Proof . Note that the 3-trampoline is the tent. Hence, if G is strongly chordal and claw-free, then G is clearly chordal, claw-free, and tent-free. Conversely, if G is chordal, claw-free, and tent-free, then G is n-trampoline-free since every n-trampoline with $n> 3$ contains an induced claw, and so, G is strongly chordal.

Proof . We may assume that the vertices of G have been labeled so that $1, 2, \dots , n$ is a strong elimination order. Then $1$ is a simple vertex of G, and we can choose j so that $N[j]$ is maximal in the totally ordered set $\{ N[i] \mid i \in N[1]\}$ . We also note that $N[1]$ is minimal in $\{ N[i] \mid i \in N[1]\}$ . Since $1$ is in particular a simplicial vertex of G, it follows that $N[1]$ is the unique maximal clique of G containing $1$ .

We claim that the edge $e = \{1, j\}$ is simplicial. We have already shown that $N[1]$ is the unique maximal clique of G containing e, so it remains to show that $G \smallsetminus e$ is chordal. If $G \smallsetminus e$ is not chordal, then it contains an induced cycle C of length at least 4. As G is chordal, e must be the unique chord of C, and so C has length 4. Hence, G has an induced subgraph of the form

which is impossible as $i, k \in N[1]$ and 1 is a simplicial vertex. Hence, e is a simplicial edge of G as claimed.

To see that $G \smallsetminus e$ is still strongly chordal, we note that $N_{G \smallsetminus e}[1] = N_G[1] \smallsetminus \{j\}$ and that $N_{G \smallsetminus e}[i] = N_G[i]$ for all $i \neq 1, j$ , so the set

$$\begin{align*}\{N_{G \smallsetminus e}[i] \mid i \in N_{G \smallsetminus e}[1]\} = \{N_G[1] \smallsetminus \{j\}\} \cup \{N_G[i] \mid i \in N_{G}[1] \smallsetminus \{j\} \} \end{align*}$$

is still totally ordered by inclusion. Hence, $1$ is still a simple vertex of $G \smallsetminus e$ , and the graph $(G \smallsetminus e) \smallsetminus \{1\} = G \smallsetminus \{1\}$ has $2, \dots , n$ as a simple elimination order. Thus, $1, 2, \dots , n$ is a simple elimination order for $G \smallsetminus e$ , so $G \smallsetminus e$ is strongly chordal by Theorem 2.3.

Suppose in addition we know that G is claw-free. If $G \smallsetminus e$ has an induced claw, then G has an induced subgraph of the form

But this is impossible since $k \in N[1]$ implies $l \in N[k] \subseteq N[j]$ .

Proof . The statement of the proposition is precisely [Reference Mohammadi and Sharifan29, 3.7] if we omit the condition that the cycles C are induced. In the statement of that theorem, the monomials are indexed instead by paths from i to j in $G \smallsetminus e$ , but such paths correspond bijectively with cycles of G containing e. We will show that each monomial $g_{C, t}$ is divisible by a suitable monomial $g_{C', t'}$ whenever C admits a chord.

Let $C: i = i_0, i_1, \dots , i_s, i_{s+1} = j$ be any cycle of G containing e. If C has a chord $\{i_p, i_q\}$ for some $p < q$ , then we have a cycle $C': i = i_0, i_1, \dots , i_p, i_q, i_{q+1}, \dots , i_{s+1} = j$ so that

$$\begin{align*}g_{C', t} = y_{i_1} \cdots y_{i_t}x_{i_{t+1}} \cdots x_{i_p}x_{i_q} \cdots x_{i_s} \end{align*}$$

divides $g_{C, t}$ for all $t < p$ ,

$$\begin{align*}g_{C', p} = y_{i_1} \cdots y_{i_p}x_{i_q} \cdots x_{i_s} \end{align*}$$

divides $g_{C, t}$ for all $p \leq t < q$ , and

$$\begin{align*}g_{C', p + t - q + 1} = y_{i_1} \cdots y_{i_p}y_{i_q} \cdots y_{i_t}x_{i_{t+1}} \cdots x_{i_s} \end{align*}$$

divides $g_{C,t}$ for all $t \geq q$ .

