/x_{j_{i-1}})\in
I$, as $u\in G(I)$, $v$ is ${\bf t}$-spread, $k_{i-1}<j_{i-1}$ and $I$ is
${\bf t}$-spread strongly stable. Moreover $\max(v)=\max(u)$, $\ell\leq i-1$
and
$\rho=\\{k_{1},k_{2},\dots,k_{\ell-1}\\}\subseteq\tau\subseteq[\max(v)-1]\setminus{\operatorname{supp}}_{\bf
t}(v)$. Hence, we can consider the element

$\displaystyle e(v;\rho)$
$\displaystyle:=\sum_{G\subseteq\rho}(-1)^{v(\rho;G)}\varepsilon({\bf
x}_{G}(v^{\prime}/{\bf x}_{G^{(v)}}))e_{\rho\setminus G}\wedge
e_{G^{(v)}}\wedge e_{\max(u)}$ (7)

where $G^{(v)}=\\{\min\\{s\in{\operatorname{supp}}(v):s>g\\}:g\in G\\}$. For
$k_{r}\in G$, $r<\ell$, so
$s_{r}=\min\\{s\in{\operatorname{supp}}(v):s>k_{r}\\}=j_{r}$, as
$j_{r}\in{\operatorname{supp}}(u)\setminus\\{j_{i-1}\\}$, and $k_{i-1}>j_{r}$,
lest $k_{i-1}\leq j_{r}<j_{i-1}$ would imply that $j_{r}=j_{i-1}$, an absurd.
So, for all $G\subseteq\rho$, we have $G^{(u)}=G^{(v)}$. This implies, by the
definition of the coefficients, that $v(\rho;G)=u(\rho;G)$ for all
$G\subseteq\rho$.

Let $F\subseteq\sigma$ with $F\neq\varnothing$ such that the corresponding
term of $e(u;\sigma)$ is non zero.

If $k_{i-1}\notin F$, then $F\subseteq\tau$ and in (a) we have already shown
that the corresponding terms of $e(u;\sigma)$ and $-e(u;\tau)\wedge
e_{k_{i-1}}$ are equal.

Suppose now that $k_{i-1}\in F$. Set $D=\\{k_{\ell},\dots,k_{i-2}\\}$. We
assume that $D\cap F$ is empty, otherwise $e_{F^{(u)}}=0$, as $j_{r}=j_{i-1}$
for some $k_{r}\in D\cap F$. So, we can write $F=G\cup\\{k_{i-1}\\}$ for a
unique $G\subseteq\rho$. Thus, the relevant sum $T$ of terms of $e(u;\sigma)$
indexed by $\big{\\{}F=G\cup\\{k_{i-1}\\}:G\subseteq\rho\big{\\}}$ is, as
$\max(u)=\max(v)$ and $x_{k_{i-1}}(u^{\prime}/x_{j_{i-1}})=v^{\prime}$,

$\displaystyle T=$
$\displaystyle\sum_{\begin{subarray}{c}G\cup\\{k_{i-1}\\}\\\
G\subseteq\rho\end{subarray}}(-1)^{u(\sigma;G\cup\\{k_{i-1}\\})}\varepsilon({\bf
x}_{G}x_{k_{i-1}}(u^{\prime}/(x_{j_{i-1}}{\bf
x}_{G^{(u)}})))e_{\sigma\setminus(G\cup\\{k_{i-1}\\})}\wedge e_{G^{(u)}}\wedge
e_{j_{i-1}}\wedge e_{\max(u)}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}G\subseteq\rho\\\
\phantom{G\cup\\{k_{i-1}\\}}\end{subarray}}(-1)^{u(\sigma;G\cup\\{k_{i-1}\\})}\varepsilon({\bf
x}_{G}(v^{\prime}/{\bf x}_{G^{(v)}}))e_{\rho\setminus G}\wedge e_{D}\wedge
e_{G^{(v)}}\wedge e_{j_{i-1}}\wedge e_{\max(v)}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}G\subseteq\rho\\\
\phantom{G\cup\\{k_{i-1}\\}}\end{subarray}}(-1)^{u(\sigma;G\cup\\{k_{i-1}\\})+1+|D|(|G|+1)}\varepsilon({\bf
x}_{G}(v^{\prime}/{\bf x}_{G^{(v)}}))e_{\rho\setminus G}\wedge
e_{G^{(v)}}\wedge e_{\max(v)}\wedge e_{D}\wedge e_{j_{i-1}}.$

Now, for all $F=G\cup\\{k_{i-1}\\}$, with $G\subseteq\rho$, we have

$\displaystyle u(\sigma;F)$
$\displaystyle=u(\sigma\setminus\\{k_{\ell},\dots,k_{i-1}\\};F\setminus\\{k_{\ell},\dots,k_{i-1}\\})+(|\\{k_{\ell},\dots,k_{i-1}\\}|-1)(|F|+1)+1$
$\displaystyle=u(\sigma\setminus\\{k_{\ell},\dots,k_{i-1}\\};G)+(|\\{k_{\ell},\dots,k_{i-1}\\}|-1)(|F|+1)+1$
$\displaystyle=u(\rho;G)+|D|(|F|+1)+1.$

Therefore, as $|G|+1=|F|$ and $|D|=i-1-\ell$, we have

$(-1)^{u(\sigma;F)+1+|D|(|G|+1)}=(-1)^{u(\rho;G)+2+2|D||F|+|D|}=(-1)^{i-1-\ell}(-1)^{u(\rho;G)}.$

So, we have $e(u;\sigma)=-e(u;\tau)\wedge e_{k_{i-1}}+T$, with, as
$u(\rho;G)=v(\rho;G)$ for all $G\subseteq\rho$,

$\displaystyle T$
$\displaystyle=\big{(}\sum_{G\subseteq\rho}(-1)^{i-1-\ell}(-1)^{u(\rho;G)}\varepsilon({\bf
x}_{G}(v^{\prime}/{\bf x}_{G^{(v)}}))e_{\rho\setminus G}\wedge
e_{G^{(v)}}\wedge e_{\max(v)}\big{)}\wedge e_{D}\wedge e_{j_{i-1}}$
$\displaystyle=(-1)^{i-1-\ell}\big{(}\sum_{G\subseteq\rho}(-1)^{v(\rho;G)}\varepsilon({\bf
x}_{G}(v^{\prime}/{\bf x}_{G^{(v)}}))e_{\rho\setminus G}\wedge
e_{G^{(v)}}\wedge e_{\max(v)}\big{)}\wedge e_{D}\wedge e_{j_{i-1}}$
$\displaystyle=(-1)^{i-1-\ell}e(v;\rho)\wedge e_{k_{\ell}}\wedge
e_{k_{\ell+1}}\wedge\dots\wedge e_{k_{i-2}}\wedge e_{j_{i-1}},$

and equation (6) holds. $\Box$

Finally we can prove Proposition 3.4.
Proof of Proposition 3.4. For $i=1$ we have $|\sigma|=0$, so
$\sigma=\varnothing$, $u(\varnothing;\varnothing)=0$ by definition, and the
element
$e(u;\varnothing)=\varepsilon(u^{\prime})e_{\max(u)}=\varepsilon(u/x_{\max(u)})e_{\max(u)}$
is clearly a cycle of $K_{1}({\bf x})$. Let $i\geq 2$. We proceed by induction
on $i\geq 2$.

For $i=2$, $\sigma=\\{k_{1}\\}$. Let
$j_{1}=\min\\{j\in{\operatorname{supp}}(u):j>k_{1}\\}$. We have
$u(\\{k_{1}\\};\varnothing)=0$ and
$u(\\{k_{1}\\},\\{k_{1}\\})=u(\\{k_{1}\\}\setminus\\{k_{1}\\};\\{k_{1}\\}\setminus\\{k_{1}\\})+(|\\{k_{1}\\}|-1)(|\\{k_{1}\\}|+1)+1=u(\varnothing;\varnothing)+1=1$,
so

$e(u;\sigma)=\varepsilon(u^{\prime})e_{k_{1}}\wedge
e_{\max(u)}-\varepsilon(x_{k_{1}}u^{\prime}/x_{j_{1}})e_{j_{1}}\wedge
e_{\max(u)}.$

If $j_{1}=\max(u)$, then $e(u;\sigma)=\varepsilon(u^{\prime})e_{k_{1}}\wedge
e_{\max(u)}$. In such a case,

$\partial_{2}(e(u;\sigma))=\varepsilon(x_{k_{1}}u^{\prime})e_{\max(u)}-\varepsilon(x_{\max(u)}u^{\prime})e_{k_{1}}=0,$

as $x_{k_{1}}u^{\prime},x_{\max(u)}u^{\prime}=u\in I$. Otherwise, if
$j_{1}<\max(u)$, then

$\partial_{2}(e(u;\sigma))=\varepsilon(x_{k_{1}}u^{\prime})e_{\max(u)}-\varepsilon(x_{\max(u)}u^{\prime})e_{k_{1}}-\varepsilon(x_{k_{1}}u^{\prime})e_{\max(u)}+\varepsilon(x_{k_{1}}(u/x_{j_{1}}))e_{j_{1}}=0,$

as the first and third terms cancel each other, and
$x_{\max(u)}u^{\prime}=u,x_{k_{1}}(u/x_{j_{1}})\in I$.

Suppose now $i>2$. Let $\sigma=\\{k_{1}<\dots<k_{i-2}<k_{i-1}\\}$, and
$\tau=\\{k_{1}<\dots<k_{i-2}\\}$. We distinguish two cases.

(a) Suppose $j_{i-1}=\max(u)$. By induction $e(u;\tau)$ is a cycle. By Lemma
3.6 (a),

$\displaystyle\partial_{i}(e(u;\sigma))$
$\displaystyle=\partial_{i}(-e(u;\tau)\wedge e_{k_{i-1}})$
$\displaystyle=-\partial_{i-1}(e(u;\tau))\wedge
e_{k_{i-1}}-(-1)^{\deg(e(u;\tau))}x_{k_{i-1}}e(u;\tau)$
$\displaystyle=-(-1)^{\deg(e(u;\tau))}x_{k_{i-1}}e(u;\tau)=0.$

Indeed,
$x_{k_{i-1}}u/x_{j_{i-1}}=x_{k_{i-1}}u/x_{\max(u)}=x_{k_{i-1}}u^{\prime}\in
I$. Thus, each non zero term of $x_{k_{i-1}}e(u;\tau)$ vanish, as it has
coefficient $\varepsilon({\bf x}_{F}x_{k_{i-1}}u^{\prime}/{\bf x}_{F^{(u)}})$,
and ${\bf x}_{F}x_{k_{i-1}}u^{\prime}/{\bf x}_{F^{(u)}}\in I$ as $I$ is a
${\bf t}$-spread strongly stable ideal. In such a case, $e(u;\sigma)$ is a
cycle, as desired.

(b) Suppose $j_{i-1}\neq\max(u)$. Set
$\ell=\min\\{\ell\in[i-1]:j_{\ell}=j_{i-1}\\}$, $v=x_{k_{i-1}}u/x_{j_{i-1}}$,
$\rho=\sigma\setminus\\{k_{\ell},\dots,k_{i-2},k_{i-1}\\}$ and
$D=\\{k_{\ell},\dots,k_{i-2}\\}$. By Lemma 3.6 (b),

$e(u;\sigma)=-e(u;\tau)\wedge e_{k_{i-1}}+(-1)^{i-1-\ell}e(v;\rho)\wedge
e_{D}\wedge e_{j_{i-1}}.$ (8)

Let $J$ be the smallest ${\bf t}$-spread strongly stable ideal of $S$ that
contains $v$. $J$ is generated only in one degree $\deg(v)=\deg(u)$, and
$J\subseteq I$. By inductive hypothesis, as $|\rho|<|\sigma|$, $e(v;\rho)$ is
a cycle of $K_{i}({\bf x};S/J)$. So, it is also a cycle of $K_{i}({\bf
x};S/I)=K_{i}({\bf x})$.

By inductive hypothesis
$\partial_{i-1}(e(u;\tau))=\partial_{\ell}(e(v;\rho))=0$. Since
$\deg(e(u;\tau))=|\tau|$ and $\deg(e(v;\rho))=|\rho|$, by equation (8) we have

$\displaystyle\partial_{i}(e(u;\sigma))=$
$\displaystyle-\partial_{i-1}(e(u;\tau))\wedge
e_{k_{i-1}}-(-1)^{\deg(e(u;\tau))}x_{k_{i-1}}e(u;\tau)$ $\displaystyle+$
$\displaystyle(-1)^{i-1-\ell}[\partial_{\ell}(e(v;\rho))\wedge e_{D}\wedge
e_{j_{i-1}}+(-1)^{\deg(e(v;\rho))}e(v;\rho)\wedge\partial_{i-\ell}(e_{D}\wedge
e_{j_{i-1}})]$ $\displaystyle=$
$\displaystyle-(-1)^{|\tau|}\big{(}x_{k_{i-1}}e(u;\tau)-(-1)^{|\rho|-|\tau|}(-1)^{i-1-\ell}e(v;\rho)\wedge\partial_{i-\ell}(e_{D}\wedge
e_{j_{i-1}})\big{)}.$

We have $i-1-\ell+|\rho|-|\tau|=i-1-\ell+\ell-1-(i-2)=0$. So
$(-1)^{|\rho|-|\tau|}(-1)^{i-1-\ell}=1$.

Set

$f:=x_{k_{i-1}}e(u;\tau)-e(v;\rho)\wedge\partial_{i-\ell}(e_{k_{\ell}}\wedge
e_{k_{\ell+1}}\wedge\dots\wedge e_{k_{i-2}}\wedge e_{j_{i-1}}).$

To show that $\partial_{i}(e(u;\sigma))=0$, it suffices to prove that $f$ is
zero. Let $F\subseteq\tau$. The set $D\cap F=\\{k_{\ell},\dots,k_{i-2}\\}\cap
F$ can have at most one element, otherwise $e_{F^{(u)}}=0$, as shown before.
Therefore, $F=G$ or $F=G\cup\\{k_{r}\\}$ for a unique $G\subseteq\rho$ and
$r\in\\{\ell,\dots,i-2\\}$. By construction we have $G^{(u)}=G^{(v)}$ and
$u(\rho;G)=v(\rho;G)$ for all $G\subseteq\rho$, as already observed in Lemma
3.6.

