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\title[Gr\"obner bases and the Lefschetz properties]{Gr\"obner bases and the Lefschetz properties for powers of a general linear form in the squarefree algebra}

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\author[Jonsson Kling, Lundqvist, Mohammadi, Orth, S\'aenz-de-Cabez\'on]{\texorpdfstring{Filip Jonsson Kling\addressmark{1}, Samuel Lundqvist\addressmark{1}, Fatemeh Mohammadi\addressmark{2}, Matthias Orth\addressmark{3}, and Eduardo S\'aenz-de-Cabez\'on\addressmark{4}}{Filip Jonsson Kling, Samuel Lundqvist, Fatemeh Mohammadi, Matthias Orth, and Eduardo S\'aenz-de-Cabez\'on}} %\thanks{\href{mailto:hello@world.c}{hello@world.c}. Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}\addressmark{1} \and you\addressmark{2}}

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\address{\addressmark{1}Department of Mathematics,  Stockholm University, Sweden\\ \addressmark{2}Departments of Mathematics and Computer Science, KU Leuven, Belgium\\
\addressmark{3}Department of Mathematics, University of Kassel, Germany\\
\addressmark{4}Departamento de Matemáticas y Computación, Universidad de La Rioja, Spain\\
}

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\abstract{For the almost complete intersection ideals $(x_1^2, \dots, x_n^2, (x_1 + \cdots + x_n)^k)$, we compute their reduced Gröbner basis for any term ordering, revealing a combinatorial structure linked to lattice paths, elementary symmetric polynomials, and Catalan numbers. Using this structure, we classify the weak Lefschetz property for these ideals. Additionally, we provide a new proof of the well-known result that the squarefree algebra satisfies the strong Lefschetz property.}
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\resume{Pour les idéaux d'intersection presque complète $(x_1^2, \dots, x_n^2, (x_1 + \cdots + x_n)^k)$, nous calculons leur base de Gröbner réduite pour tout ordre monomial, mettant en évidence une structure combinatoire liée aux chemins de réseaux, aux polynômes symétriques élémentaires et aux nombres de Catalan. En utilisant cette structure, nous classifions la propriété de Lefschetz faible pour ces idéaux. Nous fournissons également une nouvelle démonstration du résultat bien connu selon lequel l'algèbre sans carrés satisfait la propriété de Lefschetz forte.}
%{Pour les idéaux presque intersections complètes $(x_1^2, \dots, x_n^2, (x_1 + \cdots + x_n)^k)$, nous calculons leur base de Gröbner réduite pour tout ordre de termes, mettant en évidence une structure combinatoire liée aux chemins de treillis, aux polynômes symétriques élémentaires et aux nombres de Catalan. En utilisant cette structure, nous classifions la propriété de Lefschetz faible pour ces idéaux. De plus, nous fournissons une nouvelle démonstration du résultat bien connu selon lequel l'algèbre sans carrés satisfait la propriété de Lefschetz forte.}
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\keywords{Gr\"obner basis, Catalan numbers, lattice paths, almost complete intersections, general linear forms, elementary symmetric polynomials, %weak and strong 
Lefschetz properties}

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\begin{document}

\maketitle
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\section{Introduction}
Let $\mathbf{k}$ be a field of characteristic zero and let $R = \mathbf{k}[x_1, \ldots, x_n]$ be the polynomial ring in $n$ variables. 
A monomial order $\prec$ on $R$ is a total order on the monomials of $R$ where any subset of monomials has a smallest element, and such that $m_1 \prec m_2$ implies that $m \cdot m_1 \prec m \cdot m_2$ for all monomials $m$ in $R$. A Gr\"obner basis for an ideal $I$ with respect to $\prec$ is a subset $\{g_1,\ldots,g_s\}$ of $I$ 
such that the largest monomial of any $f$ in $I$ is divisible by the largest monomial of one of the $g_i$. Gr\"obner bases are a key tool in computer algebra systems for performing computations. However, they are also known for their lack of respect for symmetry, strong dependence on the monomial ordering, and often involved explicit description. 
In contrast, our main result gives, for any term ordering, a Gr\"obner basis for
$$ I_{n,k} = (x_1^2, \ldots, x_n^2, (x_1 + \cdots + x_n)^k) $$  which exhibits a different behavior.
First, all Gr\"obner basis elements, except for the squares, are elementary symmetric polynomials in a subset of the variables, and in particular, they have  $0,1$-coefficients.
Second, the sequence formed by the number of Gr\"obner basis elements of degree greater than 2 corresponds to the $(k-1)$-fold convolution of Catalan numbers.


Notice that we can present our ideals
$I_{n,k}$ as $(x_1^2,\ldots,x_n^2,e_k(x_1,\ldots,x_n))$, where $e_k$ denotes the $k$'th elementary symmetric polynomial.  This gives a connection to earlier work by Haglund, Rhoades, and Shimozon \cite{HAGLUND2018851} on the Delta conjecture \cite{haglund2018delta} where they determine the Gr\"obner basis with respect to the lexicographical monomial order for ideals of the form $(x_1^k,\ldots,x_n^k,e_n(x_1,\ldots,x_n), \ldots, e_{n-k+1}(x_1,\ldots,x_n))$, and surprisingly show that its elements can be described in terms of Demazure characters, which provides an overlap with our class in the trivial cases $I_{n,n-1}$ and $I_{n,n}.$ 

The ideals $I_{n,k}$ are also well studied for their relations to the Lefschetz properties. An algebra $A=R/I$, where $I$ is an ideal generated by homogeneous polynomials, so called forms, can be written as $A=A_0 \oplus A_1 \oplus \cdots$, where $A_i$ is the vector space consisting of zero and all forms of degree $i$. The algebra $A$ has the weak Lefschetz property (WLP) if there is an $\ell$ in $A_1$ such that, for every $i$, the map $A_i \to A_{i+1}$ given by $a \mapsto \ell  \cdot a$ is either injective or surjective. If also all powers of $\ell$ give maps that are always injective or surjective, then $A$ is said to have the strong Lefschetz property (SLP). It is well known that the Lefschetz properties often serve as a bridge between different areas of mathematics, with well-established connections to commutative algebra, algebraic geometry, combinatorics, representation theory, and algebraic topology. In this spirit, one can view the present paper as strengthening the link to combinatorics, and contributing with a link to computer algebra. Indeed, we use our combinatorially described Gröbner basis to show that 
$$R/I_{n,k} \text{ has the WLP if and only if }
\begin{cases}
k\geq \frac{n-3}{2} & \text{ for } n \text{ odd,}\\
k\geq \frac{n}{2} & \text{ for } n \text{ even,}\\
\end{cases}$$
and we give a new proof of the fact the squarefree algebra $R/(x_1^2,\ldots,x_n^2)$ has the strong Lefschetz property, originally shown independently by Stanley \cite{Stanley} and Watanabe \cite{WATANABE1989194}. As a corollary, we also obtain a minor result related to the Fr\"oberg conjecture \cite{Froberg}.

The ideal $(x_1^{2}, \ldots, x_n^{2}, (x_1 + \cdots + x_n)^{k})$, which may initially appear quite specialized, is in fact a general object. After a linear change of coordinates, we obtain the isomorphism
$$ R/(x_1^{2}, \ldots, x_n^{2}, (x_1 + \cdots + x_n)^{k}) \cong R/(\ell_1^{2}, \ldots,\ell_{n}^{2},  \ell_{n+1}^{k}), $$
where the $\ell_i$ are general linear forms. This reveals a deeper geometric connection, particularly in the context of the interpolation of general fat points via Macaulay's inverse system, as described by Emsalem and Iarrobino \cite{emsalemiarrobino}.
In light of this, it is natural to further explore the structure of algebras generated by $n+1$ powers of general linear forms. 



%Finally, we study the minimal free resolution of the initial ideal of $I_{n,k}$.~We~show that the squarefree part of the initial ideal is strongly squarefree stable, allowing us to apply results by Gasharov, Hibi, and Peeva \cite{GasharovHibiPeeva2002} to obtain closed formulas for their Betti numbers. % of the squarefree part. 
%The initial ideal itself is a combination of squarefree and powers ideals, for which Murai \cite{Murai2008} has provided recursive formulas for their Betti numbers.~To~explicitly compute the Betti numbers, we construct a minimal free resolution using Mayer--Vietoris trees, as described by the fifth author \cite{Saenz-De-Cabezon2009MVT}.~Our main result regarding the homological properties of the initial ideal of $I_{n,k}$ is that the sequence of extremal Betti numbers is a $(k-1)$-fold convolution of the Catalan number sequence.


%\medskip
%We would like to note that 
We have included only the proofs of the main results and refer to \cite{jonsson2024} for the more detailed proofs. % of the lemmas and propositions.

\begin{remark}{\rm While finalizing the paper, we were informed that Booth, Singh, and Vraciu in \cite{booth2024weak} independently have provided a description of the initial ideal of $(x_1^d,\ldots,x_n^d,(x_1+\cdots+x_n)^d)$ for $d=2,3$, using different techniques. Their work overlaps with ours regarding the initial ideal for $I_{n,2}$. %, thus overlapping our work with respect to the initial ideal for $I_{n,2}$. %In the case when $d=2$, 
They also find relations for establishing failure of the weak Lefschetz property (WLP) due to injectivity of $I_{n,2}$ for certain $n$, which are similar to but distinct from special cases of the relations we derive for the failure of the WLP for $I_{n,k}$ for certain $n$ and $k$. %
%for $I_{n,k}$ for certain $n$ and $k$.
}
\end{remark}


\section{The Gr\"obner basis} \label{sec:description} 
\noindent We fix notation as follows. Let $[n] = \{1, \ldots, n\}$. For any subset $S = \{i_1, \ldots, i_s\} \subseteq [n]$, denote by $x_S$ the monomial $x_{i_1} \cdots x_{i_s}$ in $\kb[x_1, \ldots, x_n]$. We fix integers $n$ and $k\geq 2$.
\subsection{Constructing the Gr\"obner basis elements}
In this section, we define a collection of polynomials $g_{A,n,k}$ combinatorially and prove that they are in the ideal $I_{n,k}$. Later, we will show that they form a Gröbner basis for $I_{n,k}$.
%In this section, we will prove that the polynomials $g_{A,n,k}$ do belong to the ideal $I_{n,k}$, establishing one part of \Cref{thm:GBasis}. An alternative description for the polynomials $g_{A,n,k}$ will also be deduced in \Cref{thm:f_description}. We will split the proofs into several lemmas. 

\begin{definition}\label{def:SFP}
    The \emph{squarefree part} of a polynomial $f\in \mathbf{k}[x_1,\ldots,x_n]$ is %the sum of the squarefree monomials appearing in $f$, i.e., 
    its normal form with respect to the monomial ideal $(x_1^2,\ldots,x_n^2)$. We write $\mathrm{SFP}(f)$ for the squarefree part of $f$. For a subset of polynomials $P\subseteq \mathbf{k}[x_1,\ldots,x_n]$ we write $\mathrm{SFP}(P)=\{ \mathrm{SFP}(f)\mid f\in P \}$.
\end{definition}
Note that in \Cref{def:SFP}, the word squarefree does not refer to irreducible factors of a polynomial $f$, but to the exponents of the variables $x_1,\ldots,x_n$ in the terms of its support. In other words, the squarefree part of a polynomial is the part obtained by removing all terms that contain a square. For example, %$\mathrm{SFP}(x_1x_2^3 + 3x_3x_4 + x_1^2) = 3x_3x_4$.
$$\mathrm{SFP}((x_1+\cdots+x_5)^2) = 2 (x_1 x_2 + x_1 x_3 + \cdots + x_4 x_5) = 2 e_2(x_1,\ldots,x_5),$$ where $e_2$ is the elementary symmetric polynomial of degree two.
Since $(x_1^2,\ldots,x_n^2)\subset I_{n,k}$ for all $n$ and $k$, the crucial part of the analysis of the homogeneous ideals $I_{n,k}$ is the degreewise description of their squarefree parts $\mathrm{SFP}((I_{n,k})_{(d)})$.

\begin{comment}
    \begin{lemma}\label{lem:SFPofPolys_and_I_n_k}
    Let $f\in \mathbf{k}[x_1,\ldots,x_n]$ be a polynomial. Then $f-\mathrm{SFP}(f)\in I_{n,k}$. In particular, $f\in I_{n,k}$ if and only if $\mathrm{SFP}(f)\in I_{n,k}$.
\end{lemma}
\begin{proof}
    Observing that $(x_1^2,\ldots,x_n^2)\subseteq I_{n,k}$, we only need to apply \Cref{def:SFP}.
\end{proof}
\end{comment}

%\begin{definition}\label{def:g_S_n_k}
%    Let $S\subseteq [n]$ be an index set. We write 
%\begin{equation}\label{eq:f_{S,n,k}}
 %      f_{S,n,k}=\frac{1}{k!}\mathrm{SFP}(x_S(x_1+\cdots +x_n)^k).
 %   \end{equation}
%\end{definition}


%Using the polynomials $f_{S,n,k}$, we obtain generating systems of the squarefree parts in single degrees of the ideals $I_{n,k}$.

\begin{lemma}\label{lem:GenSysDegreewiseSFP}
    \begin{enumerate}
        \item 
        Let $S\subseteq [n]$ and $f_{S,n,k}=\frac{1}{k!}\mathrm{SFP}(x_S(x_1+\cdots +x_n)^k)$. %\{1,\ldots,n\}
         Then 
        $f_{S,n,k}$ %from Equation~\eqref{eq:f_{S,n,k}}
        is squarefree of degree $k+|S|$ and can be written as $$f_{S,n,k}=\sum_{\substack{u\; \mathrm{ squarefree}\\ 
            \deg(u)=k+|S| \\
            x_S \mid u}} u.$$
       % is squarefree of degree $k+|S|$.
        \item 
        Let $k\leq d\leq n$. Then $\mathrm{SFP}((I_{n,k})_{(d)})$ is generated as a vector space by the polynomials $f_{S,n,k}$ with $|S|=d-k$.
    \end{enumerate}
\end{lemma}


\begin{comment}

\begin{proof}
(1) First observe that $\mathrm{SFP}((x_1+\cdots + x_n)^k)$ is $k!$ times the sum of all squarefree terms of degree $k$. Multiplying by $x_S$, applying $\mathrm{SFP}$ again, and dividing by $k!$, we obtain the desired expression.

