Proof of Theorem 2. Let \(I\in\mathcal{F}\) and \(P_F\in\operatorname{Ass}_S(I^k)\) for some \(F\subset[n]\) and \(k\ge1\). Then \(P_F\) is the maximal ideal of \(S_F\) and \(P_F\in\operatorname{Ass}_{S_F}(I^k(P_F))=\operatorname{Ass}_{S_F}(I(P_F)^k)\). This is equivalent to the fact that \(\operatorname{depth}S_F/I(P_F)^k=0\). By property (a), \(I(P_F)\in\mathcal{F}\). Note that \(\operatorname{depth}S/I(P_F)^k=\operatorname{depth}S_F/I(P_F)^k+n-|F|\). By property (b), the depth function \(k\mapsto\operatorname{depth}S/I(P_F)^k\) is non-increasing. It follows that \(\operatorname{depth}S_F/I(P_F)^k\) is non-increasing too. Hence, \(\operatorname{depth}S_F/I(P_F)^\ell=0\), and so \(P_F\in\operatorname{Ass}_{S_F}(I(P_F)^\ell)=\operatorname{Ass}_{S_F}(I^\ell(P_F))\), for all \(\ell\ge k\). Therefore \(P_F\in\operatorname{Ass}_S(I^\ell)\) for all \(\ell\ge k\), and so \(I\) satisfies the persistence property. This argument also proves the second assertion. ◻

Proof. Recall that the support of a monomial ideal \(I\subset S\) is defined as the set \(\operatorname{supp}(I)=\bigcup_{u\in\mathcal{G}(I)}\operatorname{supp}(u)\), where the support of a monomial \(u\in S\) is defined as the set \(\operatorname{supp}(u)=\{i:\;x_i\;\mathrm{divides}\;u\}\).

Let us define the family \[\mathcal{F}\;=\;\bigg\{I\subset S\,\,\bigg|\,\, \substack{\displaystyle I\;\mathrm{is a squarefree monomial ideal}\\[3pt] \displaystyle\mathrm{generated in degrees}\;\ge\;|\operatorname{supp}(I)|-2}\,\bigg\}.\]

In view of Theorem 2, our assertion follows once we show that the properties (a) and (b) in Theorem 2 are satisfied by the family \(\mathcal{F}\).

For a non-empty subset \(A\subset[n]\), we put \({\boldsymbol{x}}_A=\prod_{i\in A}x_i\). Let \(I\in\mathcal{F}\) and put \(G=\operatorname{supp}(I)\subset[n]\). Let \(i,j\in G\) be distinct elements. Let \(F\subset[n]\). We note that \[\varphi_F({\boldsymbol{x}}_G)\;=\;{\boldsymbol{x}}_{G\cap F},\quad\quad\quad \varphi_F({\boldsymbol{x}}_{G\setminus\{i\}})\;=\;\begin{cases} \,{\boldsymbol{x}}_{G\cap F}&\mathrm{if}\;i\notin G,\\ \,{\boldsymbol{x}}_{(G\cap F)\setminus\{i\}}&\mathrm{if}\;i\in G, \end{cases}\] \[\varphi_F({\boldsymbol{x}}_{G\setminus\{i,j\}})\;=\;\begin{cases} \,{\boldsymbol{x}}_{G\cap F}&\mathrm{if}\;i,j\notin G,\\ \,{\boldsymbol{x}}_{(G\cap F)\setminus\{i\}}&\mathrm{if}\;i\in G,\,j\notin G,\\ \,{\boldsymbol{x}}_{(G\cap F)\setminus\{j\}}&\mathrm{if}\;i\notin G,\,j\in G,\\ \,{\boldsymbol{x}}_{(G\cap F)\setminus\{i,j\}}&\mathrm{if}\;i,j\in G. \end{cases}\]

Since \(\operatorname{supp}(I(P_F))\subset\operatorname{supp}(I)\cap F=G\cap F\), the above computations show that \(I(P_F)\) is a squarefree monomial ideal generated in degrees \(\ge|G\cap F|-2\ge|\operatorname{supp}(I(P_F))|-2\). Hence \(I(P_F)\in\mathcal{F}\) and the property (a) is satisfied.

The property (b) follows from statement (a) of Theorem 3. ◻

Proof of Theorem 2. Note that \(I_{\langle n-2\rangle}=I_c(G)\) is the complementary edge ideal of some graph \(G\). Since \(I\) is minimally generated in the degrees \(n-2\) and \(n-1\), we can find a minimal monomial generator \(u\in\mathcal{G}(I)\) of degree \(n-1\). Then \(u={\boldsymbol{x}}_{[n]}/x_i\) for some \(i\in[n]\). We claim that \(i\) is an isolated vertex of \(G\). Otherwise, we would have \(\{i,j\}\in E(G)\) for some \(j\), and then \({\boldsymbol{x}}_{[n]}/(x_ix_j)\in I_{\langle n-2\rangle}\) would divide \(u\) against the fact that \(u\) is a minimal generator. Let \(F=\{i:\;{\boldsymbol{x}}_{[n]}/x_i\in\mathcal{G}(I)\}\). Then \(F\) is a subset of the isolated vertices of \(G\), and therefore \({\boldsymbol{x}}_F\) divides all the monomial generators of \(I_c(G)\). Up to relabeling, we may assume that \(F=[p]\) for some \(p\ge1\). Let \(H=G|_{[n]\setminus[t]}\). Then \(I_c(G)={\boldsymbol{x}}_{[p]}I_c(H)\). For our convenience, we relabel the variables \(\{x_{p+1},\dots,x_n\}\) by \(\{y_1,\dots,y_q\}\). Our discussion then shows that we can write \[I\;=\;{\boldsymbol{x}}_{[t]}I_c(H)+{\boldsymbol{y}}_{[q]}({\boldsymbol{x}}_{[t]}/x_i:\;i\in[t]).\] We put \(I_y=I_c(H)\) and \(I_x=({\boldsymbol{x}}_{[t]}/x_i:\;i\in[t])\). Then \[I^k\;= \;\sum_{h=0}^k{\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h\] for all \(k\ge1\). Fix an integer \(k\ge1\), and set \(J_\ell=\sum_{h=0}^\ell{\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h\) for \(0\le\ell\le k\).