Suppose $F=G$, then the corresponding term of $f$ is $a-b$, where

$\displaystyle a$ $\displaystyle=(-1)^{u(\tau;G)}\varepsilon(x_{k_{i-1}}{\bf
x}_{G}(u^{\prime}/{\bf x}_{G^{(u)}}))e_{\tau\setminus G}\wedge
e_{G^{(u)}}\wedge e_{\max(u)},$ and, as $i-1-\ell=|D|$, $\displaystyle b$
$\displaystyle=(-1)^{i-1-\ell}(-1)^{v(\rho;G)}\varepsilon(x_{j_{i-1}}{\bf
x}_{G}x_{k_{i-1}}u^{\prime}/(x_{j_{i-1}}{\bf x}_{G^{(v)}}))e_{\rho\setminus
G}\wedge e_{G^{(v)}}\wedge e_{\max(v)}\wedge e_{D}$
$\displaystyle=(-1)^{u(\rho;G)+|D|}(-1)^{|D|(|G|+1)}\varepsilon(x_{k_{i-1}}{\bf
x}_{G}(u^{\prime}/{\bf x}_{G^{(u)}}))e_{\rho\setminus G}\wedge e_{D}\wedge
e_{G^{(u)}}\wedge e_{\max(u)}$
$\displaystyle=(-1)^{u(\rho;G)+|D|(|G|+2)}\varepsilon(x_{k_{i-1}}{\bf
x}_{G}(u^{\prime}/{\bf x}_{G^{(u)}}))e_{\tau\setminus G}\wedge
e_{G^{(u)}}\wedge e_{\max(u)}.$

We have, as $i-1-\ell=|D|$,

$\displaystyle u(\tau;G)$ $\displaystyle=u(\tau\setminus\\{k_{i-2}\\};G)+|G|$
$\displaystyle=u(\tau\setminus\\{k_{i-3},k_{i-2}\\};G)+2|G|$
$\displaystyle\phantom{..}\vdots$
$\displaystyle=u(\tau\setminus\\{k_{\ell},\dots,k_{i-2}\\};G)+(i-1-\ell)|G|$
$\displaystyle=u(\rho;G)+|D|\cdot|G|.$

Thus, $(-1)^{u(\rho;G)+|D|(|G|+2)}=(-1)^{u(\tau;G)}$ and $a-b=0$, in this
case.

Otherwise, if $F=G\cup\\{k_{r}\\}$, the corresponding term of $f$ is $a-b$,
where

$\displaystyle a$
$\displaystyle=(-1)^{u(\tau;G\cup\\{k_{r}\\})}\varepsilon(x_{k_{i-1}}x_{k_{r}}{\bf
x}_{G}u^{\prime}/(x_{j_{i-1}}{\bf
x}_{G^{(u)}}))e_{\tau\setminus(G\cup\\{k_{r}\\})}\wedge e_{G^{(u)}}\wedge
e_{j_{i-1}}\wedge e_{\max(u)}$
$\displaystyle=(-1)^{u(\tau;G\cup\\{k_{r}\\})}\varepsilon(x_{k_{i-1}}x_{k_{r}}{\bf
x}_{G}u^{\prime}/(x_{j_{i-1}}{\bf x}_{G^{(u)}}))e_{\tau\setminus F}\wedge
e_{G^{(u)}}\wedge e_{j_{i-1}}\wedge e_{\max(u)},$ and, setting
$c=v(\rho;G)+r-\ell$, $\displaystyle b$
$\displaystyle=(-1)^{r-\ell}(-1)^{v(\rho;G)}\varepsilon(x_{k_{r}}{\bf
x}_{G}v^{\prime}/{\bf x}_{G^{(v)}})e_{\rho\setminus G}\wedge e_{G^{(v)}}\wedge
e_{\max(v)}\wedge e_{D\setminus\\{k_{r}\\}}\wedge e_{j_{i-1}}$
$\displaystyle=(-1)^{c+(|D|-1)(|G|+1)+1}\varepsilon(x_{k_{i-1}}x_{k_{r}}{\bf
x}_{G}u^{\prime}/({\bf x}_{G^{(u)}}x_{j_{i-1}}))e_{\rho\setminus G}\wedge
e_{D\setminus\\{k_{r}\\}}\wedge e_{F^{(u)}}\wedge e_{\max(u)}$
$\displaystyle=(-1)^{c+(|D|-1)(|G|+1)+1}\varepsilon(x_{k_{i-1}}x_{k_{r}}{\bf
x}_{G}u^{\prime}/({\bf x}_{G^{(u)}}x_{j_{i-1}}))e_{\tau\setminus F}\wedge
e_{G^{(u)}}\wedge e_{j_{i-1}}\wedge e_{\max(u)}.$

Now,

$\displaystyle u(\tau;G\cup\\{k_{r}\\})$
$\displaystyle=u(\tau\setminus\\{k_{i-2}\\};G\cup\\{k_{r}\\})+(|G|+1)$
$\displaystyle=u(\tau\setminus\\{k_{i-3},k_{i-2}\\};G\cup\\{k_{r}\\})+2(|G|+1)$
$\displaystyle\phantom{..}\vdots$
$\displaystyle=u(\tau\setminus\\{k_{r+1},\dots,k_{i-2}\\};G\cup\\{k_{r}\\})+(i-r)(|G|+1)$
$\displaystyle=u(\tau\setminus\\{k_{\ell},\dots,k_{i-2}\\};G)+(|\\{k_{\ell},\dots,k_{r}\\}|-1)(|G|+2)+1+(i-r)(|G|+1)$
$\displaystyle=u(\rho;G)+(r-\ell+i-r)(|G|+1)+(r-\ell)+1$
$\displaystyle=v(\rho;G)+(i-\ell)(|G|+1)+(r-\ell)+1.$

Hence, since
$(-1)^{(i-\ell)(|G|+1)}=(-1)^{(|D|+1)(|G|+1)}=(-1)^{(|D|-1)(|G|+1)}$, we have

$(-1)^{u(\tau;G\cup\\{k_{r}\\})}=(-1)^{v(\rho;G)+(i-\ell)(|G|+1)+(r-\ell)+1}=(-1)^{c+(|D|-1)(|G|+1)+1}$

and $a-b=0$. Therefore, $f=0$ and $e(u;\sigma)$ is a cycle, as desired. $\Box$

###### Remark 3.7

Let
$\sigma=\\{k_{1}<k_{2}<\dots<k_{i-1}\\}\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)$, $i>2$. Set

$r(u;\sigma):=\sum_{\begin{subarray}{c}F\subseteq\sigma\\\ k_{1}\in
F\end{subarray}}(-1)^{u(\sigma;F)}\varepsilon({\bf x}_{F}(u^{\prime}/{\bf
x}_{F^{(u)}}))e_{\sigma\setminus F}\wedge e_{F^{(u)}}\wedge e_{\max(u)}.$

We show that $e(u;\sigma)=e_{k_{1}}\wedge
e(u;\sigma\setminus\\{k_{1}\\})+r(u;\sigma)$. For this aim, it is enough to
prove that, for all $F\subseteq\sigma$ such that $k_{1}\notin F$, we have
$(-1)^{u(\sigma;F)}=(-1)^{u(\sigma\setminus\\{k_{1}\\};F)}$. For
$\sigma=\\{k_{1}\\}$ this is clear. Let $|\sigma|=i-1\geq 2$. We distinguish
two cases.

Case 1. Suppose $\max(\sigma)\notin F$, then
$u(\sigma;F)=u(\sigma\setminus\\{k_{i-1}\\};F)+|F|$. Moreover
$\max(\sigma\setminus\\{k_{1}\\})=\max(\sigma)$ as $|\sigma|\geq 2$. Therefore
$u(\sigma\setminus\\{k_{1}\\};F)=u(\sigma\setminus\\{k_{1},k_{i-1}\\};F)+|F|$.
By induction on $|\sigma|$,
$(-1)^{u(\sigma\setminus\\{k_{1},k_{i-1}\\};F)}=(-1)^{u(\sigma\setminus\\{k_{i-1}\\};F)}$,
and the desired conclusion follows in such a case.

Case 2. Suppose $\max(\sigma)\in F$. Set $D=\\{k_{r}\in F:j_{r}=j_{i-1}\\}$
and $d=|\\{k_{r}\in\sigma:j_{r}=j_{i-1}\\}|$. Observe that, as $k_{1}\notin
F$, $d=|\\{k_{r}\in\sigma\setminus\\{k_{1}\\}:j_{r}=j_{i-1}\\}|$. Therefore,
by the definition of the coefficients, we have

$\displaystyle u(\sigma;F)\ $ $\displaystyle=\ u(\sigma\setminus
D;F\setminus\\{k_{i-1}\\})+(d-1)(|F|+1)+1,$ $\displaystyle
u(\sigma\setminus\\{k_{1}\\};F)\ $ $\displaystyle=\
u(\sigma\setminus(\\{k_{1}\\}\cup D);F\setminus\\{k_{i-1}\\})+(d-1)(|F|+1)+1.$

As $k_{i-1}\in D$ so $D\neq\varnothing$, we have $|\sigma\setminus
D|<|\sigma|$. So by inductive hypothesis, $(-1)^{u(\sigma\setminus
D;F\setminus\\{k_{i-1}\\})}=(-1)^{u(\sigma\setminus(\\{k_{1}\\}\cup
D);F\setminus\\{k_{i-1}\\})}$, and the desired conclusion follows.

Note that $e_{k_{1}}$ doesn’t appear in $r(u;\sigma)$. Hence, we have the
useful decomposition $e(u;\sigma)=e_{k_{1}}\wedge
e(u;\sigma\setminus\\{k_{1}\\})+r(u;\sigma)$. Moreover, equations (5) and (6)
give us recurrence relations for our Koszul cycles.

We are in position to state and prove the main result of this Section.

###### Theorem 3.8

Let $I\subset S$ be a ${\bf t}$-spread strongly stable ideal. Then, for all
$i\geq 1$, the $K$-vector space $H_{i}({\bf x};S/I)$ has as a basis the
homology classes of the Koszul cycles

$e(u;\sigma)\ \ \ \text{such that}\ \ \ u\in G(I),\ \ \
\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u),\ \ \
|\sigma|=i-1.$ (9)

Proof. Let us prove the following more general statement,

Claim 1. For all $i\geq 1$ and all $j=1,\dots,n$, a minimal generating set for
$H_{i}({\bf x}_{j})$, as a $S/({\bf x}_{j})$-module, is given by the homology
classes of the Koszul cycles

$e(u;\sigma)\ \ \textit{such that}\ \ u\in G(I),\ \ \
\sigma\subseteq\big{(}[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)\big{)}\cap[j,n],\ \ \ |\sigma|=i-1.$

We proceed by induction on $n-j\geq 0$. For the base case, let $n-j=0$. We
only have to consider $H_{1}({\bf x}_{n})$. Indeed, for $i\geq 2$, $H_{i}({\bf
x}_{n})=0$, since $K_{{}_{\text{\large$\bm{\cdot}$}}}(x_{n};S/I)$ has length
one. $H_{1}({\bf x}_{n})$ is generated by the elements
$[e(u;\varnothing)]=[\varepsilon(u/x_{\max(u)})e_{\max(u)}]$ with $u\in G(I)$
and $\max(u)=n$. Moreover $({\bf x}_{n})=(x_{n})$ clearly annihilates these
elements, so they form a minimal generating set of $H_{1}({\bf x}_{n})$ as a
$S/({\bf x}_{n})$-module.

For the inductive step, suppose $n-j>0$ and that the thesis holds for $j+1$.
First, we consider the case $i=1$. By the sequence (3), we have the exact
sequence

$\displaystyle H_{1}({\bf x}_{j+1})$ $\displaystyle\xrightarrow{\ \alpha_{1}\
}H_{1}({\bf x}_{j})\xrightarrow{\ \beta_{1}\ }H_{0}({\bf
x}_{j+1})\xrightarrow{\ \delta_{0}\ }H_{0}({\bf x}_{j+1}).$ (10)

By the third isomorphism theorem for commutative rings,

$\displaystyle H_{0}({\bf x}_{j+1})\ $ $\displaystyle\cong\ \frac{S/I}{({\bf
x}_{j+1},I)/I}\ \cong\ \frac{S}{(x_{j+1},x_{j+2},\dots,x_{n},I)}\ \cong\
S_{\leq j}/I_{\leq j},$

where $S_{\leq j}=K[x_{1},\dots,x_{j}]$ and $I_{\leq j}=I\cap S_{\leq j}$. We
observe that $I_{\leq j}$ is a monomial ideal of $S_{\leq j}$ with minimal
generating set $G(I_{\leq j})=\\{u\in G(I):\max(u)\leq j\\}$.

Let $\textup{Ker}(\delta_{0})=\textup{Im}(\beta_{1})$ be the kernel of the
rightmost non zero map of sequence (10), we obtain the short exact sequence of
$S/({\bf x}_{j+1})$-modules,

$0\rightarrow\textup{Im}(\alpha_{1})\xrightarrow{\ \alpha_{1}\ }H_{1}({\bf
x}_{j})\xrightarrow{\ \beta_{1}\ }\textup{Ker}(\delta_{0})\rightarrow 0.$ (11)

By inductive hypothesis, $H_{1}({\bf x}_{j+1})$ is generated by the homology
classes of the elements

$e(u;\varnothing)=\varepsilon(u/x_{\max(u)})e_{\max(u)},$

such that $\max(u)\geq j+1$ and $u\in G(I)$. These elements also generate
$\textup{Im}(\alpha_{1})$, as $\alpha_{1}$ sends these homology classes to the
corresponding homology classes in $H_{1}({\bf x}_{j})$. Whilst,
$\textup{Ker}(\delta_{0})$ has as a basis the elements
$\varepsilon(u/x_{\max(u)})$ with $\max(u)=j$ and $u\in G(I)$. Each of these
elements is pulled back in $H_{1}({\bf x}_{j})$ to the homology class of the
element $e(u;\varnothing)$, with $\max(u)=j$ and $u\in G(I)$. Moreover $({\bf
x}_{j})$ annihilates $H_{1}({\bf x}_{j})$. Indeed, consider
$x_{\ell}[e(u;\varnothing)]$, $\ell\in[j,n]$. If $\ell=\max(u)$, then
$x_{\ell}[e(u;\varnothing)]=[0]$. If $\ell\neq\max(u)$, then
$\partial_{2}(e_{\ell}\wedge e(u;\varnothing))=x_{\ell}e(u;\varnothing)$, so
$x_{\ell}[e(u;\varnothing)]=[0]$. Therefore, we see that a generating set for
$H_{1}({\bf x}_{j})$ as a $S/({\bf x}_{j})$-module is as given in Claim 1.

