Proof. The case where the number of facets of $\Delta $ is one holds trivially. Assume that the number of facets of $\Delta $ is more than one. It follows from [Reference Herzog, Hibi, Trung and Zheng22, Corollary 3.4] that there is a good leaf, say $F_r$ , in $\Delta $ , where r is the number of facets of $\Delta $ . Set $\Delta _1=\Delta \setminus \{F_r\}.$ Since $I(\Delta )$ has a linear resolution, by [Reference Zheng35, Proposition 3.9], $\Delta $ is pure and connected in codimension one. Therefore, by [Reference Zheng35, Corollary 1.15], $\Delta _1$ is a pure simplicial tree connected in codimension one, and consequently, by [Reference Zheng35, Lemma 3.11], $I(\Delta _1)$ has a linear resolution. We claim that there exists a facet $F_{r-1}$ of $\Delta _1$ such that:

• • $F_{r-1}$ has a free vertex in $\Delta _1$ ;

• • $\operatorname {\mathrm {dist}}_{\Delta }(F_{r-1},F_{r}) =1$ .

Let $H_1, \ldots , H_k$ be the facets of $\Delta $ such that $\operatorname {\mathrm {dist}}_{\Delta }(F_r, H_i) =1$ for all $1 \le i \le k.$ If possible, we assume that the claim does not hold, that is, for each $1 \le i \le k $ , $H_i$ does not have a free vertex in $\Delta _1.$ Since $\operatorname {\mathrm {dist}}_{\Delta }(F_r,H_i) =1$ , $|F_r \cap H_i|=|F_r|-1$ for each $1 \le i \le k.$ Consequently, $F_r \cap H_i=F_r\cap H_j=H_i\cap H_j$ for each $1 \le i<j \le k$ as $F_r$ is a good leaf of $\Delta .$ Now, for each $1 \le i \le k$ , there exists a vertex $f_i \in H_i \setminus F_r$ such that $H_i=\{f_i\} \cup (H_i \cap F_r).$ It is easy to note that $f_1,\ldots , f_k$ are distinct. Since $H_i$ does not have a free vertex in $\Delta _1$ , there exists a facet $M_i$ of $\Delta _1$ such that $f_i \in H_i \cap M_i.$ With the help of [Reference Zheng35, Corollary 1.13], we can even assume that $|M_i \cap H_i|=|H_i|-1$ . By construction, $F_r,H_i,M_i$ is a proper chain in $\Delta. $ Since $F_r$ is a good leaf in $\Delta $ , $M_i \cap F_r \subseteq H_i \cap F_r$ . Observe that $|M_i \cap F_r| <|F_r|-1$ as if $|M_i \cap F_r| =|F_r|-1$ , then $M_i \cap F_r = H_i \cap F_r,$ which implies $M_i=H_i$ . Consequently, $F_r,H_i,M_i$ is a proper irredundant chain in $\Delta $ and $|M_i\cap F_r|=|F_r|-2$ , which implies that $M_i\cap F_r =M_j \cap F_r$ for each $1 \le i,j \le k$ . Also, there exists $z_i \in F_r \cap H_i$ such that $z_i \not \in M_i$ . Now, we have the following cases:

1. Case 1. Suppose that $k\ge 2.$ We claim that $M_1,H_1, H_2,M_2$ is a proper chain in $\Delta _1. $ Suppose that $|M_1 \cap H_2|\ge |H_2|-1$ . As $f_1 \in M_1 \setminus H_2$ , $M_1=(M_1\cap H_2) \cup \{f_1\}$ . This will impose $f_2 \in M_1$ , which further implies that the collection of facets $\{M_1, H_1, H_2\}$ of $\Delta _1$ does not have a leaf. We have a contradiction. Therefore, $f_2 \not \in M_1$ and $|M_1 \cap H_2|<|H_2|-1.$ Similarly, $f_1 \not \in M_2$ and $|H_1 \cap M_2|<|M_2|-1.$ Consequently, $z_1,z_2,f_1,f_2 \not \in M_1 \cap M_2$ and $z_1,z_2 \not \in \{f_1,f_2\}$ , and hence, $|M_1 \cap M_2| <|M_2|-1.$ This concludes $M_1,H_1, H_2,M_2$ is a proper irredundant chain in $\Delta _1.$ Therefore, $\operatorname {\mathrm {dist}}_{\Delta _1}(M_1,M_2)=3.$

   Since $I(\Delta _1)$ has a linear resolution, $\Delta _1$ satisfies the intersection property, and hence, $|M_1 \cap M_2|= |M_1|-3.$ However, $M_1 \cap F_r =M_2 \cap F_r \subseteq M_1 \cap M_2$ , we get $|M_1 \cap M_2| \ge |M_2|-2$ that is a contradiction.

2. Case 2. Suppose that $k=1.$ Since $z_1 \in H_1\setminus M_1$ and $H_1$ has no free vertex in $\Delta _1$ , there exists a facet N of $\Delta _1$ such that $z_1\in H_1 \cap N.$ Without loss of generality, by [Reference Zheng35, Corollary 1.13], we can assume that $|H_1\cap N|=|H_1|-1.$ In $\Delta $ , we have $N\cap F_r \subseteq H_1\cap F_r$ , as $F_r$ is a good leaf of $\Delta .$ Now, $F_r,H_1,N$ is a proper irredundant chain in $\Delta $ , and thus, $\operatorname {\mathrm {dist}}_{\Delta }(F_r,N)=2.$ Observe that $F_r\cap N \not \subset F_r\cap M_1$ as $z_1\in (F_r\cap N)\setminus M_1.$ Therefore, $F_r\cap M_1 \subseteq F_r\cap N$ , which implies that $(F_r\cap M_1)\sqcup \{z_1\} \subseteq F_r\cap N$ . Thus, $|F_r\cap N|= |F_r|-1$ . Since $I(\Delta )$ has linear resolution, $\operatorname {\mathrm {dist}}_{\Delta }(F_r,N)=1,$ which is contradiction.

Thus, in both scenarios, we have a contradiction. This concludes that $H_i$ has a free vertex in $\Delta _1$ for some $1 \le i \le k$ . Assume, without loss of generality, that $H_1$ has a free vertex in $\Delta _1$ . Next, we claim that $H_1$ is a good leaf of $\Delta _1$ . Let $w \in H_1$ be a free vertex of $H_1$ in $\Delta _1$ . Now, we have the following two cases:

1. Case a. Suppose that $w \not \in F_r.$ Then, $w=f_1.$ For any facet, P of $\Delta _1$ such that $P \ne H_1$ , $P\cap H_1=P \cap (\{f_1\} \sqcup (H_1\cap F_r))=(P\cap \{f_1\}) \sqcup (P\cap H_1 \cap F_r) =P\cap H_1 \cap F_r.$ This implies that $P\cap H_1=P \cap F_r$ . Let $A, B$ be any two facets of $\Delta _1$ such that $A,B \not \in \{H_1\}.$ In $\Delta $ , either $A\cap F_r \subseteq B\cap F_r$ or $B\cap F_r \subseteq A\cap F_r$ as $F_r$ is a good leaf of $\Delta ,$ which implies that either $A\cap H_1 \subseteq B\cap H_1$ or $B\cap H_1 \subseteq A\cap H_1$ . Thus, $H_1$ is a good leaf of $\Delta _1.$

2. Case b. Suppose that $w \in F_r.$ If possible, assume that $H_1$ is not a leaf of $\Delta _1.$ Then, there exist two facets $A, B$ of $\Delta _1$ other than $H_1$ such that $A\cap H_1 \not \subset B\cap H_1$ and $B \cap H_1 \not \subset A \cap H_1.$ Since $F_r$ is a leaf in $\Delta ,$ assume, without loss of generality, that $ A\cap F_r \subseteq B\cap F_r.$ Observe that $f_1 \in A $ and $f_1\not \in B$ as if $f_1 \notin A$ or $f_1 \in B$ , then $A\cap H_1\subseteq B\cap H_1$ . Therefore, $B\cap H_1=B\cap F_r \neq \emptyset $ and $|B\cap H_1|\le |H_1|-2$ . Let $|B\cap F_r|=|B\cap H_1|=|F_r|-j$ for some $j\ge 2.$ Since $\Delta _1$ satisfies the intersection property, there exists a unique proper irredundant chain of length j, say $H_1=B_0,B_1,\ldots , B_j=B$ . Notice that $F_r,H_1=B_0,B_1,\ldots ,B_j=B$ is a proper irredundant chain in $\Delta $ between $F_r$ and $B.$ As $\Delta $ satisfies intersection property $j+1=\operatorname {\mathrm {dist}}_{\Delta }(F_r,B)=|F_r|-|B\cap F_r|=j,$ which is a contradiction. Therefore, $H_1$ is a good leaf of $\Delta _1$ .

