Proof. Let \(\mathcal{G}(I)=\{u_1,\dots,u_m\}\) with \(u_i={\boldsymbol{x}}^{{\boldsymbol{a}}_i}\) for each \(i\). Then \(\partial(I)=\sum_{i=1}^mI_{n,{\boldsymbol{a}}_i,|{\boldsymbol{a}}_i|-1}\). Using [5] (see also [6]) we obtain that \[\begin{align} \mathrm{reg}\,\partial(I)\;&\le\;\sum_{i=1}^m\mathrm{reg}(I_{n,{\boldsymbol{a}}_i,|{\boldsymbol{a}}_i|-1})-(m-1)\;=\;\sum_{i=1}^m(|{\boldsymbol{a}}_i|-1)-m+1\\ &=\;\sum_{i=1}^m\deg(u_i)-2m+1. \end{align}\] If \(I\) is a complete intersection, then \(I_{n,{\boldsymbol{a}}_i,|{\boldsymbol{a}}_i|-1}\) are monomial ideals generated in pairwise disjoint sets of variables, and the above inequalities become equalities. ◻

Proof. Let \(a\le-1\). Set \(b=2-a\ge1\), and let \[I\;=\;(x_1x_2,\,x_1x_3^b,\,x_2x_4^b)\subset S=K[x_1,x_2,x_3,x_4].\] Notice that \((x_1x_2):(x_1x_3^b)=(x_2)\) and \((x_1x_2,x_1x_3^b):(x_2x_4^b)=(x_1)\). Hence, \(I\) has linear quotients and \(\mathrm{reg}\,I=\omega(I)=b+1\). Notice that \(\partial(I)=(x_1,x_2,x_3^b,x_4^b)\) is a complete intersection. So \(\mathrm{reg}\,\partial(I)=2b-1\) and \(\mathrm{reg}\,I-\mathrm{reg}\,\partial(I)=2-b=a\le-1\).

For \(a=0\), let \[I\;=\;(x_1x_2x_3,\,x_2x_3x_4,\,x_3x_4x_5,\,x_4x_5x_6)\subset S=K[x_1,\dots,x_6].\]

Then \(I\) has linear quotients for the generators in the given order. Hence \(\mathrm{reg}\,I=3\). We have \(\mathrm{reg}\,\partial(I)=3\) as well, as one can check with Macaulay2 [8]. This is also proved in the next Theorem 4. Hence \(\mathrm{reg}\,I-\mathrm{reg}\,\partial(I)=a=0\).

Now let \(a\ge1\). Put \(x=x_1\) and \(y=x_2\). Let \(b\) and \(c\) integers \(\ge2\) such that \(b-c+1=a\). Then \(b\ge c\ge2\). We put \[I\;=\;(x^c,\,x^{c-1}y,\,\dots,\,x^{c-i}y^{i},\,\dots,\,xy^{c-1},\,y^b)\subset S=K[x,y].\]

Using [9] we have \[\begin{align} \mathrm{reg}\,I\;&=\;\max\Bigg\{\substack{\displaystyle\max\{\deg(x^{c-i}y^i),\,\deg(y^b):0\le i\le c-1\}\\[3pt] \displaystyle\max\{\deg(x^{c-j}y^{j+1})-1,\,\deg(xy^{b})-1:0\le j\le c-2\}}\Bigg\}\\[3pt] &=\;\max\{\max\{c,\,b\},\,\max\{c+1-1,\,b+1-1\}\}\;=\;b. \end{align}\]

Since \(b\ge c\), we have \(\partial(I)=(x,y)^{c-1}\) and so \(\mathrm{reg}\,\partial(I)=c-1\). Finally, we have \(\mathrm{reg}\,I-\mathrm{reg}\,\partial(I)=b-(c-1)=a\), as desired. ◻

Proof. If \(a=-1\), it is enough to consider any monomial ideal \(I\) with \(2\)-linear resolution. Then \(\partial(I)\) is a monomial prime ideal and so \(\mathrm{reg}\,\partial(I)=1\).

Now, let \(a\ge0\). Put \(d=a+3\) and let \[I\;=\;\sum_{i=1}^{d+1}(\prod_{j=0}^{d-1}x_{i+j})\;=\;(\prod_{j=0}^{d-1}x_{i+j}:\;i=1,\dots,d+1)\subset S=K[x_1,\dots,x_{2d}].\]

We claim that \(\mathrm{reg}\,I=d\) and \(\mathrm{reg}\,\partial(I)=2d-3\). Then \(\mathrm{reg}\,\partial(I)-\mathrm{reg}\,I=d-3=a\) and the statement follows.

To prove that \(\mathrm{reg}\,I=d\), we put \(u_i=x_{i}x_{i+1}\cdots x_{i+d-1}\) for all \(i=1,\dots,d+1\). Then \(I=(u_1,\dots,u_{d+1})\) and \((u_1,\dots,u_{i-1}):(u_i)=(x_{i-1})\) for \(i=2,\dots,d+1\). Hence \(I\) has linear quotients and so \(\mathrm{reg}\,I=\alpha(I)=d\).