Now, let $i>1$. By (3), we have the short exact sequence of $S/({\bf
x}_{j+1})$-modules,

$\displaystyle 0\rightarrow\textup{Im}(\alpha_{i})\xrightarrow{\ \alpha_{i}\
}H_{i}({\bf x}_{j})\xrightarrow{\ \beta_{i}\
}\textup{Ker}(\delta_{i-1})\rightarrow 0.$ (12)

By inductive hypothesis, a minimal generating set of the $S/({\bf
x}_{j+1})$-module $H_{i-1}({\bf x}_{j+1})$ is given by the homology classes of
the Koszul cycles

$e(u;\sigma)\ \ \ \text{such that}\ \ \ u\in G(I),\ \ \
\sigma\subseteq\big{(}[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)\big{)}\cap[j+1,n],\ \ \ |\sigma|=i-2.$

The map $\delta_{i-1}$ is multiplication by $\pm x_{j}$. We show that the
minimal generating set of $\textup{Ker}(\delta_{i-1})$ is given by those
elements $[e(u;\sigma)]$ of $H_{i-1}({\bf x}_{j+1})$ such that
$j\notin{\operatorname{supp}}_{\bf t}(u)$.

Let $u\in G(I)$ and let $[e(u;\sigma)]$ be an element of $H_{i-1}({\bf
x}_{j+1})$ as in Claim 1.

Case 1. Suppose that $j\in{\operatorname{supp}}_{\bf t}(u)$, then
$x_{j}(u/x_{\max(u)})\notin I$. So, $\pm x_{j}e(u;\sigma)\neq 0$, as $\pm
x_{j}\varepsilon(u^{\prime})e_{\sigma}\wedge e_{\max(u)}\neq 0$. If for absurd
$[e(u;\sigma)]\in\textup{Ker}(\delta_{i-1})$, then
$\delta_{i-1}([e(u;\sigma)])=\pm x_{j}[e(u;\sigma)]=[0]$. Since $\pm
x_{j}e(u;\sigma)\neq 0$, there exists $a\in K_{i}({\bf x}_{j+1})$,
$a=\sum\varepsilon(u_{\gamma})e_{\gamma}$, for some $\gamma\subseteq[j+1,n]$,
$|\gamma|=i$, $u_{\gamma}\in S$, such that $\partial_{i}(a)=\pm
x_{j}e(u;\sigma)$. Hence,

$\displaystyle x_{j}e(u;\sigma)\ $ $\displaystyle=\
\varepsilon(x_{j}u/x_{\max(u)})e_{\sigma}\wedge e_{\max(u)}+(\textup{smaller
terms})=\partial_{i}(a)$ $\displaystyle=\
\partial_{i}\big{(}\sum\varepsilon(u_{\gamma})e_{\gamma}\big{)}=\sum_{\gamma\
:\
\gamma\setminus\\{\ell\\}=\sigma\cup\\{\max(u)\\}}\varepsilon(x_{\ell}u_{\gamma})e_{\sigma}\wedge
e_{\max(u)}+R,$

where $R$ is a sum of other terms not involving $e_{\sigma}\wedge
e_{\max(u)}$. We have $x_{j}(u/x_{\max(u)})\notin I$, _i.e._ ,
$\varepsilon(x_{j}(u/x_{\max(u)}))\neq 0$. Hence, for some $e_{\gamma_{0}}$
occurring in $a$ and some $\ell_{0}\in\gamma_{0}$ such that
$\gamma_{0}\setminus\\{\ell_{0}\\}=\sigma\cup\\{\max(u)\\}$, we must have
$x_{j}(u/x_{\max(u)})=x_{\ell_{0}}u_{\gamma_{0}}$. We have
$\ell_{0}\neq\max(u)$, and since
$x_{j}(u/x_{\max(u)})=x_{\ell_{0}}u_{\gamma_{0}}$ and $j<j+1\leq\max(u)$, we
also have $\ell_{0}<\max(u)$. Now, $\max(u)\in\gamma_{0}$ and the term
$\pm\varepsilon(x_{\max(u)}u_{\gamma_{0}})e_{\gamma_{0}\setminus\\{\max(u)\\}}$
appears in $R$. The inequality $\ell_{0}<\max(u)$ implies that
$\gamma_{0}\setminus\\{\max(u)\\}>\gamma_{0}\setminus\\{\ell_{0}\\}=\sigma\cup\\{\max(u)\\}$,
and since each wedge product appearing in $x_{j}e(u;\sigma)$ is smaller than
$e_{\sigma}\wedge e_{\max(u)}$, we must have either
$\varepsilon(x_{\max(u)}u_{\gamma_{0}})=0$ or there exist
$\gamma_{1}\subseteq[j+1,n]$ and an integer $\ell_{1}\in\gamma_{1}$ such that
the term
$\pm\varepsilon(x_{\ell_{1}}u_{\gamma_{1}})e_{\gamma_{1}\setminus\\{\ell_{1}\\}}$
appears in $R$ and cancels with
$\pm\varepsilon(x_{\max(u)}u_{\gamma_{0}})e_{\gamma_{0}\setminus\\{\max(u)\\}}$.

Subcase 1.1. We have $x_{\max(u)}u_{\gamma_{0}}\in I$. Hence,

$x_{\max(u)}u_{\gamma_{0}}\ =\ x_{\max(u)}x_{j}(u/x_{\max(u)})/x_{\ell_{0}}\
=\ x_{j}(u/x_{\ell_{0}})\in I.$

Therefore, $x_{j}(u/x_{\ell_{0}})\in I$, absurd. Indeed, write
$u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}$, then $j\in{\operatorname{supp}}_{\bf
t}(u)$ implies that $j=j_{p}+r$, with $0\leq r\leq t_{p}-1$, $t_{p}\geq 1$.
Moreover, $\ell_{0}>j$, so $\ell_{0}=j_{q}$, $q>p$. If
$x_{j}(u/x_{\ell_{0}})=x_{j_{1}}\cdots x_{j_{p}}x_{j_{p}+r}x_{j_{p+1}}\cdots
x_{j_{q-1}}x_{j_{q+1}}\cdots x_{j_{d}}\in I$, then for some $v\in G(I)$, $v$
divides $x_{j}(u/x_{\ell_{0}})$. As $v$ is ${\bf t}$-spread but
$x_{j}(u/x_{\ell_{0}})$ is not, we have $v\neq x_{j}(u/x_{\ell_{0}})$, hence
$\deg(v)<\deg(x_{j}(u/x_{\ell_{0}}))=\deg(u)$. So, $v$ must divide
$x_{j}(u/x_{\ell_{0}}))/x_{j}=u/x_{\ell_{0}}$, and $u/x_{\ell_{0}}\in I$,
absurd as $u\in G(I)$.

Subcase 1.2. We have
$\gamma_{1}\setminus\\{\ell_{1}\\}=\gamma_{0}\setminus\\{\max(u)\\}=\sigma\cup\\{\ell_{0}\\}$
and $x_{\max(u)}u_{\gamma_{0}}=x_{\ell_{1}}u_{\gamma_{1}}$. Therefore,
$\gamma_{1}=\sigma\cup\\{\ell_{0},\ell_{1}\\}$. The term
$\pm\varepsilon(x_{\ell_{0}}u_{\gamma_{1}})e_{\sigma\cup\\{\ell_{1}\\}}$
appears in $R$ and it is bigger than $\varepsilon(u^{\prime})e_{\sigma}\wedge
e_{\max(u)}$. So, we have two cases to consider. As before, in the first case,
$\varepsilon(x_{\ell_{0}}u_{\gamma_{1}})=0$, and recalling that
$x_{j}(u/x_{\max(u)})=x_{\ell_{0}}u_{\gamma_{0}}$, we have

$\displaystyle x_{\ell_{0}}u_{\gamma_{1}}\ $ $\displaystyle=\
x_{\ell_{0}}(x_{\max(u)}u_{\gamma_{0}})/x_{\ell_{1}}\ =\
x_{\max(u)}(x_{\ell_{0}}u_{\gamma_{0}})/x_{\ell_{1}}\ =\
x_{\max(u)}x_{j}(u/x_{\max(u)})/x_{\ell_{1}}$ $\displaystyle=\
x_{j}(u/x_{\ell_{1}})\in I,$

with $\ell_{1}>j$. Arguing as in Subcase 1.1 we obtain an absurd. Otherwise
there exist $\gamma_{2}\subseteq[j+1,n]$ and an integer
$\ell_{2}\in\gamma_{2}$ such that the term
$\pm\varepsilon(x_{\ell_{2}}u_{\gamma_{2}})e_{\gamma_{2}\setminus\\{\ell_{2}\\}}$
appears in $R$ and cancels with
$\pm\varepsilon(x_{\ell_{0}}u_{\gamma_{1}})e_{\sigma\cup\\{\ell_{1}\\}}$. We
have $\gamma_{2}=\sigma\cup\\{\ell_{1},\ell_{2}\\}$ and consider the term
arising from $\gamma_{2}\setminus\\{\ell_{1}\\}$. We can distinguish two cases
as before. After a finite number of steps $s$, we have
$x_{j}(u/x_{\ell_{s}})\in I$ for some $\ell_{s}>j$, obtaining an absurd.
Hence, $\pm x_{j}e(u;\sigma)\notin\textup{Im}(\partial_{i})$, and $\pm
x_{j}[e(u;\sigma)]\notin\textup{Ker}(\delta_{i-1})$.

Case 2. Suppose now $j\notin{\operatorname{supp}}_{\bf t}(u)$. By Remark 3.7,

$e(u;\sigma\cup\\{j\\})=e_{j}\wedge e(u;\sigma)+r(u;\sigma).$

Recalling the map $\beta_{i}:K_{i}({\bf x}_{j})\rightarrow K_{i-1}({\bf
x}_{j+1})$, we have that $\beta_{i}(e(u;\sigma\cup\\{j\\}))=e(u;\sigma)$. By
Proposition 3.4, $e(u;\sigma\cup\\{j\\})$ is a cycle. We prove that
$[e(u;\sigma\cup\\{j\\})]\neq[0]$ in $H_{i}({\bf x}_{j})$. Suppose on the
contrary that there exists $a\in K_{i+1}({\bf x}_{j})$ such that
$\partial_{i+1}(a)=e(u;\sigma\cup\\{j\\})$. Now,
$a=\sum\varepsilon(u_{\gamma})e_{\gamma}$, for some $\gamma\subseteq[j,n]$,
$|\gamma|=i+1$ and $u_{\gamma}\in S$. So,

$\displaystyle e(u;\sigma\cup\\{j\\})\ $ $\displaystyle=\
\varepsilon(u/x_{\max(u)})e_{\sigma\cup\\{j\\}}\wedge
e_{\max(u)}+(\textup{smaller terms})=\partial_{i}(a)$ $\displaystyle=\
\partial_{i}\big{(}\sum\varepsilon(u_{\gamma})e_{\gamma}\big{)}=\sum_{\gamma\
:\
\gamma\setminus\\{\ell\\}=\sigma\cup\\{j,\max(u)\\}}\varepsilon(x_{\ell}u_{\gamma})e_{\sigma\cup\\{j\\}}\wedge
e_{\max(u)}+R,$

where $R$ is a sum of other terms not involving $e_{\sigma\cup\\{j\\}}\wedge
e_{\max(u)}$. For some $e_{\gamma_{0}}$ occurring in $a$ and some
$\ell_{0}\in\gamma_{0}$ such that
$\gamma_{0}\setminus\\{\ell_{0}\\}=\sigma\cup\\{j,\max(u)\\}$, we must have
$u/x_{\max(u)}=x_{\ell_{0}}u_{\gamma_{0}}$. We have $\ell_{0}\neq\max(u)$ and
$j<j+1\leq\max(u)$, so $\ell_{0}<\max(u)$. Therefore,
$\gamma_{0}=\sigma\cup\\{j,\ell_{0},\max(u)\\}$ and
$\pm\varepsilon(x_{\max(u)}u_{\gamma_{0}})e_{\gamma_{0}\setminus\\{\max(u)\\}}=\pm\varepsilon(x_{\max(u)}u_{\gamma_{0}})e_{\sigma\cup\\{j,\ell_{0}\\}}$
appears in $R$. Now $\ell_{0}<\max(u)$ implies
$\gamma_{0}\setminus\\{\max(u)\\}>\gamma_{0}\setminus\\{\ell_{0}\\}=\sigma\cup\\{j,\max(u)\\}$,
and since each wedge product appearing in $e(u;\sigma\cup\\{j\\})$ is smaller
than $e_{\sigma\cup\\{j\\}}\wedge e_{\max(u)}$, we must have either
$\varepsilon(x_{\max(u)}u_{\gamma_{0}})=0$ or there exist
$\gamma_{1}\subseteq[j,n]$ and $\ell_{1}\in\gamma_{1}$ such that the term
$\pm\varepsilon(x_{\ell_{1}}u_{\gamma_{1}})e_{\gamma_{1}\setminus\\{\ell_{1}\\}}$
appears in $R$ and cancels with
$\pm\varepsilon(x_{\max(u)}u_{\gamma_{0}})e_{\gamma_{0}\setminus\\{\max(u)\\}}$.

