2 Background

Let $\mathbb {P}^n$ denote n-dimensional projective space obtained from a $\mathbb {C}$ -vector space of dimension $n+1$ . A commutative Noetherian $\mathbb {C}$ -algebra R is said to be graded if $R= \bigoplus _{i \in \mathbb {N}} R_i$ as an Abelian group such that for all nonnegative integers i and j we have $R_iR_j \subseteq R_{i+j},$ and is standard graded if $R_0= \mathbb {C}$ and R is generated as a $\mathbb {C}$ -algebra by a finite set of degree $1$ elements. Additionally, an R-module M is called graded if R is graded and M can be written as $M= \bigoplus _{i \in \mathbb {N}} M_i$ as an Abelian group such that for all nonnegative integers i and j we have $R_iM_j \subseteq M_{i+j}.$ Note each summand $R_i$ and $M_i$ is a $\mathbb {C}$ -vector space of finite dimension. We always assume our rings are standard graded. Let S be the symmetric algebra of $R_1$ over $\mathbb {C};$ that is, S is the polynomial ring $S=\mathbb {C}[x_0,\ldots ,x_n],$ where $\mathrm {dim}(R_1)=n+1$ and $x_0,\ldots ,x_n$ is a $\mathbb {C}$ -basis of $R_1.$ We have an induced surjection $S \rightarrow R$ of standard graded $\mathbb {C}$ -algebras, and so $R \cong S/J,$ where J is a homogenous ideal and the kernel of this map. We say that J defines R and call this ideal J the defining ideal. Denote by $\mathfrak {m}_R$ the maximal homogeneous ideal of $R.$ Except when explicitly said, all rings are graded and Noetherian and all modules are finitely generated. We may view $\mathbb {C}$ as a graded R-module since $\mathbb {C} \cong R/\mathfrak {m}_R.$ The function $\mathrm {Hilb}_M:\mathbb {N} \rightarrow \mathbb {N}$ defined by $\mathrm {Hilb}_M(d) = \mathrm {dim}_{\mathbb {C}}(M_d)$ is called the Hilbert function of the R-module $M.$ Furthermore, there exists a unique polynomial $\mathrm {HilbP}(d)$ with rational coefficents, called the Hilbert polynomial such that $\mathrm {HilbP}(d) = \mathrm {Hilb}(d)$ for $d \gg 0$ .

The minimal graded free resolution $\textbf {F}$ of an R-module M is an exact sequence of homomorphisms of finitely generated free R-modules

$$ \begin{align*}\textbf{F}: \cdots \rightarrow F_n \xrightarrow{d_n} F_{n-1} \xrightarrow{d_{n-1}} \cdots \rightarrow F_1 \xrightarrow{d_1} F_{0},\end{align*} $$

such that $d_{i-1}d_i = 0$ for all $i, M \cong F_0/\mathrm {Im}(d_1),$ and $d_{i+1}(F_{i+1}) \subseteq (x_0,\ldots ,x_n)F_i$ for all $i \geq 0.$ After choosing bases, we may represent each map in the resolution as a matrix. We can write $F_i = \bigoplus _j R(-j)^{\beta _{i,j}^R(M)},$ where $R(-j)$ denotes a rank one free module with a generator in degree $j,$ and the numbers $\beta _{i,j}^R(M)$ are called the graded Betti numbers of M and are numerical invariants of M. The total Betti numbers of M are defined as $\beta _i^R(M) = \sum _{j} \beta _{i,j}^R(M)$ . When it is clear which module we are speaking about, we will write $\beta _{i,j}$ and $\beta _i$ to denote the graded Betti numbers and total Betti numbers, respectively. By construction, we have the equalities

$$ \begin{align*} \beta_i^R(M) &= \mathrm{dim}_{\mathbb{C}}\mathrm{Tor}_i^R(M,\mathbb{C}), \\ \beta_{i,j}^R(M) &= \mathrm{dim}_{\mathbb{C}}\mathrm{Tor}_i^R(M,\mathbb{C})_j. \end{align*} $$

Two more invariants of a module are its projective dimension and relative Castelnuovo–Mumford regularity. These invariants are defined for an R-module M as follows:

$$ \begin{align*}\mathrm{pdim}_R (M) = \sup\{ i\,|\, F_i \neq 0 \}= \sup \{ i \,|\, \beta_i(M) \neq 0 \}, \end{align*} $$

$$ \begin{align*}\mathop{\mathrm{reg}}\nolimits_R (M) = \sup\{ j-i\,|\, \beta_{i,j}(M) \neq 0\}. \end{align*} $$

Both invariants are interesting and measure the growth of the resolution of $M.$ For instance, if $R=S,$ then by Hilbert’s Syzygy Theorem we are guaranteed that $\mathrm {pdim}_S(M) \leq n+1,$ where $n+1$ is the number of indeterminates of S.

Certain invariants are related to one another. For example, if $\mathrm {pdim}_R(M)$ is finite, then the Auslander–Buchsbaum formula relates the projective dimension to the depth of a module (see [Reference Peeva19, Th. 15.3]), where the depth of an R-module M is the length of the largest M-regular sequence consisting of elements of $R,$ and is denoted $\mathrm {depth}(M).$ Letting $R=S,$ the Auslander–Buchsbaum formula states that the projective dimension and depth of an S-module M are complementary to one another:

(1) $$ \begin{align} \mathrm{pdim}_S(M) + \mathrm{depth}(M) = n+1. \end{align} $$

The Krull dimension, or dimension, of a ring is the supremum of the lengths k of strictly increasing chains $P_0 \subset P_1 \subset \ldots \subset P_k$ of prime ideals of $R.$ The dimension of an R-module is denoted $\mathrm {dim}(M)$ and is the Krull dimension of the ring $R/I,$ where $I = \mathrm {Ann}_R(M)$ is the annihilator of $M.$ The depth and dimension of a ring have the following properties along a short exact sequence.

Proposition 2.1. ([Reference Eisenbud11, Cor. 18.6])

Let R be a graded Noetherian ring and suppose that

$$ \begin{align*} 0 \rightarrow M' \rightarrow M \rightarrow M^{"} \rightarrow 0 \end{align*} $$

is an exact sequence of finitely generated graded R-modules. Then

1. (a) $\mathrm {depth}(M^{'}) \geq \mathrm {min}\{ \mathrm {depth}(M), \mathrm {depth}(M^{"}) +1 \}, $

2. (b) $\mathrm {depth}(M) \geq \mathrm {min}\{ \mathrm {depth}(M^{'}), \mathrm {depth}(M^{"}) \}, $

3. (c) $\mathrm {depth}(M^{"}) \geq \mathrm {min}\{ \mathrm {depth}(M), \mathrm {depth}(M^{'}) - 1 \}, $

4. (d) $\mathrm {dim}(M)= \mathrm {max}\{ \mathrm {dim}(M^{"}),\mathrm {dim}(M^{'})\}.$

Furthermore, $\mathrm {depth}(M) \leq \mathrm {dim}(M).$

An R-module M is Cohen–Macaulay, if $\mathrm {depth}(M) = \mathrm {dim}(M).$ Since R is a module over itself, we say R is a Cohen–Macaulay ring if it is a Cohen–Macaulay R-module. Cohen–Macaulay rings have been studied extensively, and the definition is sufficiently general to allow a rich theory with a wealth of examples in algebraic geometry. This notion is a workhorse in commutative algebra, and provides very useful tools and reductions to study rings [Reference Bruns and Herzog6]. For example, if one has a graded Cohen–Macaulay $\mathbb {C}$ -algebra, then one can take a quotient by generic linear forms to produce an Artinian ring. A reduction of this kind is called an Artinian reduction and provides many useful tools to work with, and almost all homological invariants of the ring are preserved [Reference Migliore and Patnott18]. Unfortunately, we will not be able to use these tools or reductions as the coordinate ring of a generic collection of lines is almost never Cohen–Macaulay, whereas the coordinate ring of a generic collection of points is always Cohen–Macaulay.

The absolute Castelnuovo–Mumford regularity, or the regularity, is denoted $\mathrm {reg}_S(M)$ and is the regularity of M as an S-module. There is a cohomological interpretation by local duality [Reference Eisenbud and Goto12]. Set $H_{\mathfrak {m}_S}^i(M)$ to be the $i^{th}$ local cohomology module with support in the graded maximal ideal of $S.$ One has $H_{\mathfrak {m}_S}^i(M) =0$ if $i < \mathrm {depth}(M)$ or $i> \mathrm {dim}(M)$ and

$$ \begin{align*} \mathrm{reg}_S(M) = \max \{j+i : H_{\mathfrak{m}_S}^i(M)_j \neq 0 \}.\end{align*} $$

In practice, bounding the regularity of M is difficult, since it measures the largest degree of a minimal syzygy of M. We have tools to help the study of the regularity of an S-module.

Proposition 2.2. ([Reference Eisenbud11, Exer. 4C.2, Th. 4.2, and Cor. 4.4])

Suppose that

$$ \begin{align*} 0 \rightarrow M' \rightarrow M \rightarrow M^{"} \rightarrow 0 \end{align*} $$

is an exact sequence of finitely generated graded S-modules. Then

1. (a) $\mathrm {reg}_S(M^{'}) \leq \mathrm {max}\{ \mathrm {reg}_S(M), \mathrm {reg}_S(M^{"}) +1 \}, $

2. (b) $\mathrm {reg}_S(M) \leq \mathrm {max}\{ \mathrm {reg}_S(M^{'}), \mathrm {reg}_S(M^{"}) \}, $

3. (c) $\mathrm {reg}_S(M^{"}) \leq \mathrm {max}\{ \mathrm {reg}_S(M), \mathrm {reg}_S(M^{'}) - 1 \}, $

and if $d_0 = \mathrm {min}\{d \, | \, \mathrm {Hilb}(d) = \mathrm {HilbP}(d) \},$ then $\mathrm {reg}(M) \geq d_0.$ Furthermore, if M is Cohen–Macaulay, then $\mathrm {reg}_S(M) = d_0.$ If M has finite length, then $\mathrm {reg}_S(M) = \mathrm {max}\{ d : M_d \neq 0 \}.$

To study these invariants, we place the graded Betti numbers of a module M into a table, called the Betti table.

The Betti table allows us to determine certain invariants easier; for example, the projective dimension is the length of the table and the regularity is the height of the table.