To prove that \(\mathrm{reg}\,\partial(I)=2d-3\), we use Hochster’s formula. Let \(\Delta\) be the simplicial complex on vertex set \([n]\) with Stanley-Reisner ideal \(I_\Delta=\partial(I)\). From Hochster’s formula ([7]), we have \[\label{eq:Hochster} \beta_{i,j}(S/\partial(I))\;=\;\sum_{\substack{W\subset[n]\\ |W|=j}}\dim_K\widetilde{H}_{j-i-1}(\Delta_W;K),\tag{1}\] where \(\Delta_W\) is the restriction of \(\Delta\) to \(W\), and \(\widetilde{H}_s(\Gamma;K)=\mathrm{Ker}(\partial_s)/\operatorname{Im}(\partial_{s+1})\) is the \(s\)-th reduced simplicial homology of the simplicial complex \(\Gamma\) computed over \(K\). Here \(\partial_s:C_s(\Delta;K)\rightarrow C_{s-1}(\Delta;K)\) is the \(K\)-linear map defined as follows: \(C_s(\Delta;K)\) is the \(K\)-vector space with basis given by the symbols \({\boldsymbol{e}}_F\) with \(F\in\Delta\) of dimension \(s-1\), and \(\partial_s\) is defined by setting \[\partial_s({\boldsymbol{e}}_F)\;=\;\sum_{p\in F}(-1)^{\mathrm{sgn}(p;F)}{\boldsymbol{e}}_{F\setminus\{p\}},\] with \(\mathrm{sgn}(p;F)=|\{q\in F:\;q<p\}|\). See also [7].

First we show that \(\mathrm{reg}\,\partial(I)\le 2d-3\). Suppose otherwise that \(\mathrm{reg}\,\partial(I)\ge2d-2\). Since \(\mathrm{reg}\,S/I=\mathrm{reg}\,I-1\), then there would exist integers \(j,i\) such that \(j-i\ge2d-3\) and \(\widetilde{H}_{j-i-1}(\Delta_W;K)\ne0\) for some subset \(W\subset[n]\) with \(|W|=j\). This implies that \(\dim(\Delta)\ge2d-3\). We show that this is impossible. To this end, let \(H\subset[2d]\) be any subset with \(|H|=2d-2\). We show that \(H\notin\Delta\). This will give the desired contradiction. We can write \(H=[2d]\setminus\{p,q\}\) with \(p<q\). If \(p\ge d\) then \(H\) contains \(H'=\{1,\dots,d-1\}\), and since \({\boldsymbol{x}}_{H'}=x_1x_2\cdots x_{d-1}=(x_1x_2\cdots x_d)/x_d\in\partial(I)\), it follows that \(H'\notin\Delta\), and consequently \(H\notin\Delta\). Similarly, if \(q\le d+1\) then \(H\) contains \(H''=\{d+2,d+3,\dots,2d\}\notin\Delta\) and so \(H\notin\Delta\). We may therefore assume that \(p\le d-1\) and \(q\ge d+2\). Hence \(d,d+1\in H\). But then \(H\) contains \(H'''=\{1,\dots,d\}\setminus\{p\}\) and \({\boldsymbol{x}}_{H'''}=(x_1\cdots x_d)/x_p\in\partial(I)\). Therefore \(H\notin\Delta\) and this proves our claim.

Hence, \(\mathrm{reg}\,\partial(I)\le 2d-3\). We claim that \(\beta_{2,2d-2}(S/\partial(I))\ne0\). This will imply that \(\mathrm{reg}\,\partial(I)=2d-3\) and conclude the proof. Let \[W=\{1,2,\dots,d-1,d+2,d+3,\dots,2d\}.\]

We claim that \(\widetilde{H}_{|W|-3}(\Delta_W;K)=\widetilde{H}_{2d-5}(\Gamma;K)\ne0\) with \(\Gamma=\Delta_W\). Using (1 ) this will show that \(\beta_{2,2d-2}(S/\partial(I))\ne0\) and conclude the proof.

We have \(\widetilde{H}_{2d-5}(\Gamma;K)=\mathrm{Ker}(\partial_{2d-5})/\operatorname{Im}(\partial_{2d-4})\) where \(\partial_{s}:C_{s}(\Gamma;K)\rightarrow C_{s-1}(\Gamma;K)\) is defined as before. We claim that \(C_{2d-4}(\Gamma;K)=0\). Let \(H\) be any subset of \(W\) of size \(2d-3\). Then \(H=W\setminus\{p\}\) for some \(p\). If \(p\in\{1,2,\dots,d-1\}\), then \(H\) contains \(H'=\{d+2,d+3\dots,2d\}\notin\Delta\), and so \(H\notin\Gamma\). Otherwise \(p\in\{d+2,d+3,\dots,2d\}\) and by symmetry \(H\notin\Gamma\). Therefore, \(\widetilde{H}_{2d-5}(\Gamma;K)=\mathrm{Ker}(\partial_{2d-5})\).

Next, notice that the elements of the form \[W\setminus\{p,q\}\;=\;(\{1,2,\dots,d-1\}\setminus\{p\})\cup(\{d+2,d+3,\dots,2d\}\setminus\{q\})\] with \(p\in\{1,\dots,d-1\}\) and \(q\in\{d+2,\dots,2d\}\) are all in \(\Gamma\) and are of dimension \(2d-5\). Suppose otherwise that \(W\setminus\{p,q\}\notin\Delta\), then \({\boldsymbol{x}}_{W\setminus\{p,q\}}\in\partial(I)\). This implies that \(W\setminus\{p,q\}\) should contain a set of the form \(\{r,r+1,\dots,r+(d-1)\}\setminus\{s\}\) for some \(1\le r\le d+1\) and some \(s=r+h\) with \(0\le h\le d-1\). However this is easily seen to be impossible. It follows that \(W\setminus\{p,q\}\in\Delta\).