Subcase 2.1. We have $x_{\max(u)}u_{\gamma_{0}}\in I$. So,
$x_{\max(u)}u_{\gamma_{0}}=x_{\max(u)}(u/x_{\max(u)})/{x_{\ell_{0}}}=u/x_{\ell_{0}}\in
I,$ but this is absurd, as $u$ is a minimal monomial generator of $I$.

Subcase 2.2. We have
$\gamma_{1}\setminus\\{\ell_{1}\\}=\gamma_{0}\setminus\\{\max(u)\\}=\sigma\cup\\{j,\ell_{0}\\}$
and $x_{\max(u)}u_{\gamma_{0}}=x_{\ell_{1}}u_{\gamma_{1}}$. Therefore,
$\gamma_{1}=\sigma\cup\\{j,\ell_{0},\ell_{1}\\}$. The term
$\pm\varepsilon(x_{\ell_{0}}u_{\gamma_{1}})e_{\sigma\cup\\{j,\ell_{1}\\}}$
appears in $R$ and it is bigger than
$\varepsilon(u^{\prime})e_{\sigma\cup\\{j\\}}\wedge e_{\max(u)}$. So, we have
two cases to consider. As before, in the first case,
$\varepsilon(x_{\ell_{0}}u_{\gamma_{1}})=0$, and recalling that
$u/x_{\max(u)}=x_{\ell_{0}}u_{\gamma_{0}}$, we have

$\displaystyle x_{\ell_{0}}u_{\gamma_{1}}\ $ $\displaystyle=\
x_{\ell_{0}}(x_{\max(u)}u_{\gamma_{0}})/x_{\ell_{1}}\ =\
x_{\max(u)}(x_{\ell_{0}}u_{\gamma_{0}})/x_{\ell_{1}}\ =\
x_{\max(u)}(u/x_{\max(u)})/x_{\ell_{1}}$ $\displaystyle=\ u/x_{\ell_{1}}\in
I,$

an absurd, as $u\in G(I)$. Otherwise, we iterate the reasoning. After a finite
number of steps $s$, we have $u/x_{\ell_{s}}\in I$, for some $\ell_{s}$, an
absurd. Hence $e(u;\sigma\cup\\{j\\})\notin\textup{Im}(\partial_{i+1})$, and
$[e(u;\sigma\cup\\{j\\})]\neq[0]$ in $H_{i}({\bf x}_{j})$. Therefore,
$\beta_{i}([e(u;\sigma\cup\\{j\\})])=[e(u;\sigma)]$, and
$[e(u;\sigma)]\in\textup{Im}(\beta_{i})=\textup{Ker}(\delta_{i-1})$, as
desired.

Finally, a basis for $\beta_{i}^{-1}(\textup{Ker}(\delta_{i-1}))$ is given by
all the elements as in Claim 1 such that $j\in\sigma$. By inductive
hypothesis, we know a basis for $H_{i}({\bf x}_{j+1})$, and as $\alpha_{i}$
sends these homology classes to the corresponding homology classes of
$H_{i}({\bf x}_{j})$, a minimal generating set for $\textup{Im}(\alpha_{i})$
is given by all the elements as in Claim 1 such that $j\notin\sigma$.

We observe that $({\bf x}_{j})$ annihilates these elements. Indeed, the
elements $[e(u;\sigma)]$ as in Claim 1 minimally generate $H_{i}({\bf x}_{j})$
as a $S/({\bf x}_{j+1})$-module. So $({\bf x}_{j+1})$ annihilates all
$[e(u;\sigma)]$. It remains to prove that $x_{j}$ annihilates all elements
$[e(u;\sigma)]$. If $j\notin\sigma$, then by definition of $e(u;\sigma)$,
$e_{j}$ doesn’t appear in the first term
$\varepsilon(u/x_{\max(u)})e_{\sigma}\wedge e_{\max(u)}$ of $e(u;\sigma)$. We
have

$\partial_{i+1}(e_{j}\wedge
e(u;\sigma))=x_{j}e(u;\sigma)+e_{j}\wedge(-1)^{\deg(e_{j})}\partial_{i}(e(u;\sigma))=x_{j}e(u;\sigma),$

so $x_{j}[e(u;\sigma)]=[0]$. Suppose now $j\in\sigma$. Then
$\beta_{i}(x_{j}[e(u;\sigma)])=x_{j}[e(u;\sigma\setminus\\{j\\})]=[0]$, as
$[e(u;\sigma\setminus\\{j\\})]\in\textup{Ker}(\delta_{i-1})$. Hence,
$x_{j}[e(u;\sigma)]\in\textup{Ker}(\beta_{i})=\textup{Im}(\alpha_{i})$. By
Remark 3.7,

$\displaystyle x_{j}[e(u;\sigma)]\ $ $\displaystyle=\ x_{j}\big{[}e_{j}\wedge
e(u;\sigma\setminus\\{j\\})+r(u;\sigma)\big{]}\ =\
x_{j}\big{[}r(u;\sigma)\big{]}\in\textup{Im}(\alpha_{i})\subseteq H_{i}({\bf
x}_{j+1}),$

the first summand vanishes, as $e_{j}\notin H_{i}({\bf x}_{j+1})$. If we set
$a=r(u;\sigma)$, $x_{j}[a]$ is a cycle, and we have
$\partial_{i+1}(e_{j}\wedge a)=-x_{j}a$, so
$x_{j}\big{[}r(u;\sigma)\big{]}=[0]$ and $x_{j}\big{[}e(u;\sigma)\big{]}=[0]$,
as desired. Hence, a minimal generating set for the $S/({\bf x}_{j})$-module
$H_{i}({\bf x}_{j})$ is as in Claim 1.

Finally for $j=1$, $S/({\bf x}_{1})=S/(x_{1},\dots,x_{n})\cong K$, and Claim 1
implies the result, as a minimal generating set of a $K$-vector space is a
basis. $\Box$

We provide an example that demonstrate our methods.

###### Example 3.9

Let ${\bf t}=(1,0,2)$, and let
$I=(x_{1},x_{2}x_{3}^{2},x_{2}x_{3}x_{4}x_{6},x_{2}x_{4}^{2}x_{6})$. We set
$w_{1}=x_{1},w_{2}=x_{2}x_{3}^{2},w_{3}=x_{2}x_{3}x_{4}x_{6},w_{4}=x_{2}x_{4}^{2}x_{6}$.
The ideal $I\subseteq S=K[x_{1},\dots,x_{6}]$ is a ${\bf t}$-spread strongly
stable ideal with minimal generating set $G(I)=\\{w_{1},w_{2},w_{3},w_{4}\\}$.
Let ${\bf x}=x_{1},x_{2},\dots,x_{6}$. The basis for the Koszul homologies of
$S/I$ are:

  1.     1.       1. $H_{1}({\bf x};S/I)$:

$e(w;\varnothing)\ =\ \varepsilon(w/x_{\max(w)})\ e_{\max(w)}$, for $w\in
G(I)$;

      2. $H_{2}({\bf x};S/I)$:

$w_{1}$ gives no rise to any basis element,

        1. $w_{2}$ gives

$e(w_{2};\\{1\\})\ =\ \varepsilon(x_{2}x_{3})\ e_{1}\wedge e_{3}$,

        2. $w_{3}$ gives

$e(w_{3};\\{1\\})\ =\ \varepsilon(x_{2}x_{3}x_{4})\ e_{1}\wedge e_{6}$,

        3. $e(w_{3};\\{3\\})\ =\ \varepsilon(x_{2}x_{3}x_{4})\ e_{3}\wedge e_{6}$,

        4. $w_{4}$ gives

$e(w_{4};\\{1\\})\ =\ \varepsilon(x_{2}x_{4}^{2})\ e_{1}\wedge e_{6}$,

        5. $e(w_{4};\\{3\\})\ =\ \varepsilon(x_{2}x_{4}^{2})\ e_{3}\wedge e_{6}-\varepsilon(x_{2}x_{3}x_{4})\ e_{4}\wedge e_{6}$;

      3. $H_{3}({\bf x};S/I)$:

$w_{1},w_{2}$ give no rise to any basis element,

        1. $w_{3}$ gives

$e(w_{3};\\{1,3\\})\ =\ \varepsilon(x_{2}x_{3}x_{4})\ e_{1}\wedge e_{3}\wedge
e_{6}$,

        2. $w_{4}$ gives

$e(w_{4};\\{1,3\\})\ =\ \varepsilon(x_{2}x_{4}^{2})\ e_{1}\wedge e_{3}\wedge
e_{6}-\varepsilon(x_{2}x_{3}x_{4})\ e_{1}\wedge e_{4}\wedge e_{6}$;

      4. $H_{j}({\bf x};S/I)$:

$\varnothing$, for all $j\geq 4$.

For instance, consider $e(w_{4};\\{1,3\\})\in K_{3}({\bf x};S/I)$. Then

$\displaystyle\partial_{3}(e(w_{4};\\{1,3\\}))\ $ $\displaystyle=\
\partial_{3}\big{(}\varepsilon(x_{2}x_{4}^{2})\ e_{1}\wedge e_{3}\wedge
e_{6}-\varepsilon(x_{2}x_{3}x_{4})\ e_{1}\wedge e_{4}\wedge e_{6}\big{)}$
$\displaystyle=\ \varepsilon(x_{1}x_{2}x_{4}^{2})\ e_{3}\wedge
e_{6}-\varepsilon(x_{2}x_{3}x_{4}^{2})\ e_{1}\wedge
e_{6}+\varepsilon(x_{2}x_{4}^{2}x_{6})\ e_{1}\wedge e_{3}$ $\displaystyle-\
\varepsilon(x_{1}x_{2}x_{3}x_{4})\ e_{4}\wedge
e_{6}+\varepsilon(x_{2}x_{3}x_{4}^{2})\ e_{1}\wedge
e_{6}-\varepsilon(x_{2}x_{3}x_{4}x_{6})\ e_{1}\wedge e_{4}$ $\displaystyle=\
0.$

In fact, the first, third, fourth and sixth terms vanish, as
$\varepsilon(x_{1}x_{2}x_{4}^{2})=\varepsilon(x_{2}x_{4}^{2}x_{6})=\varepsilon(x_{1}x_{2}x_{3}x_{4})=\varepsilon(x_{2}x_{3}x_{4}x_{6})=0$,
and the second and fifth terms are opposite.

We illustrate how to obtain some of these elements.

Consider $w_{3}=x_{2}x_{3}x_{4}x_{6}$. Then ${\operatorname{supp}}_{\bf
t}(w_{3})={\operatorname{supp}}_{(1,0,2)}(x_{2}x_{3}x_{4}x_{6})=\\{2,4,5\\}$
and $[\max(w_{3})-1]\setminus{\operatorname{supp}}_{\bf t}(w_{3})=\\{1,3\\}$.
Let $\vartheta=\\{1,3\\}$, then $\vartheta^{(w_{3})}=\\{2,4\\}$. Moreover
$\max(w_{3})=6\notin\vartheta^{(w_{3})}$. So, we can use equation (6) to
compute the relevant Koszul cycles $e(w_{3};\sigma)$. Of course, we may also
use equation (4).

The monomial $w_{3}$ gives rise to the following Koszul cycles:

$\begin{array}[]{rrrll}\displaystyle\sigma=\varnothing;&&e(w_{3};\varnothing)&=\
\phantom{-}\varepsilon(w_{3}/x_{6})e_{6}=\varepsilon(x_{2}x_{3}x_{4})e_{6},\\\
\\\ \displaystyle\sigma=\\{1\\};&&e(w_{3};\\{1\\})&=\
-e(w_{3};\varnothing)\wedge e_{1}+e(x_{1}(w_{3}/x_{2});\varnothing)\wedge
e_{2}\\\ \displaystyle\hfil&&&=\ -\varepsilon(x_{2}x_{3}x_{4})e_{6}\wedge
e_{1}+\varepsilon(x_{1}x_{3}x_{4})e_{6}\wedge e_{2}\\\ \displaystyle\hfil&&&=\
\phantom{-}\varepsilon(x_{2}x_{3}x_{4})e_{1}\wedge e_{6},\\\ \\\
\displaystyle\sigma=\\{3\\};&&e(w_{3};\\{3\\})&=\ -e(w_{3};\varnothing)\wedge
e_{3}+e(x_{3}(w_{3}/x_{4});\varnothing)\wedge e_{4}\\\ \displaystyle\hfil&&&=\
-\varepsilon(x_{2}x_{3}x_{4})e_{6}\wedge
e_{3}+\varepsilon(x_{2}x_{3}^{2})e_{6}\wedge e_{4}\\\ \displaystyle\hfil&&&=\
\phantom{-}\varepsilon(x_{2}x_{3}x_{4})e_{3}\wedge e_{6},\\\ \\\
\displaystyle\sigma=\\{1,3\\};&&e(w_{3};\\{1,3\\})&=\ -e(w_{3};\\{1\\})\wedge
e_{3}+e(x_{3}(w_{3}/x_{4});\\{1\\})\wedge e_{4}\\\ \displaystyle\hfil&&&=\
-\varepsilon(x_{2}x_{3}x_{4})e_{1}\wedge e_{6}\wedge
e_{3}+e(x_{2}x_{3}^{2}x_{6};\\{1\\})\wedge e_{4}\\\ \displaystyle\hfil&&&=\
\phantom{-}\varepsilon(x_{2}x_{3}x_{4})e_{1}\wedge e_{3}\wedge
e_{6}+\varepsilon(x_{2}x_{3}^{2})e_{1}\wedge e_{6}\wedge e_{4}\\\
\displaystyle\hfil&&&=\ \phantom{-}\varepsilon(x_{2}x_{3}x_{4})e_{1}\wedge
e_{3}\wedge e_{6}.\end{array}$