Take $F_{r-1}=H_1$ is a good leaf of $\Delta _1=\Delta \setminus \{F_r\}.$ Now, using the claim successively, we get a desired ordering of facets of $\Delta .$

Proof. (a) We use induction on $k \ge i$ . The base case $k=i$ is trivial. Assume that $k> i$ and $f \notin F_{k-1}$ . Suppose, on the contrary, that $f \in F_k$ . Then, $f \in F_j \cap F_k$ . Since $\operatorname {\mathrm {dist}}_{\Delta }(F_{k-1},F_k) =1$ , $|F_{k-1} \cap F_k|=|F_k|-1$ . Also, as $F_k$ is a good leaf of $\langle F_1,\ldots ,F_{k-1}\rangle $ , therefore, $F_j \cap F_k \subseteq F_{k-1} \cap F_k$ . Therefore, $f \in F_{k-1}$ , a contradiction.

(b) For $j<i$ , it is easy to note that $F_j,F_{j+1},\ldots ,F_i$ is a proper chain between $F_j$ and $F_i$ . Now, after removing suitable facets from this proper chain, we obtain an irredundant proper chain, $F_j,F_{j_1},\ldots ,F_{j_{l-1}},F_{j_l}=F_i$ . If $\operatorname {\mathrm {dist}}_{\Delta }(F_j,F_i)=1$ , then take $k=j.$ Otherwise, we have $ \operatorname {\mathrm {dist}}_{\Delta }(F_j,F_{j_{l-1}}) = \operatorname {\mathrm {dist}}_{\Delta }(F_j,F_i)-1>0.$ Consequently, since $\Delta $ satisfies intersection property, $|F_j \cap F_{j_{l-1}}|=|F_j \cap F_i|+1.$ Suppose that $F_j \cap F_{j_{l-1}} \subseteq F_i$ . Then, $F_j \cap F_{j_{l-1}} \subseteq F_i \cap F_j$ , which is a contradiction to the fact that $|F_j \cap F_{j_{l-1}}|=|F_j \cap F_i|+1.$ Hence, $F_j \cap F_{j_{l-1}} \not \subset F_i$ .

Proof. By Lemma 2.1, there exists an ordering of facets of $\Delta $ , say $F_1,\ldots ,F_r$ , such that:

• • $F_1,\ldots ,F_r$ is a good leaf ordering on the facets of $\Delta ;$

• • $\operatorname {\mathrm {dist}}_{\Delta }(F_i,F_{i+1}) =1$ for all $1 \le i \le r-1.$

We fix that $m_1>\cdots >m_r$ , where $m_i=\prod \limits _{j\in F_i}x_j$ for $1 \leq i \leq r$ . One can easily verify that for any $s \geq 1$ , any minimal monomial generator M of $I(\Delta )^s$ has the unique expression $M=m_1^{a_1}m_2^{a_2}\dots m_r^{a_r}$ with $\sum \limits _{i=1}^ra_i=s$ . Now, for any two minimal monomial generators ${M=m_1^{b_1}m_2^{b_2}\dots m_r^{b_r}, N=m_1^{a_1}m_2^{a_2}\dots m_r^{a_r}}$ of $I(\Delta )^s$ , we say $M> N$ if and only if $(b_1,\ldots ,b_r)>_{lex} (a_1,\ldots ,a_r)$ . In this way, we get a total order among the minimal monomial generators of $I(\Delta )^s.$ We claim that $I(\Delta )^s$ has linear quotients with respect to this total order among the minimal monomial generators of $I(\Delta )^s.$ By [Reference Herzog and Hibi21, Lemma 8.2.3], it is sufficient to prove that for any two minimal monomial generators $M>N$ of $I(\Delta )^s$ , there exists a minimal monomial generator $P>N$ such that $(P):N$ is generated by a variable and $(M):N \subseteq (P):N.$

Let $M=m_1^{b_1}\dots m_r^{b_r}$ and $N=m_1^{a_1}\dots m_r^{a_r}$ be two minimal monomial generators of $I(\Delta )^s$ such that $M>N.$ Therefore, $a_t,b_t \ge 0$ for all $1 \le t \le r$ and $\sum \limits _{t=1}^ra_t=\sum \limits _{t=1}^rb_t=s$ . We set ${p =\min \{ t \in [r] \;: b_t-a_t>0 \}}$ and $q =\min \{t \in [r]: b_t-a_t<0\}. $ Since $(b_1,\ldots ,b_r)>_{lex} (a_1,\ldots ,a_r)$ , we get $p<q.$ Observe that $ b_p> a_p$ , $b_i \geq a_i$ for all $p <i<q,$ and $b_q<a_q.$

By Lemma 2.3 (2), there exists $k \in [p,q)$ such that $F_p \cap F_k \not \subset F_q$ and $|F_k \cap F_q|=|F_q|-1.$ Let $f \in (F_p \cap F_k) \setminus F_q.$ We define $c=(c_1,\ldots ,c_r)$ as follows:

$$ \begin{align*}c_t= \begin{cases} a_t & ~\text{if}~ t \notin \{q,k\} \\ a_{t}+1 & ~\text{if} ~ t=k \\ a_{t}-1 & ~\text{if} ~t=q. \end{cases} \end{align*} $$

Clearly, $(c_1,\ldots ,c_r)>_{lex} (a_1,\ldots ,a_r),$ and $\sum \limits _{t=1}^rc_t=s.$ Take $P=m_1^{c_1}\dots m_r^{c_r}$ . It is easy to verify that $(P):N=(m_k):m_q=(x_f).$ Consider,

$$ \begin{align*} (M):N &= (m_p^{b_p}\dots m_r^{b_r}):m_p^{a_p} \dots m_r^{a_r} \\ &= m_p^{b_p-a_p}\dots m_{q-1}^{b_{q-1}-a_{q-1}} \left(\frac{m_q^{b_q}\dots m_r^{b_r}}{\gcd(m_q^{b_q}\dots m_r^{b_r}, m_q^{a_q}\dots m_r^{a_r})} \right). \end{align*} $$

Since $x_f \in (F_p \cap F_k) \setminus F_q$ and $b_p>a_p$ , using Lemma 2.3 (1), we obtain $(M):N \subseteq (x_f)=(P):N.$ Hence, the assertion follows.

Proof. It follows from [Reference Zheng35, Proposition 3.17] that $(1)$ and $(2)$ are equivalent. The implication $(2) \implies (3)$ follows from Proposition 2.4 and the implications $(3) \implies (4),$ and $(5) \implies (6)$ are immediate. Note that if $\Delta $ is pure, then for all $s \ge 1$ , $I(\Delta )^s$ is generated in the same degree. Consequently, by [Reference Herzog and Hibi21, Proposition 8.2.1], the implication $(4) \implies (5)$ follows. Next, we prove $(6) \implies (2)$ . Since $I(\Delta )^s$ has linear first syzygies, $I(\Delta )^s$ is generated in the same degree. Therefore, $\Delta $ is pure. Suppose that $I(\Delta )$ has no linear resolution. By [Reference Zheng35, Proposition 3.9(i)], $I(\Delta )$ cannot have linear quotients with respect to any order on minimal monomial generators. Let $F_1, \ldots , F_r$ be a good leaf order on the facets of $\Delta .$ Then, for some $k \le r-1$ , $(m_1,\ldots , m_k):m_{k+1}$ is not generated by monomials of degree one. Since $F_{k+1}$ is a good leaf of the subcomplex with facets $F_1,\ldots , F_{k+1}, \left \{F_i \cap F_{k+1} ~ \mid ~ 1 \le i \le k \right \}$ is a total order set, that is, there exist distinct $i_1, \ldots , i_k \in \{ 1, \ldots , k\}$ such that $F_{i_1} \cap F_{k+1} \subseteq \cdots \subseteq F_{i_k} \cap F_{k+1}$ . Let $t \in \{ 1, \ldots , k\}$ be the largest such that:

• • $(m_{i_t}) : m_{k+1}$ is not generated by a monomial of degree one;

• • $(m_{i_j} ~\mid ~ j>t): m_{k+1}$ is generated by monomials of degree one;