    (2) By \Cref{def:g_S_n_k} and item (1), we have $f_{S,n,k}\in \mathrm{SFP}((I_{n,k})_{(d)})$ if $|S|=d-k$. In order to show that these polynomials generate $\mathrm{SFP}((I_{n,k})_{(d)})$, first note that $\{(1/k!)x_S\mid x_S \;\mathrm{squarefree}\;\mathrm{and}\deg(x_S)=d-k\}$ generates $\mathrm{SFP}(\mathbf{k}[x_1,\ldots,x_n]_{(d-k)})$. So by \Cref{def:g_S_n_k}, the vector space $V:=\mathrm{SFP}(\mathbf{k}[x_1,\ldots,x_n]_{(d-k)}\cdot (x_1+\cdots +x_n)^k)$ is generated by the $f_{S,n,k}$ with $|S|=d-k$. Now, since $\mathrm{SFP}((I_{n,k})_{(d)})=\mathrm{SFP}(((x_1+\cdots +x_n)^k)_{(d)})=V$, we are done.
\end{proof}
\end{comment}

%\begin{example}\label{ex:GenSysHomogPart}
%    Consider the linear form $\ell=x_1+\cdots + x_5$ in %the ring 
%    $\mathbf{k}[x_1,x_2,x_3,x_4,x_5]$. We then have $I_{5,2}=(x_1^2,x_2^2,x_3^2,x_4^2,x_5^2,\ell^2)$. We obtain the generating system $\{f_{\{1\},n,k},\ldots,f_{\{5\},n,k}\}$ of $(I_{5,2})_{(3)}$ as a $\mathbf{k}$-vector space, where
%    \begin{align*}
%        f_{\{1\},n,k}&=\frac{1}{2}\mathrm{SFP}(x_1 \ell^2)=x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_2 x_5 + x_1 x_3 x_4 + x_1 x_3 x_5 + x_1 x_4 x_5,\\
%        f_{\{2\},n,k}&=\frac{1}{2}\mathrm{SFP}(x_2 \ell^2)=x_1 x_2 x_3 + x_1 x_2 x_4 + x_2 x_3 x_4 + x_1 x_2 x_5 + x_2 x_3 x_5 + x_2 x_4 x_5,\\
%        f_{\{3\},n,k}&=\frac{1}{2}\mathrm{SFP}(x_3 \ell^2)=x_1 x_2 x_3 + x_1 x_3 x_4 + x_2 x_3 x_4 + x_1 x_3 x_5 + x_2 x_3 x_5 + x_3 x_4 x_5,\\
%        f_{\{4\},n,k}&=\frac{1}{2}\mathrm{SFP}(x_4 \ell^2)=x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 + x_1 x_4 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5,\\
%        f_{\{5\},n,k}&=\frac{1}{2}\mathrm{SFP}(x_5 \ell^2)=x_1 x_2 x_5 + x_1 x_3 x_5 + x_2 x_3 x_5 + x_1 x_4 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5.
%    \end{align*}
%\end{example}

%The plan  is to later show that the polynomials $g_{A,n,k}$ from Equation~\eqref{eq:g_{A,n,k}-intro} are members of the ideal $I_{n,k}$ by showing that they can be written as linear combinations of the basis polynomials $f_{S,n,k}$. 

\begin{lemma}\label{lem:Combinations_g_S_n_k}
    Let $k\leq d\leq n$. For any $S\subseteq [n]$ with $|S|=d-k$ let $\lambda_S\in \mathbf{k}$ be a scalar. Then
    $$\sum_{|S|=d-k}\lambda_S f_{S,n,k}=\sum_{\substack{u\;\mathrm{squarefree}\\ \deg(u)=d}}\left(\sum_{\substack{x_S\mid u\\ |S|=d-k}} \lambda_S\right)u.$$
\end{lemma}

\begin{proof}
    Write $f=\sum_{|S|=d-k}\lambda_S f_{S,n,k}$. As a $\mathbf{k}$-linear combination of squarefree polynomials of degree $d$, $f$ is also squarefree of degree $d$. By \Cref{lem:GenSysDegreewiseSFP}, the coefficient of a squarefree term $u$ of degree $d$ in the polynomial $f_{S,n,k}$ is $\lambda_S$ if $x_S\mid u$ and zero otherwise. Thus the coefficient of $u$ in $f$ is the sum of all $\lambda_S$ where $x_S\mid u$, as claimed.
\end{proof}

\begin{definition}\label{def:f_A_n_k}
    Let $k\leq d\leq n$, and $A=\{i_1,\ldots,i_d\}\subseteq [n]$ with $\max(A)\leq 2d-k$. % Write $C=\{1,\ldots,2d-k\}\setminus A$. 
    We define
\begin{equation}\label{eq:g}
g_{A,n,k}=\sum_{\substack{u\;\mathrm{squarefree}\\ \deg(u)=d\\ \mathrm{supp}(u)\cap (\{1,\ldots,2d-k\}\setminus A))=\emptyset}} u.
\end{equation}
\end{definition}

%%Note that this is the same definition for $g_{A,n,k}$ in terms of elementary symmetric polynomials as given in Equation~\eqref{eq:g_{A,n,k}-intro} 
%since for any set $S\subseteq [n]$, we have that\[S\cap (\{1,\dots, 2d-k\}\setminus A) = \emptyset \iff S\cap \{1,\dots, 2d-k\} \subseteq A.\]

\begin{example}\label{ex:f_A_n_k}
    For the values $n=5$, $k=2$, and $d=3$, consider the set $A=\{1,3,4\}$. Note that $\max(A)=4\leq 4=2\cdot d-k$. In this case, $g_{A,n,k}$ is given by
    $$x_1 x_3 x_4 + x_1 x_3 x_5 + x_1 x_4 x_5 + x_3 x_4 x_5,$$
    the elementary symmetric polynomial $e_3(x_1,x_3,x_4,x_5)$ of degree $d$ supported on the variables indexed by $A\cup \{5\}=A\cup \big([n]\setminus \{1,\ldots,2d-k\} \big)$.
\end{example}

We now show that %With this preparation, we are now ready to establish that 
$g_{A,n,k}\in I_{n,k}$, which is a major step needed for the proof of \Cref{thm:GBasis}.

\begin{comment}
\begin{proposition}\label{prop:ExplicitCoefficients_f_A_n_k} 
    Let $k\leq d\leq n$, and let $A\subseteq [n]$ with $\max(A)\leq 2d-k$. %the sets $A$ (with $|A|=d$) and $C$ (with $|C|=d-k$) be given as in \Cref{def:f_A_n_k}. 
    Write $\mathcal{T}_i=\{ S\subseteq [n]: |S|=d-k, \text{ and } |S\cap (\{1,\dots, 2d-k\}\setminus A)|=i \}$. Then we have
    $$g_{A,n,k}=\sum_{i=0}^{d-k} \lambda_i \left(\sum_{S\in\mathcal{T}_i} f_{S,n,k}\right),$$
    where $\lambda_i=(-1)^{i}\frac{ k }{ (k+i)\binom{d}{k+i} }$ for $0\leq i \leq d-k$. %for some coefficients $\lambda_0,\ldots,\lambda_{d-k}\in \mathbf{k}$. 
    In particular, $g_{A,n,k}\in I_{n,k}$.
\end{proposition}
\end{comment}

%\begin{lemma}\label{lem:BinomialCoeffsAlternatingSums}
 %   The unique solution of the inhomogeneous system of linear equations~\eqref{eq:InhomogLinSys_3} is $\lambda_i=(-1)^{i}\frac{ k }{ (k+i)\binom{d}{k+i} }$ for $0\leq i \leq d-k$.
%\end{lemma}

%As an easy consequence of \Cref{thm:f_description}, we have the following theorem, which gives an explicit description of the elements $g_{A,n,k}$ in terms of the generators of $I_{n,k}$.

\begin{theorem}
\label{thm:f_description}
Let $f_{S,n,k}\in I_{n,k}$ for $S\subseteq [n]$ be the squarefree part of the polynomial $x_S(x_1+\cdots+x_n)^k$. Then the elements $g_{A,n,k}$ can be written as
$$g_{A,n,k}=\sum_{i=0}^{d-k} (-1)^{i}\frac{ k }{ (k+i)\binom{d}{k+i} } \sum_{S\in\mathcal{T}_i(A)} f_{S,n,k}$$
where $d=|A|$, $\max(A)\leq 2d-k$, and 
$$\mathcal{T}_i(A)=\{S\subseteq [n]: |S|=d-k \text{ and }|S\cap\{1,\ldots,2d-k\}\setminus A| =i \}.$$
\end{theorem}
\begin{proof}
Let $C=\{1,\ldots,2d-k\}\setminus A$. Write 
    $\textstyle{f=\sum_{i=0}^{d-k} \lambda_i \left(\sum_{S\in\mathcal{T}_i} f_{S,n,k}\right)}.$
    By construction, $f$ is a sum of squarefree monomials of degree $d$. Let $u$ be a squarefree term with $\deg(u)=d$. By \Cref{lem:Combinations_g_S_n_k}, its coefficient in $f$ is $\sum_{i=0}^{d-k}  |\{S\in\mathcal{T}_i : x_S\mid u \}|\cdot \lambda_i$. On the other hand, the coefficient of $u$ in $g_{A,n,k}$ is $1$  if $|\mathrm{supp}(u)\cap C|=0$, and zero otherwise.
    Thus, $f=g_{A,n,k}$ holds if and only if the coefficients $\lambda_i$ solve the inhomogeneous linear system of $\binom{n}{d}$ equations (labeled by the terms $u$)
    \begin{equation}\label{eq:InhomogLinSys_1}
        \sum_{i=0}^{d-k}  |\{S\in\mathcal{T}_i : x_S\mid u \}|\cdot \lambda_i=\begin{cases}
        1, & \mathrm{if}\; |\mathrm{supp}(u)\cap C|=0 \\
        0, & \mathrm{if}\; |\mathrm{supp}(u)\cap C|>0.
        \end{cases}
    \end{equation}
If $|\mathrm{supp}(u)\cap C|=j$ and $S\in \mathcal{T}_i$ with $i>j$, then $x_S\nmid u$. Thus, we can rewrite~\eqref{eq:InhomogLinSys_1} as 
    \begin{equation}\label{eq:InhomogLinSys_2}
        \sum_{i=0}^{j}  |\{S\in\mathcal{T}_i : x_S\mid u \}|\cdot \lambda_i=\begin{cases}
        1, & \mathrm{if}\; |\mathrm{supp}(u)\cap C|=0 \\
        0, & \mathrm{if}\; |\mathrm{supp}(u)\cap C|=j>0.
        \end{cases}
    \end{equation}
    We now claim that the system~\eqref{eq:InhomogLinSys_2} contains only $d-k+1$ distinct equations. First, it is easy to see that for $|\mathrm{supp}(u)\cap C|=0$ we have $|\{S\in\mathcal{T}_0 : x_S\mid u \}|=\binom{d}{d-k}$ independently of $u$. Now, consider $u$ with $|\mathrm{supp}(u)\cap C|=j>0$. Note that $|\mathrm{supp}(u)\setminus C|=d-j$. Furthermore, consider $S\in\mathcal{T}_i$ with $0\leq i\leq j$. Then $|S\cap C|=i$ and $|S\setminus C|=d-k-i$. The condition $x_S\mid u$ is then equivalent to satisfying both $S\cap C\subseteq \mathrm{supp}(u)\cap C$ and $S\setminus C\subseteq \mathrm{supp}(u)\setminus C$. The number of sets $S\in\mathcal{T}_i$ that fulfill this is $\binom{j}{i}\binom{d-j}{d-k-i}$ independently of $u$.% Note that $\binom{j}{i}\binom{d-j}{d-k-i}$ if and only if $d-k-i>d-j$, i.e., $j> i+k$.
    
    The system~\eqref{eq:InhomogLinSys_2} now simplifies to 
    \begin{equation}\label{eq:InhomogLinSys_3}
        \left\{\begin{array}{rl}
             \binom{d}{d-k} \lambda_0  & =1 \\
            \sum_{i=0}^1\binom{1}{i}\binom{d-1}{d-k-i} \lambda_i & =0 \\
             \vdots & \\
             \sum_{i=0}^{d-k}\binom{d-k}{i}\binom{k}{d-k-i} \lambda_i & =0
        \end{array}\right. .
    \end{equation}
    This is a lower triangular system with diagonal elements $\binom{d-j}{d-k-j}=\binom{d-j}{k}\neq 0$ (recall that $k\leq d$ and $d-j\geq d-(d-k)=k$). In particular, it is straightforward to verify that its unique solution is $\lambda_i=(-1)^{i}\frac{ k }{ (k+i)\binom{d}{k+i} }$ for $0\leq i \leq d-k$.
%Hence the system has a unique solution and $g_{A,n,k}$ can be written as desired.
\end{proof}



%\begin{proof}
%The proof follows directly by %combining \Cref{lem:BinomialCoeffsAlternatingSums} with 
%\Cref{thm:f_description}.
%\end{proof}
\begin{comment}
    
 We illustrate \Cref{thm:f_description} with an example.
 \begin{example}\label{ex:InhomogSystem_n=5_d=3_k=2}
    Let $n=5$, $k=2$, and $d=3$. Moreover, let $A=\{1,3,4\}$. Then $g_{A,n,k}=x_1 x_3 x_4 + x_1 x_3 x_5 + x_1 x_4 x_5 + x_3 x_4 x_5$. Since $d-k=1$, we look at the set $\{ f_{\{1\},n,k},\ldots, f_{\{5\},n,k}\}\subset (I_{n,k})_{(3)}$. \Cref{ex:GenSysHomogPart} gives
    \begin{align*}
        f_{\{1\},n,k}&=x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_2 x_5 + x_1 x_3 x_4 + x_1 x_3 x_5 + x_1 x_4 x_5,\\
        f_{\{2\},n,k}&=x_1 x_2 x_3 + x_1 x_2 x_4 + x_2 x_3 x_4 + x_1 x_2 x_5 + x_2 x_3 x_5 + x_2 x_4 x_5,\\
        f_{\{3\},n,k}&=x_1 x_2 x_3 + x_1 x_3 x_4 + x_2 x_3 x_4 + x_1 x_3 x_5 + x_2 x_3 x_5 + x_3 x_4 x_5,\\
        f_{\{4\},n,k}&=x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 + x_1 x_4 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5,\\
        f_{\{5\},n,k}&=x_1 x_2 x_5 + x_1 x_3 x_5 + x_2 x_3 x_5 + x_1 x_4 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5.
    \end{align*}
    Since $2d-k=4$, the set $C=\{1,\ldots,4\}\setminus A=\{2\}$. Moreover, $\mathcal{T}_0=\{ \{1\},\{3\},\{4\},\{5\} \}$ and $\mathcal{T}_1=\{ \{2\} \}$. We obtain an inhomogeneous linear system of equations
    \begin{equation}\label{eq:InhomogLinSys_example}
        \left\{\begin{array}{ll}
             \binom{3}{1} \lambda_0  & =1 \\[1em]
             \binom{2}{1} \lambda_0 + \binom{2}{0} \lambda_1 & =0
        \end{array}\right. ,
    \end{equation}
    whose solution is clearly $\lambda_0=\frac{1}{3}$, $\lambda_1=-\frac{2}{3}$. On the other hand, one can easily verify that $\lambda_0=(-1)^{0}\frac{ 2 }{ 2\binom{3}{2} }$ and $\lambda_1=(-1)^{1}\frac{ 2 }{ 3\binom{3}{3} }$ as stated in \Cref{thm:f_description}, and that 
    $$g_{A,n,k}=\frac{1}{3}\left(f_{\{1\},n,k}+f_{\{3\},n,k}+f_{\{4\},n,k}+f_{\{5\},n,k}\right)-\frac{2}{3}f_{\{2\},n,k},$$
    as stated in \Cref{thm:f_description}.
\end{example}
\end{comment}

\subsection{Hilbert series via Lattice paths}

In what follows, we will establish a Gr\"{o}bner basis with respect to any monomial order with $x_1 \succ\cdots  \succ x_n$ of $I_{n,k}$ that is the union of $\{x_1^2,\ldots,x_n^2\}$ with the set of $g_{A,n,k}$ whose leading term is minimal with respect to division. We will need a counting argument, and, to this end, we introduce a certain type of lattice paths.

\begin{definition}\label{def:LatticePaths}
 An \emph{$(N,E)$-lattice path} is a path on the lattice $\mathbb{Z}^2$ that begins at $(0,0)$ and consists only of northward steps (in the direction $(0,1)$, denoted $N$) and eastward steps (in the direction $(1,0)$, denoted $E$).

