2.1 Some key exact sequences

Let us come back to the setting of Theorem 1.3. To simplify the notation, in the sequel, we set

(2.4) $$ \begin{align} e_1 = \deg f_1 = \deg C_1 \text{ and } e_2=\deg f_2 = \deg C_2. \end{align} $$

Assume in addition that $C_2$ is smooth and all the points in $C_1 \cap C_2$ are quasi-homogeneous singularities of $C=C_1 \cup C_2$ . Then, there is an exact sequence for any integer k given by

(2.5) $$ \begin{align} 0 \to E_{C_1}(k-e_2) \stackrel{f_2} \longrightarrow E_C(k) \to i_{2*}{\mathcal{O}}_{C_2}(-K_{C_2}-R+(k-1)D) \to 0, \end{align} $$

where $E_C= \widetilde {D_0(f)} $ (resp., $E_{C_1}= \widetilde {D_0(f_1)} $ is the sheafification of the graded S-module $D_0(f)$ (resp., $D_0(f_1)$ ), which is a rank two vector bundle on $\mathbb {P}^2$ , $K_{C_2}$ is the canonical divisor on ${C_2}$ , R is the reduced scheme of $C_1 \cap C_2$ , and the divisor $D'$ corresponds to the intersection of a line in $\mathbb {P}^2$ with the curve $C_2$ (see [Reference Dimca, Ilardi and Sticlaru11, Reference Schenck, Terao and Yoshinaga26]). By taking the associated long cohomology exact sequence of the sequence (2.5), we get

(2.6) $$ \begin{align} & 0 \to D_0(f_1)_{k-e_2} \to D_0(f)_k \to H^0(C_2,{\mathcal{O}}_{C_2}(D+(k-1)D')) \to \\ & \to N(f_1)_{k-e_2+e_1-1} \to N(f)_{k+e_1+e_2-1} \to H^1(C_2,{\mathcal{O}}_{C_2}(D+(k-1)D')) \to \nonumber\\ & \to H^2(\mathbb{P}^2, E_{C_1}(k-e_2)) \to H^2(\mathbb{P}^2, E_{C}(k)) \to 0, \nonumber \end{align} $$

with $D = -K_{C_2}-R$ , and hence $\deg D=2-2g_2-r$ , where $g_{C_2}$ is the genus of the smooth curve $C_2$ and r is the number of points in the reduced scheme of $C_1 \cap C_2$ . Moreover, one has

(2.7) $$ \begin{align} \dim H^2(\mathbb{P}^2, E_{C_1}(k-e_2))=\dim D_0(f_1)_{e_1+e_2-(k-e_2)-4} \end{align} $$

and

$$ \begin{align*}\dim H^2(\mathbb{P}^2, E_{C}(k))=\dim D_0(f)_{(e_1+e_2+1)-k-4}, \end{align*} $$

see [Reference Dimca5, Reference Sernesi28].

Remark 2.2 If we identify $D_0(f)$ with the quotient S-module

$$ \begin{align*}\overline D(f)=D(f)/SE,\end{align*} $$

then the morphism $\theta $ is just given by the multiplication by $f_1$ . Here, $D(f)$ denotes the S-module of all derivations of f, that is,

$$ \begin{align*}D(f)=\{(a,b,c) \in S^3 \ : \ af_x+bf_y+cf_z \in (f)\},\end{align*} $$

where $(f)$ is the principal ideal generated by f, and $SE$ denotes the submodule in $D(f)$ generated by the Euler derivation $E=(x,y,z)$ . The first non-zero morphism in (2.5) is multiplication by $f_2$ if we replace $D_0(f_1)$ and $D_0(f)$ by $\overline D(f_1)$ and $\overline D(f),$ respectively (see, for instance, [Reference Schenck, Terao and Yoshinaga26, Theorem 1.6]). This fact is used in the proof of Theorems 1.15 and 1.19 below.

2.2 Exact sequences for line arrangements

Consider the triple of line arrangements $({\mathcal {A}}',{\mathcal {A}}, {\mathcal {A}}")$ , where ${\mathcal {A}}'={\mathcal {A}} \setminus \{L\}$ with L being a line in ${\mathcal {A}}$ , and ${\mathcal {A}}"$ is the arrangement consisting of points on the line L corresponding to the intersection set ${\mathcal {A}}' \cap L$ . By a change of coordinates, we can take $L:x=0$ , and we set

$$ \begin{align*}{\mathcal{A}}' :f' = 0, \ {\mathcal{A}}: f = 0 \text{ and } {\mathcal{A}}":f" = 0,\end{align*} $$

for some polynomial $f" \in R=\mathbb {C}[y,z]$ . Then, there is an exact sequence of graded S-modules, sometimes called the Euler exact sequence of the triple $({\mathcal {A}}',{\mathcal {A}}, {\mathcal {A}}")$ , given by

(2.8) $$ \begin{align} 0 \to D(f')(-1) \to D(f) \to D(f"), \end{align} $$

where $D(f')$ , $D(f),$ and $D(f")$ are as in Remark 2.2 and in addition, $D(f")$ is regarded as an S-module using the restriction of scalars given by the ring morphism $\psi : S \to R$ , which replaces x by $0$ in any polynomial $h \in S$ . Here, the first morphism $D(f')(-1) \to D(f)$ is the multiplication by x, and the second morphism is given by

$$ \begin{align*}\rho: D(f) \to D(f"), \ \rho(a_1,a_2,a_3)= (\psi(a_2), \psi(a_3)),\end{align*} $$

see [Reference Orlik and Terao24, Proposition 4.45] and [Reference Schenck and Tohaneanu27, Lemma 2.6] that covers a more general case when ${\mathcal {A}}'$ is a conic-line arrangement, maybe under some quasi-homogeneity conditions. If we replace the modules $D(f')$ , $D(f),$ and $D(f")$ by the modules $\overline D(f')$ , $\overline D(f),$ and $\overline D(f")$ , the sequence obtained from (2.8) is no longer exact. However, if we pass to the associated coherent sheaves on $\mathbb {P}^2$ , we get the following short exact sequence:

(2.9) $$ \begin{align} 0 \to E_{{\mathcal{A}}'}(-1) \to E_{{\mathcal{A}}} \to i_*E_{{\mathcal{A}}"} \to 0, \end{align} $$

where we followed the notation from (2.5) (see [Reference Schenck25, Theorem 3.2]). Using this exact sequence, we can give the following useful sufficient condition for the sequence (2.8) to be right-exact, i.e., for the morphism $\rho $ to be surjective.