Proof . By the preceding proposition, we know that $J_{G \smallsetminus e}: f_e = J_{G_e} + L$ , where

$$\begin{align*}L = (g_{C,t} : C \text{ is an induced cycle of } G, e \subseteq C, \text{ and } 0 \leq t \leq s).\end{align*}$$

If $e = \{i,j\}$ , then for every $k \in U \smallsetminus e$ , we have an induced $3$ -cycle $i,j,k$ so that

$$\begin{align*}(x_k,y_k \mid k \in U\smallsetminus e) \subseteq L. \end{align*}$$

On the other hand, every induced cycle of G has length 3 since G is chordal so that $L = (x_k,y_k \mid k \in U\smallsetminus e)$ .

Next, we show that $J_{G_e} + L = J_{G \smallsetminus e} + L$ . Since $G \smallsetminus e$ is a subgraph of $G_e$ , it is clear that $J_{G \smallsetminus e} + L \subseteq J_{G_e} + L$ . Every additional generator of $J_{G_e}$ but not $J_{G \smallsetminus e}$ is of the form $f_{k, l}$ for some $k,l \in N_{G \smallsetminus e}(i)$ or $k, l \in N_{G \smallsetminus e}(j)$ . If $k, l \in N_{G \smallsetminus e}(i)$ , the induced subgraph on $\{i,j,k,l\}$ has as an edge $\{j,k\}$ , $\{j,l\}$ , or $\{k,l\}$ because e does not belong to an induced claw of G. If $\{k,l\}$ is an edge, then $f_{k, l} \in J_{G \smallsetminus e}$ . On the other hand, if $\{j,l\}$ or $\{j,k\}$ is an edge of G, then either $\{i, j, l\}$ or $\{i, j, k\}$ is a clique containing e, and so, either l or k belongs to U as e is simplicial, in which case it is clear that $f_{k,l} \in L$ . Either way, we see that $f_{k, l} \in J_{G \smallsetminus e} + L$ , and the argument is completely analogous if $k, l \in N_{G \smallsetminus e}(j)$ . Thus, we have $J_{G_e} + L = J_{G \smallsetminus e} + L$ as claimed.

Proof of Theorem 2.1.

We argue by induction on the number of edges of G. If G has no edges, then $J_G = 0$ and the statement holds trivially. So, we may assume that G has at least one edge and that the theorem holds for all claw-free strongly chordal graphs with fewer edges. In that case, since G is strongly chordal and claw-free, it has a simplicial edge e such that $G \smallsetminus e$ is strongly chordal and claw-free by Lemma 2.7. Since $G \smallsetminus e$ is a claw-free strongly chordal graph with fewer edges than G, we know that $R_{G \smallsetminus e}$ is Koszul by induction. We will show that $R_G$ is Koszul as well.

Let K be the unique maximal clique of G containing e. Corollary 2.9 yields the exact sequence of $R_{G \smallsetminus e}$ -modules

$$\begin{align*}0 \to R_H(-2) \stackrel{f_e}{\to} R_{G \smallsetminus e} \to R_G \to 0, \end{align*}$$

where H is the induced subgraph of $G \smallsetminus e$ obtained by deleting the vertices of $K \smallsetminus e$ . Since H is an induced subgraph of $G \smallsetminus e$ , we have that $R_H$ is an algebra retract of $R_{G \smallsetminus e}$ ; this means that we have the homogeneous inclusion $i: \ R_H\subset R_{G \smallsetminus e}$ and homogeneous surjection $\varphi : R_{G \smallsetminus e} \longrightarrow R_H$ so that $\varphi i=id$ (see [Reference Römer and Saeedi Madani38] for other applications of algebra retracts in an algebra-graph setting). Combining this with the assumption that $R_{G \smallsetminus e}$ is Koszul yields that $R_H$ has a linear free resolution as a module over $R_{G \smallsetminus e}$ by [Reference Ohsugi, Herzog and Hibi31, 1.4]. Now, combine this with the above short exact sequence, and conclude that $\operatorname {\mathrm {reg}}_{R_{G \smallsetminus e}} R_G \leq 1$ . It follows that $R_G$ is Koszul by [Reference Conca, De Negri and Evelina Rossi12, Theorem 2].