Next, we claim that the element \[z\;=\;\sum_{\substack{1\le p\le d-1\\[2pt] d+2\le q\le 2d}}(-1)^{p+q}\,{\boldsymbol{e}}_{W\setminus\{p,q\}}\in C_{2d-5}(\Gamma;K)\] is a cycle, that is \(\partial_{2d-5}(z)=0\). Then \(\mathrm{Ker}(\partial_{2d-5})\ne0\) and this will conclude the proof. To this end, we can write \[\partial_{2d-5}(z)\;=\;\sum_{\substack{1\le p<q\le d-1\\[2pt] d+2\le r\le 2d\\}}c_{pqr}\,{\boldsymbol{e}}_{W\setminus\{p,q,r\}}+\sum_{\substack{1\le p\le d-1\\[2pt] d+2\le q<r\le 2d\\}}c_{pqr}\,{\boldsymbol{e}}_{W\setminus\{p,q,r\}},\] with \(c_{pqr}\in K\).

We must prove that \(c_{pqr}=0\) for all \(p,q,r\). There are two cases to consider. We verify only one, and leave the other to the reader. Let \(\{p<q<r\}\subset W\) with \(p,q\in\{1,2,\dots,d-1\}\) and \(r\in\{d+2,d+3,\dots,2d\}\). Then, in \(z\) there are only two summands \((-1)^{p+r}{\boldsymbol{e}}_{W\setminus\{p,r\}}\) and \((-1)^{q+r}{\boldsymbol{e}}_{W\setminus\{q,r\}}\) whose derivation contributes to the sign of \({\boldsymbol{e}}_{W\setminus\{p,q,r\}}\) in the expansion of \(\partial_{2d-5}(z)\). Since \(p<q\), we have \[\begin{align} c_{pqr}\;&=\;(-1)^{p+r}(-1)^{\mathrm{sgn}(W\setminus\{p,r\};q)}+(-1)^{q+r}(-1)^{\mathrm{sgn}(W\setminus\{q,r\};p)}\\ &=\;(-1)^{p+r}(-1)^{|\{s\in W\setminus\{p,r\}\,:\,s<q\}|}+(-1)^{q+r}(-1)^{|\{s\in W\setminus\{q,r\}\,:\,s<p\}|}\\ &=\;(-1)^{p+r}(-1)^{q-2}+(-1)^{q+r}(-1)^{p-1}\\ &=\;(-1)^{p+q+r}(1-1)\;=\;0, \end{align}\] as desired. ◻

Proof. It is clear that \(\partial(\mathfrak{m}^k)=\mathfrak{m}^{k-1}\) and \(I\subset\mathfrak{m}\,\partial(I)\) for all integers \(k\ge1\). Hence \(\mathfrak{m}^{k-1}I\subset\mathfrak{m}^k\partial(I)\). Using Leibniz’s rule which is valid for the derivation of the product of monomial ideals, we have \[\partial(\mathfrak{m}^kI)\;=\;\partial(\mathfrak{m}^k)I+\mathfrak{m}^k\partial(I)\;=\;\mathfrak{m}^{k-1}I+\mathfrak{m}^k\partial(I)\;=\;\mathfrak{m}^k\partial(I)\] for all \(k\ge1\). It is proved in [10] that \(\mathfrak{m}^kJ\) has linear quotients for any monomial ideal \(J\subset S\) and all \(k\gg0\). Hence \(\partial(\mathfrak{m}^kI)=\mathfrak{m}^k\partial(I)\) has linear quotients for all \(k\gg0\). For a monomial ideal \(J\subset S\) with linear quotients its regularity is \(\omega(J)\). Since we obviously have \(\omega(\partial(\mathfrak{m}^kI))<\omega(\mathfrak{m}^kI)\), we conclude that \(\mathrm{reg}\,\partial(\mathfrak{m}^kI)<\mathrm{reg}(\mathfrak{m}^kI)\) for all \(k\gg0\). ◻

Proof. Let \(P\) be the discrete polymatroid on \([n]\) with \(B(P)=\{{\boldsymbol{u}}:{\boldsymbol{x}^u}\in\mathcal{G}(I)\}\) as the set of bases of \(P\). We have \(\mathrm{rank}\,P=\alpha(I)=d\). Let \(Q=\{{\boldsymbol{u}}\in P:|{\boldsymbol{u}}|\le d-1\}\). By [7], \(Q\) is a discrete polymatroid as well. Finally, it is clear that \(B(Q)=\{{\boldsymbol{u}}:{\boldsymbol{x}^u}\in\mathcal{G}(\partial(I))\}\), and this concludes the proof. ◻

Proof. As noticed before, we can write \(\partial(I)=\sum_{i\in[n]}(I:x_i)\). Then \[\partial(I)_{\langle j\rangle}\;=\;(\sum_{i\in[n]}(I:x_i))_{\langle j\rangle}\;=\;\sum_{i\in[n]}(I:x_i)_{\langle j\rangle}\;=\;\sum_{i\in[n]}(I_{\langle j+1\rangle}:x_i)\;=\;\partial(I_{\langle j+1\rangle}),\] as desired. ◻

Proof. We prove that if \(I\) is stable, then \(\partial(I)\) is stable as well. The proof in the case that \(I\) is strongly stable is analogous. Let \(u\in\partial(I)\) be a monomial and let \(i<\max(u)\). We must prove that \(x_i(u/x_{\max(u)})\in\partial(I)\). Since \(u\in\partial(I)\), there exists \(j\) such that \(v=x_ju\in I\). We distinguish the two possible cases.