Theorem 2.3 With the above notation, let

$$ \begin{align*}0 \rightarrow \oplus_{i=1}^{m-2} S(-c^{\prime}_i) \rightarrow \oplus_{i=1}^m S(-d_i) \rightarrow D_0({\mathcal{A}}') \rightarrow 0 \end{align*} $$

be a minimal free resolution of $D_0({\mathcal {A}}')$ . If $|{\mathcal {A}}"|>c_i'-1$ for all i, then the Euler exact sequence

$$ \begin{align*}0 \rightarrow D({\mathcal{A}}')(-1) \stackrel{\cdot x}{\rightarrow} D({\mathcal{A}}) \stackrel{\rho}{\rightarrow} D({\mathcal{A}}") \end{align*} $$

is right exact.

Proof Let $|{\mathcal {A}}"| = n$ , and note that this implies $|{\mathcal {A}}'| \geq n$ . Then, the S-module $D({\mathcal {A}}")$ has two generators, one of degree $1$ and another of degree $n-1$ . The first element is the Euler derivation, which is always in the image of $\rho $ . It follows that

$$ \begin{align*}D({\mathcal{A}}) \stackrel{\rho}{\rightarrow} D({\mathcal{A}}")\end{align*} $$

is surjective if and only if the induced morphism

$$ \begin{align*}\overline D({\mathcal{A}}) \stackrel{\overline \rho}{\rightarrow} \overline D({\mathcal{A}}")\end{align*} $$

is surjective. Moreover, it suffices to show that $ \overline \rho $ is surjective in degree $n - 1$ . Using the exact sequence (2.9), it suffices to show that $H^1(\mathbb {P}^2,E_{{\mathcal {A}}'}(n-2))=0$ . By the minimal free resolution of $D_0({\mathcal {A}}')$ , we have

$$ \begin{align*} 0&= H^1(\mathbb{P}^2, \oplus\mathcal{O}(-d_i+n-2)) \rightarrow H^1(\mathbb{P}^2,E_{{\mathcal{A}}'}(n-2)) \\ & \rightarrow H^2(\oplus\mathcal{O}(-c^{\prime}_i+n-2)) \simeq H^0(\mathbb{P}^2,\oplus \mathcal{O}(c^{\prime}_i-n+2-3))=0. \end{align*} $$

The last vanishing follows by our assumption, and the first one from the fact that $H^1(\mathbb {P}^2, \mathcal {O}(k))=0$ for any integer k. This completes the proof.

The exact sequence in Theorem 2.3 plays a key role in our proof of Theorem 1.19. When we pass from line arrangements to conic-line arrangements, there are similar exact sequences according to [Reference Schenck and Tohaneanu27], and they are discussed in detail and used in the proof of Theorem 1.15 below.

3.1 Proof of Theorem 1.3

Observe that there is a natural S-linear morphism

$$ \begin{align*}\theta: D_0(f_1)_{k-e_2} \to D_0(f)_k\end{align*} $$

for any integer k. If $\delta \in D_0(f_1)_{k-e_2}$ is a syzygy, we can think about it as being a derivation killing $f_1$ , and then we set

$$ \begin{align*}\theta (\delta)=f_2\delta -\frac{\delta(f_2)}{d}E,\end{align*} $$

where $d= \deg C=e_1+e_2$ and E denotes the Euler derivation (see [Reference Dimca, Ilardi and Sticlaru11, Theorem 5.1]. This morphism is injective. Indeed, if we suppose that $\theta (\delta )=0$ for some $\delta \in D_0(f_1)_{k-e_2}$ , then, in particular, $\theta (\delta )(f_1)=0$ . Since $E(f_1)=e_1f_1 \ne 0$ and $\delta (f_1)=0$ , we get $\delta (f_2)=0$ . It follows that

$$ \begin{align*}0=\theta (\delta)=f_2\delta\end{align*} $$

and hence, $\delta =0$ since $f_2 \ne 0$ .

We denote by $d_1,d_2$ (resp., by $d_1', d_2'$ ) the first two exponents of $C_1$ (resp., of C). By Proposition 1.2, we have

$$ \begin{align*}d_1' \leq d_1+e_2.\end{align*} $$

If $d_2' \leq d_2+e_2$ , then by adding these two inequalities, we get our claim. Suppose now that

$$ \begin{align*}d_2'> d_2+e_2.\end{align*} $$

Let $r_1$ and $r_2$ be the first two generators in a minimal set of generators for $D_0(f_1)$ and let $r_1'$ be a generator of $D_0(f)$ of minimal degree $d_1'$ . Then, $\theta (r_1) \in D_0(f)_{d_1+e_2}$ and $\theta (r_2 )\in D_0(f)_{d_2+e_2}$ are non-zero derivations. It follows that there are non-zero homogeneous polynomials $p,q \in S$ such that $\theta (r_1)=pr_1'$ and $\theta (r_2)=qr_1'$ . This implies $q\theta (r_1)-p\theta (r_2)=\theta (qr_1-pr_2)= 0$ , and hence by the injectivity of $\theta $ (see the beginning of the proof of [Reference Dimca, Ilardi and Sticlaru11, Theorem 5.1]), we get

$$ \begin{align*}qr_1-pr_2=0.\end{align*} $$

This is a contradiction since the quotient S-module $D_0(f_1)/Sr_1$ is torsion free, as it follows, for instance, from the exact sequence in [Reference Dimca and Sticlaru17, Theorem 5.1]. Indeed, this exact sequence may be rewritten as

$$ \begin{align*}0 \to S(-d_1) \to D_0(f) \to B(C,r_1) \to 0,\end{align*} $$

where the first morphism is given by $h \mapsto hr_1$ . Hence, $D_0(f_1)/Sr_1$ is isomorphic to an S-submodule I of the ideal $B(C,r_1)$ , in other words, I is an ideal in S, and hence has no torsion.

