Proof. It is known that $ \operatorname {\mathrm {depth}} S/I\geq \operatorname {\mathrm {depth}} S/\mathrm {in}_<(I)$ (see, e.g., [Reference Herzog and Hibi10, Theorem 3.3.4]). Hence, replacing I by $\mathrm {in}_<(I)$ , we may suppose that I is a monomial ideal. We use induction on $|A|$ . There is nothing to prove for $|A|=0$ , as in this case $q=0$ . Therefore, assume that $|A|\geq 1$ . If $t_i=0$ , for every $i=1, 2, \ldots , q$ , then it follows from condition (ii) that $x_{1,0}, \ldots , x_{q,0}$ do not divide the minimal monomial generators of I. In particular, they form a regular sequences on $S/I$ and the assertion follows. Thus, suppose that $t_i\geq 1$ , for some i with $1\leq i\leq q$ . Without loss of generality, suppose $i=1$ . Consider the following short exact sequence:

$$ \begin{align*} 0\longrightarrow S/(I:x_{1,t_1})\longrightarrow S/I\longrightarrow S/(I,x_{1,t_1})\longrightarrow 0. \end{align*} $$

This yields that

(1) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S/I \geq \min \big\{\operatorname{\mathrm{depth}} S/(I:x_{1,t_1}), \operatorname{\mathrm{depth}} S/(I,x_{1,t_1})\big\}. \end{array} \end{align} $$

By condition (i), the variable $x_{1,t_1}$ does not appear in the minimal monomial generators of $(I:x_{1,t_1})$ . In particular, $x_{1,t_1}$ is a regular element on $S/(I:x_{1,t_1})$ . Let $S^{\prime }$ be the polynomial ring obtained from S by deleting the variable $x_{1,t_1}$ (in other words, $S^{\prime }\cong S/(x_{1,t_1})$ ). Set $I^{\prime }:=(I:x_{1,t_1})\cap S^{\prime }$ . It follows that

$$\begin{align*}\operatorname{\mathrm{depth}} S/(I:x_{1,t_1})=\operatorname{\mathrm{depth}} S/((I:x_{1,t_1}), x_{1,t_1})+1=\operatorname{\mathrm{depth}} S^{\prime}/I^{\prime}+1.\end{align*}$$

Clearly, $I^{\prime }$ satisfies the assumptions with respect to the set $\{x_{i,j} \mid 2\leq i\leq q, 0\leq j\leq t_i\}$ of variables. Thus, the induction hypothesis implies that $ \operatorname {\mathrm {depth}} S^{\prime }/I^{\prime }\geq q-1$ . Hence, we deduce from the above equalities that

$$\begin{align*}\operatorname{\mathrm{depth}} S/(I:x_{1,t_1})\geq q.\end{align*}$$

Using inequality (1), it suffices to prove that $ \operatorname {\mathrm {depth}} S/(I,x_{1,t_1})\geq q$ . Set $I^{\prime \prime }:=I\cap S^{\prime }$ . Then $S/(I,x_{1,t_1})\cong S^{\prime }/I^{\prime \prime }$ . Put $t_1^{\prime }:=t_1-1$ and $t_i^{\prime }:=t_i$ , for $i=2, \ldots , q$ . Obviously, $I^{\prime \prime }$ satisfies the assumptions with respect to the set $\{x_{i,j} \mid 1\leq i\leq q, 0\leq j\leq t_i^{\prime }\}$ of variables. Therefore, we conclude from the induction hypothesis that

$$\begin{align*}\operatorname{\mathrm{depth}} S/(I,x_{1,t_1})=\operatorname{\mathrm{depth}} S^{\prime}/I^{\prime\prime}\geq q.\end{align*}$$

Proof. Let $x_{1,0}, \ldots , x_{q,0}$ be the centers of stars in a largest star packing of G. Moreover, for $1\leq i\leq q$ , assume that $N_G(x_{i,0})=\{x_{i,1}, \ldots , x_{i,t_i}\}$ . Then the assumptions of Proposition 2.1 are satisfied and it follows that

$$\begin{align*}\operatorname{\mathrm{depth}} S/I(G)\geq q=\alpha_2(G).\end{align*}$$