• • $(m_{i_t}) : m_{k+1} \not \subset (m_{i_j}) : m_{k+1} $ for each $j> t.$

We claim that $F_{i_j} \not \subseteq F_{i_t} \cup F_{k+1}$ for each $j \in \{1, \ldots , k\} \setminus \{t\}.$ For each $j>t,$ there exists a variable $z_j \in F_{i_j} \setminus F_{k+1}$ such that $z_j \notin F_{i_t}$ as $(m_{i_t}) : m_{k+1} \not \subset (m_{i_j}) : m_{k+1} $ . Consequently, $F_{i_j} \not \subseteq F_{i_t} \cup F_{k+1}$ for each $j>t.$ Suppose that for some $j<t$ , $F_{i_j} \subseteq F_{i_t} \cup F_{k+1}.$ Since $F_{i_j} \cap F_{k+1} \subseteq F_{i_t} \cap F_{k+1} \subseteq F_{i_t}$ , we get that $F_{i_j} \subseteq F_{i_t}$ , which is not possible as $\Delta $ is pure and $F_{i_j}, F_{i_t}$ are distinct facets. Thus, the claim follows. Let $\Delta '$ be the induced subcomplex of $\Delta $ on the vertex set $F_{i_t} \cup F_{k+1}.$ It follows from the above claim that $I(\Delta ')=(m_{i_t}, m_{k+1}).$ Set $u=\prod \limits _{l \in F_{i_t}\cap F_{k+1}} x_l$ , $v=\prod \limits _{l \in F_{i_t}\setminus F_{k+1}} x_l$ and $w=\prod \limits _{l \in F_{k+1}\setminus F_{i_t}} x_l$ . Observe that v and $ w$ are monomials of degree at least two, as $(m_{i_t}) : m_{k+1}$ is not generated by a monomial of degree one. Since v and w are monomials in a disjoint set of variables, it follows from [Reference Guardo and Van Tuyl16, Theorem 2.1] that $\beta _{1,j}^R\left ((v,w)^s\right ) \neq 0$ , where $j=\deg \left ((vw)^s\right ).$ Consequently, $\beta _{1,j}^R\left (I(\Delta ')^s\right )=\beta _{1,j}^R\left ((uv,uw)^s\right ) \neq 0$ , where $j=\deg \left ((uvw)^s\right ).$ Since $\Delta '$ is an induced subcomplex of $\Delta $ , it follows using the same arguments as in [Reference Beyarslan, Hà and Trung3, Lemma 4.2] that $\beta _{1,j}^R\left (I(\Delta )^s\right ) \neq 0$ with $j=\deg \left ((uvw)^s\right ).$ This is a contradiction to the fact that $I(\Delta )^s$ has linear first syzygies for some $s.$ Thus, $(6) \implies (2)$ follows. Hence, the assertion follows.

Proof. By Remark 3.4, it suffices to establish the assertion for $C_t(\Gamma )$ . Thus, we assume, without loss of generality, that $\Gamma = C_t(\Gamma )$ . We employ induction on $\mathrm {ht}(\Gamma )$ . Suppose that $\mathrm {ht}(\Gamma )=1$ . Then, $I_t(\Gamma ) =(x_0y\; :\; \ell _{\Gamma }(y)=1)=x_0(y \; : \; \ell _{\Gamma }(y)=1).$ Thus,

$$ \begin{align*}\mathrm{ reg}\left(\frac{R}{I_t(\Gamma)}\right)= 1+ \mathrm{reg}\left( \frac{R}{(y \; : \; \ell_{\Gamma}(y)=1)}\right)= 1.\end{align*} $$

Since $\mathrm {ht}(\Gamma )=1$ , every t-rooted star in $\Gamma $ must have root $x_0$ . Thus, every t-star packing of $\Gamma $ contains only one element. Since $t=\mathrm {ht}(\Gamma )+1=2$ , by Definition 3.2, $\zeta _t(\Gamma )=1$ , and hence, the result is true for $\mathrm {ht}(\Gamma )=1$ . Assume now that $\mathrm {ht}(\Gamma )>1$ . Consider the following short exact sequence:

$$ \begin{align*}\displaystyle 0 \rightarrow \frac{R}{I_t(\Gamma):x_0} (-1)\xrightarrow{\cdot x_0} \frac{R}{I_t(\Gamma)} \rightarrow \frac{R}{I_t(\Gamma)+(x_0)} \rightarrow 0.\end{align*} $$

Since $t=\mathrm {ht}(\Gamma )+1, x_0 $ divides each monomial generator of $I_t(\Gamma )$ , and thus, $I_t(\Gamma )+(x_0)=(x_0).$ Consequently, $\mathrm {reg}\left (\frac {R}{I_t(\Gamma )+(x_0)}\right )=0<t-1 \leq \mathrm {reg}\left (\frac {R}{I_t(\Gamma )}\right ).$ Therefore, by [Reference Dao, Huneke and Schweig12, Lemma 2.10], $ \mathrm { reg}\left (\frac {R}{I_t(\Gamma )}\right )=\mathrm {reg}\left (\frac {R}{I_t(\Gamma ):x_0}(-1)\right )=1+\mathrm { reg}\left (\frac {R}{I_t(\Gamma ):x_0}\right ).$ Let $x_{i_1},\ldots ,x_{i_k}$ be descents of $x_{0}$ in $\Gamma $ . For each $1 \le j \le k$ , let $\Gamma _j$ denote the $x_{i_j}$ -descendant rooted tree of $\Gamma .$ Then, $\Gamma \setminus \{x_0\}$ is the disjoint union of rooted trees, say, $(\Gamma _1,x_{i_1}),\ldots , (\Gamma _k,x_{i_k})$ each of height $\mathrm {ht}(\Gamma )-1$ . Observe that

$$ \begin{align*}I_t(\Gamma):x_0=I_{t-1}(\Gamma \setminus \{x_0\})=\sum\limits_{j=1}^kI_{t-1}(\Gamma_j).\end{align*} $$

Thus, $ \mathrm { reg}\left (\frac {R}{I_t(\Gamma )}\right )=1+\mathrm {reg}\left (\frac {R}{I_t(\Gamma ):x_0}\right )=1+\sum \limits _{j=1}^k\mathrm { reg}\left (\frac {R}{I_{t-1}(\Gamma _j)}\right ).$ Since for each j, $(\Gamma _j,x_{i_j})$ is a rooted tree with $t-1=\mathrm { ht}(\Gamma )=\mathrm {ht}(\Gamma _j)+1$ , and $\Gamma _{j}=C_{t-1}(\Gamma _{j}) $ , by induction,

$$ \begin{align*}\mathrm{ reg}\left(\frac{R}{I_{t-1}(\Gamma_j)}\right) = \zeta_{t-1}(\Gamma_j)=\begin{cases} 1 & \text{if } \mathrm{ht}(\Gamma_j)=1, \\ 1+ \sum\limits_{\ell_{\Gamma_j}(x)=0}^{\mathrm{ht}(\Gamma_j)-2} \deg_{\Gamma_j}^+(x) &\text{if } \mathrm{ht}(\Gamma_j) \ge 2.\end{cases}\end{align*} $$

Note that for $x \in V(\Gamma \setminus \{x_0\})$ and for each j, $\ell _{\Gamma _j}(x) =\ell _{\Gamma }(x)-1$ and $\mathrm {ht}(\Gamma _j) =\mathrm {ht}(\Gamma )-1.$ Therefore,

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I_t(\Gamma)}\right)&= 1+\sum\limits_{j=1}^k\mathrm{ reg}\left(\frac{R}{I_{t-1}(\Gamma_j)}\right)=1+\sum_{j=1}^k\zeta_{t-1}(\Gamma_j)\\ &= \begin{cases} 1+k & \text{if } \mathrm{ht}(\Gamma)=2, \\ 1+\sum\limits_{j=1}^k \left( 1+ \sum\limits_{\ell_{\Gamma_j}(x)=0}^{\mathrm{ht}(\Gamma_j)-2} \deg_{\Gamma_j}^{+}(x) \right) & \text{if } \mathrm{ht}(\Gamma)>2 \end{cases}\\ &= \begin{cases} 1+k & \text{if } \mathrm{ht}(\Gamma)=2,\\ 1+k+\sum\limits_{j=1}^k\sum\limits_{\ell_{\Gamma_j}(x) = 0}^{\mathrm{ht}({\Gamma_j})-2}\deg_{\Gamma_j}^+(x) & \text{if } \mathrm{ht}(\Gamma)>2 \end{cases} \\ &= 1+\sum\limits_{\ell_{\Gamma}(x)=0}^{\mathrm{ht}(\Gamma)-2} \deg_{\Gamma}^{+}(x) \end{align*} $$

as $\deg _{\Gamma }^+(x_0)=k.$ Notice that $x_0$ is the only vertex for which there exists a vertex w such that $d_{\Gamma }(x_0,w)=t-1.$ So, by definition, every t-rooted star in $\Gamma $ must have root $x_0$ ; thus, every t-star packing of $\Gamma $ contains only one element. Since $\Gamma $ is a perfect rooted tree of height $t-1$ , $\{\Gamma \}$ forms a t-star packing of $\Gamma $ . Furthermore, for any t-star packing $\mathcal {E}=\{\Gamma _1\}$ , $\Gamma _1$ is an induced subtree of $\Gamma $ , therefore, $\zeta _t(\mathcal {E}) =\zeta _t(\{\Gamma _1\}) \le \zeta _t(\{\Gamma \})= 1+\sum \limits _{\ell _{\Gamma }(x)=0}^{\mathrm {ht}(\Gamma )-2} \deg _{\Gamma }^{+}(x).$ Therefore, we conclude that

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I_t(\Gamma)}\right)= 1+\sum\limits_{\ell_{\Gamma}(x)=0}^{\mathrm{ht}(\Gamma)-2} \deg_{\Gamma}^{+}(x) =\zeta_t(\Gamma), \end{align*} $$

which establishes the assertion.