3.2 Proof of Theorem 1.6 and Corollaries 1.7 and 1.9

We use the exact sequence (2.6) above, where

$$ \begin{align*}g_2= \operatorname{\mathrm{genus}}(C_2)=\frac{(e_2-1)(e_2-2)}{2}\end{align*} $$

and $r=e_1e_2$ . Moreover, $D'$ consists of $e_2$ points, and hence,

$$ \begin{align*}\deg (D+(k-1)D')=2-(e_2-1)(e_2-2)-e_1e_2+(k-1)e_2=e_2(k-e_1-e_2+2).\end{align*} $$

For $k < e_1+e_2-2$ , we have $H^0(C_2,{\mathcal {O}}_{C_2}(D+(k-1)D'))=0$ , and this yields an isomorphism

$$ \begin{align*}D_0(f_1)_{k-e_2} \to D_0(f)_k.\end{align*} $$

It follows that all the generators of $D_0(f)$ of degree $k < e_1+e_2-2$ come from generators of $D_0(f_1)$ of degree $k-e_2$ . In this way, we get two generators of degrees

$$ \begin{align*}d_1' = d_1+e_2 \text{ and } d_2' = d_2+e_2.\end{align*} $$

A new generator, possibly coming from $H^0(C_2,{\mathcal {O}}_{C_2}(D+(k-1)D'))=\mathbb {C}$ , arises when $k=e_1+e_2-2$ . Since

$$ \begin{align*}d_2'=d_2+e_2 \leq e_1+e_2 - 2\end{align*} $$

by our assumption, we get that $d_1'$ and $d_2'$ are the first two exponents of the curve C, and hence, C has type

$$ \begin{align*}d_1'+d_2'-(e_1+e_2-1)=(d_1+d_2-(e_1-1))+e_2=t_1+e_2\end{align*} $$

as claimed.

The proof above implies that if the curve $C_1$ satisfies the condition $d_1 \leq e_1-2$ , then the resulting curve C also satisfies this condition, since

$$ \begin{align*}d_2'=d_2+e_2 \leq e_1+e_2-2.\end{align*} $$

This completes the proof of Theorem 1.6.

Example 3.2 The condition $d_2 \leq \deg (C_1)-2$ is necessary in Theorem 1.6 as the following example shows. Let $C_1:f_1=x^2+y^2+z^2=0$ and $C_2:f_2=x=0$ . Then, $d_1=d_2=1$ and hence, $t(C_1)=1$ . A direct computation shows that $C=C_1 \cup C_2$ is a 3-syzygy curve with exponents $(1,2,2)$ and hence of type

$$ \begin{align*}1+2-2=1<t(C_1)+\deg(C_2)=2.\end{align*} $$

Now, we prove Corollary 1.7.

Proof The first claim follows from (1.3), since

$$ \begin{align*}t({\mathcal{A}})=d_1+d_2-(d-1) \leq 2d-4-(d-1)=d-3.\end{align*} $$

Moreover, (1.3) tells us that we can apply Theorem 1.6 to any line arrangement $C_1$ . Starting with the integers t and d as in the second claim of Corollary 1.7, we choose $C_1$ to be a free line arrangement of $e_1=d-t \geq 3$ lines, for instance, $x^{e_1}+y^{e_1}=0$ . Then, we add t generic lines to this arrangement and we get by Theorem 1.6 a line arrangement ${\mathcal {A}}$ of degree d and type t. This completes the proof of Corollary 1.7.

Now, we present our proof of Corollary 1.9.

Proof We first notice that our additional assumption on ${\mathcal {C}}$ implies

$$ \begin{align*}t({\mathcal{C}})=d_1+d_2-\deg({\mathcal{C}})+1 \leq 2d_2-\deg({\mathcal{C}})+1\leq \deg({\mathcal{C}})-3.\end{align*} $$

Then, for the second claim, it is enough as above to construct free conic-line arrangements ${\mathcal {C}}_1$ for any $e_1=\deg ({\mathcal {C}}_1)=d-t \geq 3$ , containing at least one conic and such that $d_2 \leq \deg ({\mathcal {C}}_1) - 2$ . When $e_1=3$ , we take ${\mathcal {C}}_1$ to be a smooth conic plus one of its tangents. It is known that then ${\mathcal {C}}_1$ is free with exponents $(1,1)$ and we are done.

For $e_1 \geq 4$ , we consider the arrangement

$$ \begin{align*}{\mathcal{C}}_1:f_1=x(xz+y^2)(x^{e_1-3}+y^{e_1-3})=0.\end{align*} $$

It is easy to check the following are two syzygies for $f_1$ :

$$ \begin{align*}r_1=(x^2,xy,-e_1y^2-(e_1-1)xz)\end{align*} $$

and

$$ \begin{align*}r_2=(-(e_1-1)xy^{e_1-4},e_1x^{e_1-3}+y^{e_1-3},-2e_1x^{e_1-4}y+(e_1+1)y^{e_1-4}z).\end{align*} $$

It follows that

$$ \begin{align*}d_1 \leq \deg(r_1)=2 \text{ and } d_2 \leq \deg(r_2)=e_1-3.\end{align*} $$

since $r_1$ and $r_2$ are not multiple of strictly lower degree syzygy. Using Theorem 1.1, it follows that ${\mathcal {C}}_1$ is a free curve and the above inequalities are in fact equalities. In particular,

$$ \begin{align*}d_2=\max\{2,e_1-3\} \leq e_1-2\end{align*} $$

for $e_1 \geq 4$ . To complete the proof of Corollary 1.9, it remains to add t generic lines to the arrangement ${\mathcal {C}}_1$ and get in this way the required arrangement ${\mathcal {C}}$ by Theorem 1.6.