Our computations yield the Betti table of $S/I$,

$\begin{matrix}&0&1&2&3\\\ \text{total:}&1&4&5&2\\\
\text{0:}&1&1&\text{-}&\text{-}\\\
\text{1:}&\text{-}&\text{-}&\text{-}&\text{-}\\\
\text{2:}&\text{-}&1&1&\text{-}\\\ \text{3:}&\text{-}&2&4&2\\\ \end{matrix}$

###### Remark 3.10

The expression of our Koszul cycles is not so nice. Indeed, a basis element
$e(u;\sigma)$ of $H_{i}({\bf x};S/I)$, $I$ a ${\bf t}$-spread strongly stable
ideal, is a sum of $2^{i-1}$ wedge products! However, if ${\bf
t}=(1,\dots,1,0,\dots,0)\in{\mathbb{Z}}_{\geq 0}^{d-1}$, $d\geq 2$, the
element

$z(u;\sigma):=\varepsilon(u/x_{\max(u)})e_{\sigma}\wedge e_{\max(u)}.$

with $u\in G(I)$ and
$\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$ is
easily seen to be a cycle. Indeed,

$\displaystyle\partial_{i}(z(u;\sigma))\ $ $\displaystyle=\
\sum_{j=1}^{i-1}(-1)^{j+1}\varepsilon(x_{k_{j}}(u/x_{\max(u)}))e_{\sigma\setminus\\{k_{j}\\}}\wedge
e_{\max(u)}+(-1)^{i+1}\varepsilon(u)e_{\sigma}$ $\displaystyle=\ 0,$

as $x_{k_{j}}(u/x_{\max(u)})\in I$ for all $j$ and $u\in I$, since ${\bf
t}=(1,\dots,1,0,\dots,0)$. It’s easy to see that the homology classes
$[z(u;\sigma)]$ are non zero and $K$-independent. Hence, they form a basis for
$H_{i}({\bf x})$, as the map $z:z(u;\sigma)\mapsto e(u;\sigma)$ is a bijection
and the elements $e(u;\sigma)$ form a basis of $H_{i}({\bf x})$ by Theorem
3.8. These Koszul cycles have been considered in the papers [9, 10]. But in
general they are cycles only when the vector ${\bf t}$ has the form ${\bf
t}=(1,\dots,1,0,\dots,0)$.

###### Example 3.11

Let
$I=(x_{1}x_{2}\,,x_{1}x_{3}\,,x_{1}x_{2}^{2}\,,x_{1}x_{2}x_{3}\,,x_{1}x_{2}x_{4}\,,x_{1}x_{3}^{2}\,,x_{1}x_{3}x_{4}\,,x_{1}x_{4}^{2})$
be a $(1,0)$-spread strongly stable ideal of $K[x_{1},x_{2},x_{3},x_{4}]$. By
Remark 3.10, since ${\bf t}=(1,0)$ and
$G(I)=\big{\\{}x_{1}x_{2},x_{1}x_{3},x_{1}x_{4}^{2}\big{\\}}$, the relevant
basis for the Koszul homologies of $S/I$ are:

  1.     1.       1. $H_{1}({\bf x};S/I)$:

$\varepsilon(x_{1}x_{2}/x_{2})\ e_{2}$, $\varepsilon(x_{1}x_{3}/x_{3})\
e_{3}$, $\varepsilon(x_{1}x_{4}^{2}/x_{4})\ e_{4}$;

      2. $H_{2}({\bf x};S/I)$:

$\varepsilon(x_{1}x_{3}/x_{3})\ e_{2}\wedge e_{3}$,
$\varepsilon(x_{1}x_{4}^{2}/x_{4})\ e_{2}\wedge e_{4}$,
$\varepsilon(x_{1}x_{4}^{2}/x_{4})\ e_{3}\wedge e_{4}$;

      3. $H_{3}({\bf x};S/I)$:

$\varepsilon(x_{1}x_{4}^{2}/x_{4})\ e_{2}\wedge e_{3}\wedge e_{4}$;

      4. $H_{j}({\bf x};S/I)$:

$\varnothing$, for all $j\geq 4$.

Therefore, using _Macaulay2_ [22] the Betti table of $I$ is

$\begin{matrix}&0&1&2\\\ \text{total:}&3&3&1\\\ \text{2:}&2&1&\text{-}\\\
\text{3:}&1&2&1\end{matrix}$

## 4\. The minimal free resolution of
vector-spread strongly stable ideals

In this Section we construct the minimal free resolution of ${\bf t}$-spread
strongly stable ideals of $S$. This resolution will generalize that of Eliahou
and Kervaire [17], and also the squarefree lexsegment analogue in [10]. We
will follow the construction given by Aramova and Herzog in [9].

Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a ${\bf t}$-spread strongly stable
ideal. Note that, since $\operatorname{Tor}_{i}^{S}(K,S/I)\cong H_{i}({\bf
x};S/I)=H_{i}({\bf x})$, for all $i$, the minimal free resolution of $S/I$ may
be written as follows,

${\mathbb{F}}:\cdots\xrightarrow{\ d_{3}\ }S\otimes_{K}H_{2}({\bf
x})\xrightarrow{\ d_{2}\ }S\otimes_{K}H_{1}({\bf x})\xrightarrow{\ d_{1}\
}S\otimes_{K}H_{0}({\bf x})\xrightarrow{\ d_{0}\ }S/I\rightarrow 0.$

We set $F_{i}=S\otimes_{K}H_{i}({\bf x})$, for all $i$, and note that
$F_{0}=S$. By Theorem 3.8 and also [9], for all $i\geq 1$ a basis of the
graded free $S$-module $F_{i}$ is given by the elements,

$f(u;\sigma):=1\otimes(-1)^{(i-1)(i-2)/2}[e(u;\sigma)],$

such that $u\in G(I)$,
$\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$ and
$|\sigma|=i-1$. For later use, we shall make the following convention. If
$\sigma\not\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$ we
set $f(u;\sigma)=0$.

Thus, it remains to describe the differentials $d_{i}$, for all $i\geq 0$. For
this purpose, suppose the differentials $d_{0},d_{1},\dots,d_{i-1}$ have
already been constructed such that

${\mathbb{F}}_{<i}:F_{i-1}\xrightarrow{d_{i-1}}F_{i-2}\xrightarrow{d_{i-2}}\cdots\xrightarrow{\
d_{1}\ }F_{0}\xrightarrow{\ d_{0}\ }S/I\rightarrow 0$

is exact. Fix a basis element $f(u;\sigma)$ of $F_{i}$. Let
$\mathbb{K}=K_{{}_{\text{\large$\bm{\cdot}$}}}({\bf x};S/I)$ be the Koszul
complex attached to ${\bf x}$ with respect to $S/I$ whose $i$th module and
differential are, respectively, $K_{i}$ and $\partial_{i}:K_{i}\rightarrow
K_{i-1}$. We consider the double complex
$\mathbb{K}\otimes_{S}{\mathbb{F}}_{<i}$,

$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{2}}$$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{2}}$$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{2}}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K_{n}\otimes
F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{1}}$$\scriptstyle{\partial_{n}\otimes\textup{id}\ \
}$$\textstyle{K_{n-1}\otimes
F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\ \ \ \ \partial_{n-1}\otimes\textup{id}}$$\scriptstyle{\textup{id}\otimes
d_{1}}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{1}\otimes\textup{id}\
\ \ \ }$$\textstyle{K_{0}\otimes
F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{1}}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K_{n}\otimes
F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{0}}$$\scriptstyle{\partial_{n}\otimes\textup{id}\ \
}$$\textstyle{K_{n-1}\otimes
F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\ \ \ \ \partial_{n-1}\otimes\textup{id}}$$\scriptstyle{\textup{id}\otimes
d_{0}}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{1}\otimes\textup{id}\
\ \ \ }$$\textstyle{K_{0}\otimes
F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\textup{id}\otimes
d_{0}}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K_{n}\otimes
S/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{n}\otimes\textup{id}\
\ }$$\textstyle{K_{n-1}\otimes
S/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\
\ \ \ \ \
\partial_{n-1}\otimes\textup{id}}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{1}\otimes\textup{id}\
\ \ \ }$$\textstyle{K_{0}\otimes
S/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0}$$\textstyle{0}$$\textstyle{0}$

where “id” denotes each time a suitable identity function.

It is known by [9, Section 1] that to describe how the differential $d_{i}$
acts on $f(u;\sigma)$ it suffices to determine elements $g_{j}\in
K_{i-j}\otimes F_{j}$, $j=0,\dots,i-1$, satisfying

$\displaystyle(\textup{id}_{K_{i}}\otimes d_{0})(g_{0})\ $ $\displaystyle=\
(-1)^{(i-1)(i-2)/2}1\otimes e(u;\sigma),\ \ \textup{and}$ (13)
$\displaystyle(\textup{id}_{K_{i-j-1}}\otimes d_{j+1})(g_{j+1})\ $
$\displaystyle=\ (\partial_{i-j}\otimes\textup{id}_{F_{j}})(g_{j})\ \
\textup{for}\ j=0,\dots,i-2.$ (14)

To construct such a sequence is a difficult combinatorial task. Thus we
restrict ourself to the case when ${\bf t}=(1,\dots,1,0,\dots,0)$, (Remark
3.10). In this case we can replace the cycles $e(u;\sigma)$ by the cycles
$z(u;\sigma)$. In order to construct the sequence of elements satisfying
equations (13) and (14) we need the following notion. We recall that the pure
lexicographic order is defined as follows: $x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{n}^{a_{n}}>_{\textup{plex}}x_{1}^{b_{1}}x_{2}^{b_{2}}\cdots x_{n}^{b_{n}}$
if and only if $a_{1}=b_{1},\ a_{2}=b_{2},\ \dots,\ a_{s-1}=b_{s-1}$ and
$a_{s}>b_{s}$ for some $s\in\\{1,\dots,n\\}$.

###### Definition 4.1

Let $I\subset S$ be a ${\bf t}$-spread strongly stable ideal, ${\bf
t}=(1,\dots,1,0,\dots,0)$. Let $M(I)$ be the set of all monomials belonging to
$I$. We define the map $g:M(I)\rightarrow G(I)$, as follows: for $w\in M(I)$,
we set $g(w):=\max_{>_{\textup{plex}}}\\{u\in G(I):u\ \textup{divides}\ w\\}$.
The map $g$ is called the ${\bf t}$-spread decomposition function of $I$.

For $u\in G(I)$, $k\in[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$,
we set

$u_{k}:=g(x_{k}u)\ \ \ \text{and}\ \ \ v_{k}:=(x_{k}u)/u_{k}.$

We shall need also the following notations. For a subset $\sigma$ of $[n]$ and
for $k\in\sigma$ we define $\alpha(\sigma;k):=|\\{s\in\sigma:s<k\\}|$. For
$\tau$ a subset of $\sigma$ we let
$\gamma(\tau):=\sum_{k\in\sigma\setminus\tau}\alpha(\sigma;k)$. In what
follows, we denote $\sigma\setminus\\{k\\}$ by $\sigma\setminus k$, and
$\sigma\cup\\{k\\}$ by $\sigma\cup k$, omitting the parentheses. To further
simplify the notations, we set $\textup{id}_{K_{i-j-1}}\otimes
d_{j+1}=d_{j+1}$ and
$\partial_{i-j}\otimes\textup{id}_{F_{j}}=\partial_{i-j}$, for all
$j=0,\dots,i-2$.

The next theorem gives the desired differentials of the resolution
${\mathbb{F}}$ of $S/I$ and generalize [9, Theorem 2.3]. To write the elements
$g_{j}$ more conveniently we switch the order in the tensor products, that is
we think $g_{j}$ as an element of $F_{j}\otimes K_{i-j}$.

###### Theorem 4.2

Let $g_{0}=(-1)^{\frac{(i-1)(i-2)}{2}}1\otimes(u^{\prime}e_{\sigma}\wedge
e_{\max(u)})$, and for $j=1,\dots,i-1$ let

$\displaystyle g_{j}=(-1)^{i-j}\sum_{\begin{subarray}{c}\tau\subset\sigma\\\
|\tau|=j-1\end{subarray}}(-1)^{\gamma(\tau)}f(u;\tau)\otimes
e_{\sigma\setminus\tau}+\sum_{\begin{subarray}{c}\tau\subset\sigma\\\
|\tau|=j\end{subarray}}(-1)^{\gamma(\tau)}s_{\tau}\otimes
e_{\sigma\setminus\tau}\wedge e_{\max(u)},$

where

$s_{\tau}=\sum_{k\in\tau}(-1)^{\alpha(\tau;k)}\frac{v_{k}}{x_{\max(u)}}f(u_{k};\tau\setminus
k).$

Then the elements $g_{0},g_{1},\dots,g_{i-1}$ satisfy equations (13) and (14).
Moreover, the $i$th differential of the minimal free resolution of $S/I$
acting on $f(u;\sigma)$ is given by

$\displaystyle d_{i}(f(u;\sigma))\ $ $\displaystyle=\ \partial_{1}(g_{i-1})$
$\displaystyle=\
\sum_{k\in\sigma}(-1)^{\alpha(\sigma;k)}\big{(}-x_{k}f(u;\sigma\setminus
k)+v_{k}f(u_{k};\sigma\setminus k)\big{)}.$ (15)

Proof. Note that in the definition of $s_{\tau}$, $x_{\max(u)}$ always divide
$v_{k}$. Indeed, if we let
$k^{(u)}=\min\\{j\in{\operatorname{supp}}(u):j>k\\}$, then
$w=x_{k}(u/x_{k^{(u)}})\in I$ is again a ${\bf t}$-spread monomial. By Lemma
2.6, $w=w_{1}w_{2}$ with $w_{1}\in G(I)$ and $\max(w_{1})\leq\min(w_{2})$.
Consequently $\\{y\in G(I):y\ \textup{divides}\ x_{k}u\\}$ is non empty and
$u_{k}$ exists. Proceeding as in the proof of Lemma 2.6 we see that
$\max(u_{k})\leq\min(v_{k})$. Finally, $v_{k}\neq 1$ otherwise
$u_{k}=x_{k}u\in G(I)$, which is absurd. Hence $x_{\max(u)}$ divides $v_{k}$
as wanted.