Proof . By our main theorem 1.2, the ring $R_{G \smallsetminus e}$ is Koszul. Then the proof above implies that $R_G$ is also Koszul. Applying our main theorem once more yields that G is strongly chordal and claw-free.

Proof . ( $\Rightarrow $ ): We argue by induction on the number of edges of G. The statement holds trivially if G has no edges, so we may assume that G has at least one edge and that all claw-free strongly chordal graphs with fewer edges have a claw-avoiding perfect edge elimination order. By Lemma 2.7, there is a simplicial edge $e_1$ of G such that $G \smallsetminus e_1$ is strongly chordal and claw-free. Note that $e_1$ is claw-avoiding as G is claw-free. By induction $G \smallsetminus e_1$ has a claw-avoiding perfect edge elimination order $e_2, \dots , e_m$ , and so $e_1, e_2, \dots , e_m$ is a claw-avoiding perfect edge elimination order for G.

( $\Leftarrow $ ): Let $e_1, \dots , e_m$ be a claw-avoiding perfect edge elimination order for G. Again, we argue by induction on m. The case $m = 0$ being trivial, we may assume that $m \geq 1$ and that the corollary holds for all chordal graphs having a claw-avoiding perfect edge elimination order with fewer edges. Since $e_1$ is a claw-avoiding simplicial edge of G, we know in particular that $G \smallsetminus e_1$ is chordal, and as $G \smallsetminus e_1$ has claw-avoiding perfect edge elimination order $e_2, \dots , e_m$ , it follows by induction that $G \smallsetminus e_1$ is strongly chordal and claw-free. Thus, Corollary 2.10 implies that G is also strongly chordal and claw-free.

Proof . For $n \ge 5$ , see the proof in [Reference Herzog, Hibi and Ohsugi26, 7.47] where it is shown that $J_{C_n}$ has a minimal first syzygy of degree n; for $n \ge 5$ , if $J_{C_n}$ were Koszul, this contradicts [Reference Kempf27, Lemma 4]. For $n = 4$ , consider the graph G shown below

and set $e = \{2, 3\}$ . Note that $G \smallsetminus e = C_4$ . We will show that $\beta ^S_{2,4}(R_{C_4}) = 9> \binom {4}{2}$ , so $R_{C_4}$ is not Koszul by [Reference Boocher, Hamid Hassanzadeh and Iyengar8, 3.4].

Note that $S_G=\Bbbk [x_1, \dots ,x_4, y_1, \dots , y_4].$ First, we observe that the labeling of G is closed, and so $R_G$ has a quadratic Gröbner basis with respect to the lex order with $x_1> \cdots > x_4 > y_1 > \cdots > y_4$ . In particular, we have that the initial ideal is

$$\begin{align*}\operatorname{\mathrm{in}}_>(J_G) = (x_1y_2, x_1y_3, x_2y_3, x_2y_4, x_3y_4), \end{align*}$$

so it is the edge ideal of a path with 6 vertices. Using [Reference Boocher, D’Alì, Grifo, Montaño and Sammartano7, 3.4, 3.5], one can check that the Betti table of $S_G/\operatorname {\mathrm {in}}_>(J_G)$ is:

Since G is a closed graph, $\beta _{i,j}^R(R_G) = \beta _{i,j}^{S_G}(S_G/\operatorname {\mathrm {in}}_>(J_G))$ for all $i,j$ ( see [Reference Peeva35, 1.3]).