Denote by $H_M(t)$ and $P_{M}^{R}(t),$ respectively, the Hilbert series of M and the Poincaré series of an R-module M:

$$ \begin{align*} H_M(t) = \sum_{i \geq 0} \mathrm{Hilb}_M(i) t^i \end{align*} $$

and

$$ \begin{align*} P_{M}^R(t) = \sum_{i \geq 0} \beta_i^R(M) t^i. \end{align*} $$

It is worth observing that since M is finitely generated by homogenous elements of positive degree, the Hilbert series of M is a rational function. A short exact sequence of modules has a property we use extensively in this paper. If we have a short exact sequence of graded S-modules

$$ \begin{align*} 0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0, \end{align*} $$

then

$$ \begin{align*} \mathrm{H}_B(t) = \mathrm{H}_A(t) + \mathrm{H}_C(t).\end{align*} $$

Whenever we use this property, we will refer to it as the additivity property of the Hilbert series.

A standard graded $\mathbb {C}$ -algebra R is Koszul if $\mathbb {C}$ has a linear R-free resolution; that is, $\beta _{i,j}^R(\mathbb {C}) =0$ for $i \neq j$ . Koszul algebras possess remarkable homological properties. For example,

Koszul rings possess other interesting properties as well. Fröberg [Reference Fröberg14] showed that R is Koszul if and only if $H_R(t)$ and the $P_{\mathbb {C}}^R(t)$ have the following relationship:

(2) $$ \begin{align} P_{\mathbb{C}}^R(t) H_R(-t) = 1. \end{align} $$

In general, the Poincaré series of $\mathbb {C}$ as an R-module can be irrational [Reference Anick1], but if R is Koszul, then Equation (2) tells us the Poincaré series is always rational. So a necessary condition for a coordinate ring R to be Koszul is $P_{\mathbb {C}}^R(t) = \frac {1}{H_R(-t)}$ must have nonnegative coefficients in its Maclaurin series. Another necessary condition is that if R is Koszul, then the defining ideal has a minimal generating set of forms of degree at most $2$ . This is easy to see since

$$ \begin{align*} \beta_{2,j}^R(\mathbb{C}) = \begin{cases} \beta_{1,j}^S(R), & \text{if }j \neq 2, \\ \beta_{1,2}^S(R) + \binom{n+1}{2}, & \text{if } j=2 \end{cases} \end{align*} $$

(see [Reference Conca8, Rem. 1.10]). Unfortunately, the converse does not hold, but Fröberg showed that if the defining ideal is generated by monomials of degree at most $2,$ then R is Koszul.

More generally, if J has a Gröbner basis of quadrics in some term order, then R is Koszul. If such a basis exists, we say that R is G-quadratic. More generally, R is LG-quadratic if there is a G-quadratic ring A and a regular sequence of linear forms $l_1,\ldots ,l_r$ such that $R \cong A/(l_1,\ldots ,l_r).$ It is worth noting that every G-quadratic ring is LG-quadratic, and every LG-quadratic ring is Koszul and that all of these implications are strict [Reference Conca8]. We briefly discuss in §6 if coordinate rings of generic collections of lines are G-quadratic or LG-quadratic.

We now define a very useful tool in proving rings are Koszul.

Conca, Trung, and Valla [Reference Conca, Trung and Valla9] showed that if R has a Koszul filtration, then R is Koszul. In fact, a stronger statement is true.

Conca, Trung, and Valla construct a Koszul filtration to show certain sets of points in general linear position are Koszul in [Reference Conca, Trung and Valla9]. Since we aim to generalize Theorems 1.1 and 1.2 to collections of lines, we must define what it means for a collection of lines to be generic and what it means for a collection of lines to be in general linear position.

We can interpret this definition as saying generic collections are sufficiently random since the collection of them forms a dense subset of large parameter space. Furthermore, as one should suspect, a generic collection of lines is in general linear position, since collections of lines in general linear position are characterized by the nonvanishing of certain determinants in the coefficients of the defining linear forms. The converse is not true (see Example 6.1).

3 Properties of coordinate rings of lines

This section aims to establish properties for the coordinate rings of generic collections of lines and collections of lines in general linear position and compare them to the coordinate rings of generic collections of points and collections of points in general linear position. We will see that the significant difference between the two coordinate rings is that the coordinate ring R of a collection of lines in general linear position is never Cohen–Macaulay, unless R is the coordinate ring of a single line, while the coordinate rings of points in general linear position are always Cohen–Macaulay. The lack of the Cohen–Macaulay property presents difficulty since many techniques are not available to us, such as Artinian reductions.

Proof. We prove the claim by induction on $|\mathcal {M}|$ . Let $m=|\mathcal {M}|$ and let J be the defining ideal of $\mathcal {M}$ . If $m=1,$ then by Remark 2.8 the ideal J is minimally generated by $n-1$ linear forms. So, R is isomorphic to a polynomial ring in two indeterminates. Now, suppose that $m \geq 2,$ and write $J = K \cap I,$ where K is the defining ideal for $m-1$ lines and I is the defining ideal for the remaining single line. By induction, $\mathrm {depth}(S/K) \leq 2$ and $\mathrm {dim}(S/K)=\mathrm {dim}(S/I)=2.$ Furthermore, $S/(K+I)$ is Artinian, since the variety K defines intersects trivially with the variety I defines. Hence, $\mathrm {dim}(S/(I+K)) =0.$ So, by Proposition 2.1 the $\mathrm {depth}(S/(I+K))=0.$

Using the short exact sequence

and Proposition 2.1, we have two inequalities

$$ \begin{align*} \mathrm{min}\left\{ \mathrm{depth}(S/K \oplus S/I),\mathrm{depth}(S/(I+K))+1 \right\} \leq \mathrm{depth}(S/J) ,\end{align*} $$

and

$$ \begin{align*} \mathrm{min}\left\{ \mathrm{depth}(S/K \oplus S/I), \mathrm{depth}(S/J) - 1 \right\} \leq \mathrm{depth}(S/(I+K)).\end{align*} $$

Regardless if $\mathrm {depth}(S/K)$ is $1$ or $2,$ our two inequalities yield $\mathrm {depth}(S/J)=1.$ By the Auslander–Buchsbaum formula, we have $\mathrm {pdim}_S(S/J)=n.$ Lastly, Proposition 2.1, yields $\mathrm {dim}(S/J) =2.$

In [Reference Conca, Trung and Valla9], Conca, Trung, and Valla used the Hilbert function of points in $\mathbb {P}^n$ in general linear position to prove the corresponding coordinate ring is Koszul, provided the number of points is at most $2n+1.$ There is a generalization for the Hilbert function to a generic collection of points. We present both together as a single theorem for completeness, we do not use the Hilbert function for a generic collection of points.

Theorem 3.3. ([Reference Carlini, Catalisano and Geramita7], [Reference Conca, Trung and Valla9])

Suppose that $\mathcal {P}$ is a collection of p points in $\mathbb {P}^n.$ If $\mathcal {P}$ is a generic collection, or $\mathcal {P}$ is a collection in general linear position with $p \leq 2n+1,$ then the Hilbert function of R is

$$ \begin{align*} \mathrm{Hilb}_R(d) = \mathrm{min} \left\{ \binom{n+d}{d}, p \right\}. \end{align*} $$

In particular, if $p \leq n+1,$ then

$$ \begin{align*} H_R (t)= \frac{(p-1)t+1}{1-t}. \end{align*} $$

Since we aim to generalize Theorems 1.1 and 1.2, we would like to know the Hilbert series of the coordinate ring of a generic collection of lines. The famous Hartshorne–Hirschowitz Theorem provides an answer.

Theorem 3.4. (Hartshorne–Hirschowitz [Reference Hartshorne and Hirschowitz16])

Let $\mathcal {M}$ be a generic collection of m lines in $\mathbb {P}^n$ , and let R be the coordinate ring of $\mathcal {M}.$ The Hilbert function of R is

$$ \begin{align*}\mathrm{Hilb}_R(d) = \mathrm{min} \Bigg\{\binom{n+d}{d}, m(d+1) \Bigg\}.\end{align*} $$

This theorem is very difficult to prove. One could ask if any generalization holds for planes, and unfortunately, this is not known and is an open problem. Interestingly, this theorem allows us to determine the regularity for the coordinate ring R of a generic collection of lines.

Proof. If $m=1,$ then by Remark 2.8 and a change of basis we can write the defining ideal as $J=(x_0,\ldots ,x_{n-2}).$ The coordinate ring R is minimally resolved by the Koszul complex on $x_0,\ldots ,x_{n-2}$ . So, $\mathrm {reg}_S(R)=0,$ and this satisfies the inequality. Suppose that $m \geq 2$ and let $\alpha $ be the smallest nonnegative integer satisfying $\binom {n+\alpha }{\alpha } \geq m(\alpha +1).$ By Theorem 3.4 and Proposition 2.2, $\mathrm {reg}_S(R) \geq \alpha .$

We show the reverse inequality by induction on $m.$ Let J be the defining ideal for the collection $\mathcal {M}.$ Note, removing a line from a generic collection of lines maintains the generic property for the new collection. Let K be the defining ideal for $m-1$ lines, and let I be the defining ideal for the remaining line such that $J = K \cap I.$ By induction $\mathrm {reg}_S(S/K)=\beta ,$ and $\beta $ is the smallest nonnegative integer satisfying the inequality $\binom {n+\beta }{\beta } \geq (m-1)(\beta +1).$

Now, we claim that $\mathrm {reg}_S(S/K) = \beta \in \{ \alpha , \alpha -1 \}.$ To prove this, we need two inequalities: $ m-2 \geq \beta $ and $\binom {n+\beta }{\beta +1} \geq n(m-1).$ We have the first inequality since

$$ \begin{align*} \binom{n+m-2}{m-2}-(m-1)(m-2+1) &= \frac{(n+m-2)!}{n!(m-2)!} - (m-1)^2 \\ &= \frac{(m+1)!}{3!(m-2)!} - (m-1)^2 \\ &= \frac{(m-3)(m-2)(m-1)}{3!} \\ &\geq 0. \end{align*} $$

Thus, $m-2 \geq \beta .$ We have the second inequality, since by assumption

$$ \begin{align*} \binom{n+\beta}{\beta} &\geq (m-1)(\beta+1), \end{align*} $$

and rearranging terms gives

$$ \begin{align*} \binom{n+\beta}{\beta+1} &\geq n(m-1). \end{align*} $$

These inequalities together yield the following:

$$ \begin{align*} \binom{n+\beta+1}{\beta+1} &= \binom{n+\beta}{\beta} + \binom{n+\beta}{\beta+1} \\ &\geq (m-1)(\beta+1) + n(m-1) \\ &= (m-1)(\beta+1) + m+(m-1)(n-1) - 1 \\ &\geq (m-1)(\beta+1) + m +(m-1)2 -1 \\ &\geq (m-1)(\beta+1) + m +\beta + 1 \\ &= m (\beta+2). \end{align*} $$