Notation 3.7. Let $(\Gamma ,x_0)$ be a rooted tree with $\mathrm {ht}(\Gamma ) \geq 1$ , and t be a positive integer so that $2 \le t \le \mathrm {ht}(\Gamma )+1$ . Let z be a leaf such that $\ell _{\Gamma }(z)=\mathrm {ht}(\Gamma )$ . Then, there exists a unique path of length $t-1$ in $\Gamma $ that terminates at $z:=x_t(z),$ say, $P(z):= x_1(z),\ldots ,x_t(z)$ . Let $x_{0}(z)$ be the parent of $x_{1}(z),$ if exists. For $j = 0,\ldots ,t,$ let $\Gamma _j(z)$ be the $x_j(z)$ -descendant rooted tree of $\Gamma $ rooted at $x_{j}(z),$ and let $\Delta ^{\Gamma }_j(z) = \Gamma _j(z) \setminus (\Gamma _{j+1}(z) \cup \{x_{j}(z)\})$ . Set

$$ \begin{align*}\Gamma(z) = \begin{cases} \Gamma \setminus \Gamma_0(z) & \text{if } x_0(z) \text{ exists}\\ \Gamma \setminus \Gamma_1(z) & \text{if } x_0(z) \text{ does not exist}. \end{cases} \end{align*} $$

Note that, in both cases, $\Gamma (z)$ is either empty or a rooted tree.

Proof. We proceed by induction on $|V(\Gamma )|\ge 2.$ If $|V(\Gamma )|=2$ , then $\mathrm {ht}(\Gamma )=1$ , and hence, $t=2=\mathrm { ht}(\Gamma )+1.$ So, the assertion follows from Theorem 3.6. Now assume that $|V(\Gamma )|\ge 3$ and the assertion is true for any rooted tree $(\Gamma ',x')$ with $|V(\Gamma ')| < |V(\Gamma )|.$ Note that if $t=\mathrm {ht}(\Gamma )+1$ , then the assertion follows immediately from Theorem 3.6. So, we assume $2 \le t \le \mathrm {ht}(\Gamma ).$ Let z be a leaf such that $\ell _{\Gamma }(z)=\mathrm {ht}(\Gamma ).$ Using Notation 3.7, there is a path $P(z):=x_1(z),\ldots ,x_t(z)$ of length $t-1$ that terminates at $z=x_t(z)$ , and since $t \le \mathrm {ht}(\Gamma ), x_0(z)$ , the parent of $x_1(z)$ exists. Again, using Notation 3.7, for each $j=0,\ldots ,t$ , $\Gamma _j(z)$ is the $x_j(z)$ -descendant rooted tree of $\Gamma $ rooted at $x_j(z)$ , and $\Delta _j^{\Gamma }(z)=\Gamma _j(z) \setminus \left (\Gamma _{j+1}(z) \cup \{x_j(z)\}\right ).$

By Lemma 3.8 part $(2)$ , we know that

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I_t(\Gamma)} \right)=& \max \left\{\mathrm{reg}\left(\frac{R}{I_t(\Gamma \setminus \{z\})}\right),\mathrm{reg}\left(\frac{R}{I_t(\Gamma(z))}\right) + \sum\limits_{j=0}^{t-1}\mathrm{ reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right)+(t-1)\right\}. \end{align*} $$

Notice that $\Gamma \setminus \{z\}$ is a rooted tree with $\mathrm {ht}(\Gamma \setminus \{z\}) \ge \mathrm {ht}(\Gamma )-1$ and $|V(\Gamma \setminus \{z\})|<|V(\Gamma )|.$ Therefore, by induction,

$$ \begin{align*}\mathrm{ reg}\left( \dfrac{R}{I_t(\Gamma \setminus \{z\})}\right)=\zeta_t(\Gamma\setminus \{z\}).\end{align*} $$

We now claim that $\zeta _t(\Gamma \setminus \{z\}) \le \zeta _t(\Gamma ).$ Note that if T is a t-rooted star with root w in $\Gamma \setminus \{z\}, $ then T is also a t-rooted star with root w in $\Gamma .$ Therefore, if $\mathcal {E}$ is a t-star packing in $\Gamma \setminus \{z\}, $ then $\mathcal {E}$ is a t-separated set in $\Gamma $ . By Definition 3.2, it follows that $\zeta _t(\Gamma \setminus \{z\}) \leq \zeta _t(\Gamma ).$

Next, we claim that $\sum \limits _{j=0}^{t-1}\mathrm {reg}\left (\frac {R}{I_{t-j}(\Delta ^{\Gamma }_j(z) )}\right )+(t-1) =\zeta _t\left ({\Gamma }_0(z) \setminus \{x_0(z)\}\right ).$ By Remark 3.4, we know that for each $j=0,\ldots ,t-1$ ,

$$ \begin{align*}\mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right)=\mathrm{ reg}\left(\frac{R}{I_{t-j}(C_{t-j}(\Delta^{\Gamma}_j(z)) )}\right).\end{align*} $$

Observe that for each $j=0,\ldots ,t-1$ , a connected component of $\Delta _j^{\Gamma }(z)$ is a rooted tree of height at most $t-j-1$ . Therefore, for each $j=0,\ldots ,t-1$ , $C_{t-j}(\Delta _j^{\Gamma }(z))$ is either empty or each connected component of $C_{t-j}(\Delta _j^{\Gamma }(z))$ is a perfect rooted tree of height $t-j-1$ . Thus, for $j=0,\ldots ,t-2,$ applying Theorem 3.6 to each connected component H of $C_{t-j}(\Delta ^{\Gamma }_j(z)) $ when $C_{t-j}(\Delta ^{\Gamma }_j(z)) \neq \emptyset $ , we get

$$ \begin{align*}\displaystyle \mathrm{ reg}\left(\frac{R}{I_{t-j}(H)}\right)=\zeta_{t-j}(H)= \begin{cases} 1 &\text{if } \mathrm{ht}(H)=1,\\ 1 + \sum\limits_{\ell_{H}(x) = 0}^{\mathrm{ht}(H)-2}\deg_{H}^+(x) & \text{if }\mathrm{ht}(H) \ge 2.\end{cases}\end{align*} $$

For $j\in \{0,\ldots ,t-2\}$ , let $a_j$ denote the number of connected components of $C_{t-j}(\Delta ^{\Gamma }_j(z)) $ . Note that $a_j=0$ when $C_{t-j}(\Delta ^{\Gamma }_j(z)) = \emptyset $ . Therefore, if $C_{t-j}(\Delta ^{\Gamma }_j(z)) \neq \emptyset $ and $0 \le j<t-2$ , then

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right) &= \mathrm{reg}\left(\frac{R}{I_{t-j}(C_{t-j}(\Delta^{\Gamma}_j(z)) )}\right)= \sum_{H} \left (1 + \sum_{\ell_{H}(x) =0}^{\mathrm{ ht}(H)-2}\deg_{H}^+(x) \right) \\& =a_j + \sum_{\ell_{C_{t-j}(\Delta^{\Gamma}_j(z))}(x) =0}^{\mathrm{ ht}(C_{t-j}(\Delta^{\Gamma}_j(z)))-2}\deg_{C_{t-j}(\Delta^{\Gamma}_j(z))}^+(x), \end{align*} $$

where H runs over the connected components of $C_{t-j}(\Delta ^{\Gamma }_j(z)).$ Also, when $j=t-2$ and $C_{t-j}(\Delta ^{\Gamma }_j(z)) \neq \emptyset $ ,

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right) &= \mathrm{reg}\left(\frac{R}{I_{t-j}(C_{t-j}(\Delta^{\Gamma}_j(z)) )}\right)= \sum_{H} 1 =a_j, \end{align*} $$

where H runs over the connected components of $C_{t-j}(\Delta ^{\Gamma }_j(z)).$

Set $A=\{j~:~ 0 \le j \le t-2, \text { and } C_{t-j}(\Delta ^{\Gamma }_j(z)) \neq \emptyset \}.$ Now, it follows from the discussion in the previous two paragraphs that

$$ \begin{align*} \displaystyle & \sum\limits_{j=0}^{t-1}\mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right)+(t-1)\\ &\quad = \sum_{j \in A} \left (a_j + \sum_{\ell_{C_{t-j}(\Delta^{\Gamma}_j(z))}(x) =0}^{\mathrm{ ht}(C_{t-j}(\Delta^{\Gamma}_j(z)))-2}\deg_{C_{t-j}(\Delta^{\Gamma}_j(z))}^+(x) \right)+0+t-1\\&\quad =\sum_{j \in A} \left (a_j + 1+\sum_{\ell_{C_{t-j}(\Delta^{\Gamma}_j(z))}(x) =0}^{\mathrm{ ht}(C_{t-j}(\Delta^{\Gamma}_j(z)))-2}\deg_{C_{t-j}(\Delta^{\Gamma}_j(z))}^+(x) \right) + (t-1-|A|). \end{align*} $$