Proof. Let $\mathcal {S}$ be the set of the centers of stars in a largest star packing of G. In particular, $|\mathcal {S}|=\alpha _2(G)$ . Since every vertex in A belongs to the closed neighborhood of a vertex in W, it follows from the definition of star packing that $|\mathcal {S}\cap A|\leq | W|$ . Then the stars in $G\setminus A$ centered at the vertices in $\mathcal {S}\setminus A$ form a star packing in $G\setminus A$ of size at least $\alpha _2(G)-| W|$ . Therefore, $\alpha _2(G\setminus A)\geq \alpha _2(G)-|W|$ .

Proof. Let $\mathcal {S}$ be the set of the centers of stars in a largest star packing of G. Thanks to (i), similar to the proof of Lemma 3.1, we have $|\mathcal {S}\cap A|\leq d$ . If $|\mathcal {S}\cap A|\leq d-1$ , then the stars in $G\setminus A$ centered at the vertices in $\mathcal {S}\setminus A$ form a star packing in $G\setminus A$ of size at least $\alpha _2(G)-d+1$ . Thus, the assertion follows in this case. Therefore, suppose $|\mathcal {S}\cap A|=d$ . In this case, we have

$$\begin{align*}x_1, \ldots, x_d\in \bigcup_{x\in \mathcal{S}\cap A}N_G(x).\end{align*}$$

It again follows from the definition of star packing that

$$\begin{align*}x_1, \ldots, x_d\notin \bigcup_{x\in \mathcal{S}\setminus A}N_{G\setminus A}(x).\end{align*}$$

Therefore, we conclude from condition (ii) that

$$ \begin{align*} N_{G\setminus A}[x_1]\cap \big(\bigcup_{x\in \mathcal{S}\setminus A}N_{G\setminus A}(x)\big)\subseteq \{x_1, \ldots, x_d\}\cap \big(\bigcup_{x\in \mathcal{S}\setminus A}N_{G\setminus A}(x)\big)=\emptyset. \end{align*} $$

As a consequence, the stars in $G\setminus A$ centered at the vertices in $(\mathcal {S}\setminus A)\cup \{x_1\}$ form a star packing in $G\setminus A$ of size $\alpha _2(G)-d+1$ . This completes the proof of the lemma.

Proof. As the isolated vertices have no effect on edge ideals, we assume that $V(H)=V(H^{\prime })=V(G)$ (i.e., we extend the vertex sets of H and $H^{\prime }$ to $V(G)$ ). We use induction on $s+|E(H)|$ . For $s=1$ , we have $I(H)^{(s)}+I(H^{\prime })=I(G)$ and the assertion follows from Corollary 2.2. Therefore, suppose $s\geq 2$ . If $E(H)=\emptyset $ , then $I(H^{\prime })=I(G)$ and again we have the required inequality by Corollary 2.2. Hence, we assume $|E(H)|\geq 1$ .

To simplify the notations, we set $I:=I(H)^{(s)}+I(H^{\prime })$ . Since H is a chordal graph, it has a simplicial vertex, say $x_1$ , with nonzero degree. Without loss of generality, suppose $N_H(x_1)=\big \{x_2, \ldots , x_d\big \}$ , for some integer $d\geq 2$ . Consider the following short exact sequence:

$$ \begin{align*} 0\longrightarrow \frac{S}{(I:x_1\cdots x_d)}\longrightarrow \frac{S}{I}\longrightarrow \frac{S}{(I,x_1\cdots x_d)}\longrightarrow 0. \end{align*} $$

Using depth lemma [Reference Bruns and Herzog2, Proposition 1.2.9], we have

(2) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S/I \geq \min \big\{\operatorname{\mathrm{depth}} S/(I:x_1\cdots x_d), \operatorname{\mathrm{depth}} S/(I,x_1\cdots x_d)\big\}. \end{array} \end{align} $$