Notice that \(J_k=I^k\). For each \(0<\ell\le k\), we claim that \[\label{eq:Iell-bs} J_\ell\;=\;J_{\ell-1}+{\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell}\tag{2}\] is a Betti splitting of \(J_\ell\). First, we note that \[\mathcal{G}(J_\ell)\;=\;\mathcal{G}(J_{\ell-1})\sqcup\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell}).\]

To this end, let \(u\in\mathcal{G}(J_{\ell-1})\) and \(v\in\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})\). Then \(u\in{\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^{h}I_y^{k-h}I_x^{h}\) for some \(0\le h<\ell\). A quick computation shows that \[\begin{align} \sum_{i\in[p]}\deg_{x_i}(u)\;=\;pk-h\;&>\;pk-\ell\;=\;\sum_{i\in[p]}\deg_{x_i}(v),\\ \sum_{j\in[q]}\deg_{y_j}(u)\;=\;(q-2)k+2h\;&<\;(q-2)k+2\ell\;=\;\sum_{j\in[q]}\deg_{y_j}(v). \end{align}\] Hence \(u\) can not divide \(v\) and \(v\) can not divide \(u\). This proves our claim. Next, we compute the intersection \[\begin{align} L\;=\;J_{\ell-1}\cap({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})\;&=\;(\sum_{h=0}^{\ell-1}{\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^{h}I_y^{k-h}I_x^{h})\cap({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})\\ &=\phantom{\;(}\sum_{h=0}^{\ell-1}[({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^{h}I_y^{k-h}I_x^{h})\cap({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})]\\ &=\phantom{\;(}\sum_{h=0}^{\ell-1}[({\boldsymbol{x}}_{[p]}^{k-h}I_x^{h})\cap({\boldsymbol{x}}_{[p]}^{k-\ell}I_x^{\ell})][({\boldsymbol{y}}_{[q]}^{h}I_y^{k-h})\cap({\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell})]. \end{align}\]

Note that \(({\boldsymbol{x}}_{[p]})\subset I_x\) and \(({\boldsymbol{y}}_{[q]})\subset I_y\). Given \(h<\ell\), we have \(k-h>k-\ell\) and so \({\boldsymbol{x}}_{[p]}^{k-h}I_x^h={\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{x}}_{[p]}^{\ell-h}I_x^h\subset{\boldsymbol{x}}_{[p]}^{k-\ell}I_x^{\ell-h}I_x^h={\boldsymbol{x}}_{[p]}^{k-\ell}I_x^\ell\). Similarly, we have \({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell}\subset{\boldsymbol{y}}_{[q]}^h I_y^{k-h}\). Consequently, \[\begin{align} L\;=\;J_{\ell-1}\cap({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})\;&=\; \sum_{h=0}^{\ell-1}({\boldsymbol{x}}_{[p]}^{k-h}I_x^h{\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\;=\, (\sum_{h=0}^{\ell-1}{\boldsymbol{x}}_{[p]}^{k-h}I_x^h)({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\\[3pt] &=\;{\boldsymbol{x}}_{[p]}^{k-\ell+1}I_x^{\ell-1}{\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell}, \end{align}\] where we used that \(({\boldsymbol{x}}_{[p]}^k)\subset{\boldsymbol{x}}_{[p]}^{k-1}I_x\subset\cdots\subset{\boldsymbol{x}}_{[p]}^{k-(\ell-1)}I_x^{\ell-1}\).

To conclude that (2 ) is indeed a Betti splitting we must prove that the inclusion maps \(L\rightarrow J_{\ell-1}\) and \(L\rightarrow{\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell}\) are \(\operatorname{Tor}\)-vanishing.

We show that the first map is \(\operatorname{Tor}\)-vanishing, the other case is similar. First, suppose for a moment that \(H\) does not contain isolated vertices. Under this assumption, we claim that \(\partial({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\subset{\boldsymbol{y}}_{[q]}^{\ell-1} I_y^{k+1-\ell}\). Indeed, take \(u={\boldsymbol{y}}_{[q]}^\ell u_1\cdots u_{k-\ell}\in\mathcal{G}({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\) with \(u_s\in\mathcal{G}(I_y)\) for all \(s=1,\dots,k-\ell\), and let \(y_i\) divide \(u\). Since we assumed that \(H\) does not contain isolated vertices, we have \({\boldsymbol{y}}_{[q]}/(y_iy_j)\in\mathcal{G}(I_y)\) for some \(j\). Then \[{\boldsymbol{y}}_{[q]}^{\ell-1}u_1\cdots u_{k-\ell}({\boldsymbol{y}}_{[q]}/(y_iy_j))\;\;\mathrm{divides}\;\;u/y_i.\]

This shows that \(u/y_i\in {\boldsymbol{y}}_{[q]}^{\ell-1} I_y^{k+1-\ell}\). Hence \(\partial({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\subset{\boldsymbol{y}}_{[q]}^{\ell-1} I_y^{k+1-\ell}\). Lemma 4 implies that there exists a map \(\varphi:\mathcal{G}({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\rightarrow\mathcal{G}({\boldsymbol{y}}_{[q]}^{\ell-1} I_y^{k+1-\ell})\) such that for any non-empty subset \(\Omega\subset\mathcal{G}({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\) we have \[\label{eq:lcm} \operatorname{lcm}(u:u\in\Omega)\;\in\;\mathfrak{m}_y\cdot(\operatorname{lcm}(\varphi(u):u\in\Omega)),\tag{3}\] where \(\mathfrak{m}_y=(y_1,\dots,y_q)\). We claim that such a map \(\Phi\) exists also when \(H\) contains isolated vertices.