Case 1. Let \(j<\max(v)\). Then \(i<\max(u)=\max(v)\). Since \(I\) is stable, it follows that \(x_i(v/x_{\max(v)})\in I\). Since \(j<\max(v)\) we have that \(x_j\) divides \(x_i(v/x_{\max(v)})\). Hence \((x_i(v/x_{\max(v)}))/x_j=(x_i(v/x_j))/x_{\max(u)}=x_i(u/x_{\max(u)})\in\partial(I)\), as desired.

Case 2. Let \(j=\max(v)\). Then \(i<\max(u)<j=\max(v)\). Since \(I\) is stable, we have that \(x_i(v/x_{\max(v)})=x_i(v/x_j)=x_iu\in I\). Using that \(x_{\max(u)}<j\), it follows that \(x_{\max(u)}\) divides \(x_iu\in I\). Hence \(x_i(u/x_{\max(u)})\in\partial(I)\), as desired. ◻

Proof. It is enough to prove the statement in the case \(\ell=1\). Using Leibniz’s rule which is valid for the derivation of the product of monomial ideals, we have that \(\partial(x_iI)=x_i\partial(I)+I\). Since \(i\notin\mathrm{supp}(I)\), \(I\subset\partial(I)\) and both ideals are vertex splittable it follows that \(\partial(x_iI)\) is vertex splittable too. ◻

Proof of Theorem 12. We can write \(I=I_c(G)\) where \(G\) is a finite simple graph on the vertex set \([n]\). We may assume that \(G\) does not contain isolated vertices. In fact, assume that \(n\) is an isolated vertex of \(G\). Then \(I=x_nJ\) where \(J=I_c(H)\) has \((n-3)\)-linear resolution and \(H=G\setminus\{n\}\) is the graph on vertex set \([n-1]\). If the statement holds for \(J\), by Lemma 13 it holds for \(I\) as well. Therefore, we may assume that \(G\) does not contain isolated vertices. By [14] it follows that \(G\) is connected. Let \(n,n-1,\dots,1\) be a labeling of the vertices of \(G\) such that \(G\setminus\{n,n-1\dots,n-i\}\) is connected for all \(i=1,\dots,n\). Then \[I\;=\;x_nJ+L\] where \(J=I_c(G_n)\) and \(L=({\boldsymbol{x}}_{[n]}/(x_ix_n):\;\{i,n\}\in E(G)).\)

Since \(G_n\) is connected on \(n-1\) vertices, by induction \(I_c(G_n)\) is vertex splittable. Notice that since \(L\) can be regarded as a squarefree monomial ideal in \(K[x_1,\dots,x_{n-1}]\) generated in degree \(n-2\), it is polymatroidal and therefore it is vertex splittable. We claim that \(L\subset J\). In fact, let \(u={\boldsymbol{x}}_{[n]}/(x_ix_n)\in\mathcal{G}(L)\). Since \(G_n\) is connected, we can find an edge \(\{i,j\}\in E(G_n)\). Then \({\boldsymbol{x}}_{[n-1]}/(x_ix_j)\in I_c(G_n)\) divides \(u\). Hence \(u\in I_c(G_n)\) and so \(L\subset I_c(G_n)\). We conclude that \(\partial^0(I)=I=x_nI_c(G_n)+L\) is in fact vertex splittable.

We have \[\partial(I)\;=\;\partial(x_1J+L)\;=\;x_1\partial(J)+(J+\partial(L)).\]

Let \(H\) be the graph on vertex set \([n-1]\) with edge set \[E(H)\;=\;E(G_n)\cup\{\{i,j\}:\;\{i,n\}\in E(G),\;j\in[n-1]\setminus\{i\}\}.\]

We claim that

1. \(J+\partial(L)=I_c(H)\), and

2. \(H\) is connected.

Since \(G_n\) is connected and \(E(H)\) contains \(E(G_n)\) it follows that \(H\) is connected too and statement (b) holds. Statement (a) follows because \[\begin{align} \mathcal{G}(J&+\partial(L))=\mathcal{G}(J)\cup\mathcal{G}(\partial(L))\\ &=\{{\boldsymbol{x}}_{[n-1]\setminus\{i,j\}}:\{i,j\}\in E(G_n)\}\cup\{{\boldsymbol{x}}_{[n]\setminus\{i,n\}}/x_j:\{i,n\}\in E(G),j\ne i,j\ne n\}\\ &=\{{\boldsymbol{x}}_{[n-1]\setminus\{i,j\}}:\{i,j\}\in E(G_n)\}\cup\{{\boldsymbol{x}}_{[n-1]\setminus\{i,j\}}:\{i,n\}\in E(G),j\in[n-1]\setminus\{i\}\}\\ &=\{{\boldsymbol{x}}_{[n-1]\setminus\{p,q\}}:\{p,q\}\in E(H)\}=\mathcal{G}(I_c(H)). \end{align}\]

Now, let \(\ell\in[n-2]\). A quick calculation shows that \[\partial^\ell(I)\;=\;x_1\partial^{\ell}(J)+(\partial^{\ell-1}(J)+\partial^\ell(L)).\]

Since we have \(I_c(H)=J+\partial(L)\), and so \(\partial^{\ell-1}(I_c(H))=\partial^{\ell-1}(J)+\partial^\ell(J)\), the previous formula becomes \[\label{eq:vs-ind} \partial^\ell(I)\;=\;x_1\partial^{\ell}(J)+\partial^{\ell-1}(I_c(H)).\tag{2}\]

Now, we claim that \(\partial^{\ell-1}(I_c(H))\subset\partial^\ell(J)\). Since \(\partial^{\ell-1}(I_c(H))=\partial^{\ell-1}(J)+\partial^\ell(L)\) and obviously \(\partial^{\ell-1}(J)\subset\partial^\ell(J)\), it remains to prove that \(\partial^\ell(L)\) is contained in \(\partial^{\ell}(J)\).