By assumption, for every pair of integers $i\neq j$ , with $1\leq i,j\leq d$ we have $x_ix_j\in E(H)$ . Therefore, $x_ix_j$ is not an edge of $H^{\prime }$ . Set

$$\begin{align*}U:=\bigcup_{i=1}^dN_{H^{\prime}}[x_i]\end{align*}$$

and

$$\begin{align*}U^{\prime}:=\bigcup_{i=1}^dN_{H^{\prime}}(x_i).\end{align*}$$

Then using [Reference Seyed Fakhari14, Lemma 2], we have

$$ \begin{align*} (I:x_1\cdots x_d)&=\big((I(H)^{(s)}+I(H^{\prime})):x_1\cdots x_d\big)\\ & =I(H)^{(s-d+1)}+I(H^{\prime}\setminus U) +\big(\text{the ideal generated by}\ U^{\prime}\big)\\ & =I(H\setminus U^{\prime})^{(s-d+1)}+I(H^{\prime}\setminus U)+\big(\text{the ideal generated by }\ U^{\prime}\big). \end{align*} $$

This yields that

$$\begin{align*}\operatorname{\mathrm{depth}} S/(I:x_1\cdots x_d)=\operatorname{\mathrm{depth}} S^{\prime}/(I(H\setminus U^{\prime})^{(s-d+1)}+I(H^{\prime}\setminus U)),\end{align*}$$

where $S^{\prime }=\mathbb {K}[x_i: 1\leq i\leq n, i\notin U^{\prime }]$ . Let $G^{\prime }$ be the union of $H\setminus U^{\prime }$ and $H^{\prime }\setminus U$ . In fact, $G^{\prime }$ is the induced subgraph of G on $V(G)\setminus U^{\prime }$ . Clearly, $N_G(x_1)\setminus \{x_2, \ldots , x_d\}$ is contained in $U^{\prime }$ . Then the above equality together with Lemma 3.2 and the induction hypothesis implies that

(3) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S/(I:x_1\cdots x_d)\geq \alpha_2(G\setminus U^{\prime})-(s-d+1)+1\geq \alpha_2(G)-s+1. \end{array} \end{align} $$

Using inequalities (2) and (3), it is enough to prove that

$$\begin{align*}\operatorname{\mathrm{depth}} S/(I,x_1\cdots x_d)\geq \alpha_2(G)-s+1.\end{align*}$$

For every integer k with $1\leq k\leq d-1$ , let $J_k$ be the ideal generated by all the squarefree monomials of degree k on variables $x_2, \ldots , x_d$ . We continue in the following steps.

Step 1. Let $1\leq k\leq d-2$ be a fixed integer and assume that $\{u_1, \ldots , u_t\}$ is the set of minimal monomial generators of $x_1J_k$ . In particular, every $u_j$ is divisible by $x_1$ and $\mathrm {deg}(u_j)=k+1$ . For every integer j with $1\leq j\leq t$ , we prove that

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/(I+x_1J_{k+1}+(u_1, \ldots, u_{j-1}))\\ & \geq\min\big\{\operatorname{\mathrm{depth}} S/(I+x_1J_{k+1}+(u_1, \ldots, u_j)), \alpha_2(G)-s+1\big\}. \end{align*} $$

(Note that for $j=1$ , we have $I+x_1J_{k+1}+(u_1, \ldots , u_{j-1})=I+x_1J_{k+1}$ .)

Consider the following short exact sequence.

$$ \begin{align*} 0 & \longrightarrow \frac{S}{(I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j}\longrightarrow \frac{S}{I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})}\\ & \longrightarrow \frac{S}{I+x_1J_{k+1}+(u_1, \ldots, u_j)}\longrightarrow 0. \end{align*} $$

As a consequence,

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/(I+x_1J_{k+1}+(u_1, \ldots, u_{j-1}))\geq\\& \min\big\{\operatorname{\mathrm{depth}} S/((I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j), \operatorname{\mathrm{depth}} S/(I+x_1J_{k+1}+(u_1, \ldots, u_j))\big\}. \end{align*} $$

Therefore, to complete this step, it is sufficient to show that

$$\begin{align*}\operatorname{\mathrm{depth}} S/((I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j)\geq \alpha_2(G)-s+1.\end{align*}$$

Set

$$\begin{align*}U_j:=\{x_i\mid 1\leq i \leq d \ \mathrm{and} \ x_i\ \text{does not divide }u_j\}.\end{align*}$$

For any $x_i\in U_j$ , the monomial $x_iu_j$ is a squarefree monomial of degree $k+2$ . Hence, $x_iu_j$ belongs to $x_1J_{k+1}$ . This shows that

$$\begin{align*}(\text{the ideal generated by }\ U_j)\subseteq \big((x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j\big).\end{align*}$$

We show the reverse inclusion holds too.