To this end, assume that \(H\) contains isolated vertices. Say that these vertices correspond to the variables \(y_1,\dots,y_r\) for some \(r<q\). Then \({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell}={\boldsymbol{y}}_{[r]}^{\ell}{\boldsymbol{y}}_{[q]\setminus[r]}^{\ell}(I_y')^{k-\ell}\) where \(I_y'=I_c(H')\) and \(H'\) is the graph obtained from \(H\) by removing its isolated vertices. Similarly, \({\boldsymbol{y}}_{[q]}^{\ell-1} I_y^{k-\ell+1}={\boldsymbol{y}}_{[r]}^{\ell-1}{\boldsymbol{y}}_{[q]\setminus[r]}^{\ell-1}(I_y')^{k-\ell+1}\). Repeating the argument as before, we obtain a map \(\Phi':\mathcal{G}({\boldsymbol{y}}_{[q]\setminus[r]}^{\ell}(I_y')^{k-\ell})\rightarrow\mathcal{G}({\boldsymbol{y}}_{[q]\setminus[r]}^{\ell-1}(I_y')^{k-\ell+1})\) satisfying the property (3 ). Then, we define the map \(\varphi:\mathcal{G}({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\rightarrow\mathcal{G}({\boldsymbol{y}}_{[q]}^{\ell-1} I_y^{k+1-\ell})\) by setting \(\Phi(u)={\boldsymbol{y}}_{[r]}^{\ell-1}\Phi(u/{\boldsymbol{y}}_{[r]}^{\ell})\). It follows immediately that the property (3 ) is again satisfied.

Having acquired the existence of the map \(\Phi\) for any graph \(H\), we define a map \[\Phi:\mathcal{G}(L)\rightarrow\mathcal{G}(J_{\ell-1})\] using the map \(\varphi\). Let \(u\in\mathcal{G}(L)\). Then \(u=u_xu_y\), where \(u_x\in\mathcal{G}({\boldsymbol{x}}_{[p]}^{k+1-\ell}I_x^{\ell-1})\) and \(u_y\in\mathcal{G}({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell})\). We put \(\Phi(u)=\varphi(u_y)u_x\). Note that \(\Phi\) is well-defined, because \(\Phi(u)\in\mathcal{G}(J_{\ell-1})\). Finally, let \(\Omega\subset\mathcal{G}(L)\) be a non-empty subset. We can write \(\operatorname{lcm}(u:u\in\Omega)= \operatorname{lcm}(u_x:u\in\Omega)\cdot\operatorname{lcm}(u_y:u\in\Omega)\) and by using (3 ) we have \[\begin{align} \operatorname{lcm}(u:u\in\Omega)\;\in&\;\mathfrak{m}_y\cdot(\operatorname{lcm}(\varphi(u_y):u\in\Omega))\cdot\operatorname{lcm}(u_x:u\in\Omega)\\ =&\;\mathfrak{m}_y\cdot(\operatorname{lcm}(\varphi(u_y)u_x:u\in\Omega))\\ \subset&\;\mathfrak{m}\cdot(\operatorname{lcm}(\Phi(u):u\in\Omega)), \end{align}\] where \(\mathfrak{m}=(x_1,\dots,x_p,y_1,\dots,y_q)\). Lemma 3 implies that the inclusion \(L\rightarrow J_{\ell-1}\) is indeed \(\operatorname{Tor}\)-vanishing, as desired.

Let \(S_x=K[x_1,\dots,x_p]\) and \(S_y=K[y_1,\dots,y_q]\). Next, we claim that \[\begin{align} \tag{4}\operatorname{reg}I^k=\max&\left\{\substack{\displaystyle\operatorname{reg}I_y^k+pk,(n-1)k,\\[5pt]\displaystyle\max_{0<h<k}\{\operatorname{reg}I_y^{k-h}\!+pk+(q-1)h\}}\right\},\\[4pt] \tag{5}\operatorname{depth}\frac{S}{I^k}=\min&\left\{\substack{\displaystyle\!\operatorname{depth}S_y/I_y^k\!+p,\operatorname{depth}S_x/I_x^k\!+q,\operatorname{depth}S_x/I_x^{k-1}\!+q-1,\\[5pt]\displaystyle\min_{0<h<k}\{\operatorname{depth}S_y/I_y^{k-h}+\operatorname{depth}S_x/I_x^h+1\},\\[3pt]\displaystyle\min_{0<h<k}\{\operatorname{depth}S_y/I_y^{k-h}+\operatorname{depth}S_x/I_x^{h-1}\}}\right\}. \end{align}\]

Since (2 ) is a Betti splitting and \(I^k=J_k\), we have \[\begin{align} \operatorname{reg}I^k&=\max\{\operatorname{reg}J_{k-1},\,\operatorname{reg}({\boldsymbol{y}}_{[q]}^kI_x^k),\,\operatorname{reg}({\boldsymbol{x}}_{[p]}{\boldsymbol{y}}_{[q]}^kI_x^{k-1})-1\},\\ \operatorname{depth}S/I^k&=\min\{\operatorname{depth}S/J_{k-1},\,\operatorname{depth}S/({\boldsymbol{y}}_{[q]}^kI_x^k),\,\operatorname{depth}S/({\boldsymbol{x}}_{[p]}{\boldsymbol{y}}_{[q]}^kI_x^{k-1})-1\}. \end{align}\]

Since \(I_x\subset S_x\) is the squarefree Veronese ideal of \(S_x\) generated in degree \(p-1\) and \(\dim(S_x)=p\), we have \(\operatorname{reg}I_x^h=(p-1)h\) and the function \(k\mapsto\operatorname{depth}S_x/I_x^h\) is non-increasing, for all \(h\ge1\). Hence \(\operatorname{reg}({\boldsymbol{y}}_{[q]}^kI_x^k)=qk+(p-1)k=(n-1)k\) for all \(k\ge1\). Analogously, \(\operatorname{reg}({\boldsymbol{y}}_{[q]}{\boldsymbol{x}}_{[p]}^kI_x^{k-1})-1=(n-1)k\). Similar computations can be performed for the depth. Hence, \[\begin{align} \operatorname{reg}I^k&=\max\{\operatorname{reg}J_{k-1},\,(n-1)k\},\\ \operatorname{depth}S/I^k&=\min\{\operatorname{depth}S/J_{k-1},\,\operatorname{depth}S_x/I_x^k+q,\,\operatorname{depth}S_x/I_x^{k-1}+q-1\}. \end{align}\]

Iterating these computations to \(J_{k-1},\dots,J_0\) by using the Betti splittings (2 ), we see that the formulas (4 ) and (5 ) indeed hold.