Remark 3.3 The curve ${\mathcal {C}}$ constructed above has a complicated singularity at the point $p=(0:0:1)$ . Indeed, looking at $({\mathcal {C}},p)$ as the union of an $(e_1-2)$ -ordinary point with the smooth germ at p given by the conic $xz+y^2=0$ and using the known formula for the Milnor number of a union (see, for instance, [Reference Wall29, Theorem 6.5.1]), one can show that the Milnor number $\mu ({\mathcal {C}},p)$ is given by

$$ \begin{align*}\mu({\mathcal{C}},p)=e_1^2-4e_1+6.\end{align*} $$

On the other hand, using the formula for the total Tjurina number of a free curve, which can be obtained by setting $\nu (C)=0$ in Theorem 2.1 (1), we get

$$ \begin{align*}\tau({\mathcal{C}})=(e_1-1)^2-2(e_1-3)=e_1^2-4e_1+7.\end{align*} $$

Now, the curve ${\mathcal {C}}$ has $e_1-3$ nodes $A_1$ as singularities except p, and hence,

$$ \begin{align*}\tau({\mathcal{C}},p)=\tau({\mathcal{C}})-(e_1-3)=e_1^2-5e_1+10.\end{align*} $$

One can alternatively compute $\tau ({\mathcal {C}},p)$ by using only local computations, based on [Reference Hefez and Hernandes20, Proposition 3.2 and Theorem 5.1]. It follows that the invariant

$$ \begin{align*}\epsilon({\mathcal{C}},p)=\mu({\mathcal{C}},p)-\tau({\mathcal{C}},p),\end{align*} $$

which was introduced in [Reference Dimca7] and measures how far $({\mathcal {C}},p)$ is from being quasi-homogeneous, is given by

$$ \begin{align*}\epsilon({\mathcal{C}},p)=e_1-4.\end{align*} $$

Hence, for all $e_1> 4,$ the singularity $({\mathcal {C}},p)$ is not quasi-homogeneous.

4.1 Proof of Proposition 1.11

To see that the number m of syzygies can be only $3$ or $4$ , and to get the corresponding values for the $\varepsilon _j$ ’s, use the equality (1.7). In case (1), the value for $\tau (C)$ follows from [Reference Dimca and Sticlaru16, Proposition 2.1] using that $d_1+d_2=d+1$ , where $d= \deg C$ . In case (2), we use (1.6) to get the following resolution of the Milnor algebra $M(f)$ :

$$ \begin{align*}0 \to S(-d-d_3)\oplus S(-d-d_4) \to \oplus_{j=1}^{4}S(1-d-d_j) \to S^3(1-d) \to S.\end{align*} $$

This resolution implies that in this case one has

$$ \begin{align*} \tau(C)=\binom{k+2}{2}-3 \binom{k+3-d}{2}+\sum_{j=1}^{4}\binom{k+3-d-d_j}{2}- \\ - \binom{k+2-d-d_3}{2}- \binom{k+2-d-d_4}{2}, \end{align*} $$

for any large k. The last claim follows from the equality (1.7).

Here are some examples of curves C having type $2$ .

Example 4.1 (i) Consider the Bolza curve $C : x^{5} -y^{2}z^{3}-xz^{4} = 0$ . It is well-known that C has a unique singularity of type $E_{8}$ and one can check that C has type $2A$ with exponents $(2,4,4)$ .

(ii) Consider a conic-line arrangement $\mathcal {CL}_1\subset \mathbb {P}^{2}_{\mathbb {C}}$ given by

$$ \begin{align*}f_1 = (x^2 + y^2 - z^2 )(y-z)(x^2 - z^2 )=0.\end{align*} $$

This arrangement is known to be free and it has three singularities of type $A_{1}$ and three singularities of type $A_{3}$ , so this gives us $\tau (\mathcal {CL}_1)=12$ (see [Reference Măcinic and Pokora23, Example 2.9]). If we add the conic $Q \, : y^2-xz=0$ to $\mathcal {CL}_1$ , then the resulting arrangement $\mathcal {CL}$ has quasi-homogeneous singularities and has type $2B$ with exponents $(4,4,5,5)$ .

(iii) Consider a conic arrangement $\mathcal {C}_1\subset \mathbb {P}^{2}_{\mathbb {C}}$ given by

$$ \begin{align*}f_1= (-3x^2+xy+yz+xz)(-3y^2 + xy+yz+zx)(-3z^2 + xy+yz+zx)=0.\end{align*} $$

This arrangement is known to be free with exponents $(2,3)$ and it has three singularities of type $A_5$ and just one ordinary triple point $D_4$ . If we add the conic $Q:f_2=x^2+y^2+z^2=0$ , we get new conic arrangement $\mathcal {C}$ which has type $2A$ and exponents $(4,5,6)$ . Note that the conic Q is generic with respect to the arrangement $\mathcal {C}_1$ , namely, the intersection $\mathcal {C}_1\cap Q$ consists of $12$ points, which give rise to $12$ nodes for the arrangement $\mathcal {C}$ .