Hence, $ \beta + 1 \geq \alpha .$ Furthermore, the inequality

$$ \begin{align*} \binom{n+\beta-1}{\beta-1} < (m-1)(\beta-1+1) \leq m \beta \end{align*} $$

implies that $ \alpha \geq \beta .$ So, $\mathrm {reg}_S(S/K) = \beta $ where $\beta \in \{ \alpha , \alpha -1 \}.$

Consider the short exact sequence

$$ \begin{align*} 0 \longrightarrow S/J \longrightarrow S/K \oplus S/I \longrightarrow S/(K+I) \longrightarrow 0.\end{align*} $$

If $\beta = \alpha ,$ then Theorem 3.4 and the additive property of the Hilbert series yields the following:

$$ \begin{align*} H_{S/(K+I)}(t) &= \left( H_{S/K}(t) + H_{S/I}(t)\right) - H_{S/J}(t) \\ &= \left(\sum\limits_{k=0}^{\alpha-1} \binom{n+k}{k} t^k + \sum\limits_{k=\alpha}^{\infty} (m-1)(k+1)t^k + \sum\limits_{k=0}^{\infty} (k+1)t^k \right)\\ &\hspace{1cm}- \sum\limits_{k=0}^{\alpha-1} \binom{n+k}{k} t^k - \sum\limits_{k=\alpha}^{\infty} m(k+1)t^k \\ &= \sum_{k=0}^{\alpha-1}(k+1)t^k. \end{align*} $$

and similarly if $\beta = \alpha -1,$ then

$$ \begin{align*} H_{S/(K+I)}(t) &= \left( H_{S/K}(t) + H_{S/I}(t)\right) - H_{S/J}(t) \\ &= \left(\sum\limits_{k=0}^{\alpha-2} \binom{n+k}{k} t^k + \sum\limits_{k=\alpha-1}^{\infty} (m-1)(k+1)t^k + \sum\limits_{k=0}^{\infty} (k+1)t^k \right)\\ &\hspace{1cm}- \sum\limits_{k=0}^{\alpha-1} \binom{n+k}{k} t^k - \sum\limits_{k=\alpha}^{\infty} m(k+1)t^k \\ &= \sum_{k=0}^{\alpha-2}(k+1)t^k + \left( m\alpha - \binom{n+\alpha-1}{\alpha-1} \right)t^{\alpha-1}. \end{align*} $$

Note that $m\alpha - \binom {n+\alpha -1}{\alpha -1}$ is positive since $\alpha $ is the smallest nonnegative integer such that $\binom {n+\alpha }{\alpha } \geq m(\alpha +1)$ . So, $S/(K+I)$ is Artinian. By Proposition 2.2, $\mathrm {reg}_S(S/(K+I))=\alpha -1.$ Since $\mathrm {reg}_S(S/K)=\alpha $ or $\mathrm {reg}_S(S/K)=\alpha -1$ and $\mathrm {reg}_S(S/I)=0,$ then $\mathrm {reg}_S(S/J) \leq \alpha $ . Thus, $\mathrm {reg}_S(R)=\alpha .$

Compare the previous result with the following general regularity bound for intersections of ideals generated by linear forms.

The assumption that R is a coordinate ring of a generic collection of lines tells us the regularity exactly, which is much smaller than the Derksen–Sidman bound for a fixed n. By way of comparison, we compute the following estimate.

Corollary 3.8. Let $\mathcal {M}$ be a generic collection of m lines in $\mathbb {P}^n$ with $n \geq 3$ , and let R be the coordinate ring of $\mathcal {M}.$ Then

$$ \begin{align*}\mathrm{reg}_S(R) \leq \left \lceil \sqrt[n-1]{n!} \left( \sqrt[n-1]{m} -1\right)\right \rceil.\end{align*} $$

Proof. Let $p(x) = (x+n)\cdots (x+2) - n!m.$ The polynomial $p(x)$ has a unique positive root by the Intermediate Value Theorem, since the $(x+n)\cdots (x+2)$ is increasing on the nonnegative real numbers. Let a be this positive root, and observe that the smallest nonnegative integer $\alpha $ satisfying the inequality $\binom {n+\alpha }{\alpha } \geq m(\alpha +1)$ is precisely the ceiling of the root $a.$

We now use an inequality of Minkowski (see [Reference Frenkel and Horváth13, (1.5)]). If $x_k$ and $y_k$ are positive for each $k,$ then

$$ \begin{align*}\sqrt[n-1]{\prod_{k=1}^{n-1} (x_k + y_k)} \geq \sqrt[n-1]{\prod_{k=1}^{n-1} x_k} + \sqrt[n-1]{\prod_{k=1}^{n-1} y_k}. \end{align*} $$

Thus,

$$ \begin{align*} \sqrt[n-1]{n!m} &= \sqrt[n-1]{ (a+n)\cdots(a+2)} \\ &\geq a + \sqrt[n-1]{n!} \\ \end{align*} $$

Therefore,

$$ \begin{align*} \sqrt[n-1]{n!m} - \sqrt[n-1]{n!} \geq a. \end{align*} $$

Taking ceilings gives the inequality.

We would like to note that $\mathrm {reg}_S(R)$ is roughly asymptotic to the upper bound. Proposition 3.1 and Theorem 3.5 tell us the coordinate ring R of a non-trivial generic collection of lines in $\mathbb {P}^n$ is not Cohen–Macaulay, $\mathrm {pdim}_S(R)=n,$ and the regularity is the smallest nonnegative integer $\alpha $ satisfying $\binom {n+\alpha }{\alpha } \geq m(\alpha +1)$ . So, the resolution of R is well-behaved, in the sense that if n is fixed and we allow m to vary we may expect the regularity to be low compared to the number of lines in our collection.

4 Koszul filtration for a collection of lines

In this section, we determine when a generic collection of lines, or a collection of lines in general linear position, will yield a Koszul coordinate ring. To this end, most of the work will be in constructing a Koszul filtration in the coordinate ring of a generic collection of lines.

Proof. We use $ \widehat {\cdot }$ to denote a term removed from a sequence. Let R be the coordinate ring of $\mathcal {M}$ with defining ideal $J.$ Through a change of basis and Remark 2.8, we may assume the defining ideal for each line has the following form:

$$ \begin{align*} L_{i} &= (x_0,\ldots, \widehat{x}_{n-2i+1} , \widehat{x}_{n-2i+2},\ldots,x_{n-1},x_{n}), \end{align*} $$

for $i = 1,\ldots ,m.$ Since every $L_i$ is monomial, so is J. Furthermore, since $n+1 \geq 2m,$ the $\mathrm {reg}_S(R) \leq 1.$ Thus, J is generated by monomials of degree at most $2.$ Theorem 2.4 guarantees R is Koszul.

Unfortunately, the simplicity of the previous proof does not carry over for larger generic collections of lines. We need a lemma.

Lemma 4.2. Let $\mathcal {M}$ be a generic collection of m lines in $\mathbb {P}^n$ , and let R be the coordinate ring of $\mathcal {M}.$ If $\mathrm {reg}_S(R) = 1$ , then the Hilbert series of R is

$$ \begin{align*} H_{S/J}(t) = \frac{(1-m)t^2+(m-2)t+1}{(1-t)^2}.\end{align*} $$

If $\mathrm {reg}_S(R) = 2$ , then the Hilbert series of R is

$$ \begin{align*} H_{S/J}(t) = \frac{(1+n-2m)t^3+(3m-2n-1)t^2+(n-1)t+1}{(1-t)^2}. \end{align*} $$

Proof. By Theorem 3.5, the regularity is the smallest nonnegative integer $\alpha $ satisfying $\binom {n+\alpha }{\alpha } \geq m(\alpha +1).$ Suppose $\mathrm {reg}_S(R)=1$ . By Theorem 3.4, the Hilbert series for R is

$$ \begin{align*} H_{R}(t) &= 1 + 2mt + 3mt^2 +4mt^3 + \cdots \\ &= 1 - m\left( \frac{t(t-2)}{(1-t)^2}\right) \\ &= \frac{t^2-2t+1-mt^2+2mt}{(1-t)^2}\\ &= \frac{(1-m)t^2+2(m-1)t+1}{(1-t)^2}. \end{align*} $$

Now, suppose $\mathrm {reg}_S(R)=2$ . By Theorem 3.4, the Hilbert series for R is

$$ \begin{align*} H_{R}(t) &= 1 +(n+1)t + 3mt^2 +4mt^3 + \cdots \\ &= 1+(n+1)t-m\left( \frac{t^2(2t-3)}{(1-t)^2} \right)\\ &= \frac{(n+1)t^3-(2n+1)t^2+(n-1)t+1-2mt^3+3mt^2}{(1-t)^2} \\ &= \frac{(n+1-2m)t^3+(3m-2n-1)t^2+(n-1)t+1}{(1-t)^2}. \end{align*} $$

We can now construct a Koszul filtration for the coordinate ring of certain larger generic collections of lines.

Proof. We only prove $(a)$ due to the length of the proof and note that $(b)$ is done identically except for the Hilbert series computations. In both cases, we may assume that $n \leq 2(m-1),$ otherwise Proposition 4.1 and Remark 2.8 prove the claim. By Remark 2.8 and a change of basis, we may assume the defining ideals for our m lines have the following form:

$$ \begin{align*} L_1 &= (x_0,\ldots,x_{n-4},x_{n-3},x_{n-2})\\ L_2 &= (x_0,\ldots,x_{n-4},x_{n-1},x_{n})\\ &\hspace{2.5cm}\vdots\\ L_i &= (x_0,\ldots, \widehat{x}_{n-2i+1}, \widehat{x}_{n-2i+2},\ldots,x_{n})\\ &\hspace{2.5cm}\vdots\\ L_k &= (x_0,\ldots, \widehat{x}_{n-2k+1}, \widehat{x}_{n-2k+2},\ldots,x_{n})\\ L_{k+1} &= (l_0,\ldots,l_{n-4},l_{n-3},l_{n-2})\\ L_{k+2} &= (l_0,\ldots,l_{n-4},l_{n-1},l_{n})\\ &\hspace{2.5cm}\vdots\\ L_{k+i} &= (l_0,\ldots, \widehat{l}_{n-2i+1} , \widehat{l}_{n-2i+2},\ldots,l_{n})\\ &\hspace{2.5cm}\vdots\\ L_{2k} &= (l_0,\ldots, \widehat{l}_{n-2k+1}, \widehat{l}_{n-2k+2},\ldots,l_{n}), \end{align*} $$

where $l_i$ are general linear forms in $S.$ Denote the ideals

$$ \begin{align*} J = \displaystyle{\bigcap_{i=1}^{2k}} L_i, \hspace{1cm} K = \displaystyle{\bigcap_{i=1}^{k}} L_i, \hspace{1cm} I = \displaystyle{\bigcap_{i=k+1}^{2k}} L_i,\end{align*} $$

so that $J = K \cap I.$ Let $R=S/J;$ to prove that R is Koszul, we will construct a Koszul filtration. To construct the filtration, we need the two Hilbert series $H_{\left ( J+(x_0) \right ):(x_1)}(t)$ and $H_{\left ( J+(l_0) \right ):(l_1)}(t).$ We first calculate the former. Observe $(x_0,x_1) \subseteq L_i$ and $(l_0,l_1) \subseteq L_{k+i}$ for $i= 1,\ldots ,k.$ Using the modular law [Reference Atiyah and Macdonald2, Chapter 1], we have the equality