There exists a bijection $\tau$ that maps an $(N,E)$-lattice path of length $n$ to the monomial $\prod_{j \in J} x_j$, where the subset $J \subseteq [n]$ contains $j$ if and only if the $j$-th step in the path is north.
\end{definition}

\begin{lemma}\label{lem:LatticePathBijections}
	Let $k\leq d\leq n$. Consider polynomials $g_{A,n,k}$ for the sets $A=\{i_1,\ldots,i_d\}\subseteq[n]$ with $\max(A)\leq 2d-k$ as in \eqref{eq:g}. %\Cref{def:f_A_n_k}. 
 Then, we have that %the following statements hold true:
    \begin{enumerate}
	   \item $\ini(g_{A,n,k})=x_{i_1}\cdots x_{i_d}$,
	   \item the $(N,E)$-lattice path $\tau^{-1}(\ini(g_{A,n,k}))$ touches the line $y=x+k$,
	   \item if $P$ is an $(N,E)$-lattice path of length $n$ that intersects the line $y=x+k$, then there exists a polynomial $g_{A,n,k} \in I_{n,k}$ such that $\ini(g_{A,n,k})$ divides $\tau(P)$.
    \end{enumerate}
\end{lemma}
\begin{comment}
    \begin{proof}
{\color{red} Proof can be removed}
(1) By the definition of $g_{A, n, k}$, the term $x_{i_1} \cdots x_{i_d}$ appears as one of the summands in $g_{A, n, k}$, and all other terms in its support must include at least one variable with an index greater than $\max(A)$. This completes the proof.
       
(2) Since $\max(A) \leq 2d-k$, there are $d$ north steps and $d-k$ east steps among the first $2d-k$ steps in $\tau^{-1}(\ini(g_{A,n,k}))$. This difference of $k$ additional north steps forces the path to intersect the line $y=x+k$ after $2d-k$ steps.

(3) In a path $P$ that intersects the line $y=x+k$, there exists a minimal starting segment (of length at least $k$) that touches the line. Within this segment, there must be exactly $k$ more steps north than east. Let $\ell$ denote the length of this segment; then there exists a positive integer $d$ such that $\ell = 2d - k$. Consider the path $\tilde{P}$, which consists of the first $\ell$ steps of $P$ followed by $n - \ell$ steps east. By construction, there is a polynomial $g_{A,n,k} \in I_{n,k}$ such that $\ini(g_{A,n,k}) = \tau(\tilde{P})$. Since $\tau(\tilde{P})$ divides $\tau(P)$, the claim follows.
\end{proof}
\end{comment}
We now proceed to compute the number of elements $g_{A,n,k}$, where $A$ is minimal with respect to inclusion. Their leading terms minimally generate $\mathrm{SFP}(\ini(I_{n,k}))$. %We begin with the following definitions.

\begin{definition}
Let $(a_i)_{i=0}^{\infty}$ be a sequence of integers. The $k$-fold \emph{self-convolution} of $(a_i)_{i=0}^{\infty}$, denoted $(a_i^k)_{i=0}^{\infty}$, is defined as the sequence of coefficients of the power series 
$\textstyle{\left(\sum_{i=0}^{\infty}a_it^i \right)^{k+1}
}$.
\end{definition}

One sequence that is central to our discussion is the Catalan numbers.
\begin{definition}
The \emph{Catalan numbers} $(C_n)_{n=0}^{\infty}$ are given by
$C_n = \binom{2n}{n} - \binom{2n}{n-1}$.
The $r$-th number of the Catalan $(k-1)$-fold \emph{convolution} is denoted by $C_r^{k-1}$. %$c(r,k-1)$.
\end{definition}

\begin{corollary}\label{cor:ConvolutionCatalanNumbers}
The number of polynomials $g_{A,n,k}$ such that $A$ is minimal with respect to inclusion, and of degree $k+r$, is zero if $n < 2r + k$; otherwise, it is given by $C_r^{k-1}$. %$c(r,k-1)$. 
In particular, for $k=1$ and $k=2$, it is the $r$-th and $(r+1)$-th Catalan number, respectively.
\end{corollary}

%{\color{red} Might be possible to shorten this proof}
\begin{proof}
By \Cref{lem:LatticePathBijections}, the %number of 
polynomials $g_{A,n,k}$ of degree $k+r$ are in bijection with $(N,E)$-lattice paths that take at most $n$ steps and touch the line $y=x+k$ exactly once, specifically at the last step after $k+r$ steps north. Since such a path comprises a total of $2r+k$ steps, there can be no such paths if $n < 2r+k$, resulting in no Gröbner basis elements of degree $r+k$. For $n \geq 2r+k$, we find that, disregarding the last step, these paths are also in bijection with paths that remain below the line $y=x+k-1$ and terminate on that same line after $k-1+r$ steps north.


Next, by \cite[Corollary 16]{Tedford}, $C_i^j$ represents the number of $(N,E)$-lattice paths that start at $(i,0)$, do not cross the line $y=x$, and end at $(i+j,i+j)$. By shifting all such paths left by $i$ steps, we can also interpret $C_i^j$ as counting $(N,E)$-paths starting at the origin that remain below the line $y=x+i$ and terminate at $(j,i+j)$. By combining this count with the previously established enumeration for the number of polynomials $g_{A,n,k}$ of degree $k+r$, we arrive at the complete enumeration. Specifically, for $k=1$ we get $C_r^0 = C_r$ polynomials in degree $r$ and for $k=2$, the claim follows from the well-known fact that $C_n^1=C_{n+1}$.%; see e.g.~\cite[Equation (1.1)]{Catalan_Stanley}.  
\end{proof}

The following lemma presents our main counting argument and serves as the final step needed to complete the main part of the proof of \Cref{thm:GBasis}.

\begin{lemma}\label{lem:CountingArgument}
Let $d$ be an integer with $0 \leq d \leq n$, and consider the terms of degree $d$ outside the degree reverse lexicographic initial ideal $\ini(I_{n,k})$. If $2d-k\geq n$, there are no such terms. Otherwise, these terms are squarefree, and their number is at most $\binom{n}{d} - \binom{n}{d - k}$.
\end{lemma}
\begin{comment}
    
\begin{proof}
First, we observe that all such terms must be squarefree, since $x_i^2 \in I_{n,k}$ for all $i = 1, \dots, n$. We will show that the number of squarefree monomials of degree $d$ that are not divisible by any $\ini(g_{A,n,k})$ when $2d-k<n$ is precisely $\binom{n}{d} - \binom{n}{d - k}$. Since $g_{A,n,k} \in I_{n,k}$ by \Cref{thm:f_description}, this establishes the desired bound on the number of terms outside $\ini(I_{n,k})$ in that case.

By \Cref{lem:LatticePathBijections}, the number of squarefree monomials of degree $d$ not divisible by any $\ini(g_{A,n,k})$ corresponds to the number of $(N, E)$-lattice paths of length $n$ taking $d$ steps north, remaining strictly below the line $y = x + k$. If $2d-k\geq n$, then all path of length $n$ taking $d$ steps north do touch the line $y=x+k$. Hence there are no monomials outside $\ini(I_{n,k})$ in that case. Assume $2d-k<n$ from now on. Given that there are $\binom{n}{d}$ such $(N, E)$-lattice paths of length $n$ with $d$ north steps, it remains to prove that the number of paths touching the line $y = x + k$ in this case is exactly $\binom{n}{d - k}$. We will establish this by induction on $n$.

The case for $n = 1$ is straightforward, so assume $n > 1$ and that the formula holds for $n - 1$. Further, suppose the point $(n - d, d)$ lies below the line $y = x + k$, ensuring that not all paths of length $n$ with $d$ north steps are forced to touch this line.
The key observation is that $(N, E)$-paths ending at $(n - d, d)$ correspond bijectively to paths ending one step earlier at either $(n - d - 1, d)$ or $(n - d, d - 1)$. This correspondence remains valid even if we enforce the condition of touching the line $y = x + k$. By the induction hypothesis, the number of such paths is given by
\[
\binom{n - 1}{d - k} + \binom{n - 1}{d - 1 - k} = \binom{n}{d - k},
\]
as required.
The one case needing extra care is when $(n - d - 1, d)$ lies exactly on the line $y = x + k$, since we cannot directly apply the induction hypothesis in this situation. However, in this case, all $\binom{n - 1}{d}$ paths necessarily touch the line. Since $\binom{n - 1}{d} = \binom{n - 1}{d - k}$ when $(n - d - 1, d)$ lies on $y = x + k$, the formula still holds.
\end{proof}
\end{comment}

\begin{corollary}\label{cor:Hilbert_series}
The Hilbert series of $R/I_{n,k}$ is given by
$\left[(1+t)^n(1-t^k)\right]$, 
where the brackets indicate truncation at the first non-positive coefficient.
\end{corollary}
\begin{comment}
    \begin{proof}
Since 
\[
\left[(1+t)^n(1-t^k)\right] = \left[\frac{(1-t^2)^n}{(1-t)^n}(1-t^k)\right],
\]
Fr\"oberg's result \cite{Froberg} implies that this is the smallest the Hilbert series of $R/I_{n,k}$ can be. Next, the coefficient of $t^d$ of this polynomial is given by 
\[
\max \left\{\binom{n}{d}-\binom{n}{d-k},0\right\} = \begin{cases}
\binom{n}{d}-\binom{n}{d-k} & \text{ if } 2d-k< n \\
0 & \text{ otherwise}.
\end{cases}
\]
However, by \Cref{lem:CountingArgument}, the Hilbert series of $R/I_{n,k}$ can not be larger than this. Therefore, equality holds.
\end{proof}
\end{comment}

%\subsection{Main result}
Before we state our main result, recall that a  Gr\"obner basis is \emph{reduced} if it is in bijection with the minimal generating set of $\ini(I)$, all its elements have $1$ as leading coefficient, and the monomials appearing after the leading monomial of every polynomial in it are not in the initial ideal.


% Proof of \Cref{thm:GBasis}.
\begin{theorem}\label{thm:GBasis} 
Consider the family $\mathcal{A}$ of subsets $A \subseteq [n]$ satisfying $\max(A) = 2|A| - k$ for some $k\geq 2$, and minimal with respect to inclusion.
Then, the reduced Gröbner basis of the ideal $I_{n,k} = (x_1^2, \dots, x_n^2, (x_1 + \cdots + x_n)^k)$ with respect to degree reverse lexicographic ordering %any term ordering that, after any permutation of the first $k$ variables and a suitable set of transpositions of the form $(k+2i - 1,k+ 2i)$ for $i\geq 1$, satisfies $x_1 \succ \cdots \succ x_n$, 
is given by
$$G_{n,k}=\{x_1^{2},\ldots,x_n^2 \}\cup \bigcup_{d=k}^{k+\lfloor (n-k)/2\rfloor}\left\{ g_{A,n,k}\mid A\in\mathcal{A}, |A|=d \right\},$$
where for $|A|=d$,
\begin{equation*}\label{eq:g_{A,n,k}}
    g_{A,n,k} = e_d(x_{i_1},\dots, x_{i_{n-d+k}})
\end{equation*}
is the elementary symmetric polynomial of degree $d$ in the variables indexed by the set \[ \{i_1,\dots, i_{n-d+k}\}=A\cup \{2d-k+1,\dots, n\} \] with leading term $x_A=\prod_{a\in A}x_a$. In particular, $g_{A,n,k}$ is supported on $\binom{n-d+k}{d}$ terms.
Moreover, for fixed $k$ the sequence of cardinalities $|\{A\in\mathcal{A}: |A|=d\}|$ is a $(k-1)$-fold convolution of the sequence of Catalan numbers.
\end{theorem}
\begin{proof}
%We start by using any term ordering with $x_1 \succ \cdots \succ x_n$.
First, by \Cref{lem:LatticePathBijections} and \Cref{cor:Hilbert_series}, the squarefree leading terms of polynomials in $I_{n,k}$ are in bijection with $n$-step $(N,E)$-lattice paths that touch the line $y = x + k$, and each of these terms is divisible by some leading term $\ini(g_{A,n,k})$. Among these leading terms, the minimal ones with respect to division are those that touch the line $y = x + k$ exactly once, occurring after their last step north. Moreover, the penultimate step cannot be east; if it were, there would be an earlier intersection with the line. This establishes the defining conditions of the set $\mathcal{A}$. Regarding non-squarefree leading terms, clearly they are divisible by some $x_i^2$. By \Cref{thm:f_description}, $G_{n,k}\subset I_{n,k}$. Thus, $G_{n,k}$ is a minimal Gr\"{o}bner basis of $I_{n,k}$.

\medskip
To show that $G_{n,k}$ is reduced, first observe that all its elements are monic. Now, consider \(g_{A,n,k} \in G_{n,k}\). Let \((a,b) \in \mathbb{Z}^2\) be the lattice point where \(\tau^{-1}(\ini(g_{A,n,k}))\) touches the line \(y = x + k\). Any term \(u \in g_{A,n,k} - \ini(g_{A,n,k})\) is of the form 
\[ u=x_{j_1}\cdots x_{j_m}\ini(g_{A,n,k})/(x_{\ell_1}\cdots x_{\ell_m}),\]
where \(0 < m \leq |A|\), \(j_1, \ldots, j_m > \max(A)\), and \(\ell_1, \ldots, \ell_m \in A\). Consequently, the lattice path \(\tau^{-1}(u)\) reaches the point \((a + m, b - m)\) after \(\max(A)\) steps. The closest it could approach the line \(y = x + k\) is at the point \((a + m, b)\) after \(\max(A) + m\) steps. However, this point lies below the line since \(m > 0\) and \((a, b)\) is on the line. Therefore, \(u \notin \ini(I_{n,k})\).


The last claim follows %on the sequence of cardinalities $|\{A\in\mathcal{A}: |A|=d\}|$ for fixed $k$ follows 
from \Cref{cor:ConvolutionCatalanNumbers}.
%\Cref{lem:Invariance_G_n_k} below will state the conditions a term ordering $\prec'$ must satisfy for $\ini_{\prec'}(I_{n,k})=(\ini(G_{n,k}))$ to hold. It shows that only the blockwise ranking $\{x_1,\ldots,x_k\}\succ \{x_{k+1},x_{k+2}\}\succ \cdots$ matters, and this establishes the statement on permutations. 
\end{proof}

\begin{comment}
    \begin{remark}\label{rem:k=1}
The reason for omitting the case $k = 1$ from \Cref{thm:GBasis} is due to notational challenges. But it is straightforward to give the Gr\"obner basis also in this case. To this end, we introduce an extra variable $x_0$ with the property that $x_0 \succ x_i$ for $i = 1, \ldots, n$ and consider the ideal  $$I=(x_0^2,x_1^2, \ldots,x_n^2,(x_0+x_1+\cdots+x_n)).$$ After reducing $x_0^2$ with respect to the linear element we get $$I=((x_1+\cdots+x_n)^2, x_1^2,\ldots,x_n^2,(x_0+x_1+\cdots+x_n)).$$ Since $x_0$ only occurs in the last linear polynomial, and is leading, it follows that the reduced Gr\"obner basis for $I$ is equal to $$\{e_1(x_0,\ldots,x_n)\} \cup G_{n,2}.$$
\end{remark}

\begin{remark}
Worth noticing is that although the elements of the Gr\"obner basis only have $0,1$-coefficients, our result only holds in characteristic $0$ in general. This is explicit from the description of the Gr\"obner basis elements given in \Cref{thm:f_description},  and implicit from results showing that the Lefschetz properties for monomial complete intersections is dependent on the characteristic of the underlying field \cite{Kustin, LUNDQVIST2019213}.
\end{remark}
\end{comment}
\begin{example}\label{ex:GB_n=5_k=2}
    %As in \Cref{ex:InhomogSystem_n=5_d=3_k=2}, we 
    We set $n=5$ and $k=2$. The reduced Gr\"{o}bner basis of $I_{5,2}$ %with respect to the degree reverse lexicographic ordering 
    is 
    \[ G_{5,2}=\{x_1^2,x_2^2,x_3^2,x_4^2,x_5^2,g_{\{1,2\},5,2},g_{\{1,3,4\},5,2},g_{\{2,3,4\},5,2}  \}, \]
    where
    \begin{align*}
        g_{\{1,2\},5,2}&= x_1 x_2 + x_1 x_3 + x_2 x_3 + x_1 x_4 + x_2 x_4 + x_3 x_4 + x_1 x_5 + x_2 x_5 + x_3 x_5 + x_4 x_5,\\
        g_{\{1,3,4\},5,2}&= x_1 x_3 x_4 + x_1 x_3 x_5 + x_1 x_4 x_5 + x_3 x_4 x_5,\\
        g_{\{2,3,4\},5,2}&= x_2 x_3 x_4 + x_2 x_3 x_5 + x_2 x_4 x_5 + x_3 x_4 x_5.
    \end{align*}
\begin{comment}
    
  The lattice path associated to $\ini(g_{\{1,3,4\},5,2})=x_1 x_3 x_4$ is illustrated in \Cref{fig:LatticePath_134_n=5_k=2}.
\begin{figure}[h!]
    \centering
    \begin{tikzpicture}
        % Grid and diagonal line
        \draw[step=1cm,gray!30,thin] (0,0) grid (5,5);
        \draw[thick, red, opacity=0.7] (0,2) -- (2.5,4.5) node[above] {$y = x + 2$};

        % Lattice path
        \draw[thick, black] (0,0) -- (0,1) -- (1,1) -- (1,3) -- (2,3);

        % Labels
        \node[below left] at (0,0) {$(0,0)$};
        \node[above right] at (1,1) {Path $\tau^{-1}(\text{in}(g_{\{1,3,4\},5,2}))$};
    \end{tikzpicture}
    
    \caption{The $(N,E)$-lattice path associated to $x_1 x_3 x_4$.}
    \label{fig:LatticePath_134_n=5_k=2}
\end{figure}
\end{comment}  
\end{example}

\begin{comment}
\subsection{The Gr\"obner basis for a general term ordering}

In the remainder of this section, we investigate all the Gr\"{o}bner bases of $I_{n,k}$, showing that they have the same general structure as $G_{n,k}$, but depend on the way the variables are ranked by a given term ordering. 

\begin{remark}\label{rem:G_n_k_NotUniversalInGeneral}
Note that $G_{5,2}$, and indeed any Gr\"{o}bner basis of the form $G_{n,k}$ is also a reduced Gr\"{o}bner basis with respect to the lexicographic ordering with $x_1\succ x_2\succ\cdots$. This is because the leading terms of the elements $g_{A,n,k}$ are the same for the degree reverse lexicographic and lexicographic orderings.