Now, we proceed to prove the statements (a), (b), (c), (d), (e) and (f).

(a) Let \(c=c(H)\) be the number of connected components of \(H\) which are not isolated vertices. By [8] we have \[\label{IndI1} \operatorname{reg}I_y^k\;=\;\begin{cases} (q-1)k&\textit{if}\;\;1\le k\le c-2,\\ (q-2)k+c-1&\textit{if}\;\;k\ge c-1. \end{cases}\tag{6}\]

From this formula it follows that \(\operatorname{reg}I_y^k\le(q-1)k\) for all \(k\ge1\). Hence, for all \(k\ge1\) and all \(0<h<k\), we have \[\begin{align} \operatorname{reg}I_y^k+pk\;&\le\;(q-1)k+pk\;=\;(n-1)k,\\ \operatorname{reg}I_y^{k-h}+pk+(q-1)h\;&\le\;(q-1)(k-h)+pk+(q-1)h\;=\;(n-1)k. \end{align}\]

These inequalities and equation (4 ) imply that \(\operatorname{reg}I^k=(n-1)k\) for all \(k\ge1\).

(b) By [8], the function \(k\mapsto\operatorname{depth}S_y/I_y^k\) is non-increasing. Moreover \(k\mapsto\operatorname{depth}S_x/I_x^k\) is non-increasing as well, because \(I_x\) is matroidal. It follows at once from formula (5 ) that \(k\mapsto\operatorname{depth}S/I^k\) is non-increasing too.

(c) By [8], the graded Betti numbers of \(I_y^k\) do not depend on the ground field \(K\) for all \(k\ge1\). The same independence occurs for the powers of \(I_x\) because it is a matroidal ideal. Consider the Betti splitting (2 ). We computed before that \(L=J_{\ell-1}\cap({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})={\boldsymbol{x}}_{[p]}^{k-\ell+1}I_x^{\ell-1}{\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell}\). This ideal is the product of the ideals \({\boldsymbol{x}}_{[p]}^{k-\ell+1}I_x^{\ell-1}\) and \({\boldsymbol{y}}_{[q]}^\ell I_y^{k-\ell}\) which have disjoint supports and whose graded Betti numbers do not depend on \(K\). Hence the graded Betti numbers of \(L\) do not depend on \(K\). Finally, the Betti splitting (2 ) for \(\ell=k\) shows that the graded Betti numbers of \(I^k\) do not depend on \(K\) for all \(k\ge1\).

(d) We denote by \(b(H)\) the number of bipartite connected components of the graph \(H\), we remark that an isolated vertex is regarded as a bipartite connected component. By [11], \(\operatorname{depth}S_y/I_y^k=\operatorname{depth}S_y/I_c(H)^k=b(H)\) for all \(k\ge\operatorname{dstab}(I_y)\). Next, note that \(\operatorname{depth}S_x/I_x^{p-1}=0\). To this end, consider the monomial \(u={\boldsymbol{x}}_{[p]}^{p-2}\). It does not belong to \(I_x^{p-1}\), because \(\deg(u)=p(p-2)\) is strictly less than \((p-1)^2\) which is the initial degree of \(I_x^{p-1}\). On the other hand, for all \(i\in[p]\) we have \[x_iu\;=\;x_i{\boldsymbol{x}}_{[p]}^{p-2}\;=\;\prod_{\substack{j\in[p]\\ j\ne i}}({\boldsymbol{x}}_{[p]}/x_j)\;\in\;I_x^{p-1}.\] Hence \((I_x^{p-1}:u)=(x_1,\dots,x_p)\in\operatorname{Ass}_{S_x}(I_x^{p-1})\) and so \(\operatorname{depth}S_x/I_x^{p-1}=0\). Since \(I_x\) is a matroidal ideal, we have \(\operatorname{depth}S_x/I_x^{k}=0\) as well for all \(k\ge p-1=\mu_{n-1}(I)-1\).

Let \(k\ge\operatorname{dstab}(I_y)+\mu_{n-1}(I)\). The previous discussion and equation (5 ) yield that \[\operatorname{depth}S/I^k\;=\;\min\left\{\substack{\displaystyle b(H)+p,\,q,\,q-1,\\[5pt]\displaystyle b(H)+1,\,b(H)}\right\}\;=\;\min\{b(H),\,q-1\}\;=\;b(H),\] because \(q-1=|V(H)|-1\ge b(H)\) since \(H\) has at least one edge.

By [11], \(\operatorname{depth}S/I_{\langle n-2\rangle}^k=\operatorname{depth}S/({\boldsymbol{x}}_{[p]}I_y)^k=\operatorname{depth}S/I_c(G)^k=b(G)\) for all \(k\ge\operatorname{dstab}(I_y)+\mu_{n-1}(I)\). Recall that \(H\) is the graph obtained from \(G\) by removing the isolated vertices \(1,\dots,p\). Hence \(b(H)=b(G)-p\). Altogether, \[\operatorname{depth}S/I^k\;=\;b(H)\;=\;b(G)-p\;=\;\operatorname{depth}S/I_{\langle n-2\rangle}^k-\mu_{n-1}(I)\] for all \(k\ge\operatorname{dstab}(I_{\langle n-2\rangle})+\mu_{n-1}(I)\).

(e) Part (d) implies that \(\operatorname{dstab}(I)\le\operatorname{dstab}(I_y)+p\). By [11], we have \(\operatorname{dstab}(I_y)\le q-2\). Since \(p+q-2=n-2\), the assertion follows.

