Proof. Let ${\mathcal {C}}$ be plus-one generated. Denote by ${\mathcal {E}}_{{\mathcal {C}}}$ the sheafification of the S-module of logarithmic derivations $D_0({\mathcal {C}})$ . It is known that ${\mathcal {E}}_{{\mathcal {C}}}$ is in fact a locally free sheaf, that is, a vector bundle. Consider the exact sequence of vector bundles induced by the sheafification of the exact sequence of graded modules from Definition 2.1:

(3) $$ \begin{align} 0 \rightarrow \mathcal{O}_{\mathbb{P}^2}(-1-\ell) \longrightarrow \mathcal{O}_{\mathbb{P}^2}(-\ell)\oplus \mathcal{O}_{\mathbb{P}^2}(-b)\oplus \mathcal{O}_{\mathbb{P}^2}(-a) \rightarrow {\mathcal{E}}_{{\mathcal{C}}} \rightarrow 0. \end{align} $$

There is a well-known formula that connects the Chern polynomials of the terms of an exact sequence of vector bundles (see, for instance, [Reference Fulton15, Section 3]). Using this formula, we compute the first Chern number of ${\mathcal {E}}_{{\mathcal {C}}}$ , namely, $c_1({\mathcal {E}}_{{\mathcal {C}}}) = 1-a-b.$ Since, for reduced curves in general, $c_1({\mathcal {E}}_{{\mathcal {C}}}) = 1 - \deg ({\mathcal {C}})$ , it follows that the plus-one generated property implies $a+b=\deg ({\mathcal {C}}).$

Conversely, let us assume that ${\mathcal {C}}$ is a curve with generalized exponents $(a,b,\ell )$ such that $a+b=\deg ({\mathcal {C}})$ . By (1), a minimal free resolution for $D_0({\mathcal {C}})$ is of the following type:

$$ \begin{align*}0 \rightarrow S(-e) \rightarrow S(-\ell) \oplus S(-b) \oplus S(-a) \rightarrow D_0({\mathcal{C}}) \rightarrow 0.\end{align*} $$

Just as before, take the exact sequence of sheaves associated with the above exact sequence of graded S-modules. This allows us to compute $c_1({\mathcal {E}}_{{\mathcal {C}}})$ . Then we use the equality $a+b=\deg ({\mathcal {C}})$ to conclude $e = \ell +1.$

Proof. Immediately from [Reference Dimca10, Theorem 1.4] and [Reference Măcinic and Pokora17, Theorem 3.2].

Proof. Recall that, by [Reference Dimca10, Theorem 2.3] (extending and based on [Reference Schenck, Terao and Yoshinaga22, Theorem 1.6]), we have an exact sequence of vector bundles:

$$ \begin{align*}0 \rightarrow \mathcal{E}_{{\mathcal{C}}'}(-1) \overset{\alpha_L}\longrightarrow \mathcal{E}_{{\mathcal{C}}} \rightarrow \mathcal{O}_L(1-|{\mathcal{C}}' \cap L| -\epsilon({\mathcal{C}}',L)) \rightarrow 0. \end{align*} $$

In particular, we have a surjective map $ \mathcal {E}_{{\mathcal {C}}} \rightarrow \mathcal {O}_L(1-|{\mathcal {C}}' \cap L| -\epsilon ({\mathcal {C}}',L)).$ Tensor this map by $\mathcal {O}_L$ , to obtain again a surjection

(4) $$ \begin{align} \mathcal{E}_{{\mathcal{C}}} \otimes \mathcal{O}_L \rightarrow \mathcal{O}_L(1-|{\mathcal{C}}' \cap L| -\epsilon({\mathcal{C}}',L)). \end{align} $$