4.2 Proofs of Propositions 1.12 and 1.13

We start with Proposition 1.12. Here, we use Theorem 2.1 and Proposition 1.11. Note that the condition $d_1 <d_2-2$ is exactly the condition $d_1<(d-1)/2,$ which occurs in Theorem 2.1 (1). Hence, to prove the claim $(1)$ in Proposition 1.12, we use the formula for $\nu (C)$ given in Theorem 2.1 (1) and the formula for $\tau (C)$ given in Proposition 1.11, $(1)$ . To prove the claims (2)–(4), we use the formula for $\nu (C)$ given in Theorem 2.1 (2) and the formula for $\tau (C)$ given in Proposition 1.11 $(1)$ . This completes the proof of Proposition 1.12.

The proof of Proposition 1.13 is analogous.

5.1 Proof of Theorem 1.14

We use the exact sequence (2.6) above with $C_1={\mathcal {A}}':f_1=0$ and $C_2=L:f_2=0$ . Hence, $e_1=m_1+m_2=r$ , $e_2=1$ , $g_2= \operatorname {\mathrm {genus}}(C_2)=0$ . Moreover, $D'$ is just a point, and hence,

$$ \begin{align*}{\mathcal{O}}_{C_2}(D+(k-1)D')={\mathcal{O}}_L(k+1-m_1-m_2).\end{align*} $$

Let $(d_1,d_2, \ldots ,d_m)$ be the exponents of the curve $C={\mathcal {A}}$ .

First, consider the case $k <m_1+m_2-1=e_1-1$ . Then, $H^0(L,{\mathcal {O}}_L(k+1-m_1-m_2))=0$ and we get an isomorphism

$$ \begin{align*}D_0(f_1)_{k-1} \to D_0(f)_k.\end{align*} $$

Note that both $k=m_1+1$ and $k=m_2+1$ satisfy the condition $k <m_1+m_2-1$ , and hence, we get that $d_1=m_1+1$ and $d_2=m_2+1$ . It follows that $d_1+d_2=\deg {\mathcal {A}}+1$ , and hence, ${\mathcal {A}}$ has type $2$ . To see what happens for the other exponents of ${\mathcal {A}}$ , we look now at the case $k=e_1-1$ . Since ${\mathcal {A}}'$ is a $3$ -syzygy curve, using (2.3), we get

$$ \begin{align*}\sigma({\mathcal{A}}')=\operatorname{\mathrm{indeg}} N(f_1)=3(e_1-1)-(m_1+m_2+m_3)=2e_1-m_3-3.\end{align*} $$

Let $T_1=3(e_1-2)$ and recall that $\dim N(f_1)_j=\dim N(f_1)_{T_1-j}$ for any integer j by (2.1). It follows that $N(f_1)_j=0$ for any $j>T_1-\sigma ({\mathcal {A}}')=e_1+m_3-3$ . Since $m_3 \leq e_1-2$ by (1.3), this yields an exact sequence

(5.1) $$ \begin{align} 0 \to D_0(f_1)_{e_1-2} \to D_0(f)_{e_1-1} \to H^0(L,{\mathcal{O}}_{L}) \to 0. \end{align} $$

When $m_3=e_1-2$ , then there are two new generators in $D_0(f)_{e_1-1}$ , one corresponding to the generator of $D_0(f_1)$ of degree $m_3$ , and the second one corresponding to the generator of $H^0(L,{\mathcal {O}}_{L})=\mathbb {C}$ . In this case,

$$ \begin{align*}(d_1,d_2,d_3,d_4)=(m_1+1, m_2+1, e_1-1, e_1-1)\end{align*} $$

as claimed. When $m_3 <e_1-2$ , then the exact sequence (5.1) shows that there is exactly one new generator for $D_0(f)$ in degree $d_3=m_3+1 < e_1-1$ , and then a new generator in degree $d_4 = e_1-1$ corresponding to the generator of $H^0(L,{\mathcal {O}}_{L})=\mathbb {C}$ .

5.2 Proof of Theorem 1.15

This rather long proof is divided into two steps to help the reader understand it more easily.

Step 1. We use again the exact sequence (2.6) above with $C_1={\mathcal {C}}' : f_1=0$ and $C_2=Q : f_2=0$ . Hence, we have $e_1=m_1+m_2+1$ , $r=2e_1$ , $e_2=2$ , $g_2= \operatorname {\mathrm {genus}}(C_2)=0$ . Moreover, $D'$ consists of two points and hence

$$ \begin{align*}{\mathcal{O}}_{C_2}(D+(k-1)D')={\mathcal{O}}_L(2k-2e_1).\end{align*} $$

Let $(d_1,d_2, \ldots ,d_m)$ be the exponents of the curve $C={\mathcal {C}}$ .

First, consider the case $k <e_1$ . Then, $H^0(L,{\mathcal {O}}_L(2k-2e_1))=0$ and we get an isomorphism

$$ \begin{align*}D_0(f_1)_{k-1} \to D_0(f)_k.\end{align*} $$

Note that both $k=m_1+2$ and $k=m_2+2$ satisfy the condition $k < m_1+m_2+1$ , and hence, we get that $d_1=m_1+2$ and $d_2=m_2+2$ . It follows that $d_1+d_2=e_1+3=\deg {\mathcal {C}}+1$ , and hence, ${\mathcal {C}}$ has type $2$ . To see what happens for the other exponents of ${\mathcal {C}}$ , we look now at the case $k=e_1$ . Since ${\mathcal {C}}'$ is a free curve, one has $N(f_1)=0$ and this yields an exact sequence

(5.2) $$ \begin{align} 0 \to D_0(f_1)_{e_1-2} \to D_0(f)_{e_1} \to H^0(L,{\mathcal{O}}_{L}) \to 0. \end{align} $$

This exact sequence shows that there is exactly one new generator, call it ${\alpha }$ , for $D_0(f)$ in degree $d_3=e_1$ corresponding to the generator of $H^0(L,{\mathcal {O}}_{L})=\mathbb {C}$ .

Step 2. If the conic-line arrangement ${\mathcal {C}}$ has type $2A$ , then the exponents of ${\mathcal {C}}$ are $(d_1,d_2,d_3)=(m_1+2,m_2+2,m_1+m_2+1)$ , and our result is proved. To show that this is the case, we have to show that there is no need for an additional generator for $D_0(f)$ beyond ${\alpha }$ .