(f) The implication (i) \(\Rightarrow\) (ii) is true for any monomial ideal. Assume (ii). Since \(n-2\) is the initial degree of \(I\), \(J=I_c(G)\) has linear powers. By [10] we have \(c(G)=1\) and (iii) follows. Finally, we show that (iii) \(\Rightarrow\) (i). Assume that \(J=I_c(G)\) with \(c(G)=1\). Again by [10], \(J^k\) has linear quotients for all \(k\ge1\). Since (2 ) is a Betti splitting, by what we proved before we have \[\label{eq:G-disjoint} \mathcal{G}(I^k)\;=\;\bigsqcup_{h=0}^k\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h),\quad\mathrm{for all}\;k\ge1.\tag{7}\] Notice that \(c(H)=1\) since \(c(G)=1\). Hence \(I_y^k=I_c(H)^k\) has linear quotients for all \(k\ge1\). Since \(I_x\) is a matroidal ideal, then also \(I_x^k\) has linear quotients for all \(k\ge1\). Since \(\operatorname{supp}(I_x)\cap\operatorname{supp}(I_y)=\emptyset\), [22] implies that \(I_x^hI_y^{k-h}\) has linear quotients for all \(k\ge1\) and all \(0\le h\le k\). Multiplying an ideal with linear quotients by a monomial preserves the linear quotients property. Hence \({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h\) has linear quotients for all \(k\ge1\) and all \(0\le h\le k\). Fix \(k\ge1\) and for each \(0\le h\le k\) fix a linear quotients order \(>_h\) of \({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h\). In view of equation (7 ), we order the monomials in \(\mathcal{G}(I^k)\) as follows. Let \(u,v\in\mathcal{G}(I^k)\). Then \(u\in\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h)\) and \(v\in\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^{\ell}I_y^{k-\ell}I_x^{\ell})\) for some \(h\) and \(\ell\). We put \[u>v\quad\Longleftrightarrow\quad h<\ell\quad\text{or}\quad h=\ell\quad\text{and}\quad u>_hv.\]

Let \(u,v\in\mathcal{G}(I^k)\) with \(u>v\) as described before. If \(h=\ell\), since \({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h\) has linear quotients, there exists \(w\in\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-h}{\boldsymbol{y}}_{[q]}^hI_y^{k-h}I_x^h)\subset\mathcal{G}(I^k)\) with \(w>_hv\) and so also \(w>v\) such that \(w:v=\operatorname{lcm}(w,v)/v\) is a variable that divides \(u:v\). Now, assume that \(h<\ell\). As we computed before, we have \(\sum_{i\in[p]}\deg_{x_i}(u)=pk-h>pk-\ell=\sum_{i\in[p]}\deg_{x_i}(v)\). Hence, there exists \(i\in[p]\) such that \(x_i\) divides \(u:v\). We can write \(v={\boldsymbol{x}}_{[p]}^{k-\ell}{\boldsymbol{y}}_{[q]}^\ell v_1\cdots v_{k-\ell}v_1'\cdots v_\ell'\) with \(v_j\in\mathcal{G}(I_y)\) and \(v_j'\in\mathcal{G}(I_x)\) for all \(j\). Since \(x_i\) divides \(u:v\) and \(I_x=({\boldsymbol{x}}_{[p]}/x_j:\;j\in[p])\), we must have \(v_t'={\boldsymbol{x}}_{[p]}/x_i\) for some \(t\). Up to relabeling, we may assume that \(t=\ell\). Let \({\boldsymbol{y}}_{[q]}/(y_ry_s)\in\mathcal{G}(I_y)\) be a monomial generator. We consider the following monomial \[\begin{align} w\;&=\;x_i(v/(y_ry_s))\\ &=\;{\boldsymbol{x}}_{[p]}^{k-(\ell-1)}{\boldsymbol{y}}_{[q]}^{\ell-1}v_1\cdots v_{k-\ell}{\boldsymbol{y}}_{[q]}/(y_ry_s)v_1'\cdots v_{\ell-1}'\in\mathcal{G}({\boldsymbol{x}}_{[p]}^{k-(\ell-1)}{\boldsymbol{y}}_{[q]}^{\ell-1}I_y^{k-(\ell-1)}I_x^{\ell-1}). \end{align}\] Then \(w>v\) by definition of the order \(>\), and \(w:v=x_i\) divides \(u:v\). We conclude that \(I^k\) has indeed linear quotients and (i) follows. ◻

Proof. Let \(I=I_c(G)\), and take a monomial \(u\in\mathcal{G}(I)\). Then, \(u={\boldsymbol{x}}_{[n]}/{\boldsymbol{x}}_e\), for some edge \(e=\{i,j\}\in E(G)\). Let \(\varphi=\varphi_F\) be the map defined in the introduction. We have \[\label{eq:calculation} \varphi(u)\;=\;\begin{cases} \,{\boldsymbol{x}}_F&\mathrm{if}\;i,j\notin F,\\ \,{\boldsymbol{x}}_F/x_i&\mathrm{if}\;i\in F,\,j\notin F,\\ \,{\boldsymbol{x}}_F/x_j&\mathrm{if}\;i\notin F,\,j\in F,\\ \,{\boldsymbol{x}}_F/{\boldsymbol{x}}_e&\mathrm{if}\;i,j\in F. \end{cases}\tag{8}\]

Now, we distinguish the two possible cases.

Case 1. Assume that \((\bigcup_{e\in E(G)}e)\cap F=\emptyset\). Then the calculation (8 ) shows that \(\varphi(v)={\boldsymbol{x}}_F\) for all \(v\in\mathcal{G}(I)\). Hence \(I(P_F)=({\boldsymbol{x}}_F)\).