However, $G_{n,k}$ is in general not a universal Gr\"{o}bner basis: For instance, in $G_{5,2}$ (see \Cref{ex:GB_n=5_k=2}), the polynomials $g_{\{1,3,4\},5,2}$ and $g_{\{2,3,4\},5,2}$ share the same leading term, $x_3 x_4 x_5$, with respect to the degree reverse lexicographic ordering with $x_5\succ x_4\succ x_3\succ x_2 \succ x_1$.
\end{remark}

We aim to compute the number of different reduced Gr\"{o}bner bases of the ideal $I_{n,k}$. We write $N_{n,k}$ for this number. It turns out that certain features of the variable ranking induced by a given term order $\prec$ suffice to determine the Gr\"{o}bner basis of $I_{n,k}$ with respect to $\prec$.

\begin{definition}\label{def:BlockVariableRanking}
 Let $\prec$ be a term ordering on $\mathbf{k}[x_1,\ldots,x_n]$ and let $X_1,\ldots,X_m$ be subsets of the variables $X=\{x_1,\ldots,x_n\}$ that partition $X$. We say that $\prec$ induces the \emph{block variable ranking} $X_1\succ X_2\succ \cdots\succ X_m$ if, for any integers $1 \leq i < j \leq m$, with $x \in X_i$ and $y \in X_j$, it holds that $x \succ y$. We call an ordering an \emph{$(\ell_1,\ell_2,\ldots,\ell_m)$-block variable ranking} if $|X_i|=\ell_i$ for $1 \leq i \leq m$. In particular, a $(k,2,\ldots,2,1)$-block variable ranking is any term ordering such that:
\begin{itemize}
    \item $\{x_1,\ldots,x_k\}\succ \{x_{k+1},x_{k+2}\}\succ\cdots \succ \{x_{n-1},x_n\}$ if $n-k \equiv 0 \mod 2$, and
    \item $\{x_1,\ldots,x_k\}\succ \{x_{k+1},x_{k+2}\}\succ\cdots\succ \{x_{n-2},x_{n-1}\}\succ \{x_n\}$ if $n-k \equiv 1 \mod 2$.
\end{itemize}
\end{definition}

\begin{lemma}\label{lem:Invariance_G_n_k}
    Let $\prec$ be a $(k,2,\ldots,2,1)$-block variable ranking. 
    Then the reduced Gr\"{o}bner basis of $I_{n,k}$ with respect to $\prec$ is $G_{n,k}$.
\end{lemma}
\begin{proof}
   We need to show that for each polynomial $g \in G_{n,k}$, its leading term with respect to the term ordering $\prec$ coincides with its leading term with respect to the standard degree reverse lexicographic ordering. This is straightforward for the monomials in $G_{n,k}$.
Now, consider any polynomial $g_{A,n,k} \in G_{n,k}$. Let $A = \{i_1, \ldots, i_d\}$ as usual, where we assume $i_1 < \cdots < i_d$ and $\max(A) = i_d = 2d - k$. Then we have $\max(A) - k = 2d - k - k = 2(d - k) \equiv 0 \mod 2$. This implies that any variable with an index less than or equal to $\max(A)$ is ranked higher by $\prec$ than any variable with an index greater than $\max(A)$. 
Recalling \Cref{def:f_A_n_k}, it follows that $\ini_{\prec}(g_{A,n,k}) = x_{i_1} \cdots x_{i_d}$, as desired.
\end{proof}

\begin{definition}\label{def:G_n_k_sigma}
 Consider a permutation $\sigma \in S_n$ and let $x_{\sigma(1)}, \ldots, x_{\sigma(n)}$ be a permutation of the variables $x_1,\ldots,x_n$. With respect to the degree reverse lexicographic ordering defined by $x_{\sigma(1)} \succ \cdots \succ x_{\sigma(n)}$, we obtain a Gröbner basis for $I_{n,k}$ as outlined in \Cref{thm:GBasis}. We denote this Gr\"obner basis by $G_{n,k}^{\sigma}$. The Gröbner basis $G_{n,k}$ from \Cref{thm:GBasis} can be expressed as $G_{n,k}^{\mathrm{id}}$, for the identity permutation `id'. % refers to the identity permutation.
\end{definition}
\end{comment}

\begin{remark}
{\rm The Gr\"obner basis of $I_{n,k}$ is independent of the monomial ordering once we fix the ordering $x_1 \succ \cdots \succ x_n$. It is also invariant under permutations of the first $k$ variables and the action of transpositions of the form $(k+2i - 1,k+ 2i)$ for $i \geq 1$. In particular, the number of distinct Gröbner bases of $I_{n,k}$ is given by the multinomial coefficient $\binom{n}{k, 2, \ldots, 2, 1}$, where the entry $1$ is included if and only if $n - k \equiv 1 \mod 2$. See \cite[Proposition~2.29]{jonsson2024} for details.
}\end{remark}

%\begin{proposition}\label{prop:N_n_k_viaMultinomialCoeffs}
%The number of distinct Gröbner bases of $I_{n,k}$ is given by the multinomial coefficient $\binom{n}{k, 2, \ldots, 2, 1}$, where the final entry $1$ is included if and only if $n - k \equiv 1 \mod 2$.
%\end{proposition}
\begin{comment}
    \begin{proof}
Since the multinomial coefficient $\binom{n}{k, 2, \ldots, 2, 1}$ counts the number of ordered partitions of an $n$-element set into subsets of cardinalities $k$, $2$, \ldots, $2$, and (possibly) $1$, it can be interpreted as the number of distinct $(k, 2, \ldots, 2, 1)$-block variable rankings of the variables $x_1, \ldots, x_n$. Therefore, it suffices to demonstrate that the reduced Gröbner bases of $I_{n,k}$ with respect to the term orderings $\prec_1$ and $\prec_2$ are equal if and only if $\prec_1$ and $\prec_2$ induce the same $(k, 2, \ldots, 2, 1)$-block variable ranking.

The `if' direction is established in \Cref{lem:Invariance_G_n_k}.

For the `only if' direction, consider term orderings $\prec_1$ and $\prec_2$ that induce different $(k, 2, \ldots, 2, 1)$-block variable rankings. There exist permutations $\sigma_1, \sigma_2 \in S_n$ such that the Gröbner bases for $\prec_1$ and $\prec_2$ are $G_{n,k}^{\sigma_1}$ and $G_{n,k}^{\sigma_2}$, respectively.
Then, there exists an earliest block from the left where the rankings differ. Let $Y$ be the union of the common blocks to the left in both rankings. If $Y$ is empty, the rankings differ at the first sets, which we denote as $\{w_1,\ldots,w_k\}$ for $\prec_1$ and $\{z_1,\ldots,z_k\}$ for $\prec_2$. Since the degree $k$ squarefree components of $\ini_{\prec_1}(I_{n,k})$ and $\ini_{\prec_2}(I_{n,k})$ are given by $x_{w_1}\cdots x_{w_k}$ and $x_{z_1}\cdots x_{z_k}$, respectively, and these are different, we conclude that $G_{n,k}^{\sigma_1} \neq G_{n,k}^{\sigma_2}$.

Now, assume that $Y$ is not empty. Let the next block in the ranking induced by $\prec_1$ be $\{w_1,w_2\}$ and the corresponding block for $\prec_2$ be $\{z_1,z_2\}$. We know that $\{w_1,w_2\} \neq \{z_1,z_2\}$. By \Cref{thm:GBasis}, there exists a degree $d$ (determined by $|Y|$) such that the leading terms $\ini(g_{A,n,k})$ of degree $d$ in $G_{n,k}^{\sigma_1}$ are divisible by $w_1 w_2$ and otherwise only by variables in $Y$. In $G_{n,k}^{\sigma_2}$, a similar statement holds, with $z_1 z_2$ replacing $w_1 w_2$. Thus, the degree $d$ squarefree components of $\ini_{\prec_1}(I_{n,k})$ and $\ini_{\prec_2}(I_{n,k})$ differ, hence $G_{n,k}^{\sigma_1} \neq G_{n,k}^{\sigma_2}$, which completes the proof.
\end{proof}
\end{comment}
\begin{comment}
\begin{remark}\label{rem:InitialIdealsBettiNumbers}
For any given term ordering, \Cref{lem:Invariance_G_n_k} states that the ordering $x_{\sigma(1)} \succ \cdots \succ x_{\sigma(n)}$ for some permutation $\sigma \in S_n$ uniquely determines the Gröbner basis of $I_{n,k}$ for that ordering. In fact, knowledge of the induced block variable ranking is sufficient. Therefore, all Gröbner bases of $I_{n,k}$ take the form $G_{n,k}^{\sigma}$ for some permutation $\sigma$, and all initial ideals of $I_{n,k}$ are equivalent up to renumbering of the variables. This implies that they share the same (multigraded) Betti numbers.
\end{remark}

\begin{example}\label{ex:N_n_k_for_n=5,6,7_k=2,3}
The number of distinct reduced Gröbner bases for the ideals $I_{n,k}$, denoted by $N_{n,k}$, is given as 
    \begin{align*}
        N_{5,2}=&\binom{5}{2,2,1}%=\binom{5}{2}\binom{3}{2}%10\cdot3
        =30,\quad\quad\ 
        N_{6,2}=\binom{6}{2,2,2}%=\binom{6}{2}\binom{4}{2}=%15\cdot6=
        =90,\quad
        N_{7,2}=\binom{7}{2,2,2,1}=%\binom{7}{2}\binom{5}{2}\binom{3}{2}%=21\cdot 10\cdot 3
        630,\\
        N_{5,3}=&\binom{5}{3,2}%=\binom{5}{3}
        =10,\quad\quad\quad
        N_{6,3}=\binom{6}{3,2,1}%=\binom{6}{3}\binom{3}{2}%=20\cdot 3
        =60,\quad
        N_{7,3}=\binom{7}{3,2,2}=%\binom{7}{3}\binom{4}{2}=%35\cdot 6=
        210.\\
    \end{align*}
\end{example}

\end{comment}

\section{The Lefschetz properties}

We now apply the results from the previous section to study the Lefschetz properties. 
%Recall that an Artinian standard graded algebra $A$ has the weak Lefschetz property (WLP) if there exists a linear element $\ell$ such that the multiplication map $\cdot \ell: A_i \to A_{i+1}$ has maximal rank for all $i$, that is, the map is either surjective or injective. Similarly, $A$ has the strong Lefschetz property (SLP) if there exists a linear element $\ell$ such that the multiplication map $\cdot \ell^k: A_i \to A_{i+k}$ has maximal rank for all $i$ and $k$.
We refer to \cite{book, atour} for more background and details on the Lefschetz properties. %For an introduction to the Lefschetz properties, see \cite{book, atour}.

\subsection{Monomial complete intersections have the SLP}

As a first application of our Gr\"obner basis results, specifically \Cref{cor:Hilbert_series}, to the Lefschetz properties, we establish the well-known result that the squarefree algebra possesses the SLP.

\begin{corollary}
The squarefree algebra $A=R/(x_1^2, \dots, x_n^2)$ has the SLP. 
\end{corollary}

\begin{proof}
Note that multiplication by a form $f$ of degree $d$ on an algebra $B$ has full rank if and only if 
\[
\mathrm{HS}(B/(f);t) = [(1 - t^d) \mathrm{HS}(B;t)],
\]
where the square brackets indicate truncation of the polynomial at the first non-positive coefficient. Since $\mathrm{HS}(A;t)=(1+t)^n$, \Cref{cor:Hilbert_series} says that $\mathrm{HS}(A/(\ell^d);t)=[(1 - t^d) \mathrm{HS}(A;t)]$ for the linear form $\ell=x_1+\dots + x_n$ and any $d$. Thus any power of $\ell$ gives a map that has full rank on $A$, so $A$ has the SLP.
\end{proof}

\begin{comment}
    \begin{proof}
Recall that an algebra defined by an ideal of forms of degree $a_1,\dots, a_m$ is thin if it has a Hilbert series given by
$$\left[\frac{\prod_{i=1}^{m}(1-t^{a_{i}})}{(1-t)^n}\right].$$ 
Moreover, the squarefree algebra $R/(x_1^2,\dots, x_n^2)$ has the strong Lefschetz property if and only if the algebras $R/I_{n,k}$ are thin for all $k$. That is, we want the Hilbert series of $R/I_{n,k}$ to be given by
\[
\left[\frac{\prod_{i=1}^{n}(1-t^{2})}{(1-t)^n}(1-t^k)\right] = \left[(1+t)^n(1-t^k) \right].
\]
However, this is precisely the statement of \Cref{cor:Hilbert_series}. Therefore, all algebras $R/I_{n,k}$ are thin, and the squarefree algebra has the SLP.
\end{proof}
\end{comment}

%As mentioned in the introduction, the SLP for an Artinian monomial complete intersection can be derived from the squarefree algebra, see the short argument by Hara and Watanabe in \cite{Watanabe_Boolean}.
%proceeds as follows. Suppose we wish to show that the algebra
%$$B=\mathbf{k}[y_1,\dots, y_m]/(y_1^{d_1+1},\dots, y_m^{d_m+1})$$ has the SLP. The idea is then to embed $B$ into the squarefree algebra $$A=\mathbf{k}[x_1,\dots, x_n]/(x_1^2,\dots, x_n^2)$$ where $n=d_1+\cdots + d_m$ is chosen so that $A$ and $B$ have the same socle degree. Now define a map $\phi:\mathbf{k}[y_1,\dots, y_m]\to A$ by \[
%\phi(y_i) = x_{d_1+d_2+\cdots + d_{i-1} + 1} + x_{d_1+d_2+\cdots + d_{i-1} + 2} + \cdots + x_{d_1+d_2+\cdots + d_{i-1} + d_i}.
%\]
%One can then verify that $\ker(\phi)=(y_1^{d_1+1},\dots, y_m^{d_m+1})$, so $\phi$ induces an inclusion of $B$ into $A$. Thus, if $\ell=y_1+\cdots + y_r$ is a linear form on $B$, then $\phi(\ell)=x_1+\cdots + x_n = \ell'$ is a linear form on $A$ where we know that $\cdot (\ell')^{n-2i}:A_i \to A_{n-i}$ is bijective. The following commutative diagram %in \Cref{fig:com_diagram} 
%\begin{figure}[ht]
%\begin{center}   
%\begin{tikzcd}
%A_i \arrow[r, "(\ell')^{n-2i}"]                      & A_{n-i}                         \\
%B_i \arrow[u, "\phi", hook] \arrow[r, "\ell^{n-2i}"] & B_{n-1} \arrow[u, "\phi", hook]
%\end{tikzcd}
%\end{center}
%\caption*{}    
%\label{fig:com_diagram}
%\end{figure}
%then forces $\cdot \ell^{n-2i}:A_i \to A_{n-i}$ to be injective. Hence, it has full rank and $A$ has the SLP.
%{\color{red} Will shorten the text below to one sentence}
The SLP for artinian monomial complete intersections can be derived from the squarefree algebra, as shown in a short argument by Hara and Watanabe \cite{Watanabe_Boolean}, so in this regard the squarefree case serves as the building block for the SLP for this class of algebras. 