To see that $\beta ^S_{2,4}(R_{C_4}) = 9$ , we note that Theorem 2.8 yields the exact sequence

$$\begin{align*}0 \to {S_G}/(x_1, y_1, x_4, y_4)(-2) \xrightarrow{f_{2,3}} R_{C_4} \to R_G \to 0, \end{align*}$$

from which we obtain the exact sequence

$$ \begin{align*} 0 = \operatorname{\mathrm{Tor}}_3^{S_G}(R_G, \Bbbk)_4 &\to \operatorname{\mathrm{Tor}}_2^{S_G}({S_G}/(x_1,y_1, x_4, y_4), \Bbbk)_2 \to \operatorname{\mathrm{Tor}}_2^{S_G}(R_{C_4}, \Bbbk)_4 \\ &\to \operatorname{\mathrm{Tor}}_2^{S_G}(R_G, \Bbbk)_4 \to \operatorname{\mathrm{Tor}}_1^{S_G}({S_G}/(x_1,y_1, x_4, y_4), \Bbbk)_2 = 0. \end{align*} $$

Since $x_1,y_1,x_4,y_4$ is a regular sequence on ${S_G}$ , we have

$$\begin{align*}\beta_{2,4}^S(R_{C_4}) = \beta_{2,2}^{S_G}({S_G}/(x_1, y_1, x_4, y_4)) + \beta^{S_G}_{2,4}(R_G) = 6 + 3 = 9 \end{align*}$$

as desired.

Proof . Consider the claw G labeled as shown below

Thus, the binomial edge ideal is $J_G = (f_{1,2}, f_{1,3}, f_{1,4}) \subseteq S_G = \Bbbk [x_1, y_1, \dots , x_4, y_4]$ . If $e = \{1,4\}$ , then Theorem 2.8 shows that $J_{G \smallsetminus e} : f_e = J_{G_e} = (f_{1,2}, f_{1,3}, f_{2,3})$ , since $G \smallsetminus e$ has no induced cycles. Hence, $J_{G \smallsetminus e} : f_e = I_2(M)$ , where

$$\begin{align*}M = \begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{pmatrix}, \end{align*}$$

which has a Hilbert-Burch resolution. Applying Theorem 2.8 once more to $G' = G \smallsetminus \{4\}$ and the edge $e' = \{1, 3\}$ shows that $(f_{1,2}) : f_{1,3} = J_{G'\smallsetminus e'} : f_{e'} = J_{G^{\prime }_{e'}} = (f_{1,2})$ , so $J_{G \smallsetminus e} = (f_{1,2}, f_{1,3})$ is a complete intersection. We can construct an explicit chain map lifting the multiplication by $f_{1,4}$ map $S_G/I_2(M)(-2) \to R_{G \smallsetminus e}$ as shown below.

The first column of the middle vertical map comes from the Plücker relation $f_{1,2}f_{3,4} - f_{1,3}f_{2,4} + f_{2,3}f_{1,4} = 0$ (see [Reference Bruns and Herzog10, 7.2.3]). Hence, the mapping cone of the above chain map yields the minimal free resolution of $R_G$ , which has Betti table:

In particular, because $\beta ^S_{2,4}(R_G) = 4> \binom {3}{2}$ , we see that $R_G$ is not Koszul by [Reference Boocher, Hamid Hassanzadeh and Iyengar8, 3.4].

Proof . Consider the tent T with labeling as shown in Figure 3.

Figure 3

The tent.

Setting $G = T \smallsetminus \{6\}$ , we note that $R_T \cong \operatorname {\mathrm {Sym}}_{R_G}(M)$ , where M is the cokernel of the matrix

$$\begin{align*}\partial_1 = \begin{pmatrix} -y_2 & -y_4 \\ x_2 & x_4 \end{pmatrix}. \end{align*}$$

If $R_T$ is Koszul, then M has a linear free resolution over $R_G$ by Proposition 4.2. However, we will show that M has a minimal nonlinear fourth syzygy, and so the ring $R_T$ is not Koszul.