Case 2. Assume that \((\bigcup_{e\in E(G)}e)\cap F\ne\emptyset\). Let \(H=G|_F\). Then, for any \(e\in E(H)\), we have \(e\in E(G)\), and equation (8 ) shows that \({\boldsymbol{x}}_F/{\boldsymbol{x}}_e\in I(P_F)\). Equation (8 ) also shows that any monomial in \(I(P_F)\) has degree at least \(|F|-2\). Hence, all monomials \({\boldsymbol{x}}_F/{\boldsymbol{x}}_e\in\mathcal{G}(I_c(H))\) are minimal monomial generators of \(I(P_F)\). Since, by assumption, \({\boldsymbol{x}}_F/{\boldsymbol{x}}_e\in I(P_F)\) for at least one edge \(e\in E(H)\), then \({\boldsymbol{x}}_F\notin\mathcal{G}(I(P_F))\). Hence, besides the minimal monomial generators coming from \(I_c(H)\), the only possibly extra minimal monomial generators of \(I(P_F)\) have degree \(|F|-1\), and therefore are of the form \(w={\boldsymbol{x}}_F/x_i\) for some \(i\in F\). Equation (8 ) implies that \(w\) belongs to \(I(P_F)\) if and only if there exists \(j\in[n]\) such that \(\{i,j\}\in E(G)\). Moreover, \(w\) is a minimal generator of \(I(P_F)\) if and only if \(\{i,j\}\notin E(H)\) for all \(j\in F\setminus\{i\}\). That is, if and only if \(\deg_{G|_F}(i)=0\). Altogether, \(w={\boldsymbol{x}}_F/x_i\in\mathcal{G}(I(P_F))\) if and only if \(i\in A_F\). ◻

Proof of Theorem 1. First of all, note that if \(i\in V(G)\) is an isolated vertex of \(G\), and we let \(H=G|_{[n]\setminus\{i\}}\), then \(I_c(G)^k=x_i^kI_c(H)^k=(x_i^k)\cap I_c(H)^k\) for all \(k\ge1\). Hence \(\operatorname{Ass}(I_c(G)^k)=\{(x_i)\}\cup\operatorname{Ass}(I_c(H)^k)\) for all \(k\ge1\). We may therefore assume without loss of generality that \(G\) does not contain isolated vertices. Now, let \(F\subset[n]\) with \(|F|\ge2\). By Theorem 2, \(P_F\in\operatorname{Ass}^\infty(I_c(G))\) if and only if \(\operatorname{depth}S_F/(I_c(G)^k(P_F))=0\) for all \(k\gg0\). By Proposition 7, since we assumed that \(G\) does not contain isolated vertices (hence \(\bigcup_{e\in E(G)}e=[n]\)), we have \[I_c(G)^k(P_F)\;=\;(I_c(G)(P_F))^k\;=\;\Big(I_c(G|_F)+({\boldsymbol{x}}_F/x_i:\;\deg_{G|_F}(i)=0)\Big)^k.\]

Now, we distinguish the two possible cases.

Case 1. Suppose that \(\widetilde{b}(G|_F)=b(G|_F)\). Then \(G|_F\) does not contain isolated vertices, and so \(I_c(G)^k(P_F)=I_c(G|_F)^k\). By [11], \(\operatorname{depth}S_F/I_c(G|_F)^k=b(G)=\widetilde{b}(G)\) for all \(k\ge|F|-2\). Hence, \(P_F\in\operatorname{Ass}^\infty(I_c(G))\) if and only if \(\widetilde{b}(G)=0\).

Case 2. Suppose that \(\widetilde{b}(G|_F)\ne b(G|_F)\). Then \(G|_F\) contains some isolated vertex. In this case, Theorem 2(d) and [11] imply that \[\begin{align} \operatorname{depth}S_F/(I_c(G)^k(P_F))\;&=\;\operatorname{depth}S_F/I_c(G|_F)^k-\mu_{|F|-1}(I_c(G)^k(P_F))\\[2pt] &=\;b(G)-|\{i\in F:\;\deg_{G|_F}(i)=0\}|\\ &=\;\widetilde{b}(G). \end{align}\] Hence, once again \(P_F\in\operatorname{Ass}^\infty(I_c(G))\) if and only if \(\widetilde{b}(G)=0\).

It remains to prove that \(\operatorname{astab}(I_c(G))\le n-2\). To this end, let \(P_F\in\operatorname{Ass}^\infty(I_c(G))\). If \(F=\{i\}\) is a singleton, then \(i\) is an isolated vertex of \(G\), and so \(P_F\in\operatorname{Ass}(I_c(G)^k)\) for all \(k\ge1\). Otherwise, let \(|F|\ge2\). Then \(\operatorname{depth}S_F/(I_c(G)^k(P_F))=0\) for all \(k\gg0\). By Theorem 2(e) and [11], we have \(\operatorname{depth}S_F/(I_c(G)^k(P_F))=0\) for all \(k\ge|F|-2\). Hence \[\operatorname{astab}(I_c(G))\;\le\;\max\{1,\max_{\substack{P_F\in\operatorname{Ass}^\infty(I_c(G))\\ |F|\ge2}}\{|F|-2\}\}\;\le\;n-2,\] as desired. ◻

Proof. (a) \(\Rightarrow\) (b) is true for any ideal.

(b) \(\Rightarrow\) (c) Suppose that \(I_c(G)^k=I_c(G)^{(k)}\) for some \(k\ge2\). As we explained before, this holds if and only if \(\operatorname{Ass}(I_c(G)^k)=\operatorname{Ass}(I_c(G))\). It is a simple exercise in graph theory to show that if \(G\) does not satisfy the statement (c), then \(V(G)\) contains a subset \(F\) with \(|F|=4\) such that \(\widetilde{b}(G|_F)=0\). Suppose for a contradiction that (c) does not hold. Then we can find such a \(4\)-subset \(F\subset[n]\). By Corollary 8, \(P_F\in\operatorname{Ass}(I_c(G)^\ell)\) for all \(\ell\ge2\). Hence, \(P_F\in\operatorname{Ass}(I_c(G)^k)\). By [8] we have \(P_F\in\operatorname{Ass}(I_c(G)^k)\setminus\operatorname{Ass}(I_c(G))\). A contradiction. Hence (c) follows.