While Stanley and Watanabe established the SLP in the 1980s, the result for squarefree algebras dates back to the 1970s, studied in the context of zero-one matrices by Kantor \cite{Kantor}, Graver and Jurkat \cite{Graver}, and Wilson \cite{Wilson}, with references highlighted by Lindström.
Besides these, we note Hara and Watanabe's proof for $d=2$, Ikeda's \cite{Ikeda} from the 1990s, and Phuong and Tran's \cite{Phuong} more recent proof. So, to our knowledge, our proof is the seventh for the base case $d=2$.
%While the SLP for Artinian monomial complete intersections was established by Stanley and Watanabe in the 1980s, the corresponding result for squarefree algebras dates back to the 1970s, originally studied in the context of combinatorial properties of zero-one matrices. Fröberg and Hollman \cite{Hollman} credit this earlier work to Kantor \cite{Kantor}, Graver and Jurkat \cite{Graver}, and Wilson \cite{Wilson}, with these references initially brought to their attention by Bernt Lindström.


%Besides the results from the 1970s, we are aware of the proof by Hara and Watanabe for the case $d=2$, an argument by Ikeda \cite{Ikeda} from the 1990s, and a more recent proof by Phuong and Tran \cite{Phuong}. Therefore, to the best of our knowledge, our proof represents the seventh contribution to the specific case of $d=2$.

\subsection{The~WLP~for~powers~of %general 
linear forms in the~squarefree~algebra}


As $R/(\ell_1^{a_1},\ldots,\ell_{n}^{a_{n}}) $ has the WLP for general $\ell_i$, it is natural to ask if also $R/(\ell_1^{d},\ldots,\ell_{n+1}^{d})$ has the WLP. However, this is not the case, and was first observed by Fr\"oberg and Hollman \cite{Hollman} for the case $(n,d)=(5,2)$, which via results by  Cruz and Iarrobino \cite{CruzIarrobino} and  Sturmfels and Xu \cite{SturmfelsXu} for $d=2$,
and Migliore, Miro-R\'oig and Nagel \cite{Miglioreetal}, Nagel and Trok \cite{nageltrok} for general $d$, ended up in a classification by Boij and the second author \cite{boijlundqvist}, who showed that for $n \geq 4$, the WLP holds only in sporadic cases. The proof relies on Macaulay’s inverse system and a reduction to the case $d = 2$, where it is shown that the WLP fails due to lack of surjectivity.

In the following theorem, we fully characterize the pairs $(n, k)$ for which the algebra $R/(\ell_1^2, \dots, \ell_n^2, \ell_{n+1}^k)$, defined by powers of general linear forms $\ell_1, \dots, \ell_{n+1}$, has the WLP.
In this case, the WLP can be derived from the structure of $R/(\ell_1^2, \dots, \ell_n^2, \ell_{n+1}^2)$. Unlike the equigenerated case, however, the failure of the WLP here is due to a lack of injectivity. This requires identifying non-trivial identities within the squarefree algebra; see \Cref{prop:syzygy_relation}.

\medskip
The classification result that we will prove is stated in the following theorem.

\begin{theorem}\label{thm:WLP_linear_forms}
Let $\mathbf{k}$ be a field of characteristic zero. % and fix integers $n,k\geq 2$. 
Then for general linear forms $\ell_1,\dots, \ell_{n+1}$, the algebra $R/(\ell_1^2,\dots, \ell_n^2, \ell_{n+1}^k) \cong R/(x_1^{2}, \ldots, x_n^{2}, (x_1 + \cdots + x_n)^{k})$ has the weak Lefschetz property if and only if \[
\begin{cases}
k\geq \frac{n-3}{2} & \text{ for } n \text{ odd,}\\
k\geq \frac{n}{2} & \text{ for } n \text{ even.}\\
\end{cases}
\] 
\end{theorem}

\begin{remark}
\Cref{thm:WLP_linear_forms} implies that the Fr\"oberg conjecture is true for $a_1=\cdots =a_{n+1}=2$ and $a_{n+2}=k,$ where
$$
\begin{cases}
k \geq \frac{n-2}{2} & \text{for } n \text{ even}, \\
k \geq \frac{n+1}{2} & \text{for } n \text{ odd}.
\end{cases}
$$
The statement is trivial when $n$ is odd, as in this case, $k$ is the socle degree or higher. However, for even $n$, the case $k = \frac{n-2}{2}$ offers a new, but minor contribution for the conjecture.
\end{remark}


%\begin{remark}\label{rem:equiv}
% Note that multiplication by a form $f$ of degree $d$ on an algebra $A$ has full rank if and only if 
%\[
%\mathrm{HS}(A/(f);t) = [(1 - t^d) \mathrm{HS}(A;t)],
%\]
%where the square brackets indicate truncation of the polynomial at the first non-positive coefficient.
%\end{remark}

In the following lemmas, it is useful to work with the ideal $I_{n-1,2}$ in place of $I_{n,k}$ to streamline the analysis and simplify computations; see \cite{jonsson2024} for proofs.
Let us fix the algebras 
\[
A = \mathbf{k}[x_1, \dots, x_{n-1}] / (x_1^2, \dots, x_{n-1}^2, (x_1 + \cdots + x_{n-1})^2)\ \text{and}\  
B = \mathbf{k}[x_1, \dots, x_n] / (x_1^2, \dots, x_n^2)
\]

\begin{lemma}\label{lem:full_rank_switch}
The algebra $R/(x_1^{2},\ldots,x_n^{2}, (x_1+\cdots+x_n)^{k})$ has the WLP if and only if, for a general linear form $\ell$ in $\mathbf{k}[x_1,\ldots,x_{n-1}]$, the multiplication % map 
by $\ell^k$ yields a full rank~map~on~$A$. %$A=\mathbf{k}[x_1,\ldots,x_{n-1}]/(x_1^{2},\ldots,x_{n-1}^{2}, (x_1+\cdots+x_{n-1})^2)$.
\end{lemma}

%\begin{proof}
%For a general linear form $\ell'$ in $n$ variables, we have that
%\begin{align*}
%&\mathbf{k}[x_1,\ldots,x_n]/(x_1^{2},\ldots,x_{n}^{2}, (x_1+\cdots+x_n)^{k}, \ell') \\
%&= \mathbf{k}[x_1,\ldots,x_n]/(x_1^{2},\ldots,x_{n}^{2}, (x_1+\cdots+x_n)^{k}, a_1 x_1 + \cdots + a_n x_n)\\ 
%&\cong \mathbf{k}[x_1,\ldots,x_n]/(x_1^{2},\ldots,x_{n}^{2}, (a_1' x_1 + \cdots + a_n' x_n)^{k}, %x_1+\cdots+x_n)\\ 
%&\cong \mathbf{k}[x_1,\ldots,x_{n-1}]/(x_1^{2},\ldots,x_{n-1}^{2}, (x_1+\cdots+x_{n-1})^2, (a_1'' x_1 + \cdots + a_{n-1}'' x_{n-1})^{k})
%\end{align*}
%for some generic coefficients $a_i, a_i'$ and $a_i''$. We denote this algebra as $B$. 
%Since we know that 
%$\mathrm{HS}(\mathbf{k}[x_1,\ldots,x_n]/I_{n,k};t)=[(1-t^k)(1+t)^n]$, 
%we conclude that multiplication by $\ell^k$ on the algebra $A$ %$\mathbf{k}[x_1, \ldots, x_{n-1}]/(x_1^2, \ldots, x_{n-1}^2, (x_1 + \cdots + x_{n-1})^2)$ 
%has full rank if and only if
%\begin{align*}
%\mathrm{HS}(B;t) &= [(1-t^k)\mathrm{HS}(A
%\mathbf{k}[x_1,\ldots,x_{n-1}])/I_{n-1,2}
%;t] 
%=[(1-t^k)(1-t^2)(1+t)^{n-1}] \\
%&=[(1-t)(1-t^k)(1+t)^{n}]
%=[(1-t)\mathbf{k}[x_1,\ldots,x_n]/I_{n,k};t].
%\end{align*}
%However, $B$ has this Hilbert series if and only if multiplication by a general linear form $\ell'$ has full rank on $\mathbf{k}[x_1, \ldots, x_n] / I_{n,k}$. Hence, we conclude that $\mathbf{k}[x_1, \ldots, x_n] / I_{n,k}$ has the WLP if and only if multiplication by $\ell^k$ has full rank on $A$, 
%as required.
%\end{proof}

%{\color{red} Do we use this notation in the remainder of this Extended Abstract?}In the upcoming lemmas, we will need some notation. For a set $S=\{s_1,\dots, s_p\}$, we denote $a_S=a_{s_1}\cdots a_{s_p}$ and $x_S=x_{s_1}\cdots x_{s_p}$.For any $(k-1)$ subset $T\subset[n]$, we define $\mathcal{T}_i=\{S\subseteq [n]:\ |S|=|T|=k-1,\ |S\cap T|=i\}$. 

% \medskip
 
The following proposition will be used to establish the failure of the WLP in \Cref{thm:WLP_linear_forms}. %For cases in which we aim to establish the failure of the WLP, the following proposition is crucial.

\begin{proposition}\label{prop:syzygy_relation}
Let % $B=\mathbf{k}[x_1, \ldots, x_n] / (x_1^2, \ldots, x_n^2)$ and 
$\ell = a_1 x_1 + \cdots + a_n x_n$ be a general linear form. Then:
\begin{itemize}
    \item[{\rm (i)}] If $n = 2p + 1$ for some $p \geq 1$, then there exists another linear form $\ell'$ and a degree $(p-1)$ form $g$ such that in the algebra $B$, we have 
$\ell^p \ell' = (x_1 + \cdots + x_n)^2 g$.
\item[{\rm (ii)}] If $n=2p$ for some $p\geq 3$, then there exist a degree $2$ form $f$ and a degree $p-2$ form $g$ such that
$\ell^{p-2}f = (x_1+\cdots + x_n)^2g$
in $B$ and $f$ is not a multiple of $(x_1+\cdots + x_n)^2$.
\end{itemize}

\end{proposition}

While \Cref{prop:syzygy_relation} will be used to establish failure of the WLP, the next lemma will be used to lift this failure to more degrees and address the nontrivial cases with~the~WLP.

\begin{lemma}\label{lem:Injective_WLP}
Let %$B=\mathbf{k}[x_1,\ldots,x_n]/(x_1^{2},\ldots,x_{n}^{2}, (x_1+\cdots+x_n)^{2})$ and let 
$\ell$ be a general linear form in the algebra $B$. Then:
\begin{itemize}
    \item[{\rm (i)}] If $n\geq 2p+2$, then $\cdot \ell^p : B_1 \to B_{p+1}$ is injective.
    \item[{\rm (ii)}] If $n = 2p$ and $p \geq 3$, then for $q$, $2 < q < p$, the map $\cdot \ell^{q-2}: B_2 \to B_q$ is injective.
\end{itemize}
\end{lemma}
\begin{comment}
    \begin{proof}
We begin by noting that, by an argument of Conca \cite[Theorem 1.1]{Conca}, which was first pointed out by Wiebe \cite[Proposition 2.9]{Wiebe} to hold in this generality, it is sufficient to demonstrate that the map $\cdot \ell^p : A_1 \to A_{p+1}$ has full rank, where $A = \mathbf{k}[x_1, \ldots, x_n]/\ini(I_{n,2})$. Next, since $A$ is a monomial algebra, it suffices to consider the linear form $\ell = x_1 + \dots + x_n$, as established for the WLP by Migliore, Mir\'o-Roig, and Nagel in \cite[Proposition 2.2]{Miglioretrans} and generalized to the SLP and arbitrary characteristics by Nicklasson \cite[Theorem 2.2] {Nicklasson_SLP_two_variables}. We will prove that the map is injective by showing that if
\[
(x_1+\cdots + x_{n})^p \big(\sum_{i=1}^{n}c_ix_i \big) = 0
\]
in $A$, then all $c_i=0$. To this end, we expand and observe that
\[
(x_1+\cdots + x_{n})^p \big(\sum_{i=1}^{n}c_ix_i \big) = p!\sum_{\substack{S\subseteq [n]\\ |S|=p+1}}\big(\sum_{j\in S}c_j\big)x_S.
\]
By our restriction on $n$ and $p$, it follows from our description of $\ini(I_{n,2})$ in \Cref{thm:GBasis} that all monomials of the form $x_ix_{n-p+1}x_{n-p+2}\cdots x_{n}$ for $i=1,2,\dots, n-p$ are nonzero in $A_{p+1}$. Therefore, we have 
\[
c_i + \sum_{j=n-p+1}^{n} c_j = 0,
\]
which implies that $c_1 = c_2 = \cdots = c_{n-p}$. Additionally, the monomials
\[
x_{n-p-1} x_{n-p} \cdots x_{n-p+i-2} x_{n-p+i} \cdots x_n
\]
are nonzero in $A_{p+1}$ for $i = 1, 2, \dots, p+1$. Hence, we obtain
\[
\sum_{\substack{j=n-p-1 \\ j \neq n-p+i-1}}^{n} c_j = 0,
\]
which, combined with the relation $c_{n-p} + \cdots + c_n = 0$, implies that $c_{n-p-1} = c_{n-p} = \cdots = c_n$. Therefore, all the $c_j$'s must be equal, and since $c_{n-p} + \cdots + c_n = 0$, it follows that $c_j = 0$ for all $j$. This establishes the injectivity of the map.
\end{proof}
\end{comment}

%Similarly to how Propositions~\ref{prop:syzygy_relation} and~\ref{prop:syzygy_relation}(ii) cover both even and odd cases, an alternative version of \Cref{lem:Injective_WLP} is also necessary.


\begin{comment}
\begin{lemma}\label{lem:Injective_WLP}(ii)
Consider the algebra $B = \mathbf{k}[x_1, \dots, x_n] / (x_1^2, \dots, x_n^2, (x_1 + \cdots + x_n)^2)$, where $n = 2p$ and $p \geq 3$. Then, for $q$ such that $2 < q < p$ and $\ell$ a general linear form, the map $\cdot \ell^{q-2}: B_2 \to B_q$ is injective.
\end{lemma}\begin{proof}
As in \Cref{lem:Injective_WLP}, it suffices to show that the map $\cdot \ell^{q-2}:A_2 \to A_{q}$ is injective, where $A=\mathbf{k}[x_1,\dots, x_n]/\ini(I_{n,2})$ and $\ell=x_1+\cdots + x_n$. Let 
\[
g=\sum_{i<j}c_{i,j}x_ix_j
\]
be a quadratic form. Then
\begin{equation}\label{eq:deg2_prod}
g(x_1+\cdots + x_n)^{q-2} = (q-2)!\sum_{|T|=q}\big(\!\!\!\sum_{\{i,j\}\in T}c_{i,j} \big)x_T
\end{equation}
in $A$. We aim to show that if the expression in \eqref{eq:deg2_prod} equals zero, then all $c_{i,j} = 0$. By the bijection in \Cref{lem:LatticePathBijections}, it is clear that only the monomials corresponding to lattice paths that do not intersect the line $y = x + 2$ impose conditions on the coefficients. Focusing on the terms in \eqref{eq:deg2_prod} involving variables $x_i$ with $i \geq q$, the corresponding paths do not intersect the line and therefore provide constraints on the coefficients. These are precisely the equations that arise from multiplying by $(\ell')^{q-2}$ for $\ell' = x_q + \cdots + x_n$ in the algebra $B' = \mathbf{k}[x_k, \dots, x_n] / (x_q^2, \dots, x_n^2)$. Since $B'$ is a monomial complete intersection with the strong Lefschetz property, we know that $\cdot (\ell')^{q-2}: B_2' \to B_q'$ has full rank. Given that the socle degree of $B'$ is $n - (q-1) = 2p - q + 1$, we have $\dim_{\mathbf{k}} B_q' \geq \dim_{\mathbf{k}} B_2'$ because $2(p-q) + 1 \geq 2$. Thus, $\cdot (\ell')^{q-2}: B_2' \to B_q'$ is injective, and hence $c_{i,j} = 0$ for all $i, j \geq q$.