Let \(u={\boldsymbol{x}}_{[n-1]}/(x_px_{q_1}\cdots x_{q_\ell})\in\mathcal{G}(\partial^\ell(L))\) be a squarefree monomial generator with \(\{p,n\}\in E(G)\). Since \(G_n\) is connected, there exists an edge \(\{p,r\}\in E(G_n)\). We distinguish the two possible cases.

Case 1. Let \(r=q_i\) for some \(i\). Say \(i=1\). Then \(v={\boldsymbol{x}}_{[n-1]}/(x_px_{q_1})\in I_c(H)\) and consequently \(u=v/(x_{q_2}\cdots x_{q_{\ell}})\in\partial^{\ell-1}(J)\subset\partial^\ell(J)\), as desired.

Case 2. Let \(r\ne q_i\) for all \(i\). Then \(u/x_r=({\boldsymbol{x}}_{[n-1]}/(x_px_r))/(x_{q_1}\cdots x_{q_\ell})\in\partial^\ell(J)\). This shows that \(u\in\partial^\ell(J)\), as desired.

Hence \(\partial^{\ell-1}(I_c(H))\subset\partial^\ell(J)\). By induction on \(|V(G)|\), \(J=I_c(G_n)\) and \(I_c(H)\) satisfy the statement because the graphs \(G_n\) and \(H\) are connected on a smaller number of vertices than \(G\). Therefore \(\partial^\ell(J)\) and \(\partial^{\ell-1}(I_c(H))\) are vertex splittable. Equation (2 ) implies that \(\partial^\ell(I)\) is indeed vertex splittable. ◻

Proof. We may assume that \(d\ge3\) otherwise the statement holds vacuously.

Let \(\mathcal{M}=\mathcal{G}(I)\). Recall that the shadow of \(\mathcal{M}\) is defined as the set \[\partial(\mathcal{M})\;=\;\{u/x_i:\;u\in\mathcal{M},\,i\in\mathrm{supp}(u)\}.\]

Then \(\partial(I)\) is the ideal generated by \(\partial(\mathcal{M})\). Let \(|\mathcal{M}|=a=\binom{a_d}{d}+\binom{a_{d-1}}{d-1}+\dots+\binom{a_t}{t}\) be the \(d\)-th binomial expansion of \(|\mathcal{M}|\). This expression is uniquely determined by the inequalities \(a_d>a_{d-1}>\dots>a_t\ge t\), see [7].

By the Kruskal-Katona’s theorem ([20], [21]) we have \[|\partial(\mathcal{M})|\;\ge\;a^{(d-1)}\;=\;\binom{a_d}{d-1}+\binom{a_{d-1}}{d-2}+\dots+\binom{a_t}{t-1}.\]

If we show that \(a^{(d-1)}\ge\binom{n}{d-1}-2(d-1)+1\), then applying again [18] (or [19]), it follows that \(\partial(I)\) is vertex splittable. Since the assumptions are again satisfied for \(\partial(I)\), it is enough to prove the statement for \(I=\partial^0(I)\). Hence, it remains to prove that \(a^{(d-1)}\ge\binom{n}{d-1}-2(d-1)+1\).

The operator \(b\mapsto b^{(d-1)}\) is an increasing function. Therefore, we may also assume that \(|\mathcal{M}|=a=\binom{n}{d}-2d+1\) and it is enough to show that \[a^{(d-1)}\;\ge\;\binom{n}{d-1}-2(d-1)+1.\]

To this end, we claim that the \(d\)-th binomial expansion of \(a\) is: \[\label{eq:Macaulay} a\;=\;\sum_{i=1}^{d-2}\binom{n-i}{d-i+1}+\begin{cases} \binom{n-d}{2}+\binom{2n-4d+2}{1}&\mathrm{if}\;2d\le n\le 3d-3,\\[3pt] \binom{n-d+1}{2}&\mathrm{if}\;n=3d-2,\\[3pt] \binom{n-d+1}{2}+\binom{n-3d+2}{1}&\mathrm{if}\;n\ge 3d-1. \end{cases}\tag{3}\]

Using Pascal’s rule, we have \(\binom{n}{d}=\binom{n-1}{d}+\binom{n-1}{d-1}\). Hence, \[a\;=\;\binom{n}{d}-2d+1\;=\;\binom{n-1}{d}+\binom{n-1}{d-1}-2d+1.\]

Applying again Pascal’s rule to \(\binom{n-1}{d-1}\) and iterating this procedure yields that \[\label{eq:a} a\;=\;\binom{n-1}{d}+\binom{n-2}{d-1}+\dots+\binom{n-d+2}{3}+\binom{n-d+2}{2}-2d+1.\tag{4}\]

Since \(n\ge 2d\), we have \(\binom{n-d+2}{2}\ge\binom{d+2}{2}=(d+2)(d+1)/2>2d-1\). Hence \(b=\binom{n-d+2}{2}-2d+1>0\). Now we distinguish two cases.

Case 1. Let \(2d\le n\le 3d-3\). Notice that \[\label{eq:b} \begin{align} b-\binom{n-d}{2}\;&=\;\binom{n-d+2}{2}-\binom{n-d}{2}-2d+1\\ &=\;\frac{(n-d+2)(n-d+1)-(n-d)(n-d-1)-4d+2}{2}\\ &=\;\frac{4n-8d+4}{2}\;= \binom{2n-4d+2}{1}\;\ge\;1, \end{align}\tag{5}\] in the given range of values of \(n\). Combining (5 ) with (4 ) we obtain the expression given in the first case of equation (3 ). Such expression is indeed the \(d\)-th binomial expansion of \(a\), because \[n-1>n-2>\dots>n-d+2>n-d>2n-4d+2\ge1,\] where the second to last inequality follows because \(n\le 3d-3\).