By [Reference Măcinic and Pokora17, Theorem 2.8], we know the possible splitting types onto an arbitrary line of the vector bundle $ \mathcal {E}_{{\mathcal {C}}}$ associated with a plus-one generated curve ${\mathcal {C}}$ . More precisely, $\mathcal {E}_{{\mathcal {C}}} \otimes \mathcal {O}_L$ must be one of the three: $ \mathcal {O}_L(-a) \otimes \mathcal {O}_L(1-b)$ , $ \mathcal {O}_L(1-a) \otimes \mathcal {O}_L(-b),$ or $ \mathcal {O}_L(-l) \otimes \mathcal {O}_L(l+1-a-b)$ . If $\mathcal {E}_{{\mathcal {C}}} \otimes \mathcal {O}_L = \mathcal {O}_L(-a) \otimes \mathcal {O}_L(1-b)$ , then the surjection (4) implies that either $|{\mathcal {C}}' \cap L| + \epsilon ({\mathcal {C}}', L) \leq a+1$ or $|{\mathcal {C}}' \cap L| + \epsilon ({\mathcal {C}}', L) = b$ . If $\mathcal {E}_{{\mathcal {C}}} \otimes \mathcal {O}_L = \mathcal {O}_L(1-a) \otimes \mathcal {O}_L(-b)$ , the surjection (4) implies that either $|{\mathcal {C}}' \cap L| + \epsilon ({\mathcal {C}}', L) = b+1$ or $|{\mathcal {C}}' \cap L| + \epsilon ({\mathcal {C}}', L) \leq a$ . Finally, if $\mathcal {E}_{{\mathcal {C}}} \otimes \mathcal {O}_L = \mathcal {O}_L(-l) \otimes \mathcal {O}_L(l+1-a-b)$ , the surjection (4) implies that either $|{\mathcal {C}}' \cap L| + \epsilon ({\mathcal {C}}', L) = l+1$ or $|{\mathcal {C}}' \cap L| + \epsilon ({\mathcal {C}}', L) \leq a$ .

Proof. The proof follows from [Reference Schenck and Tohăneanu23, Lemmas 3.6 and 3.8 and Proposition 3.7]. Let us recall briefly the steps of this proof, which is based on the existence of the short exact sequence of vector bundles (5). Notice that (5) can also be recovered as a particular case of the short exact sequence from [Reference Dimca10, Theorem 2.3].

The first step is to take the long exact cohomology sequence associated with the short exact sequence of vector bundles (5). Since ${\mathcal {C} \mathcal {L}}'$ is free, it follows that $H^1({\mathcal {E}}_{{\mathcal {C} \mathcal {L}}'}(t))=0$ , for any integer t. So the long exact cohomology sequence associated with (5) gives a short exact sequence of S-graded modules, namely,

(8)

(here, for a coherent sheaf ${\mathcal {E}}$ on $\mathbb {P}^2$ , one denotes .

Furthermore, we know that, if ${\mathcal {C}}$ is a reduced plane curve in $\mathbb {P}^2$ , there is an isomorphism $\Gamma _*({\mathcal {E}}_{{\mathcal {C}}}) = D_0({\mathcal {C}})$ (see [Reference Dimca and Sernesi12]). In particular, $\Gamma _*({\mathcal {E}}_{{\mathcal {C} \mathcal {L}}'}) = D_0({\mathcal {C} \mathcal {L}}')$ , $\Gamma _*({\mathcal {E}}_{{\mathcal {C} \mathcal {L}}}) = D_0({\mathcal {C} \mathcal {L}})$ . Recall that, for projective lines and for smooth conics, a divisor is completely determined by its degree. Consider the k reduced points $i^{-1}(C \cap {\mathcal {C} \mathcal {L}}')$ on $\mathbb {P}^1$ , seen as a divisor on $\mathbb {P}^1$ . The ideal sheaf of this divisor on $\mathbb {P}^1$ is and its associated ideal is an ideal in $\mathbb {C}[s,t]$ . The ideal sheaf of the divisor defined by the reduced k points $C \cap {\mathcal {C} \mathcal {L}}'$ on C is . is the associated ideal in $S / \langle y^2-xz \rangle $ , and its resolution as a graded S-module depends on the parity of k. More precisely, a minimal resolution for $I_k$ is of type

(9) $$ \begin{align} \begin{cases} 0 \rightarrow S(-2-m) \rightarrow S(-m) \rightarrow I_k \rightarrow 0, & \textrm{ if } k = 2m\\ 0 \rightarrow S^2(-2-m) \rightarrow S^2(-1-m) \rightarrow I_k \rightarrow 0, & \textrm{if } k=2m+1. \end{cases} \end{align} $$