To prove that all $c_{i,j} = 0$ without restrictions on $i, j$, we will show that no smallest index $r$ exists for which $c_{i,j} = 0$ whenever $i, j \geq r$. Suppose, for contradiction, that $c_{i,j} = 0$ for all $i, j \geq r > 1$. We will demonstrate that this implies $c_{i,j} = 0$ for all $i, j \geq r - 1$, as well.
Considering all terms in \eqref{eq:deg2_prod} involving variables with index at least $r-1$, we obtain, by assumption, the expression
\[
\sum_{\substack{T \subset \{r, \dots, n\} \\ |T| = q-1}} \left(\sum_{j \in T} c_{r-1,j} \right) x_{r-1} x_T.
\]
Factoring out $x_{r-1}$, this expression resembles the one in \Cref{lem:Injective_WLP}. Similar to that case, examining monomials of the form $x_T = x_i x_{n-q+3} x_{n-q+4} \cdots x_n$ for $i = r, r+1, \dots, n-q+2$ yields
\[
c_{r-1, i} + \sum_{j=n-q+3}^n c_{r-1, j} = 0,
\]
which implies $c_{r-1, r} = \cdots = c_{r-1, n-q+2}$. Additionally, the monomials
\[
x_{n-q+1} x_{n-q+2} \cdots x_{n-q+j-1} x_{n-q+j+1} \cdots x_n
\]
for $j = 1, 2, \dots, q$ imply that $c_{r-1, n-q+1} = c_{r-1, n-q+2} = \cdots = c_{r-1, n}$. Thus, all $c_{r-1, j}$ are equal, and any one of these equations implies that $c_{r-1, j} = 0$ for all $j \geq r$, as desired.
Note that the case $r = 2$ is slightly different since $x_1 x_2 x_{n-q+3} \cdots x_n = 0$ in $A$, which means we obtain no information about $c_{1,2}$. However, as $x_1 x_2 = 0$ in $A$, this poses no issue. Thus, the proof is complete.
\end{proof}
\end{comment}

With this preparation, we are now ready to prove \Cref{thm:WLP_linear_forms}.
\begin{proof}[{\bf Proof of \Cref{thm:WLP_linear_forms}}]
By \Cref{lem:full_rank_switch}, this is equivalent to showing that the $k$-th power of a general linear form has full rank on $A$. %$\mathbf{k}[x_1, \ldots, x_{n-1}]/(x_1^{2}, \ldots, x_{n-1}^{2}, (x_1 + \cdots + x_{n-1})^2)$ 
for the same values of $n$ and $k$. We begin by examining the ideals for which we want to show the WLP.

%Let $A = \mathbf{k}[x_1, \ldots, x_{n-1}]/(x_1^{2}, \ldots, x_{{n-1}}^{2}, (x_1 + \cdots + x_{n-1})^{2})$. 
Note that the socle degree of $A$ is given by the smallest integer $d$ for which $\binom{n-1}{d} - \binom{n-1}{d-2} > 0$, which is $(n-1)/2$ if $n$ is odd, and $n/2$ if $n$ is even. Since any multiplication by a power of a linear form from $A_0$, or into a zero-dimensional vector space, has full rank, it follows that any $k$ greater than $(n-1)/2$ (for $n$ odd) or greater than $n/2$ (for $n$ even) works. Additionally, when $n = 2k + 3$, the socle degree is $k + 1$, so the only map potentially lacking full rank is from degree 1 to degree $k + 1$. However, this map is injective by \Cref{lem:Injective_WLP}(i), establishing all the required full-rank maps.



We are now left to show that the remaining values of $n,k$ give algebras that do not have the WLP. Assume first that $n=2p+2$. %$ is even. 
Then we claim that for any general linear form $\ell$ in $A$, the map 
$
\cdot \ell^k:A_{p+1-k} \to A_{p+1}
$
does not have full rank. For $k = p$, \Cref{prop:syzygy_relation}(i) provides a linear form $\ell' \in A_1$ such that $\ell^p \ell' = 0$ in $A$. Additionally, by \Cref{lem:Injective_WLP}(i), we have $\ell^k \ell' \neq 0$ for any $k < p$. Therefore, $\ell^{p-k} \ell'$ is a nonzero element in the kernel of $\ell^k$ for all $k = 2, \dots, p-1$, demonstrating that $\cdot \ell^k: A_{p+1-k} \to A_{p+1}$ cannot be injective. 
Moreover, this map cannot be surjective either. Using the theory of inverse systems, Boij and the second author in \cite[Theorem 5.2]{boijlundqvist} have shown that $A/(\ell^k)$ is always nonzero in degree $p + 1$ for any $k \geq 2$. This proves all required failures of maximal rank for the case where $n$ is even.


The case where $n = 2p + 1$ is odd follows a similar proof. First, the map 
$\cdot \ell^k: A_{p-k} \to A_p$
is not injective for $k = p - 2$ by \Cref{prop:syzygy_relation}(ii), as there exists a nonzero form $f$ of degree 2 with $\ell^{p-2} f = 0$ in $A$. Furthermore, by \Cref{lem:Injective_WLP}(ii), we have $\ell^{p-k-2} f \neq 0$ in $A$ for $k < p - 2$, which means that $\ell^{p-k-2} f$ is a nonzero element in the kernel of $\cdot \ell^k: A_{p-k} \to A_p$, demonstrating that this map is never injective.
Additionally, by 
\cite[Theorem 5.2]{boijlundqvist}, 
setting the last variable to zero shows that $A/(\ell^k)$ is nonzero in degree $p$ for any $k \geq 2$. This completes the verification of all required failures of full rank, proving the theorem.
\end{proof}

%\subsection{A remark on the Fr\"oberg conjecture}
%We have the isomorphism $$\mathbf{k}[x_1,\dots, x_n]/(\ell_1^2,\dots, \ell_n^2, \ell_{n+1}^k,\ell_{n+2}) \cong 
%\mathbf{k}[x_1,\dots, x_{n-1}]/(\ell_1^2,\dots, \ell_n^2, \ell_{n+1}^k),$$
%which implies, by \Cref{thm:WLP_linear_forms}, that the latter algebra is thin when $$
%\begin{cases}
%k\geq \frac{n-3}{2} & \text{ for } n \text{ odd,}\\
%k\geq \frac{n}{2} & \text{ for } n \text{ even}.\\
%\end{cases}
%$$
%From this, it follows that the Fr\"oberg conjecture is true for $a_1 = \cdots = a_{n+1} = 2$ and $a_{n+2} = k$, where
%$$
%\begin{cases}
%k \geq \frac{n-2}{2} & \text{for } n \text{ even}, \\
%k \geq \frac{n+1}{2} & \text{for } n \text{ odd}.
%\end{cases}
%$$
%The statement is trivial when $n$ is odd, as in this case, $k$ is the socle degree or higher. However, for even $n$, the case $k = \frac{n-2}{2}$ offers a new result for the conjecture.

\printbibliography


\end{document}
\section{A minimal free resolution of the initial ideal}
The initial ideal of $I_{n,k}$ belongs to the class of squarefree stable ideals plus powers, as studied in \cite{Murai2008}. The structure of the minimal free resolution of such ideals and formulas for computing their Betti numbers are well-established, often relying on subsets of the generators. In this section, we provide an explicit description of the Betti numbers of $I_{n,k}$ using the combinatorial properties of its generators. This allows us to directly extract the extremal Betti numbers, regularity, and projective dimension in terms of $n$ and $k$.

We briefly recall the notion of Betti numbers and refer to \cite{eisenbud2013commutative} for more details.
Consider the ideal $I$ in the polynomial ring $R=\kb[x_1, \ldots, x_n]$. Assume that
\[
F : 0 \to \cdots \to F_i \to F_{i-1} \to \cdots \to F_0 \to I \to 0
\]
is a minimal free resolution of $I$ (i.e., a  free resolution such that $\varphi_{i+1}(F_{i+1}) \subseteq m F_i$ for all $i \geq 0$) and $m=(x_1,\ldots,x_n)$. The $i$-th Betti number $\beta_i(I)$ of $I$ is defined as the rank of $F_i$. The $i$-th graded Betti number in degree $j$ 
denoted by $\beta_{i,j}(I)$, is the rank of the degree $j$ part of $F_i$.


Finding the Betti numbers of ideals is, in general, challenging. However, by semi-upper semicontinuity, we have the inequality $\beta_{i,j}(I) \leq \beta_{i,j}(\ini(I))$ for all $i \geq 0$ and all $j$. Thus, the Betti numbers of the initial ideal provide an upper bound for the Betti numbers of the ideal itself. Our goal in this section is to compute the Betti numbers of the initial ideal $\ini(I_{n,k})$, hence providing an upper bound for those of the ideal $I_{n,k}$.

\subsection{The squarefree part of 
$\ini(I_{n,k})$}
\begin{definition}
Let $I$ be a squarefree monomial ideal with the minimal generating set $G(I)$. Then $I$ is called \emph{strongly squarefree stable} if, for any monomial $ux_j \in G(I)$ and any index $i < j$ such that $x_i$ does not divide $u$, the monomial $ux_i$ is also in $I$.
\end{definition}

\begin{proposition}\label{prop:Strongly_stable}
Let $J_{n,k}$ denote the squarefree part of $\ini(I_{n,k})$. Then $J_{n,k}$ is a strongly squarefree stable ideal.
\end{proposition}
\begin{proof}
By \Cref{lem:LatticePathBijections}, the generators of $J_{n,k}$ correspond to $(N,E)$-lattice paths that touch the line $y = x + k$ for the first time after their last step. Let $P$ be a path corresponding to a generator $ux_j$. If $x_i$ does not divide $u$ and $i < j$, we need to show that $ux_i$ is also in $J_{n,k}$. 
In terms of the path, $ux_i$ corresponds to a new path $P'$ constructed from $P$ by changing the $j$-th step from going north to east, and the $i$-th step from going east to north. Since both $P$ and $P'$ take the same number of total steps and the same number of steps north, one of two cases must occur: 

(1) If the first time $P'$ touches the line $y = x + k$ is at the last step, then $P'$ corresponds to a monomial that is a minimal generator of $J_{n,k}$.

(2) If $P'$ touches the line $y = x + k$ at an earlier step, then the path up to that point corresponds to another generator $v$ of $J_{n,k}$ in an earlier degree. In this case, $ux_i$ is a multiple of $v$ and thus is in $J_{n,k}$, as desired.
\end{proof}

We now recall a formula for the Betti numbers of a strongly stable ideal from Gasharov, Hibi, and Peeva \cite{GasharovHibiPeeva2002}, as presented in \cite{Murai2008}. 
\begin{lemma}[\cite{GasharovHibiPeeva2002,Murai2008}]\label{lem:Murai}
   Let $G(I)$ be the set of minimal generators of a strongly squarefree stable ideal $I\subset R$. Then the Betti numbers $\beta_{p,p+s}(I)$
   are given by
\begin{equation}\label{eq:Gasharov_betti_formula}
\beta_{p,p+s}(I) = %\textcolor{red}{\cancel{\beta_{s,p}(I)}} = 
\sum_{\substack{u\in G(I)\\ \deg u = s}}\binom{\max u - s}{p}. 
\end{equation}
%}
\end{lemma}



In~\eqref{eq:Gasharov_betti_formula}, we used that $|\{j\in [n]:\; j<\max u \text{ and } x_ju \text{ is squarefree}\}|=\max u - \deg u$ for any squarefree monomial $u$. Combining \Cref{prop:Strongly_stable} and \Cref{lem:Murai}, we can determine the Betti~numbers of the squarefree part of $\ini(I_{n,k})$. % of $J_{n,k}$.

\begin{proposition}
Let $J_{n,k}$ denote the squarefree part of $\ini(I_{n,k})$. Then the Betti numbers $\beta_{p,p+s}(J_{n,k})$ are zero for $n<2s-k$ and otherwise are given by
\[
\beta_{p,p+s}(J_{n,k}) = C_{s - k}^{k-1}\binom{s - k}{p}.
\]
\end{proposition}

\begin{proof}
By \Cref{prop:Strongly_stable}, the ideal $J_{n,k}$ is strongly squarefree stable. Hence, the Betti numbers are given by \Cref{lem:Murai}. By \Cref{thm:GBasis}, we know that 
there are no minimal generators of the Gr\"obner basis of degree $s$ if $n < 2s - k$. Consequently, the Betti numbers are zero for these values of $n$. Otherwise, for any minimal generator $u$ of degree $s$, we have that $\max u = 2s - k$. Moreover, by \Cref{cor:ConvolutionCatalanNumbers}, there are $C_{s-k}^{k-1}$ minimal generators of degree $s$ in $J_{n,k}$. Thus, the result follows by applying \eqref{eq:Gasharov_betti_formula}.
%}
\end{proof}

We now apply Proposition~2.1 in \cite{Murai2008} to compute the Betti numbers of the ideal $\ini(I_{n,k})=J_{n,k}+(x_1^2,\ldots,x_n^2)$. Before presenting the formula, however, we need to fix the following notation.
For each squarefree monomial $u$ of $\ini(I_{n,k})$, we define
\[
A_p(u) = \sum_{j=1}^{t} \binom{i_j - 1}{p - j},\quad \text{if } u=x_{i_1}\cdots x_{i_t} \text{ with } i_1>\cdots > i_t.
\]
\begin{corollary}\label{cor:MuraiFormula}
Consider the ideal $\ini(I_{n,k})$ and its squarefree part $J_{n,k}$. Then, the Betti numbers of $\ini(I_{n,k})=J_{n,k}+(x_1^2,\ldots,x_n^2)$ are given by
\begin{equation}\label{eq:Murai_formula}
\beta_{p,p+s}(\ini(I_{n,k})) = \sum_{\substack{u\in \mathrm{SFP}(J_{n,k})\\ \deg(u)=s}}A_{p+1}(u)-\sum_{\substack{u\in \mathrm{SFP}(J_{n,k})\\ \deg(u)=s-1}}\big(\binom{n}{p + 1}-A_{p+2}(u)\big)+\delta_{s-1, p+1}\binom{n}{p+1},
\end{equation}
where $\delta_{s-1, p+1}$ equals $1$ if $s - 1 = p + 1$ and $0$ otherwise.
\end{corollary}


\begin{example}\label{ex:Murai_formula}
Consider the ideal $\ini(I_{4,2})=(x_1 x_2, x_1 x_3 x_4, x_2 x_3 x_4)+(x_1^2,\ldots,x_4^2)$ and its squarefree part $J_{4,2}=(x_1 x_2, x_1 x_3 x_4, x_2 x_3 x_4)$.
We compile the table 
$$
\begin{array}{r||c|c|c|c|}
u & A_1(u) & A_2(u) & A_3(u) & A_4(u) \\
\hline 
x_1 x_2         & 1 & 2 & 0  & 0 \\
\hline 
x_1 x_2 x_3     & 1 & 3 & 3  & 0 \\
\hline 
x_1 x_2 x_4     & 1 & 4 & 5  & 1 \\
\hline 
x_1 x_3 x_4     & 1 & 4 & 6  & 2 \\
\hline 
x_2 x_3 x_4     & 1 & 4 & 6  & 3 \\
\hline 
x_1 x_2 x_3 x_4 & 1 & 4 & 6  & 4 \\
\end{array}
$$
of the nonzero values of $A_p(u)$.
Now we apply \Cref{cor:MuraiFormula} to obtain the Betti numbers of $\ini(I_{4,2})$,  
gathered in the Betti diagram
$$
\begin{array}{c|cccc}
      & 0 & 1 & 2 & 3\, \\
    \hline 
    2 & 5 & 2 & 0 & 0\, \\
    3 & 2 & 15 & 16 & 5\, \\
    4 & 0 & 0 & 0 & 0\, \\
    5 & 0 & 0 & 0 & 0, \\
\end{array}
$$
where $\beta_{p,p+s}$ is the entry in row $s$ and column $p$.
\end{example}



\subsection{Mapping cone resolution}
Herzog and Takayama introduced a general mapping cone procedure in~\cite{herzog2002resolutions} for constructing free resolutions of monomial ideals. 
Mayer--Vietoris trees offer a combinatorial method to compute the support of these resolutions~\cite{Saenz-De-Cabezon2009MVT}. 
We now provide the minimal resolution of $J_{n,k}$ using~mapping~cone. 