We proceed by induction on $i$. The case $i=1$ is trivial. By induction, we
can assume that the last formula for the differential $d_{\ell}$ holds for
$\ell<i$. We need to verify the equations
$\partial_{i-j}(g_{j})=d_{j+1}(g_{j+1})$. For $j=0$ this is trivial. Let
$j>0$.

Firstly, we calculate $\partial_{i-j}(g_{j})$. Since
$|\sigma\setminus\tau|=i-j-1$, we have

$\displaystyle\partial_{i-j}(g_{j})=(-1)^{i-j}\sum_{\begin{subarray}{c}\tau\subset\sigma\\\
|\tau|=j-1\end{subarray}}(-1)^{\gamma(\tau)}f(u;\tau)\otimes\big{(}\sum_{k\in\sigma\setminus\tau}(-1)^{\alpha(\sigma\setminus\tau;k)}x_{k}e_{\sigma\setminus(\tau\cup
k)}\big{)}$ $\displaystyle+\sum_{\begin{subarray}{c}\tau\subset\sigma\\\
|\tau|=j\end{subarray}}(-1)^{\gamma(\tau)}s_{\tau}\otimes\big{(}\sum_{k\in\sigma\setminus\tau}(-1)^{\alpha(\sigma\setminus\tau;k)}x_{k}e_{\sigma\setminus(\tau\cup
k)}\wedge e_{\max(u)}+(-1)^{i-j-1}x_{\max(u)}e_{\sigma\setminus\tau}\big{)}.$

We suitably rewrite both sums.

For the first sum, note that for $\tau\subseteq\sigma$, $|\tau|=j-1$ and
$k\in\sigma\setminus\tau$, then setting $\rho=\tau\cup k$ we have that
$|\rho|=j$, $\gamma(\tau)=\gamma(\rho\setminus
k)=\sum_{s\in\sigma\setminus(\rho\cup
k)}\alpha(\sigma;s)=\gamma(\rho)+\alpha(\sigma;k)$,
$\alpha(\sigma;k)=\alpha(\sigma\setminus\tau;k)+\alpha(\tau;k)$ and also
$\alpha(\tau;k)=\alpha(\rho;k)$ for it is $k\notin\tau$. Hence,

$(-1)^{\gamma(\tau)}(-1)^{\alpha(\sigma\setminus\tau;k)}=(-1)^{\gamma(\rho)+\alpha(\sigma\setminus\tau;k)+\alpha(\tau;k)}(-1)^{\alpha(\sigma\setminus\tau;k)}=(-1)^{\gamma(\rho)}(-1)^{\alpha(\rho;k)}.$

As $\tau\subseteq\sigma$, $|\tau|=j-1$ and $k\in\sigma\setminus\tau$ are
arbitrary, $\rho=\tau\cup k\subseteq\sigma$ with $|\rho|=j$ is arbitrary too,
thus the first sum of $\partial_{i-j}(g_{j})$ can be rewritten as follows,

$\displaystyle A$
$\displaystyle=(-1)^{i-j-1}\sum_{\begin{subarray}{c}\rho\subseteq\sigma\\\
|\rho|=j\end{subarray}}(-1)^{\gamma(\rho)}\big{(}\sum_{k\in\rho}(-1)^{\alpha(\rho;k)+1}x_{k}f(u;\rho\setminus
k)\big{)}\otimes e_{\sigma\setminus\rho}.$ (16) Analogously, the second sum
can be written as $B+C$, where $\displaystyle B$
$\displaystyle=\sum_{\begin{subarray}{c}\vartheta\subseteq\sigma\\\
|\vartheta|=j+1\end{subarray}}(-1)^{\gamma(\vartheta)}\big{(}\sum_{k\in\vartheta}(-1)^{\alpha(\vartheta;k)}x_{k}s_{\vartheta\setminus
k}\big{)}\otimes e_{\sigma\setminus\vartheta}\wedge e_{\max(u)},$ (17)
$\displaystyle C$
$\displaystyle=\phantom{\sum_{\begin{subarray}{c}\rho\subseteq\sigma\\\
|\rho|=j+1\end{subarray}}}(-1)^{i-j-1}\sum_{\begin{subarray}{c}\rho\subseteq\sigma\\\
|\rho|=j\end{subarray}}(-1)^{\gamma(\rho)}s_{\rho}\otimes
x_{\max(u)}e_{\sigma\setminus\rho}.$ (18)

Taking into account equations (16), (18), the inductive hypothesis and the
definition of $s_{\rho}$ we have that

$\displaystyle A+C$
$\displaystyle=(-1)^{i-(j+1)}\sum_{\begin{subarray}{c}\rho\subset\sigma\\\
|\rho|=j\end{subarray}}(-1)^{\gamma(\rho)}\big{(}\sum_{k\in\rho}(-1)^{\alpha(\rho;k)+1}x_{k}f(u;\rho\setminus
k)+x_{\max(u)}s_{\rho}\big{)}\otimes e_{\sigma\setminus\rho}$
$\displaystyle=(-1)^{i-(j+1)}\sum_{\begin{subarray}{c}\rho\subset\sigma\\\
|\rho|=j\end{subarray}}(-1)^{\gamma(\rho)}d_{j+1}(f(u;\rho))\otimes
e_{\sigma\setminus\rho}.$

Thus, by the structure of $g_{j+1}$, to complete our proof we need to prove
that

$B=\sum_{\begin{subarray}{c}\vartheta\subseteq\sigma\\\
|\vartheta|=j+1\end{subarray}}(-1)^{\gamma(\vartheta)}d_{j+1}(s_{\vartheta})\otimes
e_{\sigma\setminus\vartheta}\wedge e_{\max(u)}.$

That is, we have to prove

$\displaystyle d_{j+1}(s_{\vartheta})$
$\displaystyle=\sum_{k\in\vartheta}(-1)^{\alpha(\vartheta;k)}x_{k}s_{\vartheta\setminus
k}$ $\displaystyle=\sum_{k\in\vartheta}\sum_{r\in\vartheta\setminus
k}(-1)^{\alpha(\vartheta;k)+\alpha(\vartheta\setminus
k;r)}x_{k}\frac{v_{r}}{x_{\max(u)}}f(u_{r};\vartheta\setminus\\{k,r\\}),$ (19)

for all $\vartheta\subseteq\sigma$, $|\vartheta|=j+1$.

Since $j<i-1$, then $j+1<i$, and by inductive hypothesis,

$\displaystyle d_{j+1}(s_{\vartheta})=$
$\displaystyle\sum_{r\in\vartheta}(-1)^{\alpha(\vartheta;r)}\frac{v_{r}}{x_{\max(u)}}\big{(}\sum_{k\in\vartheta\setminus
r}(-1)^{\alpha(\vartheta\setminus r;k)}\times$
$\displaystyle\big{(}-x_{k}f(u_{r};\vartheta\setminus\\{k,r\\})+\frac{x_{k}u_{r}}{g(x_{k}u_{r})}f(g(x_{k}u_{r});\vartheta\setminus\\{k,r\\})\big{)}.$

For $u\in G(I)$ and $\tau\subseteq[n]$, we define

$\Gamma(u;\tau):=\big{\\{}r\in\tau:\tau\setminus
r\subseteq[\max(u_{r})-1]\setminus{\operatorname{supp}}_{\bf
t}(u_{r})\big{\\}},$

where $u_{r}=g(x_{r}u)$. Note that for $r\in\tau\setminus\Gamma(u;\tau)$,
$f(u_{r};\tau\setminus r)=0$.

Now, for the first sum of terms of $d_{j+1}(s_{\vartheta})$, note that
$\alpha(\vartheta;r)=\alpha(\vartheta\setminus k;r)+\alpha(k;r)$,
$\alpha(\vartheta;k)=\alpha(\vartheta\setminus r;k)+\alpha(r;k)$ and
$\alpha(k;r)-\alpha(r;k)=1$ if $k<r$ or $-1$ if $k>r$. Thus,

$\displaystyle(-1)^{\alpha(\vartheta;r)+\alpha(\vartheta\setminus r;k)+1}$
$\displaystyle=(-1)^{\alpha(\vartheta\setminus
k;r)+\alpha(k;r)+\alpha(\vartheta;k)-\alpha(r;k)+1}=(-1)^{\alpha(\vartheta;k)+\alpha(\vartheta\setminus
k;r)}.$

Taking into account this calculation and exchanging the indices $k$ with $r$
in the second sum of terms of $d_{j+1}(s_{\vartheta})$, we can write
$d_{j+1}(s_{\vartheta})$ as $B_{1}+B_{2}$, where

$\displaystyle B_{1}$
$\displaystyle=\sum_{r\in\Gamma(u;\vartheta)}\sum_{k\in\vartheta\setminus
r}(-1)^{\alpha(\vartheta;k)+\alpha(\vartheta\setminus
k;r)}x_{k}\frac{v_{r}}{x_{\max(u)}}f(u_{r};\vartheta\setminus\\{k,r\\}),$
$\displaystyle B_{2}$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\Gamma(u;\vartheta)\\\
r\in\Gamma(u_{k};\vartheta\setminus
k)\end{subarray}}(-1)^{\alpha(\vartheta;k)+\alpha(\vartheta\setminus
k;r)}\frac{x_{r}u_{k}v_{k}}{g(x_{r}u_{k})x_{\max(u)}}f(g(x_{r}u_{k});\vartheta\setminus\\{k,r\\}).$

In all terms of the right–hand side in equation (19), for $k,r\in\vartheta$,
$k\neq r$, we have either $r\in\Gamma(u;\vartheta)$ or
$r\notin\Gamma(u;\vartheta)$ and $r\in\Gamma(u;\vartheta\setminus k)$. Let
$B_{3}$ be the sum of terms such that $r\in\Gamma(u;\vartheta)$, and let
$B_{4}$ be the sum of terms such that $r\notin\Gamma(u;\vartheta)$ and
$r\in\Gamma(u;\vartheta\setminus k)$. To finish the proof, it is enough to
show that $B_{1}=B_{3}$ and $B_{2}=B_{4}$.

It is clear that $B_{1}=B_{3}$.

Let us see that $B_{2}=B_{4}$. The hypotheses $r\notin\Gamma(u;\vartheta)$ and
$r\in\Gamma(u;\vartheta\setminus k)$ imply that
$k\notin[\max(u_{r})-1]\setminus{\operatorname{supp}}_{\bf t}(u_{r})$, where
$u_{r}v_{r}=x_{r}u$ and $\max(u_{r})\leq\min(v_{r})$. But
$k\in\vartheta\subseteq\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)$. Thus, either $k\in[\max(u_{r}),\max(u)-1]$ or
$k\in{\operatorname{supp}}_{\bf t}(u_{r})\setminus{\operatorname{supp}}_{\bf
t}(u)$. We show in both cases that $g(x_{r}u_{k})=u_{r}$.

If $k\in[\max(u_{r}),\max(u)-1]$, then $k\geq\max(u_{r})\geq r$, so $k>r$
since $k\neq r$. This implies that $r<k\leq\max(u_{k})$ too. Hence, $u_{r}$
divides $x_{r}u_{k}$. Finally, $g(x_{r}u_{k})=u_{r}$.

If $k\in{\operatorname{supp}}_{\bf
t}(u_{r})\setminus{\operatorname{supp}}_{\bf t}(u)$, then $k>r$, and so
$r<\max(u_{k})$. Since $k\in{\operatorname{supp}}_{\bf t}(u_{r})$ we have that
$k<\max(u_{r})$. Let us see that $\max(u_{r})\leq\max(u_{k})$. Suppose on the
contrary that $\max(u_{r})>\max(u_{k})$. If $u=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{d}}$, then $u_{k}=x_{k}\cdot x_{j_{1}}\cdots x_{j_{p}}$ and
$u_{r}=x_{r}\cdot x_{j_{1}}\cdots x_{j_{q}}$ are both ${\bf t}$-spread
monomials of $I$ with $p<q<d$. Then $x_{r}(u_{k}/x_{k})$ is a ${\bf t}$-spread
monomial of $I$ that divides $x_{r}u$ and
$x_{r}(u_{k}/x_{k})>_{\text{plex}}u_{r}$, an absurd. Hence
$\max(u_{r})\leq\max(u_{k})$, so $u_{r}$ divides $x_{r}u_{k}$ and again
$g(x_{r}u_{k})=u_{r}$.

Thus, $g(x_{r}u_{k})=u_{r}$ and

$\begin{array}[]{lrlrr}&\displaystyle\frac{x_{r}u_{k}v_{k}}{g(x_{r}u_{k})x_{\max(u)}}&f(g(x_{r}u_{k});\vartheta\setminus\\{k,r\\})&\phantom{aaaaa}&(\textup{as}\
u_{k}v_{k}=x_{k}u),\\\
=&\displaystyle\frac{x_{r}x_{k}u}{u_{r}x_{\max(u)}}&f(u_{r};\vartheta\setminus\\{k,r\\})&\phantom{aaaaa}&(\textup{as}\
x_{r}u=u_{r}v_{r}),\\\ =&\displaystyle
x_{k}\frac{v_{r}}{x_{\max(u)}}&f(u_{r};\vartheta\setminus\\{k,r\\}).\end{array}$

This shows that $B_{2}=B_{4}$ and completes our proof. $\Box$

We consider the ideal in Example 3.11 and construct the differentials of its
minimal free resolution. Note that in this case ${\bf t}=(1,0)$.

###### Example 4.3

Let $I\subset S=K[x_{1},\dots,x_{6}]$ be the $(1,0)$-spread strongly stable
ideal of Example 3.11 with minimal generating set

$G(I)=\big{\\{}w_{1}=x_{1}x_{2},\,\,w_{2}=x_{1}x_{3},\,\,w_{3}=x_{1}x_{4}^{2}\big{\\}}.$

By Example 3.11, ${\operatorname{pd}}(S/I)=3$. Let

${\mathbb{F}}:0\rightarrow F_{3}\xrightarrow{\ d_{3}\ }F_{2}\xrightarrow{\
d_{2}\ }F_{1}\xrightarrow{\ d_{1}\ }F_{0}=S\xrightarrow{\ d_{0}\
}S/I\rightarrow 0$

the minimal free resolution of $S/I$. We know that
$d_{0}=\varepsilon:S\rightarrow S/I$ is the canonical map. We shall describe
the differentials $d_{1},d_{2},d_{3}$ by appropriate monomial matrices.