A triple $({\mathcal {C}}',{\mathcal {C}},{\mathcal {C}}")$ of CL (short for conic-line) arrangements, where ${\mathcal {C}}'= {\mathcal {C}} \setminus \{Q\}$ with $Q \in {\mathcal {C}}$ a smooth conic and ${\mathcal {C}}"$ the zero-dimensional arrangement on Q given by the reduced intersection ${\mathcal {C}}' \cap Q$ , is called quasi-homogeneous if all the singular points of ${\mathcal {C}}'$ and of ${\mathcal {C}}$ situated on Q are quasi-homogeneous.

Proof This result is a stronger version of [Reference Schenck and Tohaneanu27, Proposition 3.7]. The authors in [Reference Schenck and Tohaneanu27] require in their result that all the singular points of ${\mathcal {C}}'$ and of ${\mathcal {C}}$ are quasi-homogeneous, but this stronger assumption is not necessary. Indeed, as noted in [Reference Dimca, Ilardi and Sticlaru11] and in Section 2.1 above, for a similar condition in [Reference Schenck, Terao and Yoshinaga26], this assumption is used in [Reference Schenck and Tohaneanu27] only to compute the difference

$$ \begin{align*}\tau({\mathcal{C}})-\tau({\mathcal{C}}')=\deg J_f -\deg J_{f'},\end{align*} $$

where we set

$$ \begin{align*}{\mathcal{C}}':f'=0 \text{ and } {\mathcal{C}}:f=0,\end{align*} $$

see, for instance, the proof of [Reference Schenck and Tohaneanu27, Propositions 2.8 and 3.7]. The difference $\tau ({\mathcal {C}})-\tau ({\mathcal {C}}')$ can be expressed in the following way:

$$ \begin{align*}\sum_{p \in {\mathcal{C}}"} (\tau({\mathcal{C}},p)-\tau({\mathcal{C}}',p)),\end{align*} $$

since for a point $q \in {\mathcal {C}} \setminus {\mathcal {C}}"$ , one has $\tau ({\mathcal {C}},q)-\tau ({\mathcal {C}}',q)=0$ , as the two singularities $({\mathcal {C}},q)$ and $\tau ({\mathcal {C}}',q)$ coincide for $q \notin Q$ , Q is smooth and we set by convention $\tau ({\mathcal {C}}',p)=0$ when the germ $({\mathcal {C}}',p)$ is the empty germ. Lemma 2.7 in [Reference Schenck and Tohaneanu27] is applied to the singularities $p \in {\mathcal {C}}"$ , since for quasi-homogeneous singularities, the Milnor number coincides with the Tjurina number. That is why it is enough to require that the singularities of ${\mathcal {C}}$ and ${\mathcal {C}}'$ situated on Q are quasi-homogeneous.

In [Reference Schenck and Tohaneanu27, Proposition 3.5], the following exact sequence of S-modules is constructed:

(5.3) $$ \begin{align} 0 \to D(f')(-2) \to D(f) \to D(f")_{\psi}, \end{align} $$

where $D(f)$ and $D(f')$ are as in Remark 2.2. To explain the equation ${\mathcal {C}}":f"=0$ and the S-module $D(f")_{\psi }$ , note that by a change of coordinates on $\mathbb {P}^2$ , we can assume that $Q:q=y^2-xz=0$ . Then, we have an isomorphism

$$ \begin{align*}v: \mathbb{P}^1 \to Q \text{ given by } v(s,t)=(s^2,st,t^2).\end{align*} $$

Using this isomorphism, we can regard the zero-dimensional arrangement ${\mathcal {C}}"$ as a zero-dimensional arrangement in $\mathbb {P}^1$ and hence it is defined by an equation $f"=~0$ , where $f" \in R=\mathbb {C}[s,t]$ is some homogeneous polynomial. Hence, the graded R-module $D(f")$ is well-defined. Moreover, the map v above induces a ring morphism

$$ \begin{align*}\psi=v^*: S \to R \text{ given by } v^*(g)=g(v(s,t)),\end{align*} $$

and the S-module $D(f")_{\psi }$ is just the module obtained from the R-module $D(f")$ by restriction of scalars using $\psi $ . Next, we consider the morphisms in the exact sequence (5.3). The first morphism $D(f')(-2) \to D(f) $ is given by the multiplication by q, the equation of Q, and hence it preserves the homogeneous components, that is,

$$ \begin{align*}q(D(f')(-2)_k)=q(D(f')_{k-2}) \subset D(f)_k.\end{align*} $$

The second morphism, call it $\rho :D(f) \to D(f")_{\psi }$ , is much more subtle. If $(a_1,a_2,a_3) \in D(f)$ , then it is shown that

$$ \begin{align*}\psi(a_1)=sQ_1, \ \psi(a_2)=\frac{tQ_1+sQ_2}{2} \text{ and } \psi(a_3)=tQ_2\end{align*} $$

for some polynomials $Q_1,Q_2 \in R$ , see [Reference Schenck and Tohaneanu27] for details. With this notation, one sets

$$ \begin{align*}\rho(a_1,a_2,a_3)=(Q_1,Q_2)\end{align*} $$

and hence

$$ \begin{align*}\rho(D(f)_k) \subset D(f")_{2k-1}.\end{align*} $$

After these preliminaries, we can continue our proof in Step 2. In our situation, ${\mathcal {C}}"$ consists of $2e_1$ points, hence, $f"$ is a homogeneous polynomial of degree $2e_1$ without multiple factors. The derivation ${\alpha } \in D_0(f)_{e_1}$ that was constructed at the end of Step 1 can be regarded as sitting in $\overline D(f)_{e_1}$ by using the identification from Remark 2.2. Then, any representative ${\alpha }_0 \in D(f)_{e_1}$ of ${\alpha }$ is clearly not in the image of the morphism $D(f')(-2) \to D(f) $ and hence $\rho ({\alpha }_0) \ne 0$ . If we assume ${\alpha }_0=(a_1,a_2,a_3)$ , then

(5.4) $$ \begin{align} za_1-xa_3 \ne 0. \end{align} $$

Indeed, $za_1=xa_3$ implies that there is $h \in S$ such that $a_1=xh$ and $a_3=zh$ . But this implies that

$$ \begin{align*}{\alpha}_1={\alpha}_0-hE=(0,a_2',0)\end{align*} $$

would be a new representative of ${\alpha }$ and clearly $\rho ({\alpha }_1)=0$ , a contradiction.