We refer to [Reference Schenck and Tohăneanu23, Lemma 3.6] for details and an explicit construction of the two resolutions. These resolutions enable us to compute the Hilbert series for $I_k$ . To do that, we only need to use the additivity of the Hilbert series with respect to exact sequences and the fact that $HS(S(-i),t) = \frac {t^i}{(1-t)^3}$ . We get $ \begin {cases} HS(I_k) = \frac {t^{m+1} + t^m}{(1-t)^2}, & \textrm { if } k = 2m,\\ HS(I_k) = \frac {2t^{m+1}}{(1-t)^2}, & \textrm {if } k=2m+1. \end {cases} $

Since ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(a',b')$ , it follows that $D_0({\mathcal {C} \mathcal {L}}') = S(-a') \oplus S(-b')$ , hence, using the additivity of the Hilbert series with respect to the exact sequence (8), the conclusion of the proposition follows.

Proof. The proof is based on [Reference Schenck and Tohăneanu23, Lemma 3.6, Proposition 3.7, and the proof of Lemma 3.10]. By dualization of the exact sequence (5), we get

Pass to sheaves in (9) to get a resolution for and use this resolution to compute . Recall that, for an arbitrary rank $2$ vector bundle ${\mathcal {E}}$ , one has ${\mathcal {E}}^{\vee } \simeq {\mathcal {E}}(-c_1({\mathcal {E}}))$ and, for a reduced curve ${\mathcal {C}}$ in $\mathbb {P}^2$ , $c_1({\mathcal {E}}_{{\mathcal {C}}}) = 1-\deg ({\mathcal {C}})$ . With this, the above exact sequence, modulo a degree shift by $1-d$ , becomes

(10)

Now the proof follows the same steps as the proof of Proposition 2.9, but using the exact sequence (10) instead of (5). More precisely, one takes the long exact cohomology sequence associated with (10), which yields a short exact sequence of graded S-modules,

(11)

Since ${\mathcal {C} \mathcal {L}}$ is free with exponents $(a,b)$ , it follows $HS(D_0({\mathcal {C} \mathcal {L}})) = \frac {t^a+t^b}{(1-t)^3}$ . A resolution for , which depends on the parity of $k+6-2d$ , see (9), enables us to compute its Hilbert series. We get

$$ \begin{align*}\begin{cases} HS(I_ {k+6-2d})= \frac{t^{d-2-m} + t^{d-3-m}}{(1-t)^2}, & \textrm{ if } k = 2m,\\ HS(I_{k+6-2d}) = \frac{2t^{d-3-m}}{(1-t)^2}, & \textrm{if } k=2m+1. \end{cases} \end{align*} $$

To compute $HS(D_0({\mathcal {C} \mathcal {L}}')),$ we use the additivity of the Hilbert series with respect to the exact sequence (11), hence the conclusion follows.

Proof. In the notations from (1), let $d_1, d_2$ be the minimal degrees of the homogeneous generators from a minimal set of generators of $D_0({\mathcal {C} \mathcal {L}}')$ , that is, the first two generalized exponents of ${\mathcal {C} \mathcal {L}}'$ .

Case $k=2m$ . In this case, we know from Proposition 2.10 that the Hilbert series of $D_0({\mathcal {C} \mathcal {L}}')$ is given by the formula

(12) $$ \begin{align} HS(D_0({\mathcal{C} \mathcal{L}}'))(t) = \frac{t^a+t^b+t^{d-3-m} - t^{d-1-m}}{(1-t)^3}. \end{align} $$

${\mathcal {C} \mathcal {L}}'$ is free only if there is a cancellation in the numerator of the above formula. If a cancellation happens, then either $a=d-1-m$ or $b=d-1-m$ . Assume first that $a=d-1-m$ . Since $d=a+b+1,$ it follows that $b = m.$ Then

$$ \begin{align*}HS(D_0({\mathcal{C} \mathcal{L}}'))(t) = \frac{t^{a-2}+t^b}{(1-t)^3}\end{align*} $$

and

$$ \begin{align*}a-2+b = d-3 = \deg({\mathcal{C} \mathcal{L}}')-1.\end{align*} $$

By Theorem 2.3, ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(a-2, b)$ . This recovers [Reference Schenck and Tohăneanu23, Theorem 3.4, case $k=2m,\; (2)\Rightarrow (1)$ ], with a slightly shortened argument. The case $b=d-1-m$ is analogous, and in this case, one gets that ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(a, b-2)$ .