(3) $$ \begin{align} \left( J+(x_0) \right):(x_1) = \left(K \cap I + K \cap (x_0) \right):(x_1) = (I+(x_0)):(x_1). \end{align} $$

So, it suffices to determine $H_{S/((I+(x_0)):(x_1))}(t)$ . To this end, we first calculate $H_{S/(I+(x_0,x_1))}(t).$ To do so, we use the short exact sequence

$$ \begin{align*} 0 \rightarrow S/\left( I+(x_0) \right) \cap \left(I+(x_1) \right) &\rightarrow S/\left( I+(x_0)\right) \oplus S/\left(I+(x_1)\right) \\ &\rightarrow S/\left(I+(x_0,x_1)\right) \rightarrow 0. \end{align*} $$

Our assumption $m+1 \leq n \leq 2(m-1)$ guarantees that $\mathop {\mathrm {reg}}\nolimits _S(S/I)=1.$ Thus, by Lemma 4.2

(4) $$ \begin{align} H_{S/I}(t) = \frac{(1-k)t^2+2(k-1)t+1}{(1-t)^2}, \end{align} $$

and since $x_0$ and $x_1$ are nonzerodivisors on $S/I,$ we have the following two Hilbert series:

$$ \begin{align*} H_{S/(I+(x_0))}(t) = H_{S/(I+(x_1))}(t) = \tfrac{(1-k)t^2+2(k-1)t+1}{1-t}.\end{align*} $$

Furthermore, the coordinate ring $S/(I+(x_0)) \cap (I+(x_1))$ corresponds precisely to a collection of $2k$ distinct points. These points must necessarily be in general linear position, since by assumption $2k=m \leq n+1$ and no three are collinear. So, by Theorem 3.3, we have

$$ \begin{align*} H_{S/\left( ((I+(x_0)) \cap (I+(x_1)) \right)}(t) = \frac{(2k-1)t+1}{1-t}. \end{align*} $$

By the additivity of the Hilbert series

$$ \begin{align*} H_{S/(I+(x_0,x_1))}(t) &= H_{S/(I+(x_0))}(t) + H_{S/(I+(x_1))}(t) - H_{S/(I+(x_0)) \cap (I+(x_1))}(t) \\ &= 2\left(\frac{(1-k)t^2+2(k-2)t+1}{1-t}\right) - \frac{(2k-1)t+1}{1-t} \\ &= 1+2(k-1)t. \end{align*} $$

Thus, by the short exact sequence

$$ \begin{align*} 0 \rightarrow S/\left( (I+(x_0)):(x_1)\right)(-1) \rightarrow S/(I+(x_0)) \rightarrow S/(I+(x_0,x_1)) \rightarrow 0, \end{align*} $$

Equation (3), and the additivity of the Hilbert series

(5) $$ \begin{align} H_{S/((J+(x_0)):(x_1))}(t) &= H_{S/(I+(x_0)):(x_1)}(t) \\ \nonumber &=\frac{1}{t} \left( H_{S/(I+(x_0))}(t) - H_{S/(I+(x_0,x_1))} \right) \\ \nonumber &= \frac{1}{t} \bigg( \frac{(1-k)t^2+2(k-1)t+1}{1-t} -1-2(k-1)t \bigg) \\ \nonumber &= \frac{(k-1)t+1}{1-t}. \end{align} $$

This gives us our desired Hilbert series. An identical argument and interchanging I with K and $x_0$ and $x_1$ with $l_0$ and $l_1$ yields

(6) $$ \begin{align} H_{S/K}(t) = \frac{(1-k)t^2+2(k-1)t+1}{(1-t)^2}, \end{align} $$

$$ \begin{align*} \hspace{1.9cm} H_{S/(K+(l_0))}(t) = H_{S/(K+(l_1))}(t) = \frac{(1-k)t^2+2(k-1)t+1}{(1-t)^2}, \end{align*} $$

and

$$ \begin{align*}H_{S/((J+(l_0)):(l_1))}(t)=H_{S/((J+(x_0)):(x_1))}(t).\end{align*} $$

We can now define a Koszul filtration $\mathcal {F}$ for R. We use $ \overline {\cdot }$ to denote the image of an element of S in $R=S/J$ for the remainder of the paper. We have already seen in Equation (5) that

$$ \begin{align*} H_{S/(J+(x_0)):(x_1)}(t) = \frac{(k-1)t+1}{(1-t)} = 1+\sum_{i=1}^{\infty} k t^i.\end{align*} $$

Hence, $n-k+1$ linearly independent linear forms are in a minimal generating set of $(J+(x_0)):(x_1).$ Clearly $l_0,\ldots ,l_{n-2k},x_0\in (J+(x_0)):(x_1),$ label $z_{n-2k+2},\ldots ,z_{n-k}$ as the remaining linear forms from a minimal generating set of $(J+(x_0)):(x_1).$ Similarly, choose $y_i$ from $(J+(l_0)):(l_1)$ so that $x_0,x_1,\ldots ,x_{n-2k},l_0,y_{n-k+2},\ldots ,y_{n-k}$ are linear forms forming a minimal generating set of $(J+(l_0)):(l_1).$

The set $\{l_0,\ldots ,l_{n-2k},x_{0},z_{n-2k+1},\ldots ,z_{n-k},x_1\}$ is a linearly independent set over $S,$ otherwise $x_1^2 \in J+(x_0).$ This means $x_1^2 \in (L_i+(x_0))$ for $i = k+1,\ldots ,2k,$ a contradiction. Similarly, $\{x_0,\ldots ,x_{n-2k},l_{0},y_{n-2k+1},\ldots ,y_{n-k},l_1\}$ is linearly independent over $S.$ Let $w_{n-k+2},\ldots ,w_{n+1},$ and $u_{n-k+2},\ldots ,u_{n+1}$ be extensions of

$$ \begin{align*}\{ \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_{0}, \overline{z}_{n-2k+1},\ldots, \overline{z}_{n-k}, \overline{x}_1\}\end{align*} $$

and

$$ \begin{align*}\{ \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+1},\ldots, \overline{y}_{n-k}, \overline{l}_1\}\end{align*} $$

to minimal systems of generators of $\mathfrak {m}_R,$ respectively. Define $\mathcal {F}$ as follows:

$$ \begin{align*} \mathcal{F} = \begin{cases} 0, ( \overline{x}_0), \hspace{.2cm } ( \overline{x}_0, \overline{x}_1), \hspace{3.3cm} ( \overline{l}_0), \hspace{.2cm} ( \overline{l}_0, \overline{l}_1),\\ \hspace{2cm}\vdots \hspace{5cm}\vdots \\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}), \hspace{2.9cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}),\\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}, \overline{l}_0), \hspace{2.5cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}, \overline{x}_0),\\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2}), \hspace{1cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2}),\\ \hspace{4cm} \vdots \\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{n-k}), \\ \hspace{1cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{n-k}), \\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{n-k}, \overline{l}_{1}), \\ \hspace{1cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{n-k}, \overline{x}_{1}),\\ \hspace{4cm}\vdots\\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{n-k}, \overline{l}_{1},u_{n-k+2}),\\ \hspace{1cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2}, \ldots, \overline{z}_{n-k}, \overline{x}_{1},w_{n-k+2}),\\ \hspace{4cm}\vdots\\ ( \overline{x}_0, \overline{x}_1,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{n-k}, \overline{l}_{1},u_{n-k+2},\ldots,u_{n}),\\ \hspace{1cm} ( \overline{l}_0, \overline{l}_1,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2}, \ldots, \overline{z}_{n-k}, \overline{x}_{1},w_{n-k+2},\ldots,w_{n}),\\ \mathfrak{m}_R. \end{cases} \end{align*} $$

We now prove $\mathcal {F}$ is a Koszul filtration. We do this by proving several claims. Throughout the process, we use the inclusion $(x_0,x_1) \cap (l_0,l_1) \subseteq J.$ Afterward, we summarize all computed colons and list the claims that prove the calculated colons.

Claim 4.4. The ideal $( \overline {x}_0, \overline {x}_1, \overline {l}_0, \overline {l}_1)$ in R has Hilbert series

$$ \begin{align*}H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1)}(t) = 1+(n-3)t\end{align*} $$

and any ideal P containing this ideal has the property that $P:(\ell ) = \mathfrak {m}_R,$ where $\ell $ is a linear form not contained in $P.$

Proof. We begin by observing that our assumption $m+1 \leq n \leq 2(m-1)$ and Proposition 3.5 yield $\mathrm {reg}_S(S/J)=2.$ Thus, by Lemma 4.2

$$ \begin{align*} H_{S/J}(t) = \frac{(n+1-4k)t^3+(6k-2n-1)t^2+(n-1)t+1}{(1-t)^2}.\end{align*} $$

Now, $L_i:(x_0) = L_i$ for $i=k+1,\ldots ,2k,$ since $x_0 \notin L_i.$ Thus,

$$ \begin{align*}J:(x_0) = \left( \bigcap_{i=1}^{2k} L_i \right) :(x_0) =\bigcap_{i=1}^{2k} \left( L_i :(x_0) \right ) = \bigcap_{i=k+1}^{2k} L_i = I.\end{align*} $$

So, $H_{S/(J:(x_0))}(t)=H_{S/I}(t).$ Using the short exact sequence

$$ \begin{align*} 0 \rightarrow S/(J:(x_0)) (-1) \rightarrow S/J \rightarrow S/(J+(x_0)) \rightarrow 0, \end{align*} $$

Equation (4), and the additivity of the Hilbert series yields

$$ \begin{align*} H_{S/(J+(x_0))}(t) &= H_{S/J}(t)-tH_{S/(J:(x_0)) }(t) \\ &=H_{S/J}(t)-tH_{S/I }(t) \\ &=\tfrac{(1+n-4k)t^3+(6k-2n-1)t^2+(n-1)t+1}{(1-t)^2} -t\left( \tfrac{(1-k)t^2+2(k-1)t+1}{(1-t)^2}\right)\\ &= \frac{(n-3k)t^3+(4k-2n+1)t^2+(n-2)t+1}{(1-t)^2}. \end{align*} $$