If ${\mathcal {C}}$ has type $2B$ , it means that there is an additional generator of degree $d_4>d_3=e_1$ . Using (1.4), we get $d_4 \leq (e_1+2)-1$ , and hence, $d_4=e_1+1$ . On the other hand, the exact sequence (5.2) which holds due to Theorem 5.1, where $e_1$ is replaced by $e_1+1$ , shows that $\overline D_0(f)_{e_1+1}$ modulo the image of $\overline D_0(f_1)_{e_1-1}$ is two-dimensional. It follows that we need a new generator of degree $e_1+1$ if and only if the vector space V spanned by $x {\alpha }_0, y {\alpha }_0,$ and $z{\alpha }_0$ in $\overline D_0(f)_{e_1+1}$ modulo the image of $\overline D(f_1)_{e_1-1}$ is one-dimensional. Note that $\rho (E)=E"$ , the Euler derivation corresponding to the ring R, and hence there is an induced morphism

$$ \begin{align*}\overline \rho: \overline D(f) \to \overline D(f")_{\psi}.\end{align*} $$

The element

$$ \begin{align*}x{\alpha}_0=(xa_1,xa_2,xa_3)\end{align*} $$

is equivalent modulo $SE$ to

$$ \begin{align*}{\beta}=x{\alpha}_0-a_1E=(0, xa_2-ya_1, xa_3-za_1).\end{align*} $$

Since $xa_3-za_1=s^2tQ_2-t^2sQ_1=st(sQ_2-tQ_1)$ , it follows that

(5.5) $$ \begin{align} {\beta}_1=\overline \rho(x{\alpha})= \overline \rho({\beta}) = (0,s(sQ_2-tQ_1)) \in \overline D(f")_{\psi}. \end{align} $$

A similar computation shows that

(5.6) $$ \begin{align} {\beta}_2=\overline \rho(z{\alpha})= (-t(sQ_2-tQ_1),0) \in \overline D(f")_{\psi}. \end{align} $$

The elements ${\beta }_1$ and ${\beta }_2$ are linearly independent in $\overline D(f")_{\psi }$ . Indeed, no linear combination $c_1{\beta }_1+c_2 {\beta }_2$ with $c_1,c_2 \in \mathbb {C}$ can have the form $hE"=(sh,th)$ for some $h \in R$ . Indeed

$$ \begin{align*}c_1{\beta}_1+c_2 {\beta}_2=(-c_2t(sQ_2-tQ_1),c_1s(sQ_2-tQ_1))=(sh,th)\end{align*} $$

would imply $(c_1s^2+c_2t^2)(sQ_2-tQ_1)=0$ , which is a contradiction by (5.4). This shows that the classes of $x {\alpha }_0$ and of $z{\alpha }_0$ in $\overline D(f)$ are linearly independent as well. Therefore, $\dim V \geq 2$ and this completes the proof of Theorem 1.15.

Example 5.2 An illustration of Theorem 1.15 occurs in Example 4.1 (iii), where $(m_1,m_2)=(2,3)$ . Note that in Example 4.1 (ii), the condition that the intersection $Q \cap {\mathcal {A}}'$ consists of $2\deg {\mathcal {A}}'=2(m_1+m_2+1)=10$ points is not fulfilled, since $\tau (\mathcal {CL})=24$ by Proposition 1.11, and hence,

$$ \begin{align*}\tau(\mathcal{CL}_1)+10 = 22\ne \tau(\mathcal{CL}).\end{align*} $$

This explains why the type $2B$ may occur here.

5.3 Proof of Proposition 1.17

The claims for $n_1=1$ and $n_1=2$ follow from [Reference Dimca, Ibadula and Măcinic9, Proposition 4.7 (4)]. Consider now the case $n_1 \geq 3$ . Let ${\mathcal {B}}:g=0$ be the line arrangement obtained from ${\mathcal {A}}'$ by adding the line $L'$ determined by the points $a_1$ and $a_2$ . Then, ${\mathcal {B}}$ is a supersolvable line arrangement, and $a_1$ is a modular point of multiplicity $m_1=n_1+1$ . Using [Reference Dimca and Sticlaru15, Proposition 3.2], it follows that

$$ \begin{align*}mdr(g)= \min \{m_1-1, (n_1+n_2+1)-m_1\}=n_1.\end{align*} $$

Since any supersolvable arrangement is free, it follows that ${\mathcal {B}}$ is a free arrangement with exponents $(n_1,n_2)$ . Next, we apply [Reference Dimca7, Theorem 1.3] to the curve $C={\mathcal {A}}'$ and the line $L=L'$ . In our case, $r=|{\mathcal {A}}' \cap L'|=2$ and $\epsilon :=\epsilon ({\mathcal {A}},L')=0$ since all the singularities of line arrangements are quasi-homogeneous. It follows that the curve ${\mathcal {A}}'$ is plus-one generated with exponents $(n_1,n_2,n_1+n_2-2)$ . Indeed, we have to use the Case $(3)$ in [Reference Dimca7, Theorem 1.3], where $d=\deg (C)=\deg ({\mathcal {A}}')=n_1+n_2$ . In the notation of [Reference Dimca7, Theorem 1.3], we have $d_1' = n_1$ and hence $d_1\geq d_1'-1=n_1-1$ . It follows that

$$ \begin{align*}d_1+1-\epsilon \geq n_1 \geq 3>r,\end{align*} $$

and hence by the last claim in [Reference Dimca7, Theorem 1.3] the curve C is not free. Therefore, we are in the case (3) of [Reference Dimca7, Theorem 1.3], and this yields our claim.