For $i=1,2,3$, the basis of the free $S$-modules $F_{i}=S\otimes_{K}H_{i}({\bf
x})$ consists of

$f(w_{j};\sigma)=(-1)^{(i-1)(i-2)/2}1\otimes[z(w_{j};\sigma)],$

for $j=1,\dots,4$,
$\sigma\subseteq[\max(w_{j})-1]\setminus{\operatorname{supp}}_{\bf t}(w_{j})$
and $|\sigma|=i-1$.

We introduce a natural order on the basis elements of $F_{i}$, as follows,

$f(w_{i};\sigma)\succ f(w_{j};\vartheta)\ \ \Longleftrightarrow\ \ i<j\ \
\textup{or}\ \ i=j\ \ \textup{and}\ \ e_{\sigma}>e_{\vartheta},$

where $e_{\sigma}>e_{\vartheta}$ with respect to the order on the wedge
products defined in Section 1.

For instance,

$f(w_{2};\\{2\\})\succ f(w_{3};\\{2\\})\succ f(w_{3};\\{3\\}).$ (20)

Then, $d_{i}$, $i=1,2,3$, may be represented by a matrix whose $j$th column is
given by the components of $d_{i}(f_{j})$ with respect to the ordered basis of
$F_{i-1}$, where $f_{j}$ is the $j$th basis element of $F_{i}$ with respect to
the order introduced.

By equation (15) we have that

$\displaystyle{\mathbb{F}}:\,\,\,\,0\,\xrightarrow{}$
$\displaystyle\underset{}{F_{3}}\,\xrightarrow{\begin{pmatrix}-x_{4}^{2}\\\
x_{3}\\\ -x_{2}\\\
\end{pmatrix}}\,\underset{}{F_{2}}\,\xrightarrow{\begin{pmatrix}x_{3}^{2}&x_{4}^{2}&0\\\
-x_{2}&0&x_{4}^{2}\\\ 0&-x_{2}&-x_{3}\\\ \end{pmatrix}}$
$\displaystyle\underset{}{F_{1}}\,\xrightarrow{\begin{pmatrix}x_{1}x_{2}&x_{1}x_{3}&x_{2}x_{4}^{2}\\\
\end{pmatrix}}\underset{}{F_{0}}\xrightarrow{\ d_{0}\
}\,\underset{}{S/I}\,\rightarrow\,0.$

For instance, taking into account the order given in (20), we have

$\displaystyle d_{3}(f(w_{3};\\{2,3\\}))\ =$ $\displaystyle\
\begin{pmatrix}-x_{4}^{2}\\\ x_{3}\\\ -x_{2}\\\
\end{pmatrix}\begin{pmatrix}1\end{pmatrix}=-x_{4}^{2}f(w_{2};\\{2\\})+x_{3}f(w_{3};\\{2\\})-x_{2}f(w_{3};\\{3\\}),$
$\displaystyle d_{2}d_{3}(f(w_{3};\\{2,3\\}))\ =$ $\displaystyle\
-x_{4}^{2}\big{(}-x_{2}f(w_{2};\varnothing)+x_{3}f(w_{1};\varnothing)\big{)}$
$\displaystyle\
+x_{3}\big{(}-x_{2}f(w_{3};\varnothing)+x_{4}^{2}f(w_{1};\varnothing)\big{)}$
$\displaystyle\
-x_{2}\big{(}-x_{3}f(w_{3};\varnothing)+x_{4}^{2}f(w_{2};\varnothing)\big{)}\
=\ 0.$

## 5\. Generalized Algebraic Shifting theory

In this final Section, we extend algebraic shifting theory to vector-spread
strongly stable ideals. From now on, $K$ is a field of characteristic zero. We
recall that by the symbol $\operatorname{Gin}(I)$ we mean the generic initial
ideal of a monomial ideal $I\subset S$, with respect to the reverse
lexicographic order, with $x_{1}>x_{2}>\cdots>x_{n}$ [23]. It is known that
$\operatorname{Gin}(I)$ is a (${\bf 0}$-spread) strongly stable ideal.

Firstly, we need some notions.

Let ${\bf t},{\bf s}\in{\mathbb{Z}}_{\geq 0}^{d-1}$, ${\bf
t}=(t_{1},\dots,t_{d-1})$, ${\bf s}=(s_{1},\dots,s_{d-1})$, with $d\geq 2$. We
can transform any ${\bf t}$-spread monomial ideal into a ${\bf s}$-spread
monomial ideal as follows: Let ${\bf 0}\in{\mathbb{Z}}_{\geq 0}^{d-1}$ be the
null vector with $d-1$ components. To denote the composition of functions

${\operatorname{Mon}}(T;{\bf t})\xrightarrow{\sigma_{{\bf t},{\bf
0}}}{\operatorname{Mon}}(T;{\bf 0})\xrightarrow{\sigma_{{\bf 0},{\bf
s}}}{\operatorname{Mon}}(T;{\bf s})$

we use the symbol $\sigma_{{\bf t},{\bf s}}$, where
$T=K[x_{1},x_{2},\dots,x_{n},\dots]$. Note that $\sigma_{{\bf t},{\bf
s}}(1)=1$, $\sigma_{{\bf t},{\bf s}}(x_{i})=x_{i}$, and for all monomials
$u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{\ell}}\in{\operatorname{Mon}}(T;{\bf t})$,
$2\leq\ell\leq d$,

$\displaystyle\sigma_{{\bf t},{\bf s}}(x_{j_{1}}x_{j_{2}}\cdots
x_{j_{\ell}})=\textstyle\prod\limits_{k=1}^{\ell}x_{j_{k}-\sum_{r=1}^{k-1}t_{r}+\sum_{r=1}^{k-1}s_{r}}.$

Finally, for $I$ a ${\bf t}$-spread monomial ideal, we let $I^{\sigma_{{\bf
t},{\bf s}}}$ the monomial ideal whose minimal generating set is
$G(I^{\sigma_{{\bf t},{\bf s}}}):=\big{\\{}\sigma_{{\bf t},{\bf s}}(u):u\in
G(I)\big{\\}}$. Note that $I^{\sigma_{{\bf t},{\bf s}}}=(I^{\sigma_{{\bf
t},{\bf 0}}})^{\sigma_{{\bf 0},{\bf s}}}$.

As mentioned in the Introduction, we define the ${\bf t}$-spread algebraic
shifting as follows: for $I$ a monomial ideal of $T$, we let $I^{s,{\bf t}}$
the following monomial ideal

$I^{s,{\bf t}}:=(\operatorname{Gin}(I))^{\sigma_{{\bf 0},{\bf t}}}.$

Note that for ${\bf t}={\bf 1}=(1,1,\dots,1)$, we obtain the classical
algebraic shifting. Indeed, for ${\bf t}={\bf 1}$, $\sigma_{{\bf 0},{\bf t}}$
is the squarefree operator defined in the Introduction.

We are going to verify the following four properties:

  1.     1. $(\textup{Shift}_{1})$

$I^{s,{\bf t}}$ is a ${\bf t}$-spread strongly stable monomial ideal;

    2. $(\textup{Shift}_{2})$

$I^{s,{\bf t}}=I$ if $I$ is a ${\bf t}$-spread strongly stable ideal;

    3. $(\textup{Shift}_{3})$

$I$ and $I^{s,{\bf t}}$ have the same Hilbert function;

    4. $(\textup{Shift}_{4})$

If $I\subseteq J$, then $I^{s,{\bf t}}\subseteq J^{s,{\bf t}}$.

###### Proposition 5.1

Let $I$ be a monomial ideal. Then, $I$ is a ${\bf t}$-spread strongly stable
ideal if and only if $I^{\sigma_{{\bf t},{\bf s}}}$ is a ${\bf s}$-spread
strongly stable ideal.

Proof. Suppose that $I$ is a ${\bf t}$-spread strongly stable ideal. Set
$I^{\prime}=I^{\sigma_{{\bf t},{\bf s}}}$. To show that $I^{\prime}$ is a
${\bf s}$-spread strongly stable ideal, it suffices to check condition (ii) of
Corollary 2.7. So, let $u\in G(I)$, $u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}$,
then

$u_{1}=\sigma_{{\bf t},{\bf
s}}(u)=\prod_{k=1}^{d}x_{j_{k}-\sum_{r=1}^{k-1}t_{r}+\sum_{r=1}^{k-1}s_{r}}=x_{j_{1}^{\prime}}x_{j_{2}^{\prime}}\cdots
x_{j_{d}^{\prime}}\in G(I^{\prime})=G(I^{\sigma_{{\bf t},{\bf s}}}).$

Let $i\in{\operatorname{supp}}(u_{1})$, $j<i$ such that
$v_{1}=x_{j}(u_{1}/x_{i})$ is ${\bf s}$-spread, we prove that $v_{1}\in
I^{\prime}$.

Now,
$i=j_{\ell}^{\prime}=j_{\ell}-\sum_{r=1}^{\ell-1}t_{r}+\sum_{r=1}^{\ell-1}s_{r}$,
for some $\ell\in\\{1,\dots,d\\}$, and $j_{p-1}^{\prime}+s_{p-1}\leq j\leq
j_{p}^{\prime}-1$, for some $p\leq\ell$, in particular for $p=1$,
$j<j_{1}^{\prime}$. Hence,

$v_{1}=x_{j}(u_{1}/x_{i})=\big{(}\prod_{k=1}^{p-1}x_{j_{k}^{\prime}}\big{)}x_{j}\big{(}\prod_{k=p}^{\ell-1}x_{j_{k}^{\prime}}\big{)}\big{(}\prod_{k=\ell+1}^{d}x_{j_{k}^{\prime}}\big{)}.$

Recall that $\sigma_{{\bf s},{\bf t}}$ is the inverse map of $\sigma_{{\bf
t},{\bf s}}$. Set $v=\sigma_{{\bf s},{\bf t}}(v_{1})$, then $v$ is ${\bf
t}$-spread, and

$v=\sigma_{{\bf s},{\bf
t}}(v_{1})=\big{(}\prod_{k=1}^{p-1}x_{j_{k}}\big{)}x_{j-\sum_{r=1}^{p-1}s_{r}+\sum_{r=1}^{p-1}t_{r}}\big{(}\prod_{k=p}^{\ell-1}x_{j_{k}-s_{k}+t_{k}}\big{)}\big{(}\prod_{k=\ell+1}^{d}x_{j_{k}}\big{)}.$

Since $j_{k+1}-j_{k}\geq t_{k}$ for all $k$ and
$j_{p}^{\prime}=j_{p}-\textstyle\sum_{r=1}^{p-1}t_{r}+\textstyle\sum_{r=1}^{p-1}s_{r}$,
we have

$\displaystyle j_{k}-s_{k}+t_{k}\leq j_{k+1}-s_{k}\ $ $\displaystyle\leq\
j_{k+1},\ \ \text{for all}\ k=p,\dots,\ell-1,\ \text{and}$ (21) $\displaystyle
j-\textstyle\sum_{r=1}^{p-1}s_{r}+\textstyle\sum_{r=1}^{p-1}t_{r}\ $
$\displaystyle<\
j_{p}^{\prime}-\textstyle\sum_{r=1}^{p-1}s_{r}+\textstyle\sum_{r=1}^{p-1}t_{r}=j_{p}.$
(22)

Setting

$\displaystyle z_{m}\ $ $\displaystyle=\
\begin{cases}x_{j-\sum_{r=1}^{p-1}s_{r}+\sum_{r=1}^{p-1}t_{r}}(u/x_{j_{p}}),&\text{for}\
m=1,\\\
x_{j_{(p+m-2)}-s_{(p+m-2)}+t_{(p+m-2)}}(z_{m-1}/x_{j_{(p+m-1)}}),&\text{for}\
m=2,\dots,\ell+1-p,\end{cases}$

we see that the monomials $z_{m}$ are ${\bf t}$-spread. Moreover, as $I$ is
${\bf t}$-spread strongly stable, $z_{1}\in I$ by (22), and inductively
$z_{m}\in I$, by (21). So, $v=z_{\ell+1-p}\in I$, and by Lemma 2.6,
$v=w_{1}w_{2}$ for unique monomials $w_{1}\in G(I)$, $w_{2}$ such that
$\max(w_{1})\leq\min(w_{2})$. Hence, $\sigma_{{\bf t},{\bf s}}(w_{1})$ divides
$v_{1}=\sigma_{{\bf t},{\bf s}}(v)$, with $\sigma_{{\bf t},{\bf s}}(w_{1})\in
G(I^{\prime})=G(I^{\sigma_{{\bf t},{\bf s}}})$. Finally, $v_{1}\in
I^{\prime}=I^{\sigma_{{\bf t},{\bf s}}}$, as desired. The converse is
trivially true as $I,{\bf t},{\bf s}$ are arbitrary. $\Box$

By virtue of this proposition, the property $(\textup{Shift}_{1})$ is
verified. Indeed, it is known that $\operatorname{Gin}(I)$ is a ${\bf
0}$-spread strongly stable ideal [23]. Consequently, $I^{s,{\bf t}}$ is a
${\bf t}$-spread strongly stable ideal, as desired.

The operators $\sigma_{{\bf t},{\bf s}}$ behave well, in fact they preserve
the graded Betti numbers. We first note that Theorem 3.8 implies a formula for
the graded Betti numbers. We remark that the next result holds whatever the
characteristic of the field $K$ is.

###### Corollary 5.2

Let $I$ be a ${\bf t}$-spread strongly stable ideal of $S$. Then,

$\beta_{i,i+j}(I)=\sum_{u\in
G(I)_{j}}\binom{\max(u)-1-\sum_{\ell=1}^{j-1}t_{\ell}}{i},\ \ \ \text{for
all}\ \ i,j\geq 0.$ (23)

In particular, the graded Betti numbers of a vector-spread strongly stable
ideal $I\subset S$ do not depend upon the characteristic of the field $K$.