Using the short exact sequence

$$ \begin{align*} 0 \rightarrow S/((J+(x_0)):(x_1))(-1) \rightarrow S/(J+(x_0)) \rightarrow S/(J+(x_0,x_1)) \rightarrow 0, \end{align*} $$

Equation (5), the previous Hilbert series, and the additivity of the Hilbert series yields

$$ \begin{align*} H_{S/(J+(x_0,x_1))}(t) &= H_{S/(J+(x_0))}(t) - tH_{S/(J+(x_0)):(x_1)}(t) \\ &= \tfrac{(n-3k)t^3+(4k-2n+1)t^2+(n-2)t+1}{(1-t)^2} - t\left( \tfrac{(k-1)t+1}{1-t} \right) \\ &= \frac{(n-2k-1)t^3+(3k-2n+3)t^2+(n-3)t+1}{(1-t)^2}. \end{align*} $$

Replacing $x_0$ and $x_1$ with $l_0$ and $l_1$ demonstrates that

$$ \begin{align*} H_{S/(J+(x_0,x_1))}(t) = H_{S/(J+(l_0,l_1))}(t).\end{align*} $$

Thus, using the short exact sequence

$$ \begin{align*} 0 \rightarrow R/(( \overline{x}_0, \overline{x}_1) \cap ( \overline{l}_0, \overline{l}_1)) &\rightarrow R/( \overline{x}_0, \overline{x}_1) \oplus R/( \overline{l}_0, \overline{l}_1) \rightarrow R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1) \rightarrow 0, \end{align*} $$

and the additivity of the Hilbert series yields

$$ \begin{align*} H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1)}(t) &= H_{R/( \overline{x_0}, \overline{x_1})}(t)+H_{R/( \overline{l_0}, \overline{l_1})}(t)-H_{R}(t) \\ \nonumber &= 2\left(\tfrac{(n-2k-1)t^3+(3k-2n+3)t^2+(n-3)t+1}{(1-t)^2}\right) -\tfrac{(1+n-4k)t^3+(6k-2n-1)t^2+(n-1)t+1}{(1-t)^2} \\ \nonumber &= \frac{(n-3)t^3+(7-2n)t^2+(n-5)t+1}{(1-t)^2} \\ \nonumber &= 1+(n-3)t. \end{align*} $$

So, $R_2 \subset ( \overline {x}_0, \overline {x}_1, \overline {l}_0, \overline {l}_1).$ This means that any ideal $P \subset R$ containing the ideal $( \overline {x}_0, \overline {x}_1, \overline {l}_0, \overline {l}_1)$ has the property that $P:(\ell ) = \mathfrak {m}_R,$ where $\ell $ is a linear form not contained in P.

Claim 4.5. For $i = 1,\ldots ,n-2k,$ we have the two Hilbert series

$$ \begin{align*} H_{R/( \overline{x}_0, \overline{x}_1, \overline{x}_2,\ldots, \overline{x}_i)}(t) &= H_{R/( \overline{l}_0, \overline{l}_1, \overline{l}_2,\ldots, \overline{l}_i)}(t) \\ &=\tfrac{(n-2k-i)t^3+(3k-2n+2i+1)t^2+(n-(i+2))t+1}{(1-t)^2}, \end{align*} $$

and the two equalities $J+(x_0,\ldots ,x_{n-2k}) = K,$ and $J+(l_0,\ldots ,l_{n-2k}) = I.$

Proof. Adding the linear forms $x_2,\ldots ,x_i$ to the ideal $( \overline {x}_0, \overline {x}_1, \overline {l}_0, \overline {l}_1)$ yields

$$ \begin{align*}H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1, \overline{x}_2,\ldots, \overline{x}_{i})}(t) = H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1, \overline{l}_2,\ldots, \overline{l}_{i})}(t) = 1+(n-(i+2))t\end{align*} $$

for $i = 2,\ldots ,n-2k$ . Using the short exact sequence

$$ \begin{align*} 0 \rightarrow R /( ( \overline{x}_0, \overline{x}_1, \overline{x}_2) \cap ( \overline{l}_0, \overline{l}_1)) &\rightarrow R/( \overline{x}_0, \overline{x}_1, \overline{x}_2) \oplus R/( \overline{l}_0, \overline{l}_1)\rightarrow R/( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{l}_0, \overline{l}_1) \rightarrow 0 \end{align*} $$

and the additivity of the Hilbert series gives

$$ \begin{align*} H_{R/( \overline{x}_0, \overline{x}_1, \overline{x}_2)}(t) &= H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1, \overline{x}_2) }(t) + H_{R}(t)- H_{R/( \overline{l}_0, \overline{l}_1)}(t)\\ &= 1+(n-4)t +\tfrac{(1+n-4k)t^3+(6k-2n-1)t^2+(n-1)t+1}{(1-t)^2} \\ &\hspace{3.5cm} -\tfrac{(n-2k-1)t^3+(3k-2n+3)t^2+(n-3)t+1}{(1-t)^2}\\ &=1+(n-4)t+\frac{(2-2k)t^3+(3k-4)t^2+2t}{(1-t)^2} \\ &= \frac{(n-2k-2)t^3+(3k-2n+5)t^2+(n-4)t+1}{(1-t)^2}. \end{align*} $$

Replacing $( \overline {x}_0, \overline {x}_1, \overline {x}_2)$ with $( \overline {x}_0, \overline {x}_1, \overline {x}_2, \overline {x}_3)$ in the above short exact sequence and using the additivity of the Hilbert series yields

$$ \begin{align*} H_{R/( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3)}(t) &= H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1, \overline{x}_2, \overline{x}_3) }(t) + H_{R}(t)- H_{R/( \overline{l}_0, \overline{l}_1)}(t)\\ &= 1+(n-5)t +\tfrac{(1+n-4k)t^3+(6k-2n-1)t^2+(n-1)t+1}{(1-t)^2} \\ &\hspace{3.5cm} -\tfrac{(n-2k-1)t^3+(3k-2n+3)t^2+(n-3)t+1}{(1-t)^2}\\ &= 1+(n-5)t + \frac{(2-2k)t^3+(3k-4)t^2+2t}{(1-t)^2} \\ &= \frac{(n-2k-3)t^3+(3k-2n+7)t^2+(n-5)t+1}{(1-t)^2} \end{align*} $$

By induction,

(7) $$ \begin{align} H_{R/( \overline{x}_0, \overline{x}_1, \overline{x}_2,\ldots, \overline{x}_i)}(t) = \tfrac{(n-2k-i)t^3+(3k-2n+2i+1)t^2+(n-(i+2))t+1}{(1-t)^2} \end{align} $$

for $i =2,\ldots ,n-2k.$ Setting $i=n-2k,$ we obtain the Hilbert series

$$ \begin{align*} H_{R/( \overline{x}_0,\ldots, \overline{x}_{n-2k})}(t) &= H_{R/( \overline{x}_0, \overline{x}_1, \overline{l}_0, \overline{l}_1, \overline{x}_2,\ldots, \overline{x}_{n-2k}) }(t) \\ \nonumber & \hspace{3cm}+H_{R}(t)- H_{R/( \overline{l}_0, \overline{l}_1)}(t)\\ \nonumber &= 1+(2k-2)t +\tfrac{(1+n-4k)t^3+(6k-2n-1)t^2+(n-1)t+1}{(1-t)^2} \\ \nonumber &\hspace{2.9cm} -\tfrac{(n-2k-1)t^3+(3k-2n+3)t^2+(n-3)t+1}{(1-t)^2}\\ \nonumber &= 1+(2k-2)t + \frac{(2-2k)t^3+(3k-4)t^2+2t}{(1-t)^2} \\ \nonumber &= \frac{(1-k)t^2+2(k-1)t+1}{(1-t)^2}. \end{align*} $$

Interchanging each $x_i$ with $l_i$ gives us the other desired Hilbert series.

The Hilbert series in Equation (7) is the same as in Equation (6). Furthermore, $J+(x_0,\ldots ,x_{n-2k})\subseteq K.$ So, we have that $J+(x_0,\ldots ,x_{n-2k})= K$ and interchanging each $x_i$ with $l_i$ gives us the other equality.

Claim 4.6. We have the equalities

$$ \begin{align*}( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0):( \overline{z}_{n-2k+2}) = \mathfrak{m}_R,\end{align*} $$

$$ \begin{align*}( \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0):( \overline{y}_{n-2k+2}) = \mathfrak{m}_R,\end{align*} $$

and

$$ \begin{align*}( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{i}):( \overline{z}_{i+1}) = \mathfrak{m}_R,\end{align*} $$

$$ \begin{align*}( \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{i}):( \overline{y}_{i+1}) = \mathfrak{m}_R,\end{align*} $$

for $i=n-2k+2,\ldots ,n-k.$ Furthermore,

$$ \begin{align*}( \overline{x}_0):( \overline{x}_1) = ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{n-k})\end{align*} $$

and

$$ \begin{align*}( \overline{l}_0):( \overline{l}_1)=( \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{n-k}).\end{align*} $$

Proof. We begin by observing

$$ \begin{align*}( \overline{l}_0, \overline{l}_{1}, \overline{x}_0, \overline{x}_1) \subseteq ( \overline{l}_0, \overline{l}_{1}, \overline{x}_0, \overline{l}_2,\ldots, \overline{l}_{n-2k}):( \overline{z}_{n-2k+2}).\end{align*} $$

So by Claim 4.4, we conclude that

$$ \begin{align*}H_{R/(( \overline{l}_0, \overline{l}_{1}, \overline{l}_2,\cdots, \overline{l}_{n-2k}, \overline{x}_0):( \overline{z}_{n-2k+2}))}(t) = 1+\alpha t,\end{align*} $$

where $\alpha \in \{0,1,\ldots ,n-3\}.$ Using the short exact sequence

(8) $$ \begin{align} 0 \rightarrow R/(( \overline{x}_0, \overline{l}_0,\ldots, \overline{l}_{n-2k}):( \overline{z}_{n-2k+2}))(-1) &\rightarrow R/( \overline{x}_0, \overline{l}_0,\ldots, \overline{l}_{n-2k}) \\ \nonumber& \hspace{-3cm} \rightarrow R/( \overline{x}_0, \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{z}_{n-2k+2}) \rightarrow 0, \end{align} $$