Since $x_1J_{k+1}+(u_1, \ldots , u_{j-1})$ is a squarefree monomial ideal, it follows that

$$\begin{align*}\big((x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j\big)\end{align*}$$

is also a squarefree monomial ideal. On the other hand, every monomial generator of $x_1J_{k+1}+(u_1, \ldots , u_{j-1})$ is a monomial over $x_1, \ldots , x_d$ . This implies that every monomial generator of $\big ((x_1J_{k+1}+(u_1, \ldots , u_{j-1})):u_j\big )$ is also a squarefree monomial over the variables $x_1, \ldots , x_d$ . Assume that v is a minimal generator of $\big ((x_1J_{k+1}+(u_1, \ldots , u_{j-1})):u_j\big )$ . If v is not equal to any of the variables belonging to $U_j$ , then by definition of $U_j$ , every variable dividing v, also divides $u_j$ . As v is a squarefree monomial, we have $v\mid u_j$ . Since

$$\begin{align*}u_jv\in x_1J_{k+1}+(u_1, \ldots, u_{j-1}),\end{align*}$$

we deduce that

$$\begin{align*}u_j^2\in x_1J_{k+1}+(u_1, \ldots, u_{j-1}),\end{align*}$$

which implies that

$$\begin{align*}u_j\in x_1J_{k+1}+(u_1, \ldots, u_{j-1}),\end{align*}$$

because $x_1J_{k+1}+(u_1, \ldots , u_{j-1})$ is a squarefree monomial ideal. This is contradiction, as the degree of $u_j$ is strictly less than the degree of any monomial in $x_1J_{k+1}$ and $u_j\notin (u_1, \ldots , u_{j-1})$ . Hence,

(4) $$ \begin{align} \begin{array}{rl} \big((x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j\big)=(\text{the ideal generated by} U_j). \end{array} \end{align} $$

Let $W_j$ be the set of variables dividing $u_j$ . In other words, $W_j=\{x_1, \ldots , x_d\}\setminus U_j$ . We remind that for any pair of integers $1\leq i, j\leq d$ , the vertices $x_i$ and $x_j$ are not adjacent in $H^{\prime }$ . Set

$$\begin{align*}U_j^{\prime}:=\bigcup_{x_i\in W_j}N_{H^{\prime}}[x_i]\end{align*}$$

and

$$\begin{align*}U_j^{\prime\prime}:=\bigcup_{x_i\in W_j}N_{H^{\prime}}(x_i).\end{align*}$$

Using equality (4), we conclude that

$$ \begin{align*} & \big((I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j\big)\\ & =\big((I(H)^{(s)}+I(H^{\prime})+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j\big)\\ & =(I(H)^{(s)}:u_j)+I(H^{\prime}\setminus U_j^{\prime})+(\text{the ideal generated by }\ U_j\cup U_j^{\prime\prime})\\ & =\big(I\big(H\setminus (U_j\cup U_j^{\prime\prime})\big)^{(s)}:u_j\big)+I\big(H^{\prime}\setminus (U_j\cup U_j^{\prime})\big)\\ & +(\text{the ideal generated by } U_j\cup U_j^{\prime\prime}). \end{align*} $$

Set $H_j:=H\setminus (U_j\cup U_j^{\prime \prime })$ and $H_j^{\prime }:=H^{\prime }\setminus (U_j\cup U_j^{\prime })$ . Then $H_j$ is a chordal graph, and $x_1$ is a simplicial vertex of $H_j$ . It is also clear that $N_{H_j}[x_1]$ is the set of variables divining $u_j$ . It thus follows from [Reference Seyed Fakhari14, Lemma 2] and the above equalities that

$$ \begin{align*} & \big((I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j\big)=I(H_j)^{(s-k)}+I(H_j^{\prime})\\ & +(\text{the ideal generated by }\ U_j\cup U_j^{\prime\prime}). \end{align*} $$