By construction (see Notation 3.7), $\Gamma _0(z)\setminus \{x_0(z)\}$ is a disjoint union of $\Delta _0^{\Gamma }(z)$ and $\Gamma _1(z)$ . Thus, using Remark 3.4, we obtain that $C_{t}(\Gamma _0(z)\setminus \{x_0(z)\})$ is an induced subforest of $\Gamma $ on the vertex set $V(C_t(\Delta _0^{\Gamma }(z))) \sqcup V(C_t(\Gamma _1(z)))=\left (\bigsqcup \limits _{j=0}^{t-1}V(C_{t-j}(\Delta _j^{\Gamma }(z))) \right )\sqcup \{x_1(z),\ldots ,x_{t}(z)\}.$ Furthermore, for each $j \in \{1, \ldots , t-2\}$ , we observe that $ \deg ^+_{C_{t}(\Gamma _0(z)\setminus \{x_0(z)\})}(x_j(z)) = a_j+1$ . Additionally, for every $x \in V(C_{t-j}(\Delta _j^{\Gamma }(z)))$ , we have $\ell _{C_{t-j}(\Delta _j^{\Gamma }(z))}(x)=\ell _{C_{t}(\Gamma _0(z)\setminus \{x_0(z)\})}(x),$ and $\deg _{C_{t-j}(\Delta ^{\Gamma }_j(z))}^+(x)= \deg _{C_{t}(\Gamma _0(z)\setminus \{x_0(z)\})}^+(x).$ Therefore, we obtain

$$ \begin{align*} \displaystyle \sum\limits_{j=0}^{t-1}\mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right)+(t-1) =(a_0+1)+\sum\limits_{\ell_{C_{t}(\Gamma_0(z)\setminus \{x_0(z)\})}(x) = 0}^{\mathrm{ht}(C_{t}({\Gamma_0(z) \setminus \{x_0(z)\}}))-2}\deg_{C_{t}(\Gamma_0(z)\setminus \{x_0(z)\})}^+(x). \end{align*} $$

Also, by construction (see Notation 3.7) and by Remark 3.4, we know that $C_{t}(\Gamma _0(z)\setminus \{x_0(z)\})$ is a disjoint union of $a_0+1$ perfect rooted trees each of height $t-1$ . Thus, applying Theorem 3.6 to each connected component of $C_{t}({\Gamma _0(z) \setminus \{x_0(z)\}})$ , we obtain

$$ \begin{align*} \displaystyle \sum\limits_{j=0}^{t-1}\mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right)+(t-1) &=(a_0+1)+\sum\limits_{\ell_{C_{t}(\Gamma_0(z)\setminus \{x_0(z)\})}(x) = 0}^{\mathrm{ht}(C_{t}({\Gamma_0(z) \setminus \{x_0(z)\}}))-2}\deg_{C_{t}(\Gamma_0(z)\setminus \{x_0(z)\})}^+(x)\\& =\zeta_t(C_{t}({\Gamma_0(z) \setminus \{x_0(z)\}})). \end{align*} $$

This concludes the claim.

Since $|V(\Gamma (z))|<|V(\Gamma )|$ , by induction,

$$ \begin{align*}\mathrm{reg}\left(\frac{R}{I_{t}(\Gamma(z) )}\right)=\begin{cases}\zeta_t({\Gamma}(z)) & \text{if } C_t(\Gamma(z)) \neq \emptyset\\ 0 & \text{if } C_t(\Gamma(z)) = \emptyset.\end{cases}\end{align*} $$

If $C_t(\Gamma (z)) = \emptyset ,$ then by Definition 3.2, any t-star packing of $\Gamma _0(z) \setminus \{x_0(z)\}$ is a t-separated set in $\Gamma $ . Therefore, $ \zeta _t(\Gamma _0(z) \setminus \{x_0(z)\}) \le \zeta _t(\Gamma )$ . Next, we assume that $C_t(\Gamma (z)) \neq \emptyset .$ Let $\mathcal {E}_1, \mathcal {E}_2$ be t-star packings of $\Gamma (z)$ and $\Gamma _0(z) \setminus \{x_0(z)\}$ , respectively, such that $\zeta _t(\mathcal {E}_1)=\zeta _t(\Gamma (z))$ and $\zeta _t(\mathcal {E}_2)=\zeta _t(\Gamma _0(z) \setminus \{x_0(z)\})$ . By Notation 3.7, $\Gamma (z)=\Gamma \setminus \Gamma _0(z)$ . This implies that $\Gamma (z) \cup \left (\Gamma _0(z) \setminus \{x_0(z)\} \right )$ is an induced subforest on the vertex set $V(\Gamma ) \setminus \{x_0(z)\}.$ Therefore, it follows from Definition 3.2 that every t-rooted star in $\mathcal {E}=\mathcal {E}_1 \cup \mathcal {E}_2$ is a t-rooted star of $\Gamma $ . Furthermore, $\mathcal {E}$ is a t-separated set in $\Gamma .$ Thus,

$$ \begin{align*} & \mathrm{reg}\left(\frac{R}{I_t(\Gamma(z))}\right) + \sum\limits_{j=0}^{t-1}\mathrm{reg}\left(\frac{R}{I_{t-j}(\Delta^{\Gamma}_j(z) )}\right)+(t-1)\\ &\quad=\zeta_t(\Gamma(z))+\zeta_t(\Gamma_0(z) \setminus \{x_0(z)\})\\ &\quad \leq \zeta_t(\Gamma). \end{align*} $$

This establishes that $ \mathrm {reg}\left ( \frac {R}{I_t(\Gamma )}\right ) \le \zeta _t(\Gamma )$ by Lemma 3.8 part $(2)$ .

For the other inequality, let $\mathcal {E}$ be a t-star packing of $\Gamma $ such that $\zeta _t(\Gamma )=\zeta _t(\mathcal {E})$ . By Definition 3.2, the parent of the root of any tree $T \in \mathcal {E}$ does not belong to any other rooted tree in $\mathcal {E}$ . Therefore, $\Gamma _1=\bigsqcup \limits _{T \in \mathcal {E}} T$ forms an induced subforest of $\Gamma $ , hence by Lemma 3.9, we get $\mathrm {reg}\left (\frac {R}{I_t(\Gamma _1)}\right ) \leq \mathrm {reg}\left (\frac {R}{I_t(\Gamma )}\right ).$ Furthermore, based on Definition 3.2 and Theorem 3.6, we find that $\mathrm { reg}\left (\frac {R}{I_t(\Gamma _1)}\right )=\sum \limits _{T \in \mathcal {E}}\mathrm {reg}\left (\frac {R}{I_t(T)}\right )=\zeta _t(\Gamma )$ . This concludes the proof.

Proof. By Theorem 3.10, we get $\mathrm {reg}\left (\frac {R}{I_t(\Gamma )}\right )=\zeta _t(\Gamma ).$ We now claim that

$$ \begin{align*}\zeta_t(\Gamma)= \sum\limits_{\ell_{\Gamma}(x)=\mathrm{ht}(\Gamma) -t}^{\mathrm{ ht}(\Gamma)-2}\deg^{+}_{\Gamma}(x).\end{align*} $$

Let $\mathcal {E}=\{(\Gamma _{i_1},x_{i_1}),\ldots ,(\Gamma _{i_k},x_{i_k})\}$ be an arbitrary t-star packing of $\Gamma $ . By Definition 3.2,

$$ \begin{align*}\zeta_t(\mathcal{E})=\sum\limits_{j=1}^k \zeta_t(\Gamma_{i_j}) \text{ and } \zeta_t(\Gamma_{i_j})=\begin{cases} 1 & \text{if } t=2,\\ 1+\sum\limits_{\ell_{\Gamma_{i_j}}(x)=0}^{\mathrm{ht}(\Gamma_{i_j})-2} \deg_{\Gamma_{i_j}}^+(x) & \text{if } t \ge 3.\end{cases}\end{align*} $$