Claim 4.5, and that the fact that $x_0$ is a nonzerodivisor on $S/I,$ we obtain

$$ \begin{align*} H_{R/( \overline{x}_0, \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{z}_{n-2k+2})}(t) &= H_{R/( \overline{x}_0, \overline{l}_0,\ldots, \overline{l}_{n-2k})}(t) \\ &\hspace{1cm}- tH_{R/( \overline{x}_0, \overline{l}_0,\ldots, \overline{l}_{n-2k}):( \overline{z}_{n-2k+2})}(t) \\ &= \frac{(1-k)t^2+2(k-1)t+1}{(1-t)} - t(1+\alpha t) \\\ &= \frac{\alpha t^3+(2-\alpha-k)t^2+(2k-3)t+1}{(1-t)} \\ &= 1+(2k-2)t+(k-\alpha)t^2 + \sum_{j=3}^{\infty} k t^j. \end{align*} $$

We also have the containment

$$ \begin{align*} ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2}) \subseteq ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0):( \overline{x}_1). \end{align*} $$

Using Claim 4.5, we obtain the equality

$$ \begin{align*}(J + (l_0,\ldots,l_{n-2k},x_0)):(x_1) = (I+(x_0)):(x_1),\end{align*} $$

which has Hilbert series computed in Equation (5). Hence,

$$ \begin{align*} H_{R/(( \overline{l}_0, \overline{l}_{1}, \overline{l}_2,\ldots, \overline{l}_{n-2k}, \overline{x}_0):( \overline{x}_{1}))}(t) &= \frac{(k-1)t+1}{1-t} = 1 + \sum_{j=1}^{\infty} k t^j. \end{align*} $$

Comparing coefficients of $t^2$ yields $k \leq k - \alpha ,$ and so $\alpha =0,$ proving the first equality. We immediately have the equality

(9) $$ \begin{align} ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{i}):( \overline{z}_{i+1}) = \mathfrak{m}_R, \end{align} $$

for each $i =n-2k+2,\ldots , n-k.$

Notice that setting $\alpha = 0,$ yields

$$ \begin{align*} H_{R/( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2})}(t) = \frac{(2-k)t^2+(2k-3)t+1}{(1-t)}. \end{align*} $$

Denote $V_i$ and $V_i'$ to be the ideals

$$ \begin{align*} V_i &= ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{i}) \end{align*} $$

and

$$ \begin{align*} V_i' &= ( \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{i}) \end{align*} $$

for $i = n-2k+2,\ldots ,n-k$ . Replacing the ideals in Equation (8) with the three ideals $V_{n-2k+3},V_{n-2k+2},$ and $V_{n-2k+2}:( \overline {z}_{n-2k+3})$ , and using the additivity of the Hilbert series yields

$$ \begin{align*} H_{R/V_{n-2k+3}}(t) &= H_{R/V_{n-2k+2}}(t) -t H_{R/V_{n-2k+2}:( \overline{z}_{n-2k+3})}(t) \\ &= \frac{(2-k)t^2+(2k-3)t+1}{(1-t)} - t \\ &= \frac{(3-k)t^2+(2k-4)t+1}{(1-t)}. \end{align*} $$

Continuing in this fashion gives

$$ \begin{align*} H_{R/V_{n-k}}(t) &= \frac{(k-1)t+1}{(1-t)}. \end{align*} $$

So, both $( \overline {x}_0):( \overline {x}_1)$ and $V_{n-k}$ have the same Hilbert series. Furthermore, $V_{n-k}\subseteq ( \overline {x}_0):( \overline {x}_1).$ So these ideals are in fact equal. Interchanging $x_0$ and $x_1$ with $l_0$ and $l_1$ yields the remaining equality.

Claim 4.7. We have the equalities

$$ \begin{align*}(0_R):( \overline{x}_0) = ( \overline{l}_0,\ldots, \overline{l}_{n-2k})\end{align*} $$

and

$$ \begin{align*}(0_R):( \overline{l}_0) = ( \overline{x}_0,\ldots, \overline{x}_{n-2k}).\end{align*} $$

Proof. The two equalities follow immediately since $(0_R):( \overline {x}_0) \subseteq ( \overline {l}_0,\ldots , \overline {l}_{n-2k})$ and $(0_R):( \overline {l}_0) \subseteq ( \overline {x}_0,\ldots , \overline {x}_{n-2k})$ and all four ideals have the same Hilbert series by Claim 4.5.

Claim 4.8. We have the equality

$$ \begin{align*}( \overline{x}_0, \overline{x}_1, \overline{x}_2,\ldots, \overline{x}_i):( \overline{x}_{i+1})=\mathfrak{m}_R,\end{align*} $$

for $i=2,\ldots ,n-2k-1.$

Proof. Using the short exact sequence

$$ \begin{align*} 0 \rightarrow R/\left(( \overline{x}_0,\ldots, \overline{x}_{i}):( \overline{x}_{i+1})\right)(-1) \rightarrow R/( \overline{x}_0,\ldots, \overline{x}_{i}) \rightarrow R/( \overline{x}_0,\ldots, \overline{x}_{i}, \overline{x}_{i+1}) \rightarrow 0, \end{align*} $$

and the Hilbert series from Equation (7), we get

$$ \begin{align*} H_{R/(( \overline{x}_0,\ldots, \overline{x}_i):( \overline{x}_{i+1}))}(t) &= \frac{1}{t} \Bigg( H_{R/( \overline{x}_0,\ldots, \overline{x}_i)}(t) - H_{R/( \overline{x}_0,\ldots, \overline{x}_i, \overline{x}_{i+1})}(t) \Bigg) \\ &= \frac{1}{t} \bigg( \tfrac{(n-2k-i)t^3+(3k-2n+2i+1)t^2+(n-(i+2))t+1}{(1-t)^2} \\ & \hspace{1cm} -\tfrac{(n-2k-i-1)t^3+(3k-2n+2i+3)t^2+(n-(i+3))t+1}{(1-t)^2}\bigg) \\ &= \frac{t^2-2t+1}{(1-t)^2} \\ &= 1, \end{align*} $$

proving the claim.

Claim 4.9. We have the four equalities

$$ \begin{align*}( \overline{x}_0,\ldots, \overline{x}_{n-2k}):( \overline{l}_0)=( \overline{x}_0,\ldots, \overline{x}_{n-2k}),\end{align*} $$

$$ \begin{align*}( \overline{l}_0,\ldots, \overline{l}_{n-2k}):( \overline{x}_0)=( \overline{l}_0,\ldots, \overline{l}_{n-2k}),\end{align*} $$

$$ \begin{align*}V_{n-k}':( \overline{l}_1) = V_{n-k}',\end{align*} $$

$$ \begin{align*}V_{n-k}:( \overline{x}_1) = V_{n-k}.\end{align*} $$

Proof. The equality

$$ \begin{align*}( \overline{x}_0,\ldots, \overline{x}_{n-2k}):( \overline{l}_0)=( \overline{x}_0,\ldots, \overline{x}_{n-2k})\end{align*} $$

follows from the genericity of $l_0.$ We now aim to show the equality

$$ \begin{align*}V_{n-k}:( \overline{x}_1) = V_{n-k}.\end{align*} $$

We always have the containment $V_{n-k} \subseteq V_{n-k}:( \overline {x}_1)$ and by Claim 4.6, we have already determined $H_{R/V_{n-k}}(t).$ So, we must only determine $H_{R/(V_{n-k}:( \overline {x}_1))}(t).$ We aim to use the additivity of the Hilbert series along the short exact sequence

(10) $$ \begin{align} 0 \rightarrow R/(V_{n-k}:( \overline{x}_1))(-1) \rightarrow R/V_{n-k} \rightarrow R/(V_{n-k}+( \overline{x}_1)) \rightarrow 0, \end{align} $$

but we first must determine $H_{R/(V_{n-k}+( \overline {x}_1))}(t).$ By Claim 4.4, adding the linear forms $ \overline {l}_2,\ldots , \overline {l}_{n-2k}, \overline {z}_{n-2k+2},\ldots , \overline {z}_{n-k}$ to the ideal $( \overline {x}_0, \overline {x}_1, \overline {l}_0, \overline {l}_1)$ yields the Hilbert series

$$ \begin{align*} H_{R/(V_{n-k}+( \overline{x}_1)))}(t) &= 1+(k-1)t. \end{align*} $$

Using Claim 4.6 and the additivity of the Hilbert series along the short exact sequence (10) yields

$$ \begin{align*} H_{R/(V_{n-k}:( \overline{x}_1))}(t) &= \frac{1}{t} \Bigg( H_{R/V_{n-k}}(t) -H_{R/(V_{n-k}+( \overline{x}_1))}(t) \Bigg) \\ &= \frac{1}{t}\bigg(\frac{(k-1)t+1}{(1-t)}- (1+(k-1)t)\bigg)\\ &= \frac{(k-1)t+1}{(1-t)}. \end{align*} $$

Interchanging $ \overline {x}_0$ and $ \overline {x}_1$ with $ \overline {l}_0$ and $ \overline {l}_1,$ proves the other two equalities.

Below is a list of calculated colons with the corresponding justification.