This yields that

$$\begin{align*}\operatorname{\mathrm{depth}} S/((I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j)=\operatorname{\mathrm{depth}} S_j/(I(H_j)^{(s-k)}+I(H_j^{\prime})),\end{align*}$$

where $S_j=\mathbb {K}[x_i: 1\leq i\leq n, i\notin U_j\cup U_j^{\prime \prime }]$ . Let $G_j$ be the union of $H_j$ and $H_j^{\prime }$ . Then $G_j$ is the induced subgraph of G on $V(G)\setminus (U_j\cup U_j^{\prime \prime })$ . We conclude from Lemma 3.2 (by considering the clique $W_j$ ) that

$$\begin{align*}\alpha_2(G_j)\geq \alpha_2(G)-| W_j| +1=\alpha_2(G)-k,\end{align*}$$

where the last equality follows from the fact that $\mathrm {deg}(u_j)=k+1$ . Hence, the induction hypothesis implies that

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/((I+x_1J_{k+1}+(u_1, \ldots, u_{j-1})):u_j)=\operatorname{\mathrm{depth}} S_j/(I(H_j)^{(s-k)}+I(H_j^{\prime}))\\ & \geq \alpha_2(G_j)-(s-k)+1\geq \alpha_2(G)-s+1, \end{align*} $$

and this step is complete.

Step 2. Let $1\leq k\leq d-2$ be a fixed integer. By a repeated use of Step 1, we have

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/(I+x_1J_{k+1}) \geq\min\big\{\operatorname{\mathrm{depth}} S/(I+x_1J_{k+1}+(u_1, \ldots, u_t)), \alpha_2(G)-s+1\big\}\\ & =\min\big\{\operatorname{\mathrm{depth}} S/(I+x_1J_k), \alpha_2(G)-s+1\big\}. \end{align*} $$

Step 3. It follows from Step 2 that

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/(I,x_1\cdots x_d)=\operatorname{\mathrm{depth}} S/(I+x_1J_{d-1})\\ & \geq\min\big\{\operatorname{\mathrm{depth}} S/(I+x_1J_{d-2}), \alpha_2(G)-s+1\big\}\\ & \geq\min\big\{\operatorname{\mathrm{depth}} S/(I+x_1J_{d-3}), \alpha_2(G)-s+1\big\}\\ & \geq \cdots \geq\min\big\{\operatorname{\mathrm{depth}} S/(I+x_1J_1), \alpha_2(G)-s+1\big\}. \end{align*} $$

In particular,

(5) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S/(I,x_1\cdots x_d)\geq\min\big\{\operatorname{\mathrm{depth}} S/\big(I+(x_1x_2, x_1x_3, \ldots, x_1x_d)\big), \alpha_2(G)-s+1\big\}. \end{array} \end{align} $$

Step 4. Let L be the graph obtained from H, by deleting the edges $x_1x_2, \ldots , x_1x_d$ . Then L is the disjoint union of $H\setminus x_1$ and the isolated vertex $x_1$ . In particular, L is a chordal graph. Also, let $L^{\prime }$ be the graph obtained from $H^{\prime }$ , by adding the edges $x_1x_2, \ldots , x_1x_d$ . Then

$$\begin{align*}E(L)\cap E(L^{\prime})=\emptyset \ \ \ and \ \ \ E(L)\cup E(L^{\prime})=E(G).\end{align*}$$

It follows from [Reference Seyed Fakhari15, Lemma 3.2] and the induction hypothesis that

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/(I+(x_1x_2, x_1x_3, \ldots, x_1x_d))\\ & =\operatorname{\mathrm{depth}} S/(I(H)^{(s)}+I(H^{\prime})+(x_1x_2, x_1x_3, \ldots, x_1x_d)\big)\\ & =\operatorname{\mathrm{depth}} S/((I(L)^{(s)}+I(L^{\prime})))\geq \alpha_2(G)-s+1. \end{align*} $$

Finally, inequality (5) implies that

(6) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S/(I,x_1\cdots x_d)\geq \alpha_2(G)-s+1. \end{array} \end{align} $$

Now, inequalities (2), (3), and (6) complete the proof of the proposition.