Case 2. Let \(n\ge 3d-2\). Using Pascal’s rule we have \[\label{eq:b1} \begin{align} b-\binom{n-d+1}{2}\;&=\;\binom{n-d+2}{2}-\binom{n-d+1}{2}-2d+1\\ &=\;\binom{n-d+1}{1}-2d+1\;=\;n-3d+2\;\ge\;0. \end{align}\tag{6}\] Combining equation (6 ) with (4 ), and distinguish between the cases \(n=3d-2\) and \(n\ge3d-1\), we obtain the expressions given in the second and third cases of equation (3 ). These expressions are indeed the \(d\)-th binomial expansion of \(a\), because \[n-1>n-2>\dots>n-d+2>n-d+1\ge 2,\] and in addition we have \(n-d+1>n-3d+2\ge1\) if \(n\ge 3d-1\).

Having acquired (3 ), using that \(\binom{N}{0}=1\) and \(\binom{N}{1}=N\) for all \(N\ge0\), we obtain that \[\label{eq:ad-1} a^{(d-1)}\;=\;\sum_{i=1}^{d-2}\binom{n-i}{d-i}+\begin{cases} \,n-d+1&\mathrm{if}\;2d\le n\le 3d-2,\\ \,n-d+2&\mathrm{if}\;n\ge 3d-1. \end{cases}\tag{7}\]

Iterations of Pascal’s rule yield that \(\binom{n}{d-1}=\sum_{j=1}^{d}\binom{n-j}{d-j}\). Therefore, \[\begin{align} \sum_{i=1}^{d-2}\binom{n-i}{d-i}\;&=\;\binom{n}{d-1}-\binom{n-(d-1)}{1}-\binom{n-d}{0}\\ &=\;\binom{n}{d-1}-n+d-2. \end{align}\]

Combining this fact with (7 ) we obtain that \[\begin{align} a^{(d-1)}\;&=\;\begin{cases} \binom{n}{d-1}-1&\mathrm{if}\;2d\le n\le 3d-2,\\[3pt] \binom{n}{d-1}&\mathrm{if}\;n\ge 3d-1. \end{cases} \end{align}\] Hence, \(a^{(d-1)}\ge\binom{n}{d-1}-2(d-1)+1\), as desired. ◻

Proof. For the proof of this result, we use some results from [22] and [23].

Let \(I\subset S\) be a quadratic squarefree monomial ideal with linear resolution. Without loss of generality, we can assume that \(I\) is fully-supported, that is \(\mathrm{supp}(I)=[n]\). Up to a suitable relabeling on the variables, by [23] we can write \(I=x_nP+J\), for monomial ideals \(P,J\subset S\) satisfying the properties:

1. \(n\notin\mathrm{supp}(P)\cup\mathrm{supp}(J)\),

2. \(P\) is a monomial prime ideal and \(J\subset P\),

3. \(P^rJ^s\) has linear quotients, and so linear resolution, for all \(r,s\ge0\).

Now, let \(k\ge1\). Then \(I^k=\sum_{r=0}^kx_n^{k-r}P^{k-r}J^r\). If \(\ell=0\), then \(\partial^0(I^k)=I^k\) has linear quotients by [22] (see also [23]).

Next, let \(\ell\ge1\). We claim that \[\label{eq:partialI40G41} \partial^\ell(I^k)\;=\;\begin{cases} \displaystyle\,(\sum_{r=0}^\ell x_n^{k-r}\mathfrak{n}^rP^{k-\ell})+(\sum_{s=\ell+1}^kx_n^{k-s}\mathfrak{n}^\ell P^{k-s}J^{s-\ell})&\mathrm{for}\;1\le\ell\le k-1,\\[2pt] \,\,\,\mathfrak{m}^{2k-\ell}&\mathrm{for}\;k\le\ell\le 2k, \end{cases}\tag{8}\] where \(\mathfrak{n}=(x_1,\dots,x_{n-1})\) and \(\mathfrak{m}=(\mathfrak{n},x_n)=(x_1,\dots,x_{n-1},x_{n})\).

For \(\ell=1\), we have \[\label{eq:HHZ1} \begin{align} \partial(I^k)\;&=\;\partial(\sum_{s=0}^kx_n^{k-s}P^{k-s}J^s)\;=\;\sum_{s=0}^k\partial(x_n^{k-s}P^{k-s}J^s)\\ &=\;\sum_{s=0}^k\partial(x_n^{k-s})P^{k-s}J^s+\sum_{s=0}^kx_n^{k-s}\partial(P^{k-s}J^{s}). \end{align}\tag{9}\]

Next, we compute both sums appearing in the above expression. We have \[\label{eq:HHZ2} \sum_{s=0}^k\partial(x_n^{k-s})P^{k-s}J^s= \sum_{s=0}^{k-1}x_{n}^{k-s-1}P^{k-s}J^s= \sum_{s=1}^kx_n^{k-s}P^{k+1-s}J^{s-1}.\tag{10}\]