Conversely, if $m \in \{a,b\}$ , then, assuming for instance $m=a$ ,

$$ \begin{align*}HS(D_0({\mathcal{C} \mathcal{L}}'))(t) = \frac{t^{b-2}+t^a}{(1-t)^3},\end{align*} $$

and, by Theorem 2.3, $(a, b-2)$ are the minimal degrees of the homogeneous generators from a minimal set of generators of $D_0({\mathcal {C} \mathcal {L}}')$ and ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(a, b-2)$ . The case $m=b$ is completely analogous and one gets that ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(a-2, b)$ .

Assume now $m \notin \{a,b\}$ , which is equivalent to $(d-m-1) \notin \{a,b\}$ . Then there is no cancellation in the numerator of the Hilbert series. This implies that ${\mathcal {C} \mathcal {L}}'$ is not free. Considering the general type (1) of resolution for $D_0({\mathcal {C} \mathcal {L}}')$ , we get

(13) $$ \begin{align} \{d_1,d_2\} \subset \{a,b,d-3-m\}, \end{align} $$

so, by the minimality of $d_1, d_2$ ,

$$ \begin{align*}d_1+d_2 \leq a+b,\end{align*} $$

that is, keeping in mind that $\deg ({\mathcal {C} \mathcal {L}}') = d-2 = a+b-1,$ $d_1+d_2 \leq \deg ({\mathcal {C} \mathcal {L}}')+1.$ Since ${\mathcal {C} \mathcal {L}}'$ is not free, by Theorem 2.3, $d_1+d_2 \geq \deg ({\mathcal {C} \mathcal {L}}').$ Summing up the two inequalities,

$$ \begin{align*}\deg({\mathcal{C} \mathcal{L}}') \leq d_1+d_2 \leq \deg({\mathcal{C} \mathcal{L}}')+1.\end{align*} $$

Consider first the case $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')$ . By Theorem 2.3, this is equivalent to ${\mathcal {C} \mathcal {L}}'$ being plus-one generated with exponents $(d_1, d_2)$ . Since at the same time $d_1+d_2 = a+b-1$ , we cannot have $\{d_1, d_2\} = \{a,b\}$ . So necessarily

$$ \begin{align*}d-3-m < b, \; \{d_1,d_2\} = \{a, d-3-m\}.\end{align*} $$

This implies $b = d-2-m$ . An inspection of the Hilbert series formula of $D_0({\mathcal {C} \mathcal {L}}')$ shows that the plus-one generated ${\mathcal {C} \mathcal {L}}'$ is of level b. To sum up, ${\mathcal {C} \mathcal {L}}'$ is plus-one generated with exponents $(a,b-1)$ and level b. Moreover, notice that $b = d-2-m$ is equivalent to $m=a-1$ .

Consider now the case $d_1+d_2 =\deg ({\mathcal {C} \mathcal {L}}')+1$ . This implies that ${\mathcal {C} \mathcal {L}}'$ is neither free nor plus-one generated, by Theorem 2.3. Since $d_1+d_2 = a+b $ , by (13), we necessarily have $\{d_1,d_2\} = \{ a, b\}.$ Moreover, since $a \leq b$ , it follows $b \leq d-3-m$ , that is, $m \leq a-2$ .

Conversely, we have to consider the cases $m=a-1$ and $m \leq a-2$ . If $m=a-1$ , then $d-3-m = b-1$ and necessarily $\{d_1,d_2\} = \{a,d-3-m\}$ , so $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')$ . Using again Theorem 2.3, it follows that ${\mathcal {C} \mathcal {L}}'$ is plus-one generated. It has exponents $(a, b-1)$ and level b, by (12). If $m \leq a-2$ , that is, $b \leq d-3-m$ , then $\{d_1,d_2\} = \{ a, b\}$ , so $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')+1.$ By Theorem 2.3, ${\mathcal {C} \mathcal {L}}'$ is neither free nor plus-one generated.

Case $k=2m+1$ . In this case, we know from Proposition 2.10 that the Hilbert series of $D_0({\mathcal {C} \mathcal {L}}')$ is given by the formula

(14) $$ \begin{align} HS(D_0({\mathcal{C} \mathcal{L}}'))(t) = \frac{t^a+t^b+2t^{d-3-m} - 2t^{d-2-m}}{(1-t)^3}. \end{align} $$

As in the previous case, a necessary condition for ${\mathcal {C} \mathcal {L}}'$ to be free is the cancellation of the term $- 2t^{d-2-m}$ in the numerator of the above formula, that is, necessarily $a=b=m+1$ . Then

$$ \begin{align*}HS(D_0({\mathcal{C} \mathcal{L}}'))(t) = \frac{2t^{m}}{(1-t)^3}\end{align*} $$

so $d_1 = d_2 = m$ , and, since $\deg ({\mathcal {C} \mathcal {L}}') = 2m+1$ , by Theorem 2.3, ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(m,m)$ .

Conversely, assume $a=b=m+1$ , then $HS(D_0({\mathcal {C} \mathcal {L}}'))(t) = \frac {2t^{m}}{(1-t)^3}$ and the same argument as above shows that ${\mathcal {C} \mathcal {L}}'$ is free with exponents $(m,m)$ . This recovers [Reference Schenck and Tohăneanu23, Theorem 3.4, case $k=2m+1,\; (1)$ ].

Assume now $a \neq m+1$ or $b \neq m+1.$ By the above argument, this is equivalent to ${\mathcal {C} \mathcal {L}}'$ being not free. Then at least one term $-t^{d-2-m}$ does not cancel in the numerator of the Hilbert series. We have

(15) $$ \begin{align} \{d_1,d_2\} \subset \{a,b,d-3-m\}, \end{align} $$

so $d_1+d_2 \leq a+b.$ Since ${\mathcal {C} \mathcal {L}}'$ is not free, by Theorem 2.3, $d_1+d_2 \geq \deg ({\mathcal {C} \mathcal {L}}')$ . But $ \deg ({\mathcal {C} \mathcal {L}}') = a+b-1$ , hence $\deg ({\mathcal {C} \mathcal {L}}') \leq d_1+d_2 \leq \deg ({\mathcal {C} \mathcal {L}}')+1$ , or, equivalently,

$$ \begin{align*}a+b-1 \leq d_1+d_2 \leq a+b.\end{align*} $$

Notice that $b=m+1$ and $a \neq m+1$ cannot happen simultaneously. If we assume the contrary, we must have $a<b, a = d-2-m$ . So $b> d-2-m$ , and this is in contradiction to the formula (14), which in this case would translate into

$$ \begin{align*}HS(D_0({\mathcal{C} \mathcal{L}}'))(t) = \frac{t^{b> d-2-m}+2t^{d-3-m} - t^{d-2-m}}{(1-t)^3}.\end{align*} $$

As we have seen, $d_1+d_2 \in \{\deg ({\mathcal {C} \mathcal {L}}'), \deg ({\mathcal {C} \mathcal {L}}')+1 \}$ . The equality $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')$ is equivalent, by Theorem 2.3, to ${\mathcal {C} \mathcal {L}}'$ being plus-one generated with exponents $(d_1, d_2)$ . Since at the same time $d_1+d_2 = a+b-1$ , then, by (15), necessarily $d-3-m < b,$ hence either

$$ \begin{align*}\{d_1,d_2\} = \{a, d-3-m\}\; \text{and}\; b = d-2-m\end{align*} $$

or

$$ \begin{align*}d_1=d_2=d-3-m.\end{align*} $$

If $\{d_1,d_2\} = \{a, d-3-m\}$ , then, by the Hilbert series formula of $D_0({\mathcal {C} \mathcal {L}}')$ (14), the level of ${\mathcal {C} \mathcal {L}}'$ is $b-1$ . Moreover, notice that $b = d-2-m$ is equivalent to $m=a-1$ . If $d_1=d_2=d-3-m$ and, moreover, $a=d-3-m$ , we are back in the previous case. Otherwise, $a> d-3-m$ , which leads to $d_1+d_2 \leq a+b-2$ , contradiction.

Let us prove now the “only if” implication of case (2)(b)(i). Hence, we assume $m=a-1$ (equivalently, $b = d-2-m$ ). Since $b \neq m+1$ , it follows $a<b$ . So $a \leq d-3-m$ . Then necessarily $\{d_1,d_2\} = \{a,d-3-m\}$ , so $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')$ . Again by Theorem 2.3, ${\mathcal {C} \mathcal {L}}'$ is plus-one generated. By (14), its exponents are $(a, b-1)$ and level $ b-1$ .

Finally, consider the case $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')+1$ . This is equivalent to ${\mathcal {C} \mathcal {L}}'$ being neither free nor plus-one generated, by Theorem 2.3. In particular, by the argument in the above paragraph, $m \neq a-1$ , since $m = a-1$ implies ${\mathcal {C} \mathcal {L}}'$ plus-one generated. So, $b \neq d-2-m$ . Then, a cancellation in the numerator of the Hilbert series formula (14) would be possible if and only if $a=d-2-m$ . But this would imply $d_1=d_2=d-3-m$ and $d_1+d_2 \leq a+b-2$ , contradiction. So, there is no cancellation in the numerator of the Hilbert series formula (14). Since $d_1+d_2 = a+b $ , after taking into account the possible cases arising from the inclusion (15), it follows that $\{d_1,d_2\} = \{ a, b\}.$ Moreover, since $a \leq b$ , it follows $b \leq d-3-m$ , that is, $m \leq a-2$ .

We have left to prove the “only if” part of case (2)(b)(ii). Assume $m \leq a-2$ (equivalently, $b \leq d-3-m$ ). Then, by (14), $\{d_1,d_2\} = \{ a, b\}$ , so $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}}')+1$ . By Theorem 2.3, ${\mathcal {C} \mathcal {L}}'$ is neither free nor plus-one generated.

Proof. From the proof of Theorem 3.1, we already know that $d_1 = a, \; d_2 = b$ . Since $\deg ({\mathcal {C} \mathcal {L}}') = \deg ({\mathcal {C} \mathcal {L}}) -2 = a+b-1$ , the conclusion follows.

Proof. Case $k=2m.$ If we assume ${\mathcal {C} \mathcal {L}}$ is free, then there must be a cancellation in the Hilbert series formula of $D_0({\mathcal {C} \mathcal {L}})$ , which happens only if $a'=m$ or $b'=m$ . In this case, ${\mathcal {C} \mathcal {L}}$ has exponents $(a', b'+2)$ or $(a'+2, b')$ . Conversely, if $a'=m$ or $b'=m$ , then by Proposition 2.9 and Theorem 2.3, ${\mathcal {C} \mathcal {L}}$ is free with exponents $(a', b'+2)$ or $(a'+2, b')$ .

If $ m \notin \{a',b'\}$ , then there is no cancellation in the Hilbert series formula of $D_0({\mathcal {C} \mathcal {L}})$ . Let $d_1, d_2$ be defined as in (1), for the resolution of the module $D_0({\mathcal {C} \mathcal {L}})$ . Then

$$ \begin{align*}\{d_1, d_2\} \subset \{a'+2, b'+2, m\}.\end{align*} $$

By (6), necessarily $m \geq b'$ . Since $m \neq b'$ , it follows $m> b'$ . If $m = b'+1,$ then $\{d_1, d_2\} = \{a'+2, b'+1\}$ , so $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}})$ . By Theorem 2.3, ${\mathcal {C} \mathcal {L}}$ is plus-one generated, and by (6), the exponents of ${\mathcal {C} \mathcal {L}}$ are $(a'+2, b'+1)$ and the level is $b'+2$ . Conversely, if ${\mathcal {C} \mathcal {L}}$ is plus-one generated, then $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}})$ , by Theorem 2.3. Since $\deg ({\mathcal {C} \mathcal {L}}) = a'+b'+3$ , this implies $\{d_1, d_2\} = \{a'+2, m\}$ and $m=b+1$ . If $m \geq b'+2,$ then $\{d_1, d_2\} = \{a'+2, b'+2\}$ , so $d_1+d_2 = \deg ({\mathcal {C} \mathcal {L}})+1$ . By Theorem 2.3, ${\mathcal {C} \mathcal {L}}$ is neither free nor plus-one generated. Conversely, if ${\mathcal {C} \mathcal {L}}$ is neither free nor plus-one generated, then $d_1+d_2> \deg ({\mathcal {C} \mathcal {L}})= a'+b'+3$ . But $\{d_1, d_2\}$ are the smallest two elements in $\{a'+2, b'+2, m\}$ , hence $m \geq b'+2$ .

Case $k=2m+1.$ If ${\mathcal {C} \mathcal {L}}$ is free, then the term $-2t^{m+2}$ from (7) must cancel out, hence necessarily $a'=b' = m$ . In this case, the exponents of ${\mathcal {C} \mathcal {L}}$ are $(m+1,m+1)$ . Conversely, if $a'=b' = m$ , then

$$ \begin{align*}HS(D_0({\mathcal{C} \mathcal{L}}))(t) = \frac{2t^{m+1}}{(1-t)^3}\end{align*} $$

and, by Theorem 2.3, since $2m+2 = a'+b'+2 = \deg ({\mathcal {C} \mathcal {L}})-1$ , ${\mathcal {C} \mathcal {L}}$ is free. From the above formula for the Hilbert series of $D_0({\mathcal {C} \mathcal {L}})$ , $\exp ({\mathcal {C} \mathcal {L}})=(m+1,m+1)$ .

Assume in what follows $a' \neq m$ or $b' \neq m$ . In any case, by (7), for $d_1, d_2$ as defined in (1), for the module $D_0({\mathcal {C} \mathcal {L}})$ , we have the inclusion

$$ \begin{align*}\{d_1, d_2\} \subset \{a'+2, b'+2, m+1\}.\end{align*} $$

First, let us prove that the case $m=a'$ and $m \neq b'$ is not possible. Assuming the contrary, we get $m=a'<b'$ . But, by (7), $m \geq b'$ , contradiction. Hence, in any case, $m \neq a'$ . This implies, again by (7), that $a'<m$ .

Consider now the case $m=b'$ . Then, still from (7), $\{d_1, d_2\} = \{a'+2, m+1\}$ . But $a'+2+m+1 = a'+b'+3 = \deg ({\mathcal {C} \mathcal {L}})$ , so, by Theorem 2.3, ${\mathcal {C} \mathcal {L}}$ is plus-one generated. By (7), ${\mathcal {C} \mathcal {L}}$ has exponents $(a'+2, b'+1)$ and level $b'+1$ . Conversely, if ${\mathcal {C} \mathcal {L}}$ is plus-one generated, then $d_1 + d_2= \deg ({\mathcal {C} \mathcal {L}}) = a'+b'+3$ , by Theorem 2.3. Since $a'+2 \leq m+1$ , $\{d_1, d_2\} = \{a'+2, m+1\},$ and $m+1 = b'+1$ , that is, $b'=m$ .

If $m \neq b'$ (in which case, we also know that $m \neq a'$ ), then, from (7), $m \geq b'+1$ . From the proof above, this is equivalent to ${\mathcal {C} \mathcal {L}}$ being neither free nor plus-one generated. Incidentally, from $m \geq b'+1$ , in this case, we get $\{d_1, d_2\} = \{a'+2, b'+2\}$ .

Proof. We know from the proof of Theorem 3.5 that $d_1 = a'+2, \; d_2 =b'+2$ . Since $ \deg ({\mathcal {C} \mathcal {L}}) = \deg ({\mathcal {C} \mathcal {L}}') +2 = a'+b'+3$ , the conclusion follows.

Proof. Immediately from Corollaries 3.4 and 3.8 and Theorem 2.6 .