We establish the claim for $t \neq 2$ , and the case when $t = 2$ follows along the same lines. Fix $1 \le j \le k$ . Let $(\Gamma _{i_j}',x_{i_j})$ denote the induced subtree of $\Gamma $ on the vertex set $V(\Gamma _{i_j}) \cup V_{i_j}$ , where $V_{i_j}=\{z: z \text { is a descendant of a leaf in } \Gamma _{i_j} \}.$ We claim that $V(\Gamma _{i_b}') \cap V(\Gamma _{i_c}')=\emptyset $ for all $b\neq c$ . Suppose, for some $b \neq c$ , $V(\Gamma _{i_b}') \cap V(\Gamma _{i_c}')\neq \emptyset $ . By the construction of $\Gamma _{i_j}'$ , it follows that either $V(\Gamma _{i_b}) \subset V(\Gamma _{i_c}')$ or $V(\Gamma _{i_c}) \subset V(\Gamma _{i_b}')$ . Without loss of generality, we assume $V(\Gamma _{i_b}) \subset V(\Gamma _{i_c}')$ . Since $\mathcal {E}$ is t-separated, the parent of $x_{i_b}$ cannot be contained in $V(\Gamma _{i_c})$ , we get

$$ \begin{align*} \mathrm{ht}(\Gamma) \geq \mathrm{ht}(\Gamma_{i_c}') &\geq \mathrm{ht}(\Gamma_{i_c})+(\mathrm{ ht}(\Gamma_{i_b})+1)+1 \\ & =(t-1)+t+1=2t. \end{align*} $$

This contradicts the given hypothesis, and thus completes the claim. Since $\Gamma $ is a perfect rooted tree, $(\Gamma _{i_j}',x_{i_j})$ is also a perfect rooted tree. Therefore, for each $0 \le m \le \mathrm {ht}(\Gamma _{i_j}')-3,$

$$ \begin{align*}\sum\limits_{\ell_{\Gamma_{i_j}'}(x)=m} \deg_{\Gamma_{i_j}'}^+(x) < \sum\limits_{\ell_{\Gamma_{i_j}'}(x)=m+1} \deg_{\Gamma_{i_j}'}^+(x).\end{align*} $$

Then,

$$ \begin{align*} \zeta_t(\Gamma_{i_j}) = 1+\sum\limits_{\ell_{\Gamma_{i_j}}(x)=0}^{\mathrm{ht}(\Gamma_{i_j})-2} \deg_{\Gamma_{i_j}}^+(x) =1+\sum\limits_{\ell_{\Gamma_{i_j}'}(x)=0}^{t-3} \deg_{\Gamma_{i_j}}^+(x) \leq 1+\sum\limits_{\ell_{\Gamma_{i_j}'}(x)=\mathrm{ ht}(\Gamma_{i_j}')-t+1}^{\mathrm{ht}(\Gamma_{i_j}')-2} \deg_{\Gamma_{i_j}'}^+(x). & \end{align*} $$

Let $x \in V(\Gamma _{i_j}')$ be a vertex so that $\ell _{\Gamma _{i_j}'}(x)=\mathrm {ht}(\Gamma _{i_j}')-a$ for some $2 \le a \le \mathrm {ht}(\Gamma _{i_j}')$ , then x is a vertex of $\Gamma $ with $\ell _{\Gamma }(x)=\mathrm {ht}(\Gamma )-a$ and $\deg _{\Gamma _{i_j}'}^+(x)=\deg _{\Gamma }^+(x)$ . Thus, we obtain

$$ \begin{align*} \zeta_t(\mathcal{E}) &=\sum_{j=1}^k \zeta_t(\Gamma_{i_j}) \le \sum\limits_{j=1}^{k} \left(1+ \sum\limits_{\ell_{\Gamma_{i_j}'}(x)=\mathrm{ht}(\Gamma_{i_j}')-t+1}^{\mathrm{ht}(\Gamma_{i_j}')-2} \deg_{\Gamma_{i_j}'}^+(x) \right) \leq \sum\limits_{\ell_{\Gamma}(x)=\mathrm{ht}(\Gamma) -t}^{\mathrm{ ht}(\Gamma)-2}\deg^{+}_{\Gamma}(x). \end{align*} $$

Next, let $A=\{x\in V(\Gamma )~:~\ell _{\Gamma }(x)=\mathrm {ht}(\Gamma )-t+1\}$ . For each $z \in A$ , let $\Gamma _z$ denote the z-descendant rooted tree of $\Gamma $ . By construction, $\mathcal {E}"=\{(\Gamma _z,z)~:~ z \in A\}$ is a t-separated set. Therefore, by Definition 3.2,

$$ \begin{align*}\zeta_t(\Gamma) \ge \zeta_t(\mathcal{E}") =|A|+ \sum\limits_{\ell_{\Gamma}(x)=\mathrm{ht}(\Gamma) -t+1}^{\mathrm{ ht}(\Gamma)-2}\deg^{+}_{\Gamma}(x)=\sum\limits_{\ell_{\Gamma}(x)=\mathrm{ht}(\Gamma) -t}^{\mathrm{ ht}(\Gamma)-2}\deg^{+}_{\Gamma}(x).\end{align*} $$

This proves the claim.

Proof. By Remark 3.4, it suffices to establish the assertion for $C_t(\Gamma )$ . Thus, we assume, without loss of generality, that $\Gamma =C_t(\Gamma ).$ We use induction on $\mathrm {ht}(\Gamma ).$ The case for ${\mathrm {ht}(\Gamma )=1}$ is trivial. So, we assume that $\mathrm {ht}(\Gamma )>1$ . As we have seen in Theorem 3.6 that $I_t(\Gamma )=(x_0)I_{t-1}(\Gamma \setminus \{x_0\})$ . Therefore, $I_t(\Gamma )^s=(x_0^s)I_{t-1}(\Gamma \setminus \{x_0\})^s.$ Let $x_{i_1},\ldots ,x_{i_k}$ be descents of $x_{0}$ in $\Gamma $ . Then, $\Gamma \setminus \{x_0\}$ is the disjoint union of perfect rooted trees, say, $(\Gamma _1,x_{i_1}),\ldots , (\Gamma _k,x_{i_k})$ each of height $\mathrm {ht}(\Gamma )-1.$ Note that $t-1=\mathrm {ht}(\Gamma )=\mathrm {ht}(\Gamma _j)+1$ for each j. Thus, by induction,

$$ \begin{align*}\mathrm{reg}\left(\frac{R}{I_{t-1}(\Gamma_j)^s}\right) = (t-1)(s-1)+\mathrm{ reg}\left(\frac{R}{I_{t-1}(\Gamma_j)}\right).\end{align*} $$

Now, applying [Reference Hà, Trung and Trung17, Lemma 2.3] and [Reference Nguyen and Vu32, Theorem 1.1] repeatedly, we get that

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I_t(\Gamma)^s}\right)&= s+(t-1)(s-1)+\sum\limits_{j=1}^k\mathrm{ reg}\left(\frac{R}{I_{t-1}(\Gamma_j)}\right)\\&= t(s-1)+1+\sum_{j=1}^k \mathrm{reg}\left(\frac{R}{I_{t-1}(\Gamma_j)}\right)\\&= t(s-1)+\mathrm{reg}\left(\frac{R}{I_t(\Gamma)}\right), \end{align*} $$

where the last equality follows from the proof of Theorem 3.6. This completes the assertion.

Proof. Let T be a t-rooted star in $\Gamma $ . By Definition 3.2, T is a perfect rooted tree of $\mathrm {ht}(T)=t-1$ . Therefore, by Theorem 4.1, $\mathrm {reg}{\left (\frac {R}{I_t(T)^s}\right )} = t(s-1)+\mathrm { reg}\left (\frac {R}{I_t(T)}\right )$ for all $s \geq 1.$ Now, let $\mathcal {E}=\{(\Gamma _{i_1},x_{i_1}),\ldots ,(\Gamma _{i_k},x_{i_k})\}$ be a t-star packing of $\Gamma $ so that $\zeta _t(\Gamma )=\zeta _t(\mathcal {E})$ . Let $\Gamma '$ be the induced subforest of $\Gamma $ on the vertex set $\bigcup \limits _{j=1}^k V(\Gamma _{i_j})$ . By Definition 3.2, $\Gamma '$ is a rooted forest with k connected components $(\Gamma _{i_1},x_{i_1}),\ldots ,(\Gamma _{i_k},x_{i_k})$ . This implies that $I_t(\Gamma ')=\sum \limits _{j=1}^kI_t(\Gamma _{i_j})$ . We claim that $\mathrm {reg}{\left (\frac {R}{I_t(\Gamma ')^s}\right )} = t(s-1)+\mathrm { reg}\left (\frac {R}{I_t(\Gamma ')}\right )$ for all $ s\ge 1.$ We proceed by induction on k. If $k=1$ , then $\Gamma '$ is a t-rooted star, and hence, the claim immediately follows. Suppose $k>1$ and the claim is true for $k-1$ . Set $\Gamma "$ to be the induced subforest of $\Gamma $ on the vertex set $\bigcup \limits _{j=1}^{k-1} V(\Gamma _{i_j})$ . Then, $I_t(\Gamma ")=\sum \limits _{j=1}^{k-1}I_t(\Gamma _{i_j})$ and $I_t(\Gamma ') =I_t(\Gamma ")+I_t(\Gamma _{i_k}).$ By induction, $\mathrm { reg}{\left (\frac {R}{I_t(\Gamma ")^s}\right )} = t(s-1)+\mathrm {reg}\left (\frac {R}{I_t(\Gamma ")}\right )$ for all $ s\ge 1$ and $\mathrm { reg}{\left (\frac {R}{I_t(\Gamma _{i_k})^s}\right )} = t(s-1)+\mathrm {reg}\left (\frac {R}{I_t(\Gamma _{i_k})}\right )$ for all $ s\ge 1.$ Now, it follows from [Reference Nguyen and Vu32, Theorem 1.1] that

$$ \begin{align*} \mathrm{reg}{\left(\frac{R}{I_t(\Gamma')^{s}}\right)}&= t(s-1)+ \mathrm{reg}{\left(\frac{R}{I_t(\Gamma")}\right)} +\mathrm{reg}{\left(\frac{R}{I_t(\Gamma_{i_k})}\right)}\\ &= t(s-1)+ \mathrm{reg}{\left(\frac{R}{I_t(\Gamma" \sqcup \Gamma_{i_k})}\right)}\\ &=t(s-1)+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma')}\right)}. \end{align*} $$