To see this, we first note that the labeling of G is closed, and so $J_G$ has a quadratic Gröbner basis with respect to the lex order with $x_1> \cdots > x_5 > y_1 > \cdots > y_5$ . In particular, we have that the initial ideal is

$$\begin{align*}\operatorname{\mathrm{in}}_>(J_G) = (x_1y_2, \, x_1y_3, \, x_2y_3, \, x_2y_4, \, x_3y_4, \, x_3y_5, \, x_4y_5). \end{align*}$$

Thus, the quotient $R_G$ has a monomial basis consisting of all monomials not contained in the above monomial ideal. We also note that $R_G$ admits a $\mathbb {Z}^5$ -grading via $\deg x_i = \deg y_i = \mathbf {e}_i$ , where $\mathbf {e}_i$ is the i-th standard basis vector of $\mathbb {Z}^5$ , and $\operatorname {\mathrm {Ker}} \partial _1$ is homogeneous with respect to this grading. We exploit this grading to simplify the computation of the first few terms of the linear strand of the minimal free resolution of M.

Let $u = (u_1, u_2) \in \operatorname {\mathrm {Ker}} \partial _1 \subseteq R_G(-\mathbf {e}_2) \oplus R_G(-\mathbf {e}_4)$ be a nonzero homogeneous element of degree $\mathbf {a} \in \mathbb {Z}^5$ . Since $(R_G)_{\mathbf {b}} = 0$ if any component of $\mathbf {b}$ is negative, we must have either $\mathbf {e}_2 \leq \mathbf {a}$ or $\mathbf {e}_4 \leq \mathbf {a}$ , where $\mathbf {b} = (b_1, \dots , b_5) \leq (a_1, \dots , a_5) = \mathbf {a}$ means that $b_i \leq a_i$ for all i. If u is a linear syzygy, then we have $\left \vert \mathbf {a} \right \vert = \sum _i a_i = 2$ , so $\mathbf {a} = \mathbf {e}_2 + \mathbf {e}_i, \mathbf {e}_4 + \mathbf {e}_j, \mathbf {e}_2 + \mathbf {e}_4$ for some $i \neq 4$ and $j \neq 2$ . In the first case, we have $u = (\alpha x_i + \beta y_i, 0)$ for some $\alpha , \beta \in \Bbbk $ , so $\partial _1u = 0$ implies $\alpha x_2x_i + \beta x_2y_i = 0$ . As $x_2x_i$ is part of the monomial basis for $R_G$ , it follows that $\alpha = 0$ , and so $\beta x_2y_i = 0$ . If $i \neq 5$ , then $0 = \beta x_2y_i = \beta x_iy_2$ . In any case, either $x_2y_i$ or $x_iy_2$ is part of the monomial basis for $R_G$ , so $\beta = 0$ , which contradicts that u is nontrivial. A similar argument shows that there are no nontrivial elements of $\operatorname {\mathrm {Ker}} \partial _1$ in degree $\mathbf {a} = \mathbf {e}_4 + \mathbf {e}_j$ for $j \neq 2$ . However, when $\mathbf {a} = \mathbf {e}_2 + \mathbf {e}_4$ , we see that $u = (\alpha _1x_4 + \beta _1y_4, \alpha _2x_2 + \beta _2y_2)$ , so $\partial _1u = 0$ implies

$$\begin{align*}0 = \alpha_1x_2x_4 + \beta_1x_2y_4 + \alpha_2x_2x_4 + \beta_2x_4y_2 = (\alpha_1 + \alpha_2)x_2x_4 + (\beta_2 + \beta_1)x_4y_2. \end{align*}$$

As $x_2x_4$ and $x_4y_2$ are part of the monomial basis for $R_G$ , it follows that $\alpha _2 = -\alpha _1$ and $\beta _2 = -\beta _1,$ so

$$\begin{align*}u = \alpha_1\begin{pmatrix} x_4 \\ -x_2 \end{pmatrix} + \beta_1\begin{pmatrix} y_4 \\ -y_2 \end{pmatrix}, \end{align*}$$

and the columns of the matrix

$$\begin{align*}\partial_2 = \begin{pmatrix} x_4 & y_4 \\ -x_2 & -y_2 \end{pmatrix} \end{align*}$$

are easily seen to generate the linear syzygies of $\partial _1$ .