Proof. Let $i,j\geq 0$. By equation (1),
$\beta_{i,i+j}(I)=\beta_{i+1,i+j}(S/I)=\dim_{K}H_{i+1}({\bf x};S/I)_{i+j}$. By
Theorem 3.8, the degree of a basis element $\big{[}e(u;\sigma)\big{]}$ of
$H_{i+1}({\bf x};S/I)_{i+j}$ is given by $|\sigma|+1+\deg(u)-1=i+j$. Thus
$u\in G(I)_{j}$. For a fixed $u\in G(I)_{j}$, we have
$\sigma\subseteq[\max(u)-1]\setminus{\operatorname{supp}}_{\bf t}(u)$. Hence,
there are $\binom{|[\max(u)-1]\setminus{\operatorname{supp}}_{\bf
t}(u)|}{i}=\binom{\max(u)-1-\sum_{\ell=1}^{j-1}t_{\ell}}{i}$ possible choices
for $\sigma$. Summing all these binomials over $u\in G(I)_{j}$, we obtain the
formula in the statement. $\Box$

Suitable choices of ${\bf t}$ return several well known formulas for the
graded Betti numbers. ${\bf t}=(0,0,\dots,0)$ returns the Eliahou–Kervaire
formula for (strongly) stable ideals [17]; ${\bf t}=(1,1,\dots,1)$ gives the
Aramova–Herzog–Hibi formula for squarefree (strongly) stable ideals [10].
Finally, in the uniform case, _i.e._ , ${\bf t}=(t,t,\dots,t)$, we have the
Ene–Herzog–Qureshi formula for uniform $t$-spread strongly stable ideals [18].

Let $P_{I}^{S}(y)=\sum_{i}\beta_{i}(I)y^{i}$ be the Poincaré series of $I$.
Equation (23) implies

###### Corollary 5.3

Let $I\subset S$ be a ${\bf t}$-spread strongly stable ideal. Then

  1. (a)

$P_{I}^{S}(y)=\sum_{u\in
G(I)}(1+y)^{\max(u)-1-\sum_{\ell=1}^{\deg(u)-1}t_{\ell}}$;

  2. (b)

${\operatorname{pd}}(I)=\max\big{\\{}\max(u)-1-\textstyle\sum_{j=1}^{\deg(u)-1}t_{j}:u\in
G(I)\big{\\}}$;

  3. (c)

${\operatorname{reg}}(I)=\max\big{\\{}\deg(u):u\in G(I)\big{\\}}$.

Let us return now to our shifting operators. As announced, we have

###### Lemma 5.4

Let $I$ be a ${\bf t}$-spread strongly stable ideal. Then $I^{\sigma_{{\bf
t},{\bf s}}}$ is a ${\bf s}$-spread strongly stable ideal, and for all
$i,j\geq 0$,

$\beta_{i,i+j}(I)=\beta_{i,i+j}(I^{\sigma_{{\bf t},{\bf s}}}).$

Proof. We have just proved that $I^{\sigma_{{\bf t},{\bf s}}}$ is a ${\bf
s}$-spread strongly stable ideal with minimal generating set
$G(I^{\sigma_{{\bf t},{\bf s}}})=\\{{\sigma_{{\bf t},{\bf s}}}(u):u\in
G(I)\\}$. Moreover, for $u\in G(I)$, we have $\max(\sigma_{{\bf t},{\bf
s}}(u))=\max(u)-\sum_{\ell=1}^{\deg(u)-1}t_{\ell}+\sum_{\ell=1}^{\deg(u)-1}s_{\ell}$.
Hence, Corollary 5.2 yields

$\displaystyle\beta_{i,i+j}(I^{\sigma_{{\bf t},{\bf s}}})\ $ $\displaystyle=\
\sum_{\sigma_{{\bf t},{\bf s}}(u)\in G(I^{\sigma_{{\bf t},{\bf
s}}})_{j}}\binom{\max({\sigma_{{\bf t},{\bf
s}}}(u))-1-\sum_{\ell=1}^{j-1}s_{\ell}}{i}$ $\displaystyle=\ \sum_{u\in
G(I)_{j}}\binom{\max(u)-\sum_{\ell=1}^{j-1}t_{\ell}+\sum_{\ell=1}^{j-1}s_{\ell}-1-\sum_{\ell=1}^{j-1}s_{\ell}}{i}$
$\displaystyle=\ \sum_{u\in
G(I)_{j}}\binom{\max(u)-1-\sum_{\ell=1}^{j-1}t_{\ell}}{i}\ =\
\beta_{i,i+j}(I).$

$\Box$

As a consequence, the property $(\textup{Shift}_{3})$ is verified too. Indeed,
it is known that $I$ and $\operatorname{Gin}(I)$ have the same Hilbert
function. Moreover, by Lemma 5.4, $\operatorname{Gin}(I)$ and
$(\operatorname{Gin}(I))^{\sigma_{{\bf 0},{\bf t}}}$ have the same graded
Betti numbers and thus the same Hilbert function.

Note that condition $(\textup{Shift}_{4})$ is trivially verified. Finally it
remains to establish condition $(\textup{Shift}_{2})$. This is accomplished in
the next theorem.

###### Theorem 5.5

Let $K$ be a field of characteristic zero. Let $I\subset S$ be a ${\bf
t}$-spread strongly stable ideal. Then

$I=(\operatorname{Gin}(I))^{\sigma_{{\bf 0},{\bf t}}}.$

Proof. We proceed by induction on the integer $\ell=\max\\{\max(u):u\in
G(I)\\}\geq 1$. If $\ell=1$, then $G(I)=\\{x_{1}^{a}\\}$, $I=(x_{1}^{a})$, and
$\operatorname{Gin}(I)=I=(x_{1}^{a})$, moreover $\sigma_{{\bf t},{\bf
0}}(x_{1}^{a})=x_{1}^{a}$, for some $a\geq 1$. So, the thesis holds for
$\ell=1$.

Let $\ell>1$. By [23, Lemma 11.2.8] we can assume $\ell=n$. So, there exists a
monomial $u\in G(I)$ with $\max(u)=n$. Let $p=\max\\{p:x_{n}^{p}\
\textup{divides}\ w\ \textup{for some}\ w\in G(I)\\}$, our hypothesis implies
that $p\geq 1$. We consider the following ideals:

$\displaystyle I^{\prime}$ $\displaystyle=I:(x_{n}^{p}),\ \ \ \
I^{\prime\prime}=\big{(}u\in G(I):\max(u)<n\big{)}.$

Both are again ${\bf t}$-spread strongly stable ideals, and
$I^{\prime\prime}\subseteq I\subseteq I^{\prime}$. By inductive hypothesis,
$\operatorname{Gin}(I^{\prime})=(I^{\prime})^{\sigma_{{\bf t},{\bf 0}}}$ and
$\operatorname{Gin}(I^{\prime\prime})=(I^{\prime\prime})^{\sigma_{{\bf t},{\bf
0}}}$. Equivalently,

$I^{\prime}=\operatorname{Gin}(I^{\prime})^{\sigma_{{\bf 0},{\bf t}}}\ \
\textup{and}\ \
I^{\prime\prime}=\operatorname{Gin}(I^{\prime\prime})^{\sigma_{{\bf 0},{\bf
t}}}.$

Therefore, $I^{\prime\prime}\subseteq\operatorname{Gin}(I)^{\sigma_{{\bf
0},{\bf t}}}\subseteq I^{\prime}$.

Claim 2. It is

$I\subseteq\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}.$ (24)

To prove Claim 2, it is enough to show that each $u\in G(I)$ with $\max(u)=n$
belongs to $\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$. Indeed, since
$I^{\prime\prime}\subseteq\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$,
all monomials $u\in G(I)$ with $\max(u)<n$ are in
$\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$.

Let $u\in G(I)$ with $\max(u)=n$. We set $a=n-1-\sum_{j=1}^{\deg(u)-1}t_{j}$
and $b=a+\deg(u)$. By Corollary 5.2 we have

$\displaystyle\beta_{a,b}(I)\ $ $\displaystyle=\ \sum_{\begin{subarray}{c}v\in
G(I)\\\
\deg(v)=\deg(u)\end{subarray}}\binom{\max(v)-1-\sum_{j=1}^{\deg(u)-1}t_{j}}{n-1-\sum_{j=1}^{\deg(u)-1}t_{j}}$
$\displaystyle=\ \big{|}\big{\\{}v\in G(I):\max(v)=n,\
\deg(v)=\deg(u)\big{\\}}\big{|}.$ Similarly, as
$\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$ is ${\bf t}$-spread
strongly stable, $\displaystyle\beta_{a,b}(\operatorname{Gin}(I)^{\sigma_{{\bf
0},{\bf t}}})\ $ $\displaystyle=\ \sum_{\begin{subarray}{c}w\in
G(\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}})\\\
\deg(w)=\deg(u)\end{subarray}}\binom{\max(w)-1-\sum_{j=1}^{\deg(u)-1}t_{j}}{n-1-\sum_{j=1}^{\deg(u)-1}t_{j}}$
$\displaystyle=\ \big{|}\big{\\{}w\in G(\operatorname{Gin}(I)^{\sigma_{{\bf
0},{\bf t}}}):\max(w)=n,\ \deg(w)=\deg(u)\big{\\}}\big{|}.$

Moreover, by [23, Corollary 3.3.3] and by Lemma 5.4, we have

$\beta_{a,b}(I)\ \leq\ \beta_{a,b}(\operatorname{Gin}(I))\ =\
\beta_{a,b}(\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}).$

Hence

$\displaystyle\big{|}\big{\\{}w\in G(\operatorname{Gin}(I)^{\sigma_{{\bf
0},{\bf t}}}):\max(w)=n,\deg(w)=\deg(u)\big{\\}}\big{|}\geq$ (25)
$\displaystyle\ \ \ \ \ \ \big{|}\big{\\{}v\in
G(I):\max(v)=n,\deg(v)=\deg(u)\big{\\}}\big{|}.$

Our aim is to prove that $u\in G(I)$ with $\max(u)=n$ belongs to
$\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$.

Let $w_{1},\dots,w_{s}$ be the monomial generators in
$G(\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}})$ such that
$\max(w_{i})=n$ and $\deg(w_{1})\leq\deg(w_{2})\leq\dots\leq\deg(w_{s})$.
Since $\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}\subseteq I^{\prime}$,
we have $w_{i}x_{n}^{p}\in I$, for all $i=1,\dots,s$. We prove that $w_{i}\in
I$ for all $i$. Since $w_{i}x_{n}^{p}\in I$, there is a monomial $v_{i}\in
G(I)$ such that $v_{i}$ divides $w_{i}x_{n}^{p}$. We have
$\deg(v_{i})\leq\deg(w_{i})+p$, for all $i=1,\dots,s$.

If $\deg(v_{1})<\deg(w_{1})$, setting $u=v_{1}$ in (25), we would have an
absurd. Hence $\deg(v_{1})\geq\deg(w_{1})$. By finite induction,
$\deg(v_{i})\geq\deg(w_{i})$, for all $i=1,\dots,s$.

Now, if $\deg(v_{s})\geq\deg(w_{s})+1$, setting $u=v_{s}$ in (25), we would
obtain an absurd. Hence, $\deg(v_{s})\leq\deg(w_{s})$, and since we have
proved that $\deg(v_{s})\geq\deg(w_{s})$, we obtain $\deg(v_{s})=\deg(w_{s})$.
Iterating this argument, $\deg(v_{i})=\deg(w_{i})$, for all $i=1,\dots,s$.

If $v_{i}=(w_{i}x_{n}^{p})/x_{n}^{p}=w_{i}$ we set $u_{i}=v_{i}$ and note that
$u_{i}=w_{i}$ divides $w_{i}$. Otherwise, $v_{i}=(w_{i}x_{n}^{p})/z_{i}$ for
some monomial $z_{i}\neq x_{n}^{p}$, we note that $v_{i}$ has bigger sorted
indexes than $w_{i}$, thus since $I$ is ${\bf t}$-spread strongly stable
$w_{i}\in I$. Hence, there is a monomial $u_{i}\in G(I)$ that divides $w_{i}$.
Finally, we have constructed monomials $u_{1},\dots,u_{s}\in G(I)$ such that
$u_{i}$ divides $w_{i}$, for all $i=1,\dots,s$. Repeating the same argument as
before, using (25), we see that $\deg(u_{i})\geq\deg(w_{i})$, for all $i$,
hence $u_{i}=w_{i}$, since $u_{i}$ divides $w_{i}$, for all $i=1,\dots,s$.

Thus, $w_{i}=u_{i}\in G(I)$, for all $i=1,\dots,s$, and we get the inclusion

$\big{\\{}w\in G(\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf
t}}}):\max(w)=n\big{\\}}\subseteq\big{\\{}u\in G(I):\max(u)=n\big{\\}}.$

This equation together with (25) yield

$\big{\\{}w\in G(\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf
t}}}):\max(w)=n\big{\\}}=\big{\\{}u\in G(I):\max(u)=n\big{\\}}.$

Hence, Claim 2 is true.

Finally, $I$ and $\operatorname{Gin}(I)$ have the same Hilbert function.
Moreover, by Lemma 5.4, $\operatorname{Gin}(I)$ and
$\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$ have the same Hilbert
function. Hence $I$ and $\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$
have the same Hilbert function. Formula (24) and this observation imply that
$I=\operatorname{Gin}(I)^{\sigma_{{\bf 0},{\bf t}}}$, or equivalently
$\operatorname{Gin}(I)=I^{\sigma_{{\bf t},{\bf 0}}}$, proving the theorem.
$\Box$

We end the paper by remarking that the operator $\sigma_{{\bf t},{\bf s}}$
establishes a bijection between ${\bf t}$-spread strongly stable ideals and
${\bf s}$-spread strongly stable ideals.

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