Proof. The assertion follows from Proposition 3.3 by substituting $H=G$ and $H^{\prime }=\emptyset $ .

Proof. Let u be a monomial in $\big (I(G)^{(k)}:xy\big )$ . Then $uxy\in I(G)^{(k)}$ . Clearly, this implies that $ux\in I(G)^{(k-1)}$ . Therefore, $u\in \big (I(G)^{(k-1)}:x\big )$ . Similarly, u belongs to $\big (I(G)^{(k-1)}:y\big )$ . Hence,

$$\begin{align*}\big(I(G)^{(k)}:xy\big)\subseteq\big(I(G)^{(k-1)}:x\big)\cap \big(I(G)^{(k-1)}:y\big).\end{align*}$$

To prove the reverse inclusion, let v be a monomial in

$$\begin{align*}\big(I(G)^{(k-1)}:x\big)\cap \big(I(G)^{(k-1)}:y\big).\end{align*}$$

We must show that $vxy\in I(G)^{(k)}$ . It is enough to prove that for any minimal vertex cover C of G, we have $vxy\in \mathfrak {p}_C^k$ . So, let C be a minimal vertex cover of G. It follows from $xy\in E(G)$ that C contains at least one of the vertices x and y. Without loss of generality, suppose $x\in C$ . Since $v\in \big (I(G)^{(k-1)}:y\big )$ , we have $vy\in I(G)^{(k-1)}\subseteq \mathfrak {p}_C^{k-1}$ . This together with $x\in \mathfrak {p}_C$ implies that $vxy\in \mathfrak {p}_C^k$ .

Proof. Set $I:=I(G)$ and let $G(I)=\{u_1, \ldots , u_m\}$ be the set of minimal monomial generators of I. For every integer i with $1\leq i\leq m$ consider the short exact sequence

$$ \begin{align*} 0 & \longrightarrow \frac{S}{(I^{(2)}+(u_1, \ldots, u_{i-1})):u_i}\longrightarrow \frac{S}{I^{(2)}+(u_1, \ldots, u_{i-1})}\\ & \longrightarrow \frac{S}{I^{(2)}+(u_1, \ldots, u_i)}\longrightarrow 0, \end{align*} $$

where for $i=1$ , the ideal $(u_1, \ldots , u_{i-1})$ is the zero ideal. It follows from depth Lemma [Reference Bruns and Herzog2, Proposition 1.2.9] that

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/(I^{(2)}+(u_1, \ldots, u_{i-1}))\\ & \geq \min\big\{\operatorname{\mathrm{depth}} S/((I^{(2)}+(u_1, \ldots, u_{i-1})):u_i), \operatorname{\mathrm{depth}} S/(I^{(2)}+(u_1, \ldots, u_i))\big\}. \end{align*} $$

Using the above inequalities inductively, we have

$$ \begin{align*} & \operatorname{\mathrm{depth}} S/I^{(2)}\\ & \geq\min\bigg\{\operatorname{\mathrm{depth}} S/(I^{(2)}+I), \min\big\{\operatorname{\mathrm{depth}} S/((I^{(2)}+(u_1, \ldots, u_{i-1})):u_i)\mid 1\leq i\leq m\big\}\bigg\}\\ & =\min\big\{\operatorname{\mathrm{depth}} S/I, \operatorname{\mathrm{depth}} S/((I^{(2)}+(u_1, \ldots, u_{i-1})):u_i)\mid 1\leq i\leq m\big\}\\ & \geq\min\big\{\alpha_2(G), \operatorname{\mathrm{depth}} S/((I^{(2)}+(u_1, \ldots, u_{i-1})):u_i)\mid 1\leq i\leq m\big\}, \end{align*} $$

where the last inequality follows from Corollary 2.2. Hence, it is enough to show that

$$\begin{align*}\operatorname{\mathrm{depth}} S/((I^{(2)}+(u_1, \ldots, u_{i-1})):u_i)\geq \alpha_2(G)-1,\end{align*}$$

for every integer i with $1\leq i\leq m$ .