If $u = (u_1, u_2) \in \operatorname {\mathrm {Ker}} \partial _2 \subseteq R(-\mathbf {e}_2-\mathbf {e}_4)^2$ is a nonzero homogeneous element of degree $\mathbf {a}$ , then $\mathbf {a} \geq \mathbf {e}_2 + \mathbf {e}_4$ . If in addition u is a linear syzygy, then $\mathbf {a} = \mathbf {e}_2 + \mathbf {e}_4 + \mathbf {e}_i$ for some i, so $u = (\alpha _1x_i + \beta _1y_i, \alpha _2x_i + \beta _2y_i)$ . Then $\partial _2u = 0$ implies:

$$ \begin{align*} 0 &= \alpha_1x_ix_4 + \beta_1x_4y_i + \alpha_2x_iy_4 + \beta_2y_iy_4 \\ 0 &= \alpha_1x_ix_2 + \beta_1x_2y_i + \alpha_2x_iy_2 + \beta_2y_iy_2 \end{align*} $$

As $x_ix_4$ and $y_iy_4$ are part of the monomial basis for $R_G$ , it follows that $\alpha _1 = \beta _2 = 0$ , so $\beta _1x_4y_i + \alpha _2x_iy_4 = 0$ and $\beta _1x_2y_i + \alpha _2x_iy_2 = 0$ . If $i = 1, 5$ , then $\alpha _2 = \beta _1 = 0$ as well since $x_4y_1, x_1y_4, x_2y_5$ , and $x_5y_2$ are also part of the monomial basis for $R_G$ , which contradicts that u is nontrivial. Hence, we must have $i = 2, 3, 4$ , in which case we have $0 = \beta _1x_4y_i + \alpha _2x_iy_4 = (\beta _1 + \alpha _2)x_iy_4 = (\beta _1 + \alpha _2)x_4y_i$ where either $x_iy_4$ or $x_4y_i$ is part of the monomial basis for $R_G$ , so $\alpha _2 = -\beta _1$ . From this, it is easily seen that the columns of the matrix

$$\begin{align*}\partial_3 = \begin{pmatrix} y_2 & y_4 & y_3 \\ -x_2 & -x_4 & -x_3 \end{pmatrix} \end{align*}$$

generate the linear syzygies of $\partial _2$ .

Let $u = (u_1, u_2, u_3) \in \operatorname {\mathrm {Ker}} \partial _3 \subseteq R(-2\mathbf {e}_2-\mathbf {e}_4) \oplus R(-\mathbf {e}_2-2\mathbf {e}_4) \oplus R(-\mathbf {e}_2-\mathbf {e}_3-\mathbf {e}_4)$ be a nonzero linear syzygy of degree $\mathbf {a}$ . If $\mathbf {a} \ngeq \mathbf {e}_2 + \mathbf {e}_3 + \mathbf {e}_4$ , then $u_3 = 0$ , so $(u_2, u_1) \in \operatorname {\mathrm {Ker}} \partial _1$ and

$$\begin{align*}u = \alpha\begin{pmatrix} -x_2 \\ x_4 \\ 0 \end{pmatrix} + \beta\begin{pmatrix} -y_2 \\ y_4 \\ 0 \end{pmatrix} \end{align*}$$

for some $\alpha , \beta \in \Bbbk $ . So, we may assume that $\mathbf {a} \geq \mathbf {e}_2 + \mathbf {e}_3 + \mathbf {e}_4$ , in which case we have either:

1. 1. $\mathbf {a} = \mathbf {e}_2 +\mathbf {e}_3 + \mathbf {e}_4 + \mathbf {e}_i$ for some $i \neq 2, 4$ ,

2. 2. $\mathbf {a} = \mathbf {e}_2 +\mathbf {e}_3 + 2\mathbf {e}_4$ , or

3. 3. $\mathbf {a} = 2\mathbf {e}_2 +\mathbf {e}_3 + \mathbf {e}_4$ .