(c) \(\Rightarrow\) (a) It is straightforward to see that if \(G\) satisfies the statement (c) then \(\operatorname{Ass}(I_c(G)^k)=\operatorname{Ass}(I_c(G))\) for all \(k\ge1\). Hence (a) follows. ◻

Proof. (a) By [27], it follows that \(\operatorname{v}(I_c(G)^k)\ge(n-2)k-1\) for all \(k\ge1\). Therefore, when \(G\ne tK_2\) to prove that equality holds for all \(k\ge1\), it suffices to find a monomial \(u_k\), for each \(k\ge1\), such that \(I_c(G)^k:u_k\in\operatorname{Ass}(I_c(G)^k)\) and \(\deg(u_k)=(n-2)k-1\).

Assume that \(G\) contains an isolated vertex \(i\). Then \((x_i)\in\operatorname{Ass}(I_c(G)^k)\) for all \(k\ge1\). Let \(e\in E(G)\), then \(i\notin e\). We consider the monomial \(u_k=({\boldsymbol{x}}_{[n]\setminus e})^k/x_i\) for all \(k\ge1\). Note that \(x_iu_k\in I_c(G)^k\) for all \(k\ge1\). Now, let \(v\) be a monomial with \(\operatorname{supp}(v)\subset[n]\setminus\{i\}\). Then \(\deg_{x_i}(vu_k)=k-1\). Hence \(vu_k\notin I_c(G)^k\). We conclude that \(I_c(G):u_k=P_{\{i\}}=(x_i)\), as desired.

Suppose now that \(G\) does not contain isolated vertices. Let \(G^c\) be the complementary graph of \(G\). That is, the graph with the same vertex set of \(G\) whose edges are the non-edges. Furthermore, let \(\mathcal{T}(G)\) be the set of triangles (also called \(3\)-cliques) of \(G\). By [8], since \(G\) does not have isolated vertices, we have \[\operatorname{Ass}(I_c(G))\;=\;\{P_e:\;e\in E(G^c)\}\cup\{P_T:\;T\in\mathcal{T}(G)\}.\]

First, suppose that \(G\) contains a triangle \(T=\{p,q,r\}\). Consider the monomial \(u_k=({\boldsymbol{x}}_{[n]}/(x_px_qx_r))({\boldsymbol{x}}_{[n]}/(x_px_q))^{k-1}\) with \(k\ge1\). Note that \(\deg(u_k)=(n-2)k-1\). We claim that \(I_c(G)^k:u_k=P_T\). It is clear that \(P_T\subset I_c(G)^k:u_k\). For instance, \[x_pu_k\;=\;({\boldsymbol{x}}_{[n]}/(x_qx_r))({\boldsymbol{x}}_{[n]}/(x_px_q))^{k-1}\in I_c(G)^k.\] To prove the reverse inclusion \(I_c(G)^k:u_k\subset P_T\), we show that if \(v\) is a monomial with \(\operatorname{supp}(v)\subset[n]\setminus\{p,q,r\}\), then \(v\notin I_c(G)^k:u_k\). Suppose on the contrary that \(v\in I_c(G)^k:u_k\). Then, there exist edges \(e_1,\dots,e_k\in E(G)\) such that \[\label{eq:divide-edges} w\;=\;\prod_{i=1}^k{\boldsymbol{x}}_{[n]\setminus e_i}\quad\text{divides}\quad vu_k\;=\;v({\boldsymbol{x}}_{[n]}/(x_px_qx_r))({\boldsymbol{x}}_{[n]}/(x_px_q))^{k-1}.\tag{9}\] It follows that \[\deg_{x_p}(w)\;=\;k-|\{j\in[k]:\;p\in e_j\}|\;\le\;\deg_{x_p}(v)+\deg_{x_p}(u_k)\;=\;0,\] because \(x_p\) does not divide \(v\) and \(u_k\). Since \(|\{j\in[k]:p\in e_j\}|\le k\), the above inequality implies that \(|\{j\in[k]:p\in e_j\}|=k\). Hence \(p\in e_j\) for all \(j\in[k]\). Similarly, \(q\in e_j\) for all \(j\in[k]\). Hence \(e_1=\dots=e_k=\{p,q\}\). It follows that \(\deg_{x_r}(w)=k\). This \(x_r\)-degree should be less than or equal to \(\deg_{x_r}(vu_k)=\deg_{x_r}(v)+\deg_{x_r}(u_k)=k-1\), because \(r\notin\operatorname{supp}(v)\). But this is not possible. Hence \(v\notin I_c(G)^k:u_k\). We conclude that \(I_c(G)^k:u_k=P_T\).

Now, assume that \(G\) contains no triangles. Since \(G\) does not have isolated vertices, either \(G\) contains an induced \(3\)-path \(P_3\) or else \(G=tK_2\) with \(t\ge2\) (since \(n\ge3\)).

Suppose first that \(G\) contains an induced \(3\)-path. Say, that \(\{p,q\},\{q,r\}\in E(G)\), but \(e=\{p,r\}\notin E(G)\). Hence \(P_e\in\operatorname{Ass}(I_c(G))\). Consider again the monomial \(u_k=({\boldsymbol{x}}_{[n]}/(x_px_qx_r))({\boldsymbol{x}}_{[n]}/(x_px_q))^{k-1}\) with \(k\ge1\). Note that \(\deg(u_k)=(n-2)k-1\). We claim that \(I_c(G)^k:u_k=P_e\). As before, \(P_e\subset I_c(G)^k:u_k\). Conversely, we show that if \(v\) is a monomial with \(\operatorname{supp}(v)\subset[n]\setminus\{p,r\}\), then \(v\notin I_c(G)^k:u_k\). Suppose on the contrary this was the case. Then, equation (9 ) would be again valid. From this equation, we derive that all edges \(e_1,\dots,e_k\) contain \(p\). Since \(\{p,r\}\notin E(G)\), it follows that \(\deg_{x_r}(w)=k\). But this \(x_r\)-degree should be less than or equal to \(\deg_{x_r}(vu_k)=k-1\), which is absurd. We conclude that \(I_c(G)^k:u_k=P_e\).