Fix an integer i with $1\leq i\leq m$ and assume that $u_i=xy$ . By [Reference Banerjee1, Theorem 4.12], for every pair of integers $1\leq j< i\leq m$ , one of the following conditions holds.

1. (i) $(u_j:u_i) \subseteq (I^2:u_i)\subseteq (I^{(2)}:u_i)$ ; or

2. (ii) there exists an integer $k\leq i-1$ such that $(u_k:u_i)$ is generated by a variable, and $(u_j:u_i)\subseteq (u_k:u_i)$ .

We know from (i) and (ii) above that

(7) $$ \begin{align} \begin{array}{rl} \big((I^{(2)}+ (u_1, \ldots, u_{i-1})):u_i\big)=(I^{(2)}:u_i)+(\text{some variables}). \end{array} \end{align} $$

Let A be the set of variables belonging to $\big ((I^{(2)}+ (u_1, \ldots , u_{i-1})):u_i\big )$ . Assume that $x\in A$ . This means that $x^2y$ belongs to the ideal $I^{(2)}+(u_1, \ldots , u_{i-1})$ . Since $u_1, \ldots , u_{i-1}$ do not divide $x^2y$ , we deduce that $x^2y\in I(G)^{(2)}$ . But this is a contradiction, as $C:=V(G)\setminus \{x\}$ is a vertex cover of G with $x^2y\notin \mathfrak {p}_C^2$ . Therefore, $x\notin A$ . Similarly, $y\notin A$ . It follows from $x,y\notin A$ and equality (7) that

$$ \begin{align*} & \big((I^{(2)}+ (u_1, \ldots, u_{i-1})):u_i\big)=\big(I^{(2)}:u_i\big)+(\text{the ideal generated by } A)\\ & =\big(I^{(2)}+ (\text{the ideal generated by } A)):u_i\big)\\ & =\big(I(G\setminus A)^{(2)}+(\text{the ideal generated by } A)):u_i\big)\\ & =\big(I(G\setminus A)^{(2)}:u_i\big)+(\text{the ideal generated by } A). \end{align*} $$

Therefore,

(8) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S/((I^{(2)}+ (u_1, \ldots, u_{i-1})):u_i)=\operatorname{\mathrm{depth}} S_A/(I(G\setminus A)^{(2)}:u_i), \end{array} \end{align} $$

where $S_A=\mathbb {K}[x_i: 1\leq i\leq n, x_i\notin A]$ . It follows from Lemma 4.1 that

$$\begin{align*}(I(G\setminus A)^{(2)}:u_i)=(I(G\setminus A):x)\cap(I(G\setminus A):y).\end{align*}$$

Consider the following short exact sequence.

$$ \begin{align*} 0 & \longrightarrow \frac{S_A}{(I(G\setminus A)^{(2)}:u_i)}\longrightarrow \frac{S_A}{(I(G\setminus A):x)}\oplus\frac{S_A}{(I(G\setminus A):y)}\\ & \longrightarrow \frac{S_A}{(I(G\setminus A):x)+(I(G\setminus A):y)}\longrightarrow 0. \end{align*} $$

Applying depth lemma [Reference Bruns and Herzog2, Proposition 1.2.9] on the above exact sequence, it suffices to prove that

1. (a) $ \operatorname {\mathrm {depth}} S_A/(I(G\setminus A):x) \geq \alpha _2(G)-1$ ,

2. (b) $ \operatorname {\mathrm {depth}} S_A/(I(G\setminus A):y) \geq \alpha _2(G)-1$ , and

3. (c) $ \operatorname {\mathrm {depth}} S_A/((I(G\setminus A):x)+(I(G\setminus A):y))\geq \alpha _2(G)-2$ .