For the second sum, we use that \(\partial(P^{r})=P^{r-1}\) for \(r\ge1\). Hence \[\label{eq:HHZ3} \begin{align} \sum_{s=0}^kx_n^{k-s}\partial(P^{k-s}J^s)\;&=\;\sum_{s=0}^kx_n^{k-s}[\partial(P^{k-s})J^s+P^{k-s}\partial(J^s)]\\ &=\;\sum_{s=0}^kx_n^{k-s}[P^{k-s-1}J^r+P^{k-s}J^{s-1}\partial(J)]. \end{align}\tag{11}\]

Combining equations (9 ), (10 ) and (11 ) we obtain \[\begin{align}\label{eq:HHZ4} \partial(I^k)\;&=\;x_n^kP^{k-1}+\sum_{s=1}^kx_n^{k-s}[P^{k+1-s}J^{s-1}+P^{k-s-1}J^s+P^{k-s}J^{s-1}\partial(J)]\\ &=\;x_n^kP^{k-1}+\sum_{s=1}^kx_n^{k-s}P^{k-s-1}J^{s-1}[P^{2}+J+P\partial(J)]. \end{align}\tag{12}\]

We claim that \[\label{eq:HHZ5} P^2+J+P\partial(J)\;=\;P(P+\partial(J)).\tag{13}\]

Clearly \(P(P+\partial(J))\subset P^2+J+P\partial(J)\). Conversely, let \(u\in\mathcal{G}(P^2+J+P\partial(J))\). If \(u\in P^2+P\partial(J)\) then \(u\in\mathcal{G}(P(P+\partial(J)))\). Else, if \(u\in\mathcal{G}(J)\), using (ii) we have \(u=x_px_q\) with \(x_p\in P\). But \(x_q\in\partial(J)\), hence \(u\in\mathcal{G}(P\partial(J))\subset\mathcal{G}(P(P+\partial(J)))\), as desired.

Since \(I\) is fully-supported, and \(n\notin\mathrm{supp}(P)\cup\mathrm{supp}(J)\) by property (i), we obtain that \(P+\partial(J)=\mathfrak{n}=(x_1,\dots,x_{n-1})\). Combining equations (12 ) and (13 ) we obtain \[\label{eq:HHZ6} \partial(I^k)\;=\;x_{n}^{k}P^{k-1}+x_{n}^{k-1}\mathfrak{n}P^{k-1}+\sum_{s=2}^kx_n^{k-s}\mathfrak{n}P^{k-s}J^{s-1}.\tag{14}\] This is the expression given in equation (8 ) for \(\ell=1\).

Now, let \(1<\ell\le k-1\), and assume by finite induction that the formula for \(\partial^\ell(I^k)\) given in equation (8 ) is correct. Then, \[\label{eq:partial-ell} \partial^{\ell+1}(I^k)\;=\;\partial(\partial^\ell(I^k))\;=\;\partial((\sum_{r=0}^\ell x_n^{k-r}\mathfrak{n}^rP^{k-\ell})+(\sum_{s=\ell+1}^kx_n^{k-s}\mathfrak{n}^\ell P^{k-s}J^{s-\ell})).\tag{15}\]

For the first sum in (15 ) we have \[\begin{align} \partial(\sum_{r=0}^\ell&x_n^{k-r}\mathfrak{n}^rP^{k-\ell})\;=\\ &=\;x_n^kP^{k-(\ell+1)}\!+\!x_n^{k-1}(P^{k-\ell}+\partial(\mathfrak{n}P^{k-\ell}))\!+\ldots+\!x_n^{k-\ell}(\mathfrak{n}^{\ell-1} P^{k-\ell}\!+\!\partial(\mathfrak{n}^\ell P^{k-\ell}))\\ &+\;x_n^{k-(\ell+1)}\mathfrak{n}^\ell P^{k-\ell}. \end{align}\]

Using that \(P\subset\mathfrak{n}\), we have \[\mathfrak{n}^{r-1}P^{k-\ell}+\partial(\mathfrak{n}^{r}P^{k-\ell})\;= \;\mathfrak{n}^{r-1}P^{k-\ell}+\mathfrak{n}^{r-1}P^{k-\ell}+\mathfrak{n}^{r}P^{k-(\ell+1)}\;=\;\mathfrak{n}^{r}P^{k-(\ell+1)}.\]

Hence \[\label{eq:partialnn} \partial(\sum_{r=0}^\ell x_n^{k-r}\mathfrak{n}^rP^{k-\ell})\;=\;\sum_{r=0}^\ell x_n^{k-r}\mathfrak{n}^r P^{k-(\ell+1)}+x_n^{k-(\ell+1)}\mathfrak{n}^\ell P^{k-\ell}.\tag{16}\]

For the second sum in (15 ) we have \[\begin{align} \partial(\sum_{s=\ell+1}^k&x_n^{k-s}\mathfrak{n}^\ell P^{k-s}J^{s-\ell})\;=\;x_n^{k-(\ell+1)}\partial(\mathfrak{n}^\ell P^{k-(\ell+1)}J)\\ &+x_n^{k-(\ell+2)}(\mathfrak{n}^\ell P^{k-(\ell+1)}J\!+\!\partial(\mathfrak{n}^\ell P^{k-(\ell+2)}J^2))\!+\ldots+\!(\mathfrak{n}^\ell PJ^{k-(\ell+1)}+\partial(\mathfrak{n}^\ell J^{k-\ell})). \end{align}\]