$$ \begin{align*}\begin{cases} (0):( \overline{x}_0) =( \overline{l}_0,\dots, \overline{l}_{n-2k}), \hspace{1cm} (0):( \overline{l}_0)=( \overline{x}_0,\ldots, \overline{x}_{n-2k}), \hspace{.2cm} 4.7 \\ ( \overline{x}_0):( \overline{x}_1) = ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0, \overline{z}_{n-2k+2},\ldots, \overline{z}_{n-k}), \hspace{.2cm} 4.6 \\ \hspace{.3cm} ( \overline{l}_0):( \overline{l}_1)=( \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0, \overline{y}_{n-2k+2},\ldots, \overline{y}_{n-k}), \hspace{.2cm} 4.6 \\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2,\ldots, \overline{x}_i):( \overline{x}_{i+1})=\mathfrak{m}_R, \hspace{.5cm} i=2,\ldots, n-2k-1, \hspace{.2cm}4.8 \\ \hspace{.5cm} ( \overline{l}_0, \overline{l}_1, \overline{l}_2,\ldots, \overline{l}_i):( \overline{l}_{i+1})=\mathfrak{m}_R, \hspace{.5cm} i=2,\ldots, n-2k-1, \hspace{.2cm}4.8 \\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2,\ldots, \overline{x}_{n-2k}):( \overline{l}_{0})=( \overline{l}_0,\ldots, \overline{l}_{n-2k}), \hspace{.2cm} 4.9\\ \hspace{.5cm} ( \overline{l}_0, \overline{l}_1, \overline{l}_2,\ldots, \overline{l}_{n-2k}):( \overline{x}_{0})=( \overline{x}_0,\ldots, \overline{x}_{n-2k}), \hspace{.2cm} 4.9 \\ ( \overline{x}_0,\ldots, \overline{x}_{n-2k}, \overline{l}_0):( \overline{y}_{n-2k+2}) = \mathfrak{m}_R, \hspace{.2cm} 4.6 \\ \hspace{.5cm} ( \overline{l}_0,\ldots, \overline{l}_{n-2k}, \overline{x}_0):( \overline{z}_{n-2k+2}) = \mathfrak{m}_R, \hspace{.2cm} 4.6 \\ V_{i}':( \overline{y}_{i+1})= \mathfrak{m}_R, \hspace{.3cm} V_{i}:( \overline{z}_{i+1})= \mathfrak{m}_R, \hspace{.1cm} i=n-2k+2,\ldots, n-k-1, \hspace{.2cm} 4.6 \\ \hspace{.3cm} V_{n-k}':( \overline{l}_1)=V_{n-k}',\hspace{.3cm} V_{n-k}:( \overline{x}_1)= V_{n-k}, \hspace{.2cm} 4.9 \\ (V_{n-k}'+( \overline{l}_1)):( \overline{u}_{n-k+1})=\mathfrak{m}_R, \hspace{.3cm} \hspace{.2cm}(V_{n-k}+( \overline{x}_1)):( \overline{w}_{n-k+1})=\mathfrak{m}_R, \hspace{.2cm} 4.4 \\ \hspace{.3cm}(V_{n-k}'+( \overline{l}_1, \overline{u}_{n-k+1},\ldots, \overline{u}_i)):( \overline{u}_{i+1})=\mathfrak{m}_R, \hspace{.01cm} i = n-k+1,\ldots,n-1, \hspace{.2cm} 4.4 \\ (V_{n-k}+( \overline{x}_1, \overline{w}_{n-k+1},\ldots, \overline{w}_i)):( \overline{w}_{i+1})=\mathfrak{m}_R, \hspace{.01cm} i = n-k+1,\ldots,n-1. \hspace{.2cm}4.4 \end{cases}\end{align*} $$

This completes the proof of Theorem 4.3.

There is at least one example of a coordinate ring with the Koszul property which is not covered by our previous theorem. Let $\mathcal {M}$ be a generic collection of five lines in $\mathbb {P}^6.$ By Remark 2.8 and a change of basis, we may assume the defining ideals for our five lines have the following form:

$$ \begin{align*} L_1 &= ( x_0,x_3,x_4,x_5,x_6), \hspace{3.1cm} L_2 = (x_0,x_1,x_4,x_5,x_2+ax_3+x_6), \\ L_3 &= (x_0,x_1,x_2,x_6,x_3+bx_4+x_5), \hspace{1.25cm} L_4 = (x_1,x_2,x_3,x_5,x_0+x_4+x_6), \\ & \hspace{3cm}L_5 = (x_2,x_3,x_4,x_6,x_0+x_1+x_5), \end{align*} $$

where $a,b \in \mathbb {C}$ are algebraically independent over $\mathbb {Q}.$ Some further explanation is needed why we may assume our five lines have this form.

By Remark 2.8, the intersection of any triple of the defining ideals of our five lines contains a single linear form in a minimal generating set. Furthermore, the intersection of any pair of defining ideals for our five contains three linear forms in a minimal generating set. Thus, after a change of basis, we may assume

$$ \begin{align*} L_1 &= ( x_0,x_3,x_4,x_5,x_6), \hspace{1.5cm} L_2 = (x_0,x_1,x_4,x_5,l_0), \\ L_3 &= (x_0,x_1,x_2,x_6,l_1), \hspace{1.6cm} L_4 = (x_1,x_2,x_3,x_5,l_2), \\ & \hspace{2.7cm}L_5 = (x_2,x_3,x_4,x_6,l_3). \end{align*} $$

where the linear forms $l_0,\ldots ,l_3$ have the form

$$ \begin{align*} l_0 &= c_{0,2}x_2+c_{0,3}x_3+c_{0,6}x_6, \hspace{1cm} l_1 = c_{1,3}x_3+c_{1,4}x_4+c_{1,5}x_5, \\ l_2 &= c_{2,0}x_0+c_{2,4}x_4+c_{2,6}x_6, \hspace{1cm} l_3 = c_{3,0}x_0+c_{3,1}x_1+c_{3,5}x_5. \end{align*} $$

It is of no loss to assume these are all monic in certain indeterminates. That is they have the form

$$ \begin{align*} l_0 &= c_{0,2}x_2+c_{0,3}x_3+x_6, \hspace{1cm} l_1 = x_3+c_{1,4}x_4+c_{1,5}x_5, \\ l_2 &= x_0+c_{2,4}x_4+c_{2,6}x_6, \hspace{1cm} l_3 = c_{3,0}x_0+c_{3,1}x_1+x_5. \end{align*} $$

Through a change of basis, we may reduce the coefficient on $x_5$ in $l_1$ to $1,$ and then normalize $l_3$ to be monic in $x_5;$ then through another change of basis, we may reduce the coefficient on $x_0$ in $l_3$ to $1,$ and then normalize $l_2$ to be monic in $x_0;$ then through another change of basis, we may reduce the coefficient on $x_6$ in $l_2$ to $1,$ and then normalize $l_0$ to be monic in $x_6;$ then through another change of basis, we may reduce the coefficient on $x_2$ in $l_0$ to $1;$ then through another change of basis, we may reduce the coefficient on $x_4$ in $l_2$ to $1;$ then through another change of basis, we may reduce the coefficient on $x_1$ in $l_3$ to $1.$ Ultimately, we obtain

$$ \begin{align*} l_0 &= x_2+c_{0,3}x_3+x_6, \hspace{1cm} l_1 = x_3+c_{1,4}x_4+x_5, \\ l_2 &= x_0+x_4+x_6, \hspace{1.6cm} l_3 = x_0+x_1+x_5. \end{align*} $$

Note the order in which we make these reductions is important.

Proof. After a change of basis, we may represent the defining ideal for our five lines as above. Below is a Koszul filtration

$$ \begin{align*}\mathcal{F}= \begin{cases} (0_R), ( \overline{x}_0), ( \overline{x}_2), ( \overline{x}_0, \overline{x}_4),( \overline{x}_0, \overline{x}_1),( \overline{x}_2, \overline{x}_3),( \overline{x}_0, \overline{x}_6),( \overline{x}_0, \overline{x}_2, \overline{x}_3),\\ ( \overline{x}_2, \overline{x}_3, \overline{x}_0+ \overline{x}_1+ \overline{x}_4+ \overline{x}_5+ \overline{x}_6),( \overline{x}_0, \overline{x}_4, \overline{x}_5),( \overline{x}_0, \overline{x}_1, \overline{x}_2),\\ ( \overline{x}_0, \overline{x}_3, \overline{x}_4), ( \overline{x}_0, \overline{x}_1, \overline{x}_4),( \overline{x}_0, \overline{x}_2, \overline{x}_4),( \overline{x}_0, \overline{x}_2, \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3+b \overline{x}_4+ \overline{x}_5+b \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3+b \overline{x}_4+ \overline{x}_5+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5), ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_2+a \overline{x}_3+a \overline{x}_5+ \overline{x}_6), \\ ( \overline{x}_0, \overline{x}_2, \overline{x}_4, \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6), ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5), ( \overline{x}_0, \overline{x}_4, \overline{x}_5, \overline{x}_6), ( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_4+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+b \overline{x}_4+b \overline{x}_6), ( \overline{x}_0, \overline{x}_2, \overline{x}_4, \overline{x}_6, \overline{x}_1+ \overline{x}_3+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_6, \overline{x}_4+\frac{1}{b} \overline{x}_5),( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4+ \overline{x}_5),( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_6), ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_4+\frac{1}{a} \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+b \overline{x}_4+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5),( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_6, \overline{x}_3+b \overline{x}_4),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_6), ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4+ \overline{x}_5), ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_6), \\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_5, \overline{x}_4+ \overline{x}_6),( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_2+ \overline{x}_6), \\ ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6), ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6), \\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_6, \overline{x}_1+ \overline{x}_5), ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_6, \overline{x}_4+\frac{1}{b} \overline{x}_5), \\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_6, \overline{x}_2+a \overline{x}_3+x_6), ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_6, \overline{x}_3+b \overline{x}_4),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_6, \overline{x}_3+ \overline{x}_5), \mathfrak{m}_R. \end{cases}\end{align*} $$

The calculated colons are

$$ \begin{align*}\begin{cases} (0_R):( \overline{x}_0) = ( \overline{x}_2, \overline{x}_3, \overline{x}_0+ \overline{x}_1+ \overline{x}_4+ \overline{x}_5+ \overline{x}_6),\\ (0_R):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_4, \overline{x}_5), \\ ( \overline{x}_0):( \overline{x}_4) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3+b \overline{x}_4+ \overline{x}_5+b \overline{x}_6),\\ ( \overline{x}_0):( \overline{x}_1) = ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_6), \\ ( \overline{x}_2):( \overline{x}_3) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3+b \overline{x}_4+ \overline{x}_5+\frac{1}{a} \overline{x}_6), \\ ( \overline{x}_0):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_4+ \overline{x}_6),\\ ( \overline{x}_2, \overline{x}_3):( \overline{x}_0)=( \overline{x}_2, \overline{x}_3, \overline{x}_0+ \overline{x}_1+ \overline{x}_4+ \overline{x}_5+ \overline{x}_6), \\ ( \overline{x}_2, \overline{x}_3):( \overline{x}_0+ \overline{x}_1+ \overline{x}_4+ \overline{x}_5+ \overline{x}_6)=( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_6, \overline{x}_4+\frac{1}{b} \overline{x}_5), \\ ( \overline{x}_0, \overline{x}_4):( \overline{x}_5) = ( \overline{x}_0, \overline{x}_2, \overline{x}_4, \overline{x}_6, \overline{x}_1+ \overline{x}_3+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5) ( \overline{x}_0, \overline{x}_4):( \overline{x}_3) = ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_2+a \overline{x}_3+a \overline{x}_5+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1):( \overline{x}_4) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3+b \overline{x}_4+ \overline{x}_5+b \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_4):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_4, \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_6):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_4, \overline{x}_5, \overline{x}_6) ( \overline{x}_0, \overline{x}_1):( \overline{x}_5) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2):( \overline{x}_3+b \overline{x}_4+ \overline{x}_5+b \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2):( \overline{x}_3+b \overline{x}_4+ \overline{x}_5+\frac{1}{a} \overline{x}_6) =( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_5, \overline{x}_4+ \overline{x}_6), \end{cases}\end{align*} $$