\smallskip

Let $I$ be a monomial ideal with a unique minimal generating set of monomials $G(I) = \{g_1, \dots, g_r\}$. Define $I_i = ( g_1, \dots, g_i )$ as the ideal generated by the first $i$ generators in a chosen ordering. For each $i$, we have the exact sequence
\begin{equation*}\label{eq:ses}
0 \longrightarrow I_{i-1} \cap ( g_i ) \stackrel{j}{\longrightarrow} I_{i-1} \oplus ( g_i )  \stackrel{l}{\longrightarrow} I_i \longrightarrow 0.
\end{equation*}
In this sequence, $g_i$ is referred to as a {\em pivot monomial}. % or {\em pivot generator}.
Assume that free resolutions $\FF_{i}'$ and $\widetilde{\FF}_i$ are known for the ideals $I'_i=I_{i-1}$ and $\widetilde{I}_i = I_{i-1} \cap ( g_i )$, respectively. Then, a (not necessarily minimal) resolution $\FF_i$ of $I_i$ can be constructed as the mapping cone of the chain complex morphism $\psi: \widetilde{\FF}_i \longrightarrow \FF_{i}'$ that lifts the inclusion $j$. 
By applying this process recursively on $i$, we obtain a free resolution $\FF = \FF_r$ of $I$, known as an {\em iterated mapping cone resolution}.

\smallskip
A Mayer--Vietoris tree of a monomial ideal $I$, denoted by MVT$(I)$ is a binary tree that represents the sequence of ideals in the iterated mapping cone process. Each node in this tree contains an ideal, given by its minimal generating set, and is labeled with a pair $(p, d)$, where $p$ is the node's position and $d$ its dimension. The root node, labeled $(1, 0)$, contains the ideal $I = I_r$. For a node labeled $(p, d)$ containing an ideal $J$ with at least two generators, there are two child nodes: the left child, labeled $(2p, d+1)$ contains the ideal $\widetilde{J}$, and the right child, labeled $(2p + 1, d)$ contains the ideal $J'$.
Nodes at position $1$ and all even positions are referred to as the {\em relevant} nodes of the tree. 

By \cite[Theorem~1]{Saenz-De-Cabezon2009MVT}, there exists a resolution of $I$ supported on the relevant nodes of MVT$(I)$. This resolution is referred to as the {\em Mayer--Vietoris resolution} of $I$. The dimension of each node in MVT$(I)$ corresponds to the homological degree of the corresponding module in the resolution. The following proposition gives a sufficient condition for a Mayer--Vietoris resolution to be minimal. % see Proposition~3 in \cite{Saenz-De-Cabezon2009MVT}.



\begin{proposition}[\cite{Saenz-De-Cabezon2009MVT}, Proposition~3]\label{prop:repeated_nodes}
Let $\mu \in \mathbb{N}^n$ be a multidegree such that $x^\mu$ is a minimal generator in exactly $k$ relevant nodes of the Mayer--Vietoris tree of $I$, and all of these nodes have the same dimension $i$. Then, $\beta_{(i, \mu)}(I) = k$. 
In particular, if $x^\mu$ appears as a minimal generator in only one relevant node, % of $MVT(I)$, 
then $\beta_{(i, \mu)}(I) = 1$. 
\end{proposition}

\begin{remark}
In any non-minimal resolution, there is at least one reduction pair of generators in consecutive homological degrees with the same multidegree. Thus, a sufficient condition for a Mayer--Vietoris resolution to be minimal is the absence of repeated multidegrees in consecutive dimensions.
\end{remark}

\begin{example}\label{ex:MVT}
{Consider the ideal $I_{4,2} = ( x_1x_2, x_1x_3x_4,x_3x_3x_4,x_1^2,x_2^2,x_3^2,x_4^2 )$}. \Cref{fig:MVT} depicts a Mayer--Vietoris tree of $I_{4,2}$, in which we display the first few nodes. Each node is labeled with its position $p$, dimension $d$, and the corresponding ideal $I$, denoted as $(p, d) \, I$. The relevant nodes are shaded darker.
\begin{figure}
\begin{center}
 \begin{tikzpicture}[scale=0.6,transform shape]
 \tikzstyle{level 1}=[sibling distance=9cm]
 \tikzstyle{level 2}=[sibling distance=5cm]
 \tikzstyle{level 3}=[sibling distance=3cm] 
 \node{$(1,0)$ $x_1x_2,x_1x_3x_4,x_2x_3x_4, x_1^2, x_2^2, x_3^2,x_4^2$}
 child{ node{$(2,1)$ $x_4^2(x_1x_2, x_1x_3,x_2x_3,x_1^2, x_2^2,x_3^2)$}
 	child{node[left=.5cm,rectangle,draw]{$(4,2)$ $x_3^2x_2^2(x_1,x_2)$}}
 	child{node[left=.3cm,xshift=.2cm,color=black!50!white]{$(5,1)$ $x_4^2(x_1x_2, x_1x_3,x_2x_3,x_1^2, x_2^2)$}
 	      child{node[xshift=-1cm,rectangle,draw]{$(10,2)$ $x_2^2x_4^2(x_1,x_3)$}}
          child{node[below=.5cm,xshift=-1cm,color=black!50!white]{$(11,1)$ $x_4^2(x_1x_2,x_1x_3,x_2x_3,x_1^2)$}
          child{node[xshift=-1cm,rectangle,draw]{$(22,2)$ $x_1^2x_4^2(x_2,x_3)$}}
	child{node[below=.6cm,xshift=-1cm,color=black!50!white,rectangle,draw]{$(23,1)$ $x_4^2(x_1x_2,x_1x_3,x_2x_3)$}}}
          }}
 child{ node[color=black!50!white]{$(3,0)$ $x_1x_2,x_1x_3x_4,x_2x_3x_4, x_1^2, x_2^2, x_3^2$}
 	child{node[xshift=-.3cm]{$(6,1)$ $x_3^2(x_1x_2,x_1x_4,x_2x_4, x_1^2, x_2^2)$}
    child{node[xshift=-1cm,rectangle,draw]{$(12,2)$ $x_2^2x_3^3(x_1,x_4)$}}
	child{node[below=.5cm,xshift=-1cm,color=black!50!white]{$(13,1)$ $x_3^2(x_1x_2,x_1x_4,x_2x_4, x_1^2)$}
    child{node[xshift=-1cm,rectangle,draw]{$(26,2)$ $x_1^2x_3^2(x_2,x_4)$}}
	child{node[below=.6cm,xshift=-1cm,color=black!50!white,rectangle,draw]{$(27,1)$ $x_3^2(x_1x_2,x_1x_4,x_2x_4)$}}}}
 	child{node[color=black!50!white]{$(7,0)$ $x_1x_2,x_1x_3x_4,x_2x_3x_4, x_1^2, x_2^2$}
	child{node[xshift=-1cm,rectangle,draw]{$(14,1)$ $x_2^2(x_1,x_3x_4)$}}
	child{node[below=.5cm,xshift=-1cm,color=black!50!white]{$(15,0)$ $x_1x_2,x_1x_3x_4,x_2x_3x_4, x_1^2$}
    child{node[xshift=-1cm,rectangle,draw]{$(30,1)$ $x_1^2(x_2,x_3x_4)$}}
	child{node[below=.6cm,xshift=-1cm,color=black!50!white,rectangle,draw]{$(31,0)$ $x_1x_2,x_1x_3x_4,x_2x_3x_4$}}}
       }};
 \end{tikzpicture}
\end{center}
\caption{Partial display of MVT$(\ini(I_{4,2}))$.}\label{fig:MVT}
\end{figure}


Since MVT$(I_{4,2})$ has no repeated multidegrees in nodes of consecutive dimensions, the corresponding resolution of $I_{4,2}$ is minimal. In particular, $\beta_0(I_{4,2}) = 7$, $\beta_1(I_{4,2}) = 17$, $\beta_2(I_{4,2}) = 16$ and $\beta_3(I_{4,2}) = 5$, with their multidegrees given by the monomials in the nodes at position $1$ for $\beta_0$, at positions $2$, $6$, $14$, $30$, $62$ and $126$ for $\beta_1$, at positions $4$, $10$, $12$, $22$, $26$, $28$ $46$, $54$, $56$, $60$, $94$ and $110$ for $\beta_2$, and at positions $8$, $20$, $24$, $44$ and $52$ for $\beta_3$.
\end{example}

We are now ready to construct a Mayer--Vietoris resolution of $\ini(I_{n,k})$. However, before proceeding, we first recall the following well-known lemmas.

\begin{lemma}\label{lem1}
Let $I \subseteq \kb[x_1, \dots, x_n]$ be a squarefree stable monomial ideal, and let $\sigma = \{x_{i_1}, \dots, x_{i_k}\}$ be a subset of the variables. The ideal $I_\sigma = ( g_\sigma \mid g \in G(I) ) \subseteq \kb[\{x_1, \dots, x_n\} \setminus \sigma]$ is squarefree stable, where $g_\sigma$ is the monomial obtained by setting each variable in $\{x_1, \dots, x_n\} \setminus \sigma$ equal to $1$ in the monomial $g$.
\end{lemma}

\begin{lemma}\label{lem2}
Let $I \subseteq \kb[x_1, \dots, x_n]$ be a monomial ideal, and let $y^\mu \in \kb[y_1, \dots, y_m]$ be a monomial in a different set of variables. Then the minimal free resolutions of $I$ and $I' = y^\mu I = ( y^\mu g \mid g \in G(I) ) \subseteq \kb[x_1, \dots, x_n, y_1, \dots, y_m]$ are isomorphic.
\end{lemma}

\begin{lemma}[\cite{Saenz-De-Cabezon2009MVT}, Proposition 7 and Remark 7]\label{lem3}
 For every squarefree stable ideal $I$ there is a Mayer--Vietoris tree whose corresponding resolution %of $I$ 
 is minimal.
\end{lemma}

\noindent{\bf A Mayer--Vietoris resolution of $\ini(I_{n,k})$.} Assume that $n = k + 2m$. Let $r$ denote the number of generators of $\ini(I_{n,k})$, which, by \Cref{cor:ConvolutionCatalanNumbers}, is given by $r = \sum_{i=0}^m C_{i}^{k-1}$.
Let $J_{n,k}$ be the squarefree part of $\ini(I_{n,k})$. 

\medskip
Consider the Mayer--Vietoris tree of $\ini(I_{n,k})$, where we select as the pivot, whenever possible, the monomial corresponding to the last squared variable among the generators of the ideal that is not a common factor of all the generators. For example, at the node in position $0$, we choose $x_n^2$ as the pivot; at the node in position $2$, we choose $x_{n-1}^2$ (since $x_n^2$ is a common factor); and at the node in position $3$, we again choose $x_{n-1}^2$. The first iteration of the iterated cone procedure yields the initial elements of MVT$(\ini(I_{n,k}))$:

\begin{center}
 \begin{tikzpicture}[scale=.8,transform shape]
 \tikzstyle{level 1}=[sibling distance=8.5cm]
 \tikzstyle{level 2}=[sibling distance=6cm]
 \node{$(1,0)$ $g_1,\dots,g_r,x_1^1,x_2^2,\dots,x_n^2$}
 child{ node{$(2,1)$ $x_n^2(g_1\vert_{x_n=1},\dots, g_r\vert_{x_n=1},\dots,x_1^2,x_2^2,\dots,x_{n-1}^2)$}}
 child{ node[color=black!50!white]{$(3,0)$ $g_1,\dots,g_r,x_1^1,x_2^2,\dots,x_{n-1}^2$}};
 \end{tikzpicture}
 \end{center}
The notation $x^\mu(x^{\nu_1}, \dots, x^{\nu_k})$ above means that each monomial $x^{\nu_j}$ is multiplied by $x^\mu$, and the monomial $g_i \vert_{x_k = 1}$ is obtained by evaluating the variable $x_k$ in $g_i$ to $1$.~This process continues until there are no more common squared variables that are not common factors. At this point, we reach nodes of the form $x_{i_1}^2 \cdots x_{i_k}^2(x^{\nu_1}, \dots, x^{\nu_l})$, where each monomial $x^{\nu_j}$ is in $\kb[\{x_1, \dots, x_n\} \backslash \{x_{i_1}, \dots, x_{i_k}\}]$. Nodes whose non-common part is squarefree are called {\em final nodes}. % of the tree. % (note that not all final nodes are leaves in the tree).
In \Cref{ex:MVT}, the final nodes are those bordered at positions $4,10,12,14,22,23,26,27,30$ and $31$.
\begin{proposition}
%With the above notation, 
The cone resolution supported on the Mayer--Vietoris tree constructed above provides a minimal free resolution for $\ini(I_{n,k})$.
\end{proposition} 

\begin{proof}
We only need to show the minimality of the associated Mayer--Vietoris resolution. Note that there is no common multidegree among relevant nodes in the subtrees attached to non-final nodes, since the common factor differs in each subtree. Moreover, by \Cref{lem1}, all ideals in relevant nodes are of the form $x_{i_1}^2 \cdots x_{i_d}^2(J)$, where $J$ is a squarefree stable ideal in the variables $\{x_1, \dots, x_n\} \backslash \{x_{i_1}, \dots, x_{i_d}\}$. Now, the only possible repeated multidegrees in MVT$(\ini(I_{n,k}))$ arise in the subtrees attached to final nodes. By Lemmas~\ref{lem2} and~\ref{lem3}, the cone resolutions of each of these subtrees are minimal, since the root of each of these subtrees have the form of a monomial times a squarefree stable ideal in a different set of variables. Thus the resolution of $\ini(I_{n,k})$ is minimal.
\end{proof}

\begin{corollary}
The projective dimension of $\ini(I_{n,k})$ is given by ${\rm pd}(\ini(I_{n,k})) = n - 1$, and its regularity is ${\rm reg}(\ini(I_{n,k})) = k + m$.
\end{corollary}

\begin{proof}
The final nodes with the highest dimension and degree in the Mayer--Vietoris tree of $\ini(I_{n,k})$ are obtained by sequentially choosing as pivots sequences of squared variables of length $k+m-1$. The final node at the end of such a sequence has the form \( x_{l_1}^2 \cdots x_{l_{k+m-1}}^2(x_{l'_1}, \dots, x_{l'_{m+1}}) \), with dimension $k+m-1$. From each of these nodes, a tree isomorphic to the one corresponding to the prime ideal \( (x_{l'_1}, \dots, x_{l'_{m+1}}) \) hangs. This prime ideal has projective dimension $m$, so the highest dimension of any node in the tree is \( (k+m-1) + m = k + 2m - 1 = n - 1 \). Since the Mayer--Vietoris tree provides a minimal free resolution of $\ini(I_{n,k})$, we conclude that the projective dimension of $\ini(I_{n,k})$ is ${\rm pd}(\ini(I_{n,k})) = n - 1$.