We have \[\begin{align} \partial(\mathfrak{n}^\ell P^{k-(\ell+1)}J)\;&=\;\mathfrak{n}^{\ell-1}P^{k-(\ell+1)}J+\mathfrak{n}^\ell P^{k-(\ell+2)}J+\mathfrak{n}^\ell P^{k-(\ell+1)}\partial(J)\\ &=\;\mathfrak{n}^\ell P^{k-(\ell+2)}J+\mathfrak{n}^\ell P^{k-(\ell+1)}\partial(J), \end{align}\] and using that \(J\subset\mathfrak{n}P\), \(\partial(J)\subset\mathfrak{n}\) and equation (13 ), we obtain \[\begin{align} \mathfrak{n}^\ell&P^{k-(s-1)}J^{s-1-\ell}\!+\!\partial(\mathfrak{n}^\ell P^{k-s}J^{s-\ell})\;=\\ &=\;\mathfrak{n}^\ell P^{k-(s-1)}J^{s-(\ell+1)}+\mathfrak{n}^{\ell-1}P^{k-s}J^{s-\ell}+\mathfrak{n}^\ell P^{k-(s+1)}J^{s-\ell}+\mathfrak{n}^{\ell}P^{k-s}J^{s-(\ell+1)}\partial(J)\\ &=\;\mathfrak{n}^{\ell-1}P^{k-(s+1)}J^{s-(\ell+1)}[\mathfrak{n}P^2+PJ+\mathfrak{n}J+\mathfrak{n}P\partial(J)]\\ &=\;\mathfrak{n}^{\ell}P^{k-(s+1)}J^{s-(\ell+1)}[P^2+J+P\partial(J)]\\ &=\;\mathfrak{n}^{\ell+1}P^{k-s}J^{s-(\ell+1)}. \end{align}\]

Consequently \[\label{eq:HHZ7} \begin{align} \partial(\sum_{s=\ell+1}^{k}&x_n^{k-s}\mathfrak{n}^\ell P^{k-s}J^{s-\ell})\;=\;x_{n}^{k-(\ell+1)}(\mathfrak{n}^{\ell}P^{k-(\ell+2)}J+\mathfrak{n}^\ell P^{k-(\ell+1)}\partial(J))\\ &+(\sum_{s=\ell+2}^kx_{n}^{k-s}\mathfrak{n}^{\ell+1}P^{k-s}J^{s-(\ell+1)}). \end{align}\tag{17}\]

Using again equation (13 ), we have \[\label{eq:HHZ8} \begin{align} \mathfrak{n}^\ell P^{k-\ell}+\mathfrak{n}^{\ell}P^{k-(\ell+2)}J+\mathfrak{n}^\ell P^{k-(\ell+1)}\partial(J)\;&=\;\mathfrak{n}^{\ell}P^{k-(\ell+2)}[P^2+J+P\partial(J)]\\ &=\;\mathfrak{n}^{\ell+1}P^{k-(\ell+1)}. \end{align}\tag{18}\]

Combining equations (15 ), (16 ), (17 ) and (18 ) we finally conclude that the formula given in equation (8 ) for \(\partial^{\ell+1}(I^k)\) is correct. In particular, for \(\ell+1=k\) we have \[\partial^k(I^k)\;=\;\sum_{r=0}^kx_n^{k-r}\mathfrak{n}^r\;=\;\mathfrak{m}^k.\]

Therefore, for each \(k\le\ell\le 2k\), we clearly have \(\partial^{\ell}(I^k)=\mathfrak{m}^{2k-\ell}\).

Having established the formula (8 ) for all values \(1\le\ell\le 2k\), it remains to prove that \(\partial^\ell(I^k)\) has linear quotients. This is clear for \(k\le\ell\le 2k\) because \(\partial^\ell(I^k)=\mathfrak{m}^{2k-\ell}\).

Now, let \(1\le\ell\le k-1\). Then we can write \(\partial^\ell(I^k)=\sum_{t=0}^k x_n^{k-t}I_t\), where \(I_t=\mathfrak{n}^tP^{k-\ell}\) for \(0\le t\le\ell\), and \(I_{t}=\mathfrak{n}^{\ell}P^{k-t}J^{t-\ell}\) for \(\ell+1\le t\le k\). Using that \(J\subset\mathfrak{n}P\) and \(P\subset\mathfrak{n}\), we obtain that \[I_0\supset I_1\supset I_2\supset\cdots\supset I_k.\]

We claim that each \(I_t\) has linear quotients. To this end, we use a result of Soleyman Jahan and Zheng [24] which says that if \(J\subset S\) is a monomial ideal with linear quotients and \(Q\subset S\) is a monomial prime ideal such that \(\mathrm{supp}(J)\subset\mathrm{supp}(Q)\), then \(QJ\) has linear quotients too. Applying this result in our case with \(Q=\mathfrak{n}\), and using the property (iii), we see that all ideals \(I_t\) have indeed linear quotients.

To finish the proof we use a result of Ficarra, Moradi and Römer [25] which says that if \(I_1,I_2\subset S\) are monomial ideals with linear quotients such that \(I_2\subset I_1\), \(i\notin\mathrm{supp}(I_2)\) and \(\mathcal{G}(x_i I_1)\subset\mathcal{G}(x_iI_1+I_2)\), then the ideal \(x_iI_1+I_2\) has again linear quotients.

Applying this result in our situation, since \(I_0\subset I_1\), both ideals have linear quotients, \(n\notin\mathrm{supp}(I_1)\) and \(\mathcal{G}(x_nI_0+I_1)=\mathcal{G}(x_nI_0)\sqcup\mathcal{G}(I_1)\), we conclude that \(x_nI_0+I_1\) has linear quotients. For the same reasons, \(x_n(x_nI_0+I_1)+I_2\) has linear quotients. Iterating this argument to the other ideals \(I_t\), we finally obtain that \(\partial^\ell(I^k)\) has indeed linear quotients, concluding the proof. ◻