This completes the claim. Since $\Gamma '$ is an induced forest of $\Gamma $ , using similar arguments as in [Reference Beyarslan, Hà and Trung3, Lemma 4.2], we get

$$ \begin{align*} \mathrm{reg}{\left(\frac{R}{I_t(\Gamma)^{s}}\right)} \geq \mathrm{reg}{\left(\frac{R}{I_t(\Gamma')^{s}}\right)}=t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma')}\right)}. \end{align*} $$

By Theorem 3.10 and Definition 3.2,

$$ \begin{align*}\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma')}\right)}=\zeta_t(\Gamma')=\sum\limits_{j=1}^k \zeta_t(\Gamma_{i_j})=\zeta_t(\Gamma)=\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma)}\right)}.\end{align*} $$

Hence, the assertion follows.

Notation 4.3. Let $\Delta $ be a simplicial forest on the vertex set $[n]$ . Let $F_1,\ldots ,F_r$ denote the facets of $\Delta $ . Without loss of generality, assume that $F_1,\ldots ,F_r$ form a good leaf ordering on the facets of $\Delta .$ For each $1 \le i \le r$ , set $m_i=\prod \limits _{j \in F_i} x_j$ For $1 \leq i \leq r-1,$ we define ${\Delta _i=\langle F_1,\ldots ,F_i \rangle }$ and $J_i=\langle m_{i+1},\ldots ,m_r \rangle .$

Proof.

1. 1) Since $F_r$ is a good leaf of $\Delta $ , it follows from the proof of [Reference Caviglia, Hà, Herzog, Kummini, Terai and Trung8, Theorem 5.1] that $I(\Delta )^{s+1} : m_r = I(\Delta )^s.$

2. 2) Fix $ 1 \leq i \leq r-1$ . Note that $\left ( I(\Delta _i)^{s+1}+J_i \right ):m_i= I(\Delta _i)^{s+1}:m_i+(J_i:m_i)$ . Since $F_1,\ldots , F_r$ is a good leaf order on the facets of $\Delta $ , $F_i$ is a good leaf of the subcomplex with facets $F_1,\ldots ,F_i.$ Therefore, by the proof of [Reference Caviglia, Hà, Herzog, Kummini, Terai and Trung8, Theorem 5.1], $I(\Delta _i)^{s+1} : m_i = I(\Delta _i)^s$ . Thus, $\left ( I(\Delta _i)^{s+1}+J_i \right ):m_i= I(\Delta _i)^s+(J_i:m_i)$ for all $1 \leq i \leq r-1$ .

3. 3) The facts that $J_1+(m_1)=I(\Delta )$ and $I(\Delta _1)=(m_1)$ give us the desired result.

4. 4) Fix $2 \leq i \leq r-1$ . Consider

   $$ \begin{align*} \left( I(\Delta_i)^{s+1}+J_i \right)+(m_i)&= \left(\left( I(\Delta_{i-1})+m_i\right)^{s+1}+J_i \right)+(m_i) \\ &= \sum\limits_{j=0}^{s+1}m_i^jI(\Delta_{i-1})^{s+1-j}+J_i+(m_i) \\ &= I(\Delta_{i-1})^{s+1}+J_i+(m_i) \\ &= I(\Delta_{i-1})^{s+1}+J_{i-1}. \end{align*} $$
   Hence, the assertion follows.

Proof. Using Lemma 4.4, we get the following short exact sequences:

$$\begin{align*}0 \rightarrow \frac{R}{I(\Delta)^{s}}(-d_r) \xrightarrow{\cdot m_r} \frac{R}{I(\Delta)^{s+1}} \rightarrow \frac{R}{I(\Delta_{r-1})^{s+1}+J_{r-1}} \rightarrow 0, \end{align*}$$

for $2 \leq i \leq r-1,$

$$\begin{align*}0 \rightarrow \frac{R}{I(\Delta_{i})^{s}+(J_{i}:m_{i})}(-d_{i}) \xrightarrow{\cdot m_i} \frac{R}{I(\Delta_{i})^{s+1}+J_{i}} \rightarrow \frac{R}{I(\Delta_{i-1})^{s+1}+J_{i-1}} \rightarrow 0, \end{align*}$$

and

$$\begin{align*}0 \rightarrow \frac{R}{ I(\Delta_1)^{s}+(J_1 :m_1)}(-d_{1}) \xrightarrow{\cdot m_1} \frac{R}{I(\Delta_{1})^{s+1}+J_1} \rightarrow \frac{R}{I(\Delta)} \rightarrow 0. \end{align*}$$

Now using [Reference Hà, Trung and Trung17, Lemma 1.2], we get

(4.1) $$ \begin{align} \mathrm{reg}{\left(\frac{R}{I(\Delta)^{s+1}}\right)} \leq \max \left\{d_r+\mathrm{reg}{\left(\frac{R}{I(\Delta)^{s}}\right)}, \mathrm{ reg}{\left(\frac{R}{I(\Delta_{r-1})^{s+1}+J_{r-1}}\right)} \right\}, \end{align} $$

for $2 \leq i \leq r-1,$

(4.2) $$ \begin{align} \mathrm{reg}{\left(\frac{R}{I(\Delta_{i})^{s+1}+J_{i}}\right)} \leq \max \left\{d_{i}+\mathrm{ reg}{\left(\frac{R}{I(\Delta_{i})^{s}+(J_{i}:m_{i})}\right)}, \mathrm{reg}{\left(\frac{R}{I(\Delta_{i-1})^{s+1}+J_{i-1}} \right)} \right\} \end{align} $$

and

(4.3) $$ \begin{align} \mathrm{reg}{\left(\frac{R}{I(\Delta_{1})^{s+1}+J_{1}}\right)} \leq \max \left\{d_{1}+\mathrm{ reg}{\left(\frac{R}{I(\Delta_{1})^{s}+(J_{1}:m_{1})}\right)},\mathrm{reg}\left(\frac{R}{I(\Delta)}\right) \right\}. \end{align} $$

Combining Equations (1)–(3), we obtain the desired result.