For the regularity, observe that the maximal degree of a generator of $\ini(I_{n,k})$ is $k + m$. Additionally, if there is more than one generator of $\ini(I_{n,k})$ of a given degree $d$, then for each such generator, there is another generator of the same degree that differs only in one variable. When taking a squared variable as a pivot, the maximal degree of the generators in a node $J$ remains unchanged in the subsequent node $J'$. While the dimension of the node stays the same, it drops by at least one in the reduced node $\widetilde{J}$, where the dimension increases by one. Therefore, the regularity does not increase. In the final nodes, the squarefree part ensures that for the generator with the highest degree, there is always another generator (of the same degree or lower) that differs only in one variable. Thus, regularity does not increase in the trees hanging from the final nodes either. Consequently, we have ${\rm reg}(\ini(I_{n,k})) = k + m$, and this value is attained in the generating set of $\ini(I_{n,k})$.
\end{proof}


\subsection{Relation between Betti numbers of $\ini(I_{n,k})$ and $\ini(I_{n-1,k})$} 
We now show that the study of $\ini(I_{n,k})$ for general $n$ can be reduced to the cases where $n = k + 2m$ for some integer $m$. Hence, in what follows, we focus on these cases.
\begin{lemma} Let $n=k+2m+1$ for some $m$. Then
\[
\beta_{i,j}(\ini(I_{n,k}))=\beta_{i,j}(\ini(I_{n-1,k}))+ \beta_{i-1,j-2}(\ini(I_{n-1,k})).
\]
\end{lemma}

\begin{proof}
    Note that $\ini(I_{n,k}) = \ini(I_{n-1,k}) + ( x_{n}^2 )$, where the ideals $\ini(I_{n-1,k})$ and $( x_{n}^2 )$ belong to  polynomial rings with disjoint sets of variables. Consequently, the minimal free resolution of $\ini(I_{n,k})$ is given by the tensor product of the minimal free resolution of $\ini(I_{n-1,k})$ and that of $( x_{n}^2 )$. Moreover, the only non-zero Betti numbers of $( x_{n}^2 )$ are $\beta_{0,0} = 1$ and $\beta_{1,2} = 1$. 
Given these facts, we derive the relation 
\begin{eqnarray*}
    \beta_{i,j}(\ini(I_{n,k})) &=& \beta_{i,j}(\ini(I_{n-1,k}))\beta_{0,0}\big( ( x_{n}^2 ) \big) + \beta_{i-1,j-2}(\ini(I_{n-1,k}))\beta_{1,2}\big( ( x_{n}^2 ) \big) \\ &=& \beta_{i,j}(\ini(I_{n-1,k}))+ \beta_{i-1,j-2}(\ini(I_{n,k})).
\end{eqnarray*}
Note that the second summand is zero when $i = 0$ or $j < 2$.
\end{proof}
\subsection{The shape of the Betti table of $\ini(I_{n,k})$}

The analysis of the structure of MVT$(\ini(I_{n,k}))$ provides insight into the structure of the Betti table of $\ini(I_{n,k})$.
The Betti table of $\ini(I_{n,k})$ encodes the Betti numbers, where the entry in column $i$ and row $j$ corresponds to $\beta_{i,i+j}(\ini(I_{n,k}))$. In the table presented below, dots represent zero values, while asterisks denote non-zero values.


\begin{proposition}
Let $n=k+2m$. Let $\beta_{i,j}$ denote $\beta_{i,j}(\ini(I_{n,k}))$. Then
\begin{enumerate}
    \item $\beta_{0,2}=n+1$ if $k=2$, $\beta_{0,2}=n$ if $k>2$, and \\ $\beta_{0,j}=C_{j-k}^{k-1}$ for all ${k}\leq j\leq {k+m}$, %; if $k=2$ then $\beta_{0,2}=n+1$.
    \item $\beta_{i,2(i+1)}=\binom{{n}}{{i+1}}$ and $\beta_{i,j}=0$ for all $j<i+k$, $j\neq 2(i+1)$ for $i=1,\dots,k-3$,
    \item $\beta_{k-1+2i,k-1+2i+k+i}=C_{i+1}^{k-1}$, for $i=\{0,\dots,m\}$, %\Fatemeh{is this only for $k=2$?}
   \item for $i=\{0,\dots,m\}$, the following hold:
    \begin{itemize}
      %  \item $\beta_{k-1+2i,j}=0$ for all $j<k+i$,
               \item $\beta_{k-1+2i,k-1+2i+j}=0$ for all $j<k+i$,        
       \item $\beta_{k-1+2i,k-1+2i+j}\neq 0$ for all $j>k+i$,
 \item 
 {$\beta_{j,j+(k+i)}=0$} for all 
 {$j>k-1+2i$},
       \item 
       {$\beta_{j,j+(k+i)}\neq 0$} for all 
       {$j<k-1+2i$}.
    \end{itemize}

\end{enumerate}
In particular, the Betti diagram of $\ini(I_{n,k})$ is 
\[
\begin{array}{c|ccccccccccc}
          & 0 & 1 & \cdots& k-1 & k & k+1 & \cdots & k-1+2i &\cdots &k-1+2m\\
      \hline
      2  & n & \cdot &\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot& \cdot\\
      3  & \cdot & * &\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot& \cdot\\
      4  & \cdot & \cdot &*&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot& \cdot\\
      \vdots  &\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot& \cdot\\
      k  & C_0^{k-1} & * & \cdots & C_1^{k-1} &\cdot&\cdot&\cdot&\cdot&\cdot& \cdot\\
      k+1  & C_1^{k-1} & * & * & * & * & C_2^{k-1} &\cdot&\cdot&\cdot& \cdot\\
      \vdots  &\vdots&*&*&*&*&*&\ddots&\cdot&\cdot& \cdot\\
    k+i  & C_i^{k-1} &*&*&*&*&*&\cdots&C_{i+1}^{k-1}&\cdot& \cdot\\
      \vdots  &\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots& \cdot\\
      k+m  & C_m^{k-1} &*&*&*&*&*& * &*&*& C_{m+1}^{k-1}
      \end{array}
\]
\end{proposition}
\begin{proof}
(1) The formula follows directly from \Cref{cor:ConvolutionCatalanNumbers} and the fact that the squares of the variables are among the minimal generators of $\ini(I_{n,k})$.
\smallskip

(2) The only generators in dimension $i$ for $i = 0, \dots, k-3$ in degree $2(i+1)$ are formed by products of $i+1$ squared variables. For each $i$, there are $\binom{{n}}{{i+1}}$ such products, as desired. 
\smallskip 

(3) Let $i \in\{0,\dots,m\}$. Take any sequence $\sigma$ consisting on $k+i-1$ variables among the first $k+2i$ variables, such that their product (which we will also denote by $\sigma$) is not in the ideal ${\rm in}(I_{n,k})$. Then, the final node corresponding to taking as pivots the elements of $\sigma$ in descending order has the form
\[
\sigma^2(x_{l_1},\dots,x_{l_{i+1}}, \xb^{\mu_1},\dots,\xb^{\mu_s}),
\]
where $\{x_{l_1},\dots,x_{l_{i+1}}\}=\{x_1\dots,x_{k+2i}\}\setminus \sigma$, and the $\xb^{\mu_j}$ are monomials of degree strictly bigger than $1$, involving also variables bigger than $x_{k+2i}$.

This final node has dimension $\vert\sigma\vert=k+i-1$, and has a descendant $\sigma^2(x_{l_1},\dots,x_{l_{i+1}})$ of the same dimension, which is also a final node. The subtree hanging from this final node has a leaf such that its only generator is $\sigma^2x_{l_1}\cdots x_{l_{i+1}}$ at dimension $(k+i-1)+i=k+2i-1$, which has degree $2(k+i-1)+(i-1)$ and is a generator of the minimal free resolution of ${\rm in}(I_{n,k})$, hence contributing in one unit to $\beta_{k-1+2i,k-1+2i+k+i}$. The number of such sequences is
\[
\binom{{k+2i}}{{k+i-1}}-\sum_{j=0}^{i-1}C^{k-1}_{j}\binom{{2i-2j}}{{i-j-1}}=C^{k-1}_{i+1}.
\]

(4) %To prove (3) 
We show that no other sequence of variables results in a final node where the corresponding hanging tree has a node at this exact dimension and degree. We consider the following cases:
\begin{itemize}
\item[-] Sequences $\sigma$ corresponding to monomials in the ideal $\ini(I_{n,k})$ do not appear in the nodes of the tree, by construction.
\item[-] Sequences $\sigma$ such that $|\sigma| < k + i - 1$ correspond to final nodes, where fewer than $i+1$ variables appear as isolated generators. Therefore, the descendants of these final nodes produce subtrees with either a projective dimension strictly less than $i + (k + i - 1 - |\sigma|)$, or a total degree greater than $2k + 3i - 1$. Thus, none of these contributes to $\beta_{k-1+2i, k-1+2i+k+i}$.
\item[-] Sequences $\sigma$ such that $|\sigma| > k + i - 1$ correspond to final nodes in dimension $|\sigma| > k + i - 1$. To reach dimension $k-1+2i$ at the correct degree, the degree increase would need to be smaller than $k-1+2i + k+i - 2|\sigma|$ in $k-1+2i - |\sigma|$ dimensional steps, meaning at least one step produces no increase. Therefore, these sequences do not contribute to $\beta_{k-1+2i, k-1+2i+k+i}$.
\end{itemize}
The data on the %zero 
Betti numbers implies the shape of the Betti diagram, as illustrated.
\end{proof}

\section{Further directions}
We conclude the paper by presenting two conjectures and discussing potential extensions of our work to ideals of the general form $$I_{a_1,\ldots,a_{n+1}}=(x_1^{a_1}, \ldots, x_n^{a_n}, (x_1 + \cdots + x_n)^{a_{n+1}})$$ for various values of $a_i$. 
The first conjecture is on the coefficients of the Gr\"obner basis for $I_{a_1,\ldots,a_{n+1}}$ and comes from experiments in Macaulay2~\cite{M2}. In the following, we fix the reverse lexicographic order.
\begin{conjecture}
Each coefficient occurring in the Gr\"obner basis of $I_{a_1,\ldots,a_{n+1}}$ 
is a (possibly empty) product of primes less than $\max \{a_1,\ldots,a_n \}.$
\end{conjecture}

In the case of $I_{n,k}$, we have $a_1=\cdots = a_n=2$ and $a_{n+1}=k$, so the conjecture states that all nonzero coefficients should be $1$, which holds by \Cref{thm:GBasis}. 

\medskip
Given a sequence of integers $(a_1, \ldots, a_n)$, there exists an extension of square-free strongly stable ideals, called $(a_1, \ldots, a_n)$-strongly stable ideals; see \cite[p.~1322]{Murai2008}. Let $J_{a_1, \ldots, a_{n+1}}$ denote the ideal generated by all monomials $x_1^{b_1} \cdots x_n^{b_n}$ in $I_{a_1, \ldots, a_{n+1}}$ such that $b_1 < \min\{a_1, a_{n+1}\}$ and $b_j < a_j$ for all $j > 1$.

Our computations suggest that $J_{a_1, \ldots, a_{n+1}}$ is $(\min\{a_1, a_{n+1} \}, a_2, \dots, a_n)$-strongly stable, and the initial ideal of $I_{a_1, \dots, a_{n+1}}$ have interesting combinatorial properties. For example, we propose the following conjectural generalization of \Cref{prop:Strongly_stable}.


\begin{conjecture}
The initial ideal $\mathrm{in}(I_{a_1,\ldots,a_{n+1}})$ %with respect to the reverse lexicographic order, 
can be decomposed as 
\[
\mathrm{in}(I_{a_1,\ldots,a_{n+1}}) = (x_1^{\min\{a_1, a_{n+1} \}}, x_2^{a_2},\dots, x_n^{a_n}) + J_{a_1,\ldots,a_{n+1}}.
\]
\end{conjecture}

We remark that the recent result by Booth, Singh, and Vraciu \cite{booth2024weak} supports this conjecture for the case $a_i=3$ for $i = 1,\ldots,n+1$. Furthermore, when $a_1 = \cdots = a_n = 3$ and $a_{n+1} = 2$, our computations show that the Motzkin numbers count the elements of each degree in the initial ideal of $(x_1^3, \ldots, x_n^3, (x_1 + \cdots + x_n)^2)$, with these numbers characterized by certain lattice paths.


\medskip

In the context of minimal free resolutions, we note that Schreyer's construction~\cite{Schreyer1980} provides a minimal free resolution for the monomial ideals of $\ini(I_{n,k})$. However, we have excluded this construction here, as it closely parallels the resolution obtained via the Mayer--Vietoris construction. We expect that this property extends to other ideals of the form $I_{a_1, \ldots, a_{n+1}}$.

\medskip
Moreover, since the resolutions of the ideals $\ini(I_{n,k})$ are given by mapping cones, we anticipate that these resolutions will be supported on a CW-complex, as discussed in~\cite{dochtermann2014cellular}. This will be explored in future work.


%\acknowledgements{
%Experimental computations in Macaulay2 have been crucial for our work, which  started during a workshop on Gr\"obner free methods held at Universidad de La Rioja (Logro\~no, Spain). We thank the local organizers for their hospitality and Victor Ufnarovski for fruitful discussions in the beginning of the project. We also thank KU Leuven for hosting us for one week during the middle part of the project. We furthermore thank Mats Boij for providing insightful comments during talks given by the first two authors in the latter part of the project. Samuel Lundqvist was supported by the Swedish Research Council grant VR2022-04009. Fatemeh Mohammadi was partially supported by the grants G0F5921N (Odysseus programme) and G023721N from the Research Foundation - Flanders (FWO), the UiT Aurora project MASCOT and the KU~Leuven~iBOF/23/064~grant. }
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\lipsum[3]

\begin{equation}\label{eqn:eq1}
\int_0^\infty \exp(-x) \, \mathrm{d}x = 1.
\end{equation}

%\begin{lem}\label{lem:lem1}
%\lipsum[11]
%\end{lem}

\lipsum[10]

\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{plot}
\caption{A plot of a function.}
\label{fig:plot}
\end{figure}

\subsection{A subsection}

\lipsum[5]

\begin{align*}
1 + 1 + 1 &= 2 + 1 \\ &= 3.
\end{align*}

\subsubsection{A subsubsection}

\lipsum[17]

Here is a citation~\cite{greenwade93} with URL. Here is another
citation~\cite{MR4}, which is the earliest-numbered Math Review with a
functioning DOI: MR0000004. Finally, a reference to
equation~\eqref{eqn:eq1}. And a reference to \Cref{fig:plot}.

\acknowledgements{\lipsum[20]}

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