To prove (a), note that

$$ \begin{align*} (I(G\setminus A):x)&=I(G\setminus(A\cup N_{G\setminus A}[x]))+(\text{the ideal generated by }\ N_{G\setminus A}(x))\\ & =I(G\setminus(A\cup N_G[x]))+(\text{the ideal generated by }\ N_{G\setminus A}(x)). \end{align*} $$

Hence,

(9) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S_A/(I(G\setminus A):x)=\operatorname{\mathrm{depth}} S^{\prime}/I(G\setminus(A\cup N_G[x])), \end{array} \end{align} $$

where $S^{\prime }=\mathbb {K}[x_i: 1\leq i\leq n, x_i\notin A\cup N_G(x)]$ . Obviously, x is a regular element of $S^{\prime }/I(G\setminus (A\cup N_G[x]))$ . Therefore, Corollary 2.2 implies that

(10) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S^{\prime}/I(G\setminus(A\cup N_G[x]))\geq \alpha_2(G\setminus(A\cup N_G[x]))+1. \end{array} \end{align} $$

Assume that $A\subseteq N_G(x)\cup N_G(y)$ . It then follows from Lemma 3.1 that

$$\begin{align*}\alpha_2(G\setminus(A\cup N_G[x]))\geq \alpha_2(G)-2.\end{align*}$$

Hence, we conclude from equality (9) and inequality (10) that

$$\begin{align*}\operatorname{\mathrm{depth}} S_A/(I(G\setminus A):x)\geq \alpha_2(G)-1.\end{align*}$$

Thus, to complete the proof of (a), we only need to show that $A\subseteq N_G(x)\cup N_G(y)$ .

Let z be an arbitrary variable in A and suppose $z\notin N_G(x)\cup N_G(y)$ . Then the only edge dividing $zu_i=zxy$ is $u_i$ . In particular,

(11) $$ \begin{align} \begin{array}{rl} z\notin \big((u_1, \ldots, u_{i-1}):u_i). \end{array} \end{align} $$

Moreover, since $\{z,x\}$ is an independent subset of G, we conclude that $C^{\prime }=V(G)\setminus \{z,x\}$ is a vertex cover of G with $zu_i=zxy\notin \mathfrak {p}_{C^{\prime }}^2$ . Thus, $zu_i\notin I^{(2)}$ . This means that $z\notin (I^{(2)}:u_i)$ . This, together with (11) implies that

$$\begin{align*}z\notin\big((I^{(2)}, u_1, \ldots, u_{i-1}):u_i\big),\end{align*}$$

which is a contradiction. Therefore, $z\in N_G(x)\cup N_G(y)$ . Hence, $A\subseteq N_G(x)\cup N_G(y)$ and this completes the proof of (a). The proof of (b) is similar to the proof of (a). We now prove (c).

Note that

$$ \begin{align*} & (I(G\setminus A):x)+(I(G\setminus A):y)\\ & =I(G\setminus(A\cup N_{G\setminus A}[x]\cup N_{G\setminus A}[y]))+(\text{the ideal generated by } N_{G\setminus A}[x]\cup N_{G\setminus A}[y])\\ & =I(G\setminus(A\cup N_G[x]\cup N_G[y]))+(\text{the ideal generated by }\ N_{G\setminus A}[x]\cup N_{G\setminus A}[y])\\ & =I(G\setminus(N_G[x]\cup N_G[y]))+(\text{the ideal generated by }\ N_{G\setminus A}[x]\cup N_{G\setminus A}[y]), \end{align*} $$

where the last equality follows from $A\subseteq N_G(x)\cup N_G(y)$ . We conclude that

(12) $$ \begin{align} \begin{array}{rl} \operatorname{\mathrm{depth}} S_A/((I(G\setminus A):x)+(I(G\setminus A):y))=\operatorname{\mathrm{depth}} S^{\prime\prime}/I(G\setminus(N_G[x]\cup N_G[y])), \end{array} \end{align} $$

where $S^{\prime \prime }=\mathbb {K}\big [x_i: 1\leq i\leq n, x_i\notin N_G[x]\cup N_G[y]\big ]$ . Using Corollary 2.2 and Lemma 3.1, we deuce that

$$\begin{align*}\operatorname{\mathrm{depth}} S^{\prime\prime}/I(G\setminus(N_G[x]\cup N_G[y]))\geq \alpha_2(G\setminus(N_G[x]\cup N_G[y]))\geq \alpha_2(G)-2.\end{align*}$$

Finally, the assertion of (c) follows from equality (12) and the above inequality. This completes the proof of the theorem.