Notation 4.7. Let $(\Gamma ,x_0)$ be a broom graph of $\mathrm {ht}(\Gamma )=h \ge 1$ and $ 2 \le t \leq \mathrm {ht}(\Gamma )+1$ . For $0 \leq i < h$ , let $l_i$ denote the number of leaves in $\Gamma $ at level i and $l_h$ denote the number of leaves at level h minus one. Assume that $l_i = 0 $ if $i<t-1$ . Then, for $0 \leq i \leq h$ , the number of vertices in $\Gamma $ at level i is $l_i+1$ . For $0 \leq i \leq h$ , let $V_i(\Gamma )=\{ x_{(i,0)},x_{(i,1)},\ldots , x_{(i,l_i)}\}$ be the set of vertices of $\Gamma $ that belongs to level i. Assume, without loss of generality, that $x_{(0,0)},\ldots ,x_{(h,0)}$ are the vertices of the handle. Now, for $0 \leq i \leq h-t+1$ and $0 \leq j \leq l_i$ , set $F_{(i,j)} =\{x_{(i,0)},x_{(i+1,0)},\ldots ,x_{(i+t-2,0)}, x_{(i+t-1,j)}\}$ and $m_{(i,j)} =\prod \limits _{x\in F_{(i,j)}} x$ . Let $\Delta $ denote the simplicial complex whose facets are $\{F_{(i,j)} \; : \; 0 \leq i \leq h-t+1, 0 \leq j \leq l_i\}$ . Then, the facet ideal $I(\Delta )$ of $\Delta $ is the t-path ideal of $\Gamma .$ Our aim is to find a good leaf ordering on the facets of $\Delta $ . Let $A= \{ (i,j) \; : \; 0 \leq i \leq h-t+1, 0 \leq j \leq l_i\}.$ For $(i,j), (k,l) \in A$ , we declare $(i,j) < (k,l)$ if either $i< k$ or if $i=k$ , then $j> l.$ We claim that the ordering on the facets of $\Delta $ induced by the order on A is a good leaf ordering. For $(i,j) \in A$ , let $\Delta _{(i,j)}$ denote the simplicial complex whose facets are $\left \{ F_{(k,l)} \; : \; (k,l)\le (i,j) \right \}.$ Then, by [Reference Herzog, Hibi, Trung and Zheng22], it is enough to prove that $F_{(i,j)}$ is a good leaf of $\Delta _{(i,j)}.$ Let $(k,l) \in A$ be such that $(k,l) < (i,j)$ . If $k=i$ , then $j < l$ , and therefore, $F_{(i,j)} \cap F_{(i,l)} =\{x_{(i,0)},x_{(i+1,0)},\ldots ,x_{(i+t-2,0)}\}$ . Suppose that $k \neq i$ , then $k <i$ . Now, we have the following cases: when $l \neq 0$ , then $F_{(i,j)} \cap F_{(k,l)} =\emptyset $ if $k \leq i-t+1$ , otherwise $F_{(i,j)} \cap F_{(k,l)} =\{x_{(i,0)},x_{(i+1,0)},\ldots ,x_{(k+t-2,0)}\} $ . When $l=0$ , then $F_{(i,j)} \cap F_{(k,l)} =\emptyset $ if $k < i-t+1$ , otherwise $F_{(i,j)} \cap F_{(k,l)} =\{x_{(i,0)},x_{(i+1,0)},\ldots ,x_{(k+t-1,0)}\} $ . Consequently, the collection $ \{ F_{(i,j)} \cap F_{(k,l)} \; : \; (k,l) < (i,j) \}$ is a totally ordered set with respect to inclusion, and therefore, $F_{(i,j)}$ is a good leaf of $\Delta _{(i,j)}.$

Proof. First, we claim that $\mathrm {reg}{\left (\frac {R}{I_t(\Gamma )^s}\right )}\leq t(s-1)+\mathrm {reg}\left (\frac {R}{I_t(\Gamma )}\right )$ for all $s \ge 1.$ We proceed by induction on s. If $s=1$ , then the result holds trivially. Assume that the claim is true for s, that is, for any broom graph $\Gamma '$ with $\mathrm {ht}(\Gamma ')=h'$ and $2 \leq t \leq h'+1,$

$$\begin{align*}\mathrm{reg}{\left(\frac{R}{I_t(\Gamma')^s}\right)}\leq t(s-1)+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma')}\right)}. \end{align*}$$

Using Notation 4.7, we know that $I_t(\Gamma ) =I(\Delta )$ , where $\Delta $ is a simplicial tree whose facets are $\{F_{(i,j)} \; : \; (i,j) \in A\}$ . Also, the ordering on the facets of $\Delta $ induced by the total order on A is a good leaf ordering. Set $J_{(i,j)}=\left ( m_{(k,l)} \; : \; (i,j) < (k,l) \right )$ for all $(i,j) \in A$ . By Lemma 4.5, we get

(4.4) $$ \begin{align} \mathrm{reg}{\left(\frac{R}{I(\Delta)^{s+1}}\right)} \leq \max \left\{t+\mathrm{reg}{\left(\frac{R}{I(\Delta)^{s}}\right)}, \alpha, \mathrm{reg}\left(\frac{R}{I(\Delta)}\right) \right\}, \end{align} $$

where

$$ \begin{align*}\alpha =\max\limits_{(i,j) \in A\setminus \{(h-t+1,0)\}} \left\{ t+ \mathrm{ reg}{\left(\frac{R}{I(\Delta_{(i,j)})^{s}+(J_{(i,j)}:m_{(i,j)})}\right)}\right\}.\end{align*} $$

Note that $I(\Delta _{(i,j)})=\left (m_{(k,l)} \; : \; (k,l) \le (i,j) \right )$ (see Notation 4.7) for all $(i,j) \in A$ . By construction and Notation 4.7, it follows that

$$ \begin{align*}J_{(i,j)}:m_{(i,j)} = ( x_{(i+t-1,l)}\;:\; l<j)+ I_t \left( \Gamma_{ {(i+t,0)}} \right)\end{align*} $$

if $ j \neq 0$ and

$$ \begin{align*}J_{(i,0)}:m_{(i,0)} = ( x_{(i+t,l)}~:~ 0 \le l\leq l_{i+t})+ I_t \left( \Gamma_{ {(i+t+1,0)}} \right),\end{align*} $$

where $\Gamma _{(i,0)}$ denote the $x_{i,0}$ -descendant rooted tree of $\Gamma .$ Consequently, for all $(i,j) \in A$ , $I(\Delta _{(i,j)})$ and $J_{(i,j)}:m_{(i,j)}$ are monomial ideals generated in a disjoint set of variables, and therefore,

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I(\Delta_{(i,j)})^s+(J_{(i,j)}:m_{(i,j)})}\right)&=\mathrm{ reg}{\left(\frac{R}{I(\Delta_{(i,j)})^s}\right)}+\mathrm{reg}{\left(\frac{R}{J_{(i,j)}:m_{(i,j)}}\right)}. \end{align*} $$

Let $\Gamma _{ \leq (i,j)}$ denote the induced subtree of $\Gamma $ such that $I(\Delta _{(i,j)})=I_t(\Gamma _{\leq (i,j)}).$ Note that by construction, $\Gamma _{ \leq (i,j)}$ is a broom graph. Therefore, by induction,

$$ \begin{align*}\mathrm{ reg}{\left(\frac{R}{I(\Delta_{(i,j)})^s}\right)}\leq t(s-1)+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma_{\leq (i,j)})}\right)}.\end{align*} $$

Note that $\Gamma _{\leq (i,j)} \sqcup \Gamma _{(i+t,0)}$ and $\Gamma _{\leq (i,0)} \sqcup \Gamma _{(i+t+1,0)}$ are disjoint union of induced subtrees of  $\Gamma .$ Thus, if $j \neq 0,$

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I(\Delta_{(i,j)})^s+(J_{(i,j)}:m_{(i,j)})}\right)&=\mathrm{ reg}{\left(\frac{R}{I(\Delta_{(i,j)})^s}\right)}+\mathrm{reg}{\left(\frac{R}{J_{(i,j)}:m_{(i,j)}}\right)} \\ & \leq t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma_{\leq (i,j)})}\right)}+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma_{(i+t,0)})}\right)} \\& =t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma_{\leq (i,j)}\sqcup \Gamma_{(i+t,0)})}\right)} \\& \leq t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma)}\right)}, \end{align*} $$

where the last inequality follows from Lemma 3.9. Similarly,

$$ \begin{align*} \mathrm{reg}\left(\frac{R}{I(\Delta_{(i,0)})^s+(J_{(i,0)}:m_{(i,0)})}\right)&=\mathrm{ reg}{\left(\frac{R}{I(\Delta_{(i,0)})^s}\right)}+\mathrm{reg}{\left(\frac{R}{J_{(i,0)}:m_{(i,0)}}\right)} \\ & \leq t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma_{\leq (i,0)})}\right)}+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma_{(i+t+1,0)})}\right)} \\& =t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma_{\leq (i,0)}\sqcup \Gamma_{(i+t+1,0)})}\right)} \\& \leq t(s-1)+\mathrm{ reg}{\left(\frac{R}{I_t(\Gamma)}\right)}. \end{align*} $$

Therefore,

$$ \begin{align*} \alpha = \max\limits_{(i,j) \in A\setminus \{(h-t+1,0)\}} \left\{ t+ \mathrm{ reg}{\left(\frac{R}{I(\Delta_{(i,j)})^{s}+(J_{(i,j)}:m_{(i,j)})}\right)}\right\} \leq ts+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma)}\right)}. \end{align*} $$

Also, by applying induction, we have

$$ \begin{align*}\mathrm{reg}{\left(\frac{R}{I(\Delta)^{s}}\right)} \leq t(s-1)+\mathrm{reg}{\left(\frac{R}{I_t(\Gamma)}\right)}.\end{align*} $$

Using Equation (4.4), $\mathrm { reg}{\left (\frac {R}{I(\Delta )^{s+1}}\right )} \leq ts+\mathrm {reg}{\left (\frac {R}{I_t(\Gamma )}\right )}.$ By induction, the claim follows. Furthermore, the assertion immediately follows from Lemma 4.2.

Proof. The equivalence of $(1)$ and $(2)$ follows from [Reference Bouchat, Hà and O’Keefe5, Theorem 4.5] or from Theorem 3.10. The rest follows from Theorem 2.5.