5.4 Proof of Theorem 1.19

This proof is similar to our proof of Theorem 1.15, so we divide it into two steps.

Step 1. We use the exact sequence (2.6) above with $C_1={\mathcal {A}}':f_1=0$ and $C_2=L:f_2=0$ . Hence, $e_1=n_1+n_2$ , $e_2=1$ , $g_2= \operatorname {\mathrm {genus}}(C_2)=0$ , $r=n_1+n_2-1$ . Moreover, $D'$ is just a point and we have

$$ \begin{align*}{\mathcal{O}}_{C_2}(D+(k-1)D')={\mathcal{O}}_L(k+2-n_1-n_2).\end{align*} $$

Let $(d_1,d_2, \ldots ,d_m)$ be the exponents of the curve $C={\mathcal {A}}$ .

Consider first the case $k <n_1+n_2-2=e_1-2$ . Then, $H^0(L,{\mathcal {O}}_L(k+2-d_1-d_2))=0$ and we get an isomorphism

$$ \begin{align*}D_0(f_1)_{k-1} \to D_0(f)_k.\end{align*} $$

Note that for $n_1\geq 4$ , both $k=n_1+1$ and $k=n_2+1$ satisfy the condition $k <n_1+ n_2-2$ , and hence, we get that $d_1=n_1+1$ and $d_2=n_2+1$ . To see what happens for $n_1=3$ , we look now at the case $k=e_1-2$ . Since ${\mathcal {A}}'$ is a 3-syzygy curve, using (2.3), we get

$$ \begin{align*}\sigma({\mathcal{A}}')=\operatorname{\mathrm{indeg}} N(f_1)=3(e_1-1)-(n_1+n_2+n_1+n_2-2)=e_1-1.\end{align*} $$

Let $T_1=3(e_1-2)$ and recall that $\dim N(f_1)_j=\dim N(f_1)_{T_1-j}$ for any integer j by (2.1). It follows that $N(f_1)_j=0$ for any $j>T_1-\sigma ({\mathcal {A}}')=2e_1-5$ . In particular, this yields an exact sequence

(5.7) $$ \begin{align} 0 \to D_0(f_1)_{e_1-3} \to D_0(f)_{e_1-2} \to H^0(L,{\mathcal{O}}_{L}) \to 0. \end{align} $$

Consider first the case $n_1=3<n_2$ , when the sequence becomes

$$ \begin{align*}0 \to D_0(f_1)_{n_2} \to D_0(f)_{n_2+1} \to H^0(L,{\mathcal{O}}_{L}) \to 0.\end{align*} $$

This shows that there are two new generators in $D_0(f)_{n_2+1}$ , one corresponding to the generator of $D_0(f_1)$ of degree $n_2$ and the other one, call it ${\alpha }$ , corresponding to the generator of $H^0(L,{\mathcal {O}}_{L})=\mathbb {C}$ . This implies that $d_2=d_3=n_2+1$ in this case. When $n_1=3=n_2$ , then there are three new generators in $D_0(f)_{n_2+1}$ , two corresponding to the two generators of $D_0(f_1)$ of degree $n_1=n_2$ , and the third one, call it again ${\alpha }$ , corresponding to the generator of $H^0(L,{\mathcal {O}}_{L})=\mathbb {C}$ . In this case $(d_1,d_2,d_3)=(4,4,4)$ as claimed. It follows that in all cases ${\mathcal {A}}$ has type $2$ . Now, consider the case $n_1 \geq 4$ and study the exponents $d_3$ and $d_4$ . The exact sequence (5.7) shows that there is exactly one new generator for $D_0(f)$ in degree $d_3=e_1-2=n_1+n_2-2$ , which we denote by ${\alpha }$ , corresponding to the generator of $H^0(L,{\mathcal {O}}_{L})=\mathbb {C}$ .

Step 2. If ${\mathcal {A}}$ is a $3$ -syzygy curve, then Theorem 1.19 is proved. We show now that this is indeed the case.

The arrangement ${\mathcal {A}}"$ consists of $n=e_1-1=n_1+n_2$ points, using the formulas (1.6) and (1.7), we see that $m=3$ , $\varepsilon _1=1,$ and $c_1=2(n_1+n_2-1).$ Hence, the integer $c^{\prime }_1$ from Theorem 2.3 is given by

$$ \begin{align*}c_1'=c_1-(e_1-1)=n_1+n_2-1.\end{align*} $$

It follows that we can apply Theorem 2.3 and get a surjective morphism $\rho $ . Now, our claim that $D_0(f)$ can be generated by three elements or, equivalently, that $D(f)$ is generated by four elements, follows from the next basic fact in commutative algebra. If we have an exact sequence of S-modules

$$ \begin{align*}0 \to M' \to M \to M" \to 0.\end{align*} $$

A a generating set for $M'$ and a set $B \subset M$ which maps onto a generating set for $M"$ , then the image of A under $M' \to M$ union B gives a generating set for M. We apply this to $M'=D(f')$ , $M=D(f),$ and $M"=D(f")$ , and A consisting of $E'=E$ the Euler derivation of ${\mathcal {A}}'$ and of ${\mathcal {A}}$ , plus the two generators of $D(f')_0$ and B consisting of E and an element ${\alpha } \in D(f)$ such that $\rho ({\alpha })$ is the generator of $D(f")$ of degree $n-1$ . The element $xE$ is discarded from the union since we have E in this union.