In case (1), we have $u = (0, 0, \alpha x_i + \beta y_i)$ for some $\alpha , \beta \in \Bbbk $ , so $\partial _3u = 0$ implies $0 = \alpha x_iy_3 + \beta y_iy_3$ and $0 = \alpha x_ix_3 + \beta y_ix_3$ . As $x_ix_3$ and $y_iy_3$ are part of the monomial basis for $R_G$ , it follows that $\alpha = \beta = 0$ , which contradicts that u is nontrivial. In case (2), we have $u = (0, \alpha _2x_3 + \beta _2 x_3, \alpha _3x_4 + \beta _3 x_4)$ for some $\alpha _i, \beta _i \in \Bbbk $ . Then $\partial _3u = 0$ implies

$$\begin{align*}0 = \alpha_2x_3y_4 + \beta_2y_3y_4 + \alpha_3x_4y_3 + \beta_3y_3y_4 = (\alpha_2 + \alpha_3)x_4y_3 + (\beta_2 + \beta_3)y_3y_4. \end{align*}$$

As $x_4y_3$ and $y_3y_4$ are part of the monomial basis for $R_G$ , we see that $\alpha _3 = -\alpha _2$ and $\beta _3 = -\beta _2$ , so

$$\begin{align*}u = \alpha_2\begin{pmatrix} 0 \\ x_3 \\ -x_4 \end{pmatrix} + \beta_2\begin{pmatrix} 0 \\ y_3 \\ -y_4 \end{pmatrix}. \end{align*}$$

By a similar argument applied to case (3), we see that the columns of the matrix

$$\begin{align*}\partial_4 = \begin{pmatrix} -x_2 & -y_2 & x_3 & y_3 & 0 & 0 \\ x_4 & y_4 & 0 & 0 & x_3 & y_3 \\ 0 & 0 & -x_2 & -y_2 & -x_4 & -y_4 \end{pmatrix} \end{align*}$$

generate the linear syzygies of $\partial _3$ .

Finally, we observe that $x_3f_{1,5} = x_1f_{3,5} + x_5f_{1,3} \in J_G$ and similarly that $y_3f_{1,5} \in J_G$ , so $q = (0, 0, f_{1,5})$ is a quadratic syzygy of $\partial _3$ . If q is a nonminimal syzygy, the above description of the linear syzygies of $\partial _3$ implies that $f_{1,5} \in J_G + (x_2, y_2, x_4, y_4)$ , which is easily seen to imply that $f_{1,5} \in J_P$ where P is the induced path of G on the vertices $\{1,3, 5\}$ . Since the vertex labeling of P is closed, we see that $\operatorname {\mathrm {in}}_>(J_P) = (x_1y_3, x_3y_5)$ , but then $\operatorname {\mathrm {in}}_>(f_{1,5}) = x_1y_5 \notin \operatorname {\mathrm {in}}_>(J_P)$ , which is a contradiction. Therefore, q is a minimal quadratic syzygy of $\partial _3$ , hence also of M. Hence, M does not have a linear resolution over $R_G$ , and $R_T$ is not Koszul.

Proof . (a) $\Leftrightarrow $ (b): This follows immediately from our main theorem 1.2 and [Reference Ene, Herzog and Hibi21, 2.6].

(b) $\Rightarrow $ (c): If G can be decomposed as described in part (b), then G is claw-free by the equivalence of (b) and (a), and the blocks of G are just the blocks of $G_v$ for all v. Since induced subgraphs of closed graphs are closed by [Reference Crupi and Rinaldo17, 2.2], it follows that every block of G is closed.

(c) $\Rightarrow $ (a): To see that G is strongly chordal when it is claw-free and all of its blocks are closed, it suffices by our main theorem 1.2 to note that G is chordal and tent-free. Since cycles and tents are 2-connected, any induced cycle or tent of G would be contained in one of its blocks. Thus, G is chordal and tent-free since the blocks of G are closed.