Now, let \(G=tK_2\) with \(t\ge2\). Say \(E(G)=\{\{1,2\},\{3,4\},\dots,\{2t-1,2t\}\}\). First we show that there is no monomial \(u\in S\) of degree \((2t-2)k-1\) such that \(I_c(G)^k:u\) is a monomial prime ideal. Suppose for a contradiction that such a monomial \(u\) exists. Then \(I_c(G)^k:u\) should be generated by variables. Up to relabeling, we may assume that \(x_1\in I_c(G)^k:u\). Then \(x_1u\in I_c(G)\). Since \(\deg(x_1u)=(2t-2)k\), then \(x_1u\in\mathcal{G}(I_c(G)^k)\). Therefore, there exist edges \(e_1,\dots,e_k\in E(G)\) such that \(x_1u=\prod_{i=1}^k{\boldsymbol{x}}_{[2t]\setminus e_i}\). Since \(x_1\) divides this monomial, we have \(1\notin e_i\) for at least one index \(i\in[k]\). Say \(e_i=e_1=\{3,4\}\). Therefore \(u={\boldsymbol{x}}_{[2t]\setminus\{1,3,4\}}(\prod_{i=2}^k{\boldsymbol{x}}_{[2t]\setminus e_i})\). Note that \(x_3x_4\in I_c(G)^k:u\) because \(x_3x_4u=x_2{\boldsymbol{x}}_{[2t]\setminus\{1,2\}}(\prod_{i=2}^{k}{\boldsymbol{x}}_{[2t]\setminus e_i})\in I_c(G)^k\). We claim that \(x_3\notin I_c(G)^k:u\) and \(x_4\notin I_c(G)^k:u\). This shows that \(x_3x_4\) is a minimal generator of \(I_c(G)^k:u\) and so \(I_c(G)^k:u\) is not generated by variables. If \(x_3\in I_c(G)^k:u\), then we should have \[x_3u\;=\;{\boldsymbol{x}}_{[2t]\setminus\{1,4\}}(\prod_{i=2}^{k}{\boldsymbol{x}}_{[2t]\setminus e_i})\;=\;{\boldsymbol{x}}_{[2t]}^k/((x_1x_4){\boldsymbol{x}}_{e_2}\cdots {\boldsymbol{x}}_{e_k})\in I_c(G)^k.\] Since \(\deg(x_3u)=(2t-2)k\), this is equivalent to \(w=(x_1x_4){\boldsymbol{x}}_{e_2}\cdots {\boldsymbol{x}}_{e_k}\in I(G)^k\). Let \(w'={\boldsymbol{x}}_{e_2}\cdots {\boldsymbol{x}}_{e_k}\). Since \(\mathcal{G}(I_c(G))=\{{\boldsymbol{x}}_{\{1,2\}},\dots,{\boldsymbol{x}}_{\{2t-1,2t\}}\}\), each minimal monomial generator of \(v\in I(G)^k\) is such that \(\deg_{x_1}(v)+\deg_{x_2}(v)\) is an even integer. On the other hand, \(\deg_{x_1}(w)+\deg_{x_2}(w)=1+\deg_{x_1}(w')+\deg_{x_2}(w')\) is an odd integer, because \(w\in\mathcal{G}(I_c(G)^{k-1})\). Thus \(w\notin I_c(G)^k\), and so \(x_3\notin I_c(G)^k:u\), as desired.

Summarizing our discussion thus far, we proved that \(\operatorname{v}(I_c(G)^k)\ge(2t-2)k\) for all \(k\ge1\). Now, we prove that equality holds. To this end, consider the monomial \(u_k={\boldsymbol{x}}_{[2t]\setminus\{1,3\}}({\boldsymbol{x}}_{[2t]\setminus\{1,2\}})^{k-1}\). Note that \(x_1u_k=x_4{\boldsymbol{x}}_{[2t]\setminus\{3,4\}}({\boldsymbol{x}}_{[2t]\setminus\{1,2\}})^{k-1}\in I_c(G)^k\). Similarly \(x_3u_k\in I_c(G)^k\). Hence \((x_1,x_3)\subset I_c(G)^k:u_k\). To prove the reverse inclusion, consider a monomial \(v\in S\) with \(\operatorname{supp}(v)\subset[2t]\setminus\{1,3\}\). Suppose for a contradiction that \(vu_k\in I_c(G)^k\), then there exist edges \(e_1,\dots,e_k\in E(G)\) such that \(w=\prod_{i=1}^k{\boldsymbol{x}}_{[2t]\setminus e_i}\) divides \(vu_k\). Since \(\deg_{x_1}(vu_k)=0\) it follows that \(\deg_{x_1}(w)=0\), and so \(e_1=\dots=e_k=\{1,2\}\). Hence \(\deg_{x_3}(w)=k\). However, this \(x_3\)-degree should be less than or equal to \(\deg_{x_3}(vu_k)=k-1\), an absurd. It follows that \(I_c(G)^k:u_k=(x_1,x_3)\) as desired. We conclude that \(\operatorname{v}(I_c(G)^k)=(2t-2)k\) for all \(k\ge1\), and statement (a) follows.

(b) Let \(c=c(G)\). By [8] or also Theorem 3(c)(ii) in this paper, it follows indeed that \(\operatorname{v}(I_c(G)^k)<\operatorname{reg}I_c(G)^k\) for all \(k\ge1\).

(c) Assume that \(I_c(G)\) has linear powers. Then \(\operatorname{reg}I_c(G)^k=(n-2)k\) for all \(k\ge1\). Moreover, [10] (or also Theorem 3(c)(ii)) yields \(c(G)=1\). Part (a) implies that \(\operatorname{v}(I_c(G)^k)=(n-2)k-1=\operatorname{reg}I_c(G)^k-1\) for all \(k\ge1\). ◻