$$ \begin{align*}\begin{cases} ( \overline{x}_0, \overline{x}_3, \overline{x}_4):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_2+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4):( \overline{x}_5) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_6, \overline{x}_3+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4):( \overline{x}_2+a \overline{x}_3+a \overline{x}_5+ \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_4):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_4+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_5, \overline{x}_4+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_5):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_4+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2):( \overline{x}_5) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_4, \overline{x}_5):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_5):( \overline{x}_2+a \overline{x}_3+ \overline{x}_4+ \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_6, \overline{x}_3+b \overline{x}_4),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5):( \overline{x}_3+b \overline{x}_4+b \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_6), \\ ( \overline{x}_0, \overline{x}_2, \overline{x}_4, \overline{x}_6):( \overline{x}_1+ \overline{x}_3+ \overline{x}_5) = ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_6):( \overline{x}_4+\frac{1}{b} \overline{x}_5 ) = ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_6, \overline{x}_1+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_6):( \overline{x}_5 ) = ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_6, \overline{x}_1+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6):( \overline{x}_3+b \overline{x}_4+ \overline{x}_5) = \mathfrak{m}_R,\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5):( \overline{x}_4 ) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+b \overline{x}_4+bx_6),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_6):( \overline{x}_4 ) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_6, \overline{x}_4+\frac{1}{b} \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5):( \overline{x}_3 ) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+b \overline{x}_4+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5):( \overline{x}_3+\frac{1}{a} \overline{x}_4+\frac{1}{a} \overline{x}_6)=( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_6, \overline{x}_3+b \overline{x}_4), \\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5):( \overline{x}_3+b \overline{x}_4+\frac{1}{a} \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_5, \overline{x}_4+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5):( \overline{x}_3 ) = ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6):( \overline{x}_3+b \overline{x}_4) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5):( \overline{x}_2+a \overline{x}_3+ \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_6):( \overline{x}_3 +b \overline{x}_4) = ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_6, \overline{x}_2+a \overline{x}_3),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5):( \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_5, \overline{x}_3+\frac{1}{a} \overline{x}_4+\frac{1}{a} \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6):( \overline{x}_4 ) = ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6, \overline{x}_3+b \overline{x}_4+ \overline{x}_5),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_6):( \overline{x}_3+b \overline{x}_4+ \overline{x}_5) = \mathfrak{m}_R,\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_5):( \overline{x}_3+\frac{1}{a} \overline{x}_6) = \mathfrak{m}_R,\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_3, \overline{x}_5):( \overline{x}_4+ \overline{x}_6) = \mathfrak{m}_R,\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5):( \overline{x}_2+ \overline{x}_6) = ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6):( \overline{x}_1) = ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_6):( \overline{x}_1+ \overline{x}_5) = ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_6, \overline{x}_4+\frac{1}{b}x_5):( \overline{x}_1)= ( \overline{x}_0, \overline{x}_2, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_6):( \overline{x}_6)= ( \overline{x}_0, \overline{x}_1, \overline{x}_4, \overline{x}_5, \overline{x}_2+a \overline{x}_3+ \overline{x}_6),\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_5, \overline{x}_6, \overline{x}_3+b \overline{x}_4):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6),\end{cases}\end{align*} $$

$$ \begin{align*}\begin{cases} ( \overline{x}_0, \overline{x}_1, \overline{x}_2, \overline{x}_4, \overline{x}_6):( \overline{x}_3+ \overline{x}_5) = \mathfrak{m}_R,\\ ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6):( \overline{x}_2) = ( \overline{x}_0, \overline{x}_1, \overline{x}_3, \overline{x}_4, \overline{x}_5, \overline{x}_6). \end{cases}\end{align*} $$

Note that $|\mathcal {F}|=|57|.$ Every colon is a non-trivial calculation and requires significant effort; the interested reader may find a link to code verifying the filtration at the author’s website (https://www.joshuaandrewrice.com).

5 Hilbert function obstruction to the Koszul property

In this section, we determine when the coordinate ring of a generic collection of lines is not Koszul. But first, we need a theorem from Complex Analysis.

Theorem 5.2. Let $\mathcal {M}$ be a generic collection of m lines in $\mathbb {P}^n$ with $n \geq 2$ , and let R be the coordinate ring of $\mathcal {M}.$ If

$$ \begin{align*}m> \frac{1}{72}\left(3(n^2+10n+13)+\sqrt{3(n-1)^3(3n+5)}\right),\end{align*} $$

then R is not Koszul.

We have at least one exceptional example of a coordinate ring of a generic collection of lines that is not Koszul that the previous theorem does not handle.

Proof. By Remark 2.8 and a change of basis, we may assume the defining ideals for our three lines have the form

$$ \begin{align*} L_1 = (x_0,x_1,x_3), \qquad L_2 = (x_0,x_2,x_4), \qquad L_3 = (x_1,x_2,l),\end{align*} $$

where $l=x_3+x_4.$ Let J be the defining ideal for R and notice that $K = L_1 \cap L_2=(x_0,x_1x_2,x_1x_4,x_3x_2,x_3x_4).$

We have the following ring isomorphism:

$$ \begin{align*} S/(K+L_3) &= \mathbb{C}[x_0,x_1,x_2,x_3,x_4]/(x_0,x_1,x_2,l,x_3x_4) \\ &\cong \mathbb{C}[x_3,x_4]/(l,x_3x_4) \\ &\cong \mathbb{C}[w]/(w^2). \end{align*} $$

Hence,

$$ \begin{align*}H_{S/(K+L_3) }(t) = \dfrac{-t^2+1}{1-t} = 1+t.\end{align*} $$

Therefore, by Proposition 2.2, the $\text {reg}_S(S/(K+L_3)) =1.$

One checks that the $\text {reg}_S(S/K) =1.$ Using the short exact sequence

$$ \begin{align*}0 \rightarrow S/J \rightarrow S/K\oplus S/L_3 \rightarrow S/(K +L_3) \rightarrow 0 \end{align*} $$

and Proposition 2.2 yields $\text {reg}(S/J) \leq 2.$ So J is generated by forms of degree at most $3$ . The previous short exact sequence, Lemma 4.2, and the additivity of the Hilbert series along the previous short exact sequence yields

$$ \begin{align*} H_{S/J}(t) &= H_{S/K}(t) + H_{S/L_3}(t) - H_{S/(K+L_3)}(t) \\ &= \frac{-t^2+2t+1}{(1-t)^2} + \frac{1}{(1-t)^2} - (1+t) \\ &= \frac{-t^3+3t+1}{(1-t)^2} \\ &= 1+5t+9t^2+12t^3 + \cdots. \end{align*} $$

Thus, J is generated by six linearly independent quadrics and possibly cubics. The cubic $x_3x_4l$ is contained in J, but is not contained in the ideal $(x_0x_1,x_0x_2,x_1x_2,x_1x_4,x_2x_3,x_0l),$ since no term divides $x_3^2x_4.$ Hence, there must be a cubic generator in a minimal generating set of $J.$ Thus, R is not Koszul.

6 Examples

Finally, it is worth observing three examples that have appeared, while studying generic lines.

Example 6.1. There are collections of lines in general linear position that are not generic collections. Consider the four lines in $\mathbb {P}^3$ :

$$ \begin{align*} \mathcal{L}_1 &= \{[0:0:\alpha:\beta] : \alpha, \beta \textrm{ not both zero} \}, \\ \mathcal{L}_2 &= \{[\alpha:\beta:0:0] : \alpha, \beta \textrm{ not both zero} \}, \\ \mathcal{L}_3 &=\{ [\alpha : \beta : -\alpha : \beta ] : \alpha, \beta \textrm{ not both zero} \}, \\ \mathcal{L}_4 &= \{[\alpha:-\beta:\alpha:\beta] : \alpha, \beta \textrm{ not both zero}\}. \end{align*} $$

These lines are in general linear position since every pair spans $\mathbb {P}^3.$ The four defining ideals in S are

$$ \begin{align*} L_1 &= (x_0,x_1), \hspace{3cm} L_2 = (x_2,x_3), \\ L_3 &= (x_0+x_2,x_1-x_3), \hspace{1.3cm} L_4 = (x_0-x_2,x_1+x_3). \end{align*} $$

The coordinate ring $S/J,$ where $J = \bigcap _{i=1}^4 L_i,$ has the following Hilbert series:

$$ \begin{align*}H_{S/J}(t) = \frac{-3t^4+2t^3+2t^2+2t+1}{(1-t)^2} = 1+4t+9t^2+\cdots,\end{align*} $$

whereas, by Theorem 3.4, the coordinate ring R for four generic lines in $\mathbb {P}^3$ has the following Hilbert series:

$$ \begin{align*}H_R(t) = \frac{-2t^3+3t^2+2t+1}{(1-t)^2}= 1+4t+10t^2 + \cdots.\end{align*} $$

So, this is not a generic collection of lines.

Example 6.2. Consider the coordinate ring R for five generic lines in $\mathbb {P}^5.$ The defining ideal J for R is minimally generated by quadrics and has the following Betti table computed via Macaulay2.

The ring R is not Koszul by Theorem 5.2. Furthermore, it is known that if R is Koszul and the defining ideal is generated by g elements, then $\beta _{i,2i} \leq \binom {g}{i}$ for $i \in \{2,\ldots ,g\}$ [Reference Avramov, Conca and Iyengar3, Cor. 3.2]. The previous inequality fails for $i=2$ . So, this ring is not Koszul for two numerical reasons.

Example 6.3. Consider the coordinate ring R for six generic lines in $\mathbb {P}^6.$ The defining ideal J for R is minimally generated by quadrics and has the following Betti table computed via Macaulay2.

The Algebra is not Koszul by Theorem 5.2, and does not fail the aforementioned inequality.

Coordinate rings with defining ideals minimally generated by quadrics are not rare, but the previous two examples are interesting since both fail for identical reasons, and one fails for an additional numerical reason. It would be interesting to determine sufficient reasons why certain numerical conditions fail and others do not. For example, why does $\beta _{i,2i} \leq \binom {g}{i}$ fail in one of the previous rings but not the other.

The Koszul property for the coordinate ring of m generic lines in $\mathbb {P}^n$ .

Furthermore, we would like to add that our theorems do not cover every coordinate ring R for every generic collection of lines in $\mathbb {P}^n.$ For the coordinate rings, we could not determine, there is a possibility these rings could be LG-quadratic or G-quadratic. In every possible case computable by Macaulay2, there exists a quadratic monomial ideal whose quotient ring gives the same Hilbert series as $R.$ There could even be some change of basis which gives a quadratic Gröbner basis. Furthermore, if we wanted to construct a Koszul filtration in these coordinate rings, then Proposition 4.10 demonstrates that there is no reason we should expect a reasonable filtration unless there is a more efficient change of basis that went unobserved. Below is a table, without $m=1$ , $n=1$ , and $n=2$ , summarizing our results: