---
address: 'Department of Mathematis, College of Humanities and Sciences, Nihon University Setagaya, Tokyo 156-8550, Japan'
author:
- 'Kei-ichi Watanabe'
title: 'Classification of 2-dimensional graded normal hypersurfaces with $a(R)\le 6$.'
---

Introduction {#introduction .unnumbered}
============

Isolated weighted homogeneous surface singularities of type $R=k[X,Y,Z]/(f)$ are extensively studied by V.I. Arnold, H. Pinkham [@P1] and K. Saito and many other authors. Especially, K. Saito [@S1] studies these in terms of regular system of weights". On the other hand, from the view point of commutative ring theory, the $a$-invariant $a(R)$ defined in [@GW] (in Saito’s paper, the notation $\varepsilon = -a(R)$ is used) and also, such singularities can be constructed by so called DPD (Dolgacev-Pinkham-Demazure) constrction [@Dol][@P2][@Dem].

The author tried the classification of all the possible weights for given $a(R)$ using DPD construction of normal graded rings and it turned out that the procedure is so simple and nearly automatic.

Although this classification is known" in the literature (cf. [@S1], [@Wag]), it seems that the algebraic or ring-theoretic classification is not done yet.

Another good point of our classification is that we can draw the graph" of the resolution of the singularity of $\Spec(R)$ instantly from the expression as $R = R(X,D)$.

Below, we present the classification of such singularities with $0<a(R)\le 6$. We prove in general that for a given $\alpha>0$, the number of types of $R$ for 2-dimensional normal graded hypersurfaces with $a(R)=\alpha$ is finite.

Preliminaries
=============

Let $R=\oplus_{n\ge 0}R_n= k[u,v,w]/(f)$ be a $2$-dimensional graded normal hypersurface, where $k$ is an algebraically closed field of any characteristic. We always assume that the grading of $R$ is given so that $R_n\ne 0$ for $n\gg 0$. We put $X=\Proj(R)$. Since $\dim R=2$ and $R$ is normal, $X$ is a smooth curve. Then by the construction of Dolgachev, Pinkham and Demazure ([@Dem], [@P2]), there is an ample $\Q$-Cartier divisor $D$ (that is, $ND$ is an ample divisor on $X$ for some positive integer $N$), such that $$R\cong R(X,D) = \oplus_{n\ge 0}H^0(X,O_X(nD))\cdot T^n\subset k(X)[T]$$ as graded rings, where $T$ is a variable over $k(X)$ and $$H^0(X,O_X(nD))=\{f\in k(X)\;|\;\divv_X(f)+nD\ge 0\}\cup\{0\}.$$

We denote by $[nD]$ the integral part of $nD$ so that $H^0(X,O_X(nD))= 
H^0(X,O_X([nD]))$.

Now, let us begin the classification. In the following, $X$ is a smooth curve of genus $g$ and $D$ is a fractional divisor on $X$ such that $ND$ is an ample integral (Cartier) divisor for some $N>0$.

We always denote $$\leqno{(1.0.1)}\quad D=E+ \sum_{i=1}^r \dfrac{p_i}{q_i} P_i \quad (p_i, q_i \;{\mathrm{are \; positive  \; integers \; and \;}}\forall i,\; (p_i,q_i)=1),$$ where $E$ is an integral divisor. In this case, we denote $$\leqno{(1.0.2)}\quad
\FRC(D)=\sum_{i=1}^r \dfrac{q_i-1}{q_i} P_i.$$

At the same time, by our assumption $R\cong k[u,v,w]/(f)$. If $\deg(u,v,w;f)=(a,b,c;h)$, then by [@GW], $$\leqno{(1.0.3)}\quad a(R)= h-(a+b+c).$$

We always assume $\deg(u,v,w;f)=(a,b,c;h)$ and also that $a\le b\le c$. We call $(a,b,c; h)$ the [*type*]{} of $R$. In this paper, we will determine all the possible types of $R$ for $0< a(R)\le 6$ [^1].

We list our tools to classify.

\[Fund\]([**Fundamental formulas**]{}) Assume that $R=R(X,D)$

$
\cong k[u,v,w]/(f)$ with $\deg(u,v,w;f)=(a,b,c;h)$ and $a(R)=h-(a+b+c)=\alpha$. Then we have the following equalities.

\(1) [@W] Since $R$ is Gorenstein with $a(R)=\alpha$, we have $$\alpha D\sim K_X + \FRC(D) = K_X +\sum_{i=1}^r \dfrac{q_i-1}{q_i} P_i,$$ where, in general, $D_1\sim D_2$ means that $D_1-D_2=\divv_X(\phi )$ for some $\phi \in k(X)$.

In particular, the genus $g$ of $X$ is given by $$g=g(X) = H^0(X, O_X(K_X))= \dim R_{\alpha}.$$

\(2) [@To] If $P(R,t)=\sum_{n\ge 0} \dim R_n t^n$ is the Poincare series of R, then $$\lim_{t\to 1} (1-t)^2P(R,t) = \deg D.$$ Since $P(R,t)=\dfrac{1-t^h}{(1-t^a)(1-t^b)(1-t^c)}$ in our case, we have $$\deg D= \dfrac{h}{abc}=\dfrac{\alpha}{abc}+\dfrac{1}{ab}+\dfrac{1}{ac}+
\dfrac{1}{bc}.$$

Note that the latter expression is a decreasing function of $a,b,c$.

\[bd-of-c\][@OW] Let $R=k[x,y,z]/(f)$ be a normal graded ring with type $(a,b,c;h)$ with $h=a+b+c+\alpha $. Then

\(1) If $h$ is not a multiple of $c$, then either $h-a$ or $h-b$ is a multiplie of $c$. The same is true for $a,b$. Namely, at least one of $h, h-a,h-c$ is a multiple of $b$ and at least one of $h,h-b,h-c$ is a multiple of $a$.

\(2) If a prime number $p$ divides two of $a,b,c$, then $p$ divides $h$. In particular, if $\alpha$ is even, then at most one among $a,b,c$ is even.

\(3) $c\le a+b+\alpha $.

If $R=k[x,y,z]/(f)$ is normal, then $f$ must contain monomial of the form $x^n$ or $x^ny$ or $x^nz$. The statement (1), (3) follows from this fact. As for (2), It is easy to see that if $p$ divides $a,b$ and not $h$, then $f$ must be divisible by $z$.

We list some properties of $R$ when $R$ is Gorenstein.

\[dual\] Let $R=R(X,D)$ be a normal graded ring with $D$ as in (1.0.1) and assume that $R$ is Gorenstein with $a(R)=\alpha$. Then we have the following formulas.

1.  For every $n$, $0\le n\le \alpha$, we have $\deg [nD]+\deg [(\alpha -n)D]
    =\deg [\alpha D] = 2g -2$, where $g$ is the genus of $X$.

2.  $[(\alpha + 1)D] = K_X + E + \sum_{i=1}^r P_i$.

3.  $[(2\alpha + 1)D] = 2K_X + E + 2\sum_{p_i \ge 2} P_i + \sum_{p_i=1} P_i$.

This follows easily from $\alpha D=K_X+\FRC(D)$.

Next we recall some fundamental property of $p_g(R)$.

If $X \to \Spec(R)$ is a resolution of singularities of $R$, then the geometric genus of $R$, $p_g(R)$ is defined by $$p_g(R)=\dim_k \HH^1(X, O_X).$$

When $R$ is a Gorenstein graded ring, it is proved in [@W] that

$$\leqno{(1.5.1)}\quad
\qquad p_g(R)=\sum_{n=0}^{a(R)} \dim_k R_n.$$

In the following, we denote $a(R)=\alpha$ to avoid confusion with $a$ (the minimal positive degree with $R_a\ne 0$).

\[resol\] [@Dem] If $R=R(X,D)$ with $D=E+ \sum_{i=1}^r \dfrac{p_i}{q_i} P_i$, the graph“ of the resolution of singularity of $\Spec(R)$ is a so called star-shaped” graph with central curve" $X$ with self intersection $X^2 = - \lceil D\rceil$ and branch of ${\Bbb P}^1$’s intersecting at $P_i\in X$ with self intersection number $-b_1,\ldots , -b_s$ if the continued-fraction expression of $\dfrac{q_i}{q_i-p_i}$ is as $\dfrac{q_i}{q_i-p_i}= b_1- \dfrac{1}{b_2- \dfrac{1}{b_3- \ldots }}$

General results for given $\alpha = a(R)>0$.
============================================

First, we will show the finiteness of the types $(a,b,c; h)$ for given $\alpha =a(R)>0$.

\[finite\] If we fix $\alpha =a(R) > 0$, the number of $(a,b,c; h)$ for a normal graded ring $R=k[x,y,z]/(f)$ with $h=
a+b+c+\alpha $ is finite [^2].

In the rest of this section, we fix $\alpha >0$, always assume that $a\le b\le c, h=a+b+c+\alpha $ and $(a,b,c)=1$. The proof is done in a series of lemmas. Note that by Lemma \[bd-of-c\], it suffices to show that the number of possible $(a,b)$ is finite. Also, for simplicity, we assume that $c\ge \alpha$.

\[degDle4ab\] (1) We have $\deg D \le \dfrac{4}{ab}$.

\(2) If $g\ge 2$, then $ab< \dfrac{4\alpha }{2g-2}$.

\(1) By our assumption $c\ge \alpha$, $\deg D=\dfrac{a+b+c+\alpha }{abc}\le \dfrac{4}{ab}$.

\(2) Since $\alpha D\sim K_X+\FRC(D)$, $\deg D  \ge \dfrac{2g-2}{\alpha }$ and the LHS is less than $\dfrac{4}{ab}$.

If $g=1$, then $ab< 8\alpha $.

As in the previous lemma, since $\alpha\; \deg D \sim \FRC(D)$ and $\deg \;\FRC(D)\ge \dfrac{1}{2}$, we have $\deg D=\dfrac{a+b+c+\alpha }{abc} \ge \dfrac{1}{2\alpha }$. Hence we have $\dfrac{4}{ab}> \dfrac{1}{2\alpha}$.

If $g=0$, then $ab< 168\alpha $.

We have $\alpha \deg D = \FRC(D) -2$ and it is easy to show that if $\sum_{i=1}^r \dfrac{q_i-1}{q_i} -2>0$, then the minimal such value of the LHS is $\dfrac{1}{42}$.

By these Lemmas, we have proved that the number of types of $R$ is finite for a fixed $\alpha=a(R)$.

\[negative\] Although the following results are somewhat known" in the literature, I include the cases of $\alpha= a(R)\le 0$ for the completeness.

If $\alpha =0$, then $K_X=0$ and $\FRC(D)=0$. Hence $g(X)=1$ and $\deg D\le 3$. We have $(a,b,c;h)= (1,2,3;6), (1,2,2;4),(1,1,1;3)$ according to $\deg D=1,2,3$ respectively.

If $\alpha <0$, then since $\deg(K_X + \FRC(D)) <0$, we have $X\cong \PP^1,  
r\le 3$ and if $r=3$, $(q_1,q_2,q_3)$ is either $(2,2,n),(2,3,3), (2,3,4)$ or $(2,3,5)$. If we put $D= 
-(K_X + \FRC(D))$ in these cases, then we have $\alpha =-1$ and we get $(a,b,c;h)=$

$(2,n,n+1; 2n+2), (3,4,6;12), 
(4,6,9;18), (6,10,15)$ respectively, corresponding to $(D_{n+2}), (E_6), (E_7), (E_8)$ singularities. Note that the number of types is infinite in this case.

Since in the case $\alpha<0$ and $r=3$, there is no integer $n<-1$ with $nD \sim K_X + \FRC(D)$, if $\alpha <-1$, then $r\le 2$ and putting $P_1=(0)$ and $P_2=(\infty)$, then $R$ is generated by monomials and is isomorphic to a ring of the form $k[u,v,w]/(uv - w^n)$. Note that in this case, the type $(a,b,c;h)$ is not uniquely determined by the ring. On the contrary, if $\alpha \ge 0$ or $r\ge 3$, the type $(a,b,c;h)$ is uniquely determined by the ring $k[X,Y,Z]/(f)$ since the resolution of $\Spec(R)$ given in \[resol\] is a minimal good resolution[^3] and conversely a minimal good resolution determines $R$ as a graded ring. Related general statement can be found in [@S0].

In the following, we consider only the case $\alpha >0$.

Assume that $R=k[X,Y,Z,W]/(f)$ is graded with type $(a,b,c,d; h)$ and has isolated singularity. For a given $\alpha >0$, is the number of $(a,b,c,d; h)$ with $\alpha = h - (a+b+c+d)$ finite ?

The following Theorem is the main result of [@S2]. We give a proof here because it is much simpler by our method than the one given there.

\[alpha+1\] For any given $\alpha >0$, we have either $R_{\alpha -1}\ne 0$ or $R_{\alpha +1}\ne 0$.

Let $R=R(X,D)$ and $g$ be the genus of $X$ and put $D= E + \sum_{i=1}^r \dfrac{p_i}{q_i} P_i$ as in (1.0.1). By Lemma \[dual\] (2), we have always $$[(\alpha + 1) D] = K_X + E + \sum_{i=1}^r P_i.$$ Since $\deg D>0$, we have $\deg E \ge 1-r $ and $\deg [ (\alpha + 1) D] 
\ge 2g-1$. Hence if $g>0$, we have always $R_{\alpha +1} \ne 0$.

If $g=0$, we have $$\deg([ (\alpha - 1) D] ) = -2 - \deg E \quad {\rm and} \quad 
\deg([ (\alpha + 1) D] ) = - 2 + \deg E + r.$$ Hence if $\deg E\le -2$, we have $R_{\alpha -1} \ne 0$ and if $\deg E \ge -1$, then $R_{\alpha +1}\ne 0$ since $r\ge 3$.

Next we will show that if $p_g(R)=1$, then $\alpha\le 7$ and that $\alpha$ is bounded if $p_g(R)$ is bounded. These result also appears in Saito’s paper [@S1], but we will show it by our method. We can easily distinguish $R$ with $p_g(R)=1$ and $\alpha\le 6$ from our table.

\[pg\] (1) If $p_g(R)=r$ is fixed, then $\alpha$ is bounded.

\(2) If $p_g(R)=1$, then $\alpha \le 7$. If $p_g(R)=1$ and $\alpha =7$, then the type of $R$ is either $(8,10,15; 40), (8,10,25; 50)$ or $(8,9,12; 36)$.

\(1) If $p_g(R)=r$, then $a\le \alpha/r$ by formula (1.5.1). Hence if $\alpha$ tends to $\infty$, then $\deg D \le c \alpha^{-2}$ for some constant $c$ by Proposition 1.1. While we have shown in Lemma 2.2 to 2.4 that $\deg D \ge c' \alpha^{-1}$ for some constant $c'$.

\(2) Put $R=R(X,D)$ and $g$ be the genus of $X$. If $g>0$, then we have $\dim R_{\alpha} =g$ and hence we have always $p_g(R) \ge g+1 \ge 2$.

Now we assume $g=0$ and $p_g(R)=1$ and as always, we write $D = E + \sum_{i=1}^r \dfrac{p_i}{q_i} P_i$. By formula (1.5.1), we have $R_n=0$ for $1\le n \le \alpha$. Hence by Theorem \[alpha+1\], $a= \alpha+1$. Since $R_n\ne 0$ if and only if $\deg [nD] \ge 0$, and $$\leqno{(\ref{pg}.1)}\quad [nD] +[(\alpha -n)D ] = [\alpha D] = -2\quad \mbox{for $1\le n \le \alpha -1$},$$ we have $$\leqno{(\ref{pg}.2)}\quad \deg [nD] = -1 \quad \mbox{for $1\le n \le \alpha -1$.}$$ In particular, if $\alpha \ge 2$ and $p_g(R)=1$, then $\deg E = -1$.

Also, since $[(\alpha +1)D] = -2 + \deg E + r$ and $\dim R_{\alpha +1}\le 2$, we get $r\le 4$. If $r =4$, then $\alpha \le 2$ by the following Lemma \[r=3\]. So, we assume $r=3$ in the following and assume $q_1\le q_2\le q_3$.

Since $a=\alpha +1$ in this case, every element of degree $\le 2 \alpha + 1$ is a member of minimal generating set of $R$. Hence we must have $$\leqno{(\ref{pg}.3)}\quad \sum_{n=\alpha +1}^{2\alpha +1} \dim R_n \le 3.$$ On the other hand, since $\deg [nD] = -1$ for $1\le n \le \alpha -1$ and $(\alpha + n)D \sim K_X + \FRC(D) + nD$ for every $n$, $1\le n\le \alpha - 1$, we have $$\leqno{(\ref{pg}.4)}\quad R_{\alpha +n}\ne 0 \LLR \mbox{no $q_i$  divides } n.$$

Thus, we conclude that every $n, 1\le n \le \alpha-1$, except at most $3$ is a multiple of some $q_i$. This excludes the possibility $\alpha =6$ since, every $q_i$ should be relatively prime to $6$ and $n=1,2,3,4$ are not multiple of any $q_i$.

Next, we assume that $\alpha$ is even and $\ge 8$. Then since every $q_i$ should be odd, we should have $(a,b,c) = (\alpha +1, \alpha +2, \alpha +4)$ by (\[pg\].4). But this contradicts the condition of \[bd-of-c\] (2) and hence this case does not occur.

Now, we assume that $\alpha$ is odd and $\alpha\ge 7$. To proceed further, we note that if $q_1=2$ and $q_2=3$, then $R_6 \ne 0$. Hence we have either $R_{\alpha +2} \ne 0$ or $R_{\alpha +3}\ne 0$.

Thus, we have either $$\leqno{(\ref{pg}.5)}\quad (a,b,c;h) = (\alpha +1, \alpha +2, c; c+3\alpha+3)\quad {\rm or}$$ $$\leqno{(\ref{pg}.6)}\quad (a,b,c;h) = (\alpha +1, \alpha +3, c; c+3\alpha+4)$$

Now we check the condition of \[bd-of-c\] (1). In case (2.8.5), $c$ should divide either $3\alpha +3, 2\alpha+2$ or $2\alpha +1$. Then we should have $(a,b,c;h) = 
(\alpha +1, \alpha +2, 2\alpha +1; 5\alpha+4)$ or $(\alpha +1, \alpha +2, (3\alpha +3)/2; 9(\alpha+1)/2+1)$. in the former case, $\alpha +2$ must divide either $5\alpha+4, 3\alpha+3$ or $4\alpha+3$ and we can see such cases do not occur. in the latter case, $\alpha+2$ should divide $9(\alpha+1)/2)$, the only solution is $\alpha =7$, then $(a,b,c;h)= (8, 9,c; c+24)$ and only possibility is and

$$(a,b,c; h) = (8,9,12; 36). \quad D = \dfrac{2}{3} P_1 + \dfrac{1}{4} P_2 + \dfrac{1}{8}P_3 -Q .$$

In case (2.8.6), $c$ should divide either $3\alpha +4, 2\alpha+1$ or $2\alpha +3$. Then we should have $(a,b,c;h) = 
(\alpha +1, \alpha +3, 2\alpha +1; 5\alpha+5), 
(\alpha +1, \alpha +3, 2\alpha +3; 5\alpha+7)$ or $(\alpha +1, \alpha +3, 3\alpha +4; 6\alpha+8)$.

As in the previous case, since we have assumed $\alpha\ge 7$, the only solutions are

$$(8,10,15; 40)\quad D = \dfrac{1}{2}P_1 + \dfrac{2}{5} P_2 + \dfrac{2}{15}P_3 -Q,$$ $$(8,10,25; 50)\quad D = \dfrac{1}{2}P_1 + \dfrac{2}{5} P_2 + \dfrac{1}{8}P_3 -Q.$$

\[r=3\] Let $R=R(X,D)$ be as in Theorem \[pg\], with $g=0$ and $r=4$. Then if $p_g(R)\le 1$, then $\alpha \le 2$.

If $\alpha \ge 3$, then by formula (\[pg\].2), $\deg E=-1$ and $\deg [2D] =-1$. Hence the number of $i$ with $q_i = 2$ is at most $1$ and $\deg \alpha D \ge \dfrac{1}{2} + 3\cdot \dfrac{2}{3} = \dfrac{5}{2}$. On the other hand, since the type of $R$ is $(\alpha +1,\alpha +1, \alpha +2: 4\alpha +4)$, $\deg D= \dfrac{4}{(\alpha +1)(\alpha +2)}$, which is smaller than $\dfrac{5}{4\alpha}$. A contradiction !

The classification of the hypersurfaces with $a(R)\le 6$.
=========================================================

Now let us begin the classification of normal graded surface $R=k[x,y,z]/(f)$ with $a(R)=\alpha \le 6$. 0.8cm

**The case $\alpha=1$**

0.8cm

Henceforce, we put $D=K_X +\sum_{i=1}^r \dfrac{q_i-1}{q_i} P_i.$ From \[Fund\] (3), the maximal value of $\deg D$ is taken when $a=b=c=1$ and $\deg D=4$ in that case.

0.5cm [**Case 1 - A. The case $g>0$.**]{} 0.3cm Assume that $g\ge 1$. Since $\deg(D)\le 4$, and $\deg D\ge \deg K_X=2g-2$, $g\le 3$ and if $g=3$, $D=K_X$. We list the cases by giving the form of $D$ and $(a,b,c;h)$. We can easily deduce the general form of the equation $f$ from this data. Also, if $f$ with the given weight has an isolated singularity, then $k[u,v,w]/(f)\cong R(X,D)$, where $D$ is a divisor of given form. 0.2cm [**(1-A-1)**]{} $g=3,D=K_X;\quad (1,1,1;4).$ 0.2cm Next, consider the case $g=2$. Note that $\dim R_1=\dim H^0(K_X)=g=2$, we have $a=b=1$ and $\deg D= 1+\dfrac{3}{c}\le \dfrac{5}{2}\quad (c\ge2)$. Since, either $\deg D=2, D=K_X$ or $\deg D\ge \dfrac{5}{2}$, we have 2 cases. 0.2cm [**(1-A-2)**]{} $g=2,D=K_X;\quad (1,1,3;6).$

[**(1-A-3)**]{} $g=2,D=K_X+\dfrac{1}{2}P;\quad (1,1,2;5).$ 0.2cm Next, assume $g=1$. In this case, $a=1$, $2\le b\le c$ and the maximal value of $\deg D$ is $\dfrac{3}{2}$. Since on the other hand, $\deg D\ge \dfrac{r}{2}$ and thus $r\le 3$ and if $r=3$, $D=\dfrac{1}{2}(P_1+P_2+P_3)$. 0.2cm [**(1-A-4)**]{} $g=1,D=\dfrac{1}{2}(P_1+P_2+P_3);\quad (1,2,2;6).$ 0.2cm Also, since $\dim R_2=r$, if $r=2$, then $a=1,b=2,c\ge 3$, $\deg(D)\le 
\dfrac{7}{6}$. 0.2cm [**(1-A-5)**]{} $g=1,D=\dfrac{1}{2}(P_1+P_2);\quad (1,2,4;8).$

[**(1-A-6)**]{} $g=1,D=\dfrac{1}{2}P_1+\dfrac{2}{3} P_2;\quad (1,2,3;7).$ 0.2cm

If $g=1$ and $D=\dfrac{q-1}q P$, we have $q-1$ new generators in degrees $1,3,\ldots, q$. Hence $q\le 4$. 0.2cm [**(1-A-7)**]{} $g=1,D=\dfrac{1}{2}P;\quad (1,4,6;12).$

[**(1-A-8)**]{} $g=1,D=\dfrac{2}{3}P;\quad (1,3,5;10).$

[**(1-A-9)**]{} $g=1,D=\dfrac{3}{4}P;\quad (1,3,4;9).$ 0.2cm We have $9$ types when $g\ge 1$. 0.5cm [**Case 1- B. The case $g=0$ and $r\ge 4$.**]{} 0.3cm Since $\deg(K_X)=-2$ and $\deg D>0$, we have $r\ge 3$. On the other hand, since $R_1=H^0(K_X)=0$, $a\ge 2, c\ge  3$, we have $\deg D\le \dfrac{2}{3}<1$. Since $\deg D\ge -2+r/2$, we have $r\le 5$. 0.2cm Now, since $\deg [ 2D ] =r-4$, $\dim R_2=2,1$, respectively, if $r=5,4$.

Thus if $r=5$, then $a=b=2$ and $c\ge 3$. Since $3.\dfrac{1}{2}+2.\dfrac{2}{3}-2 = \dfrac{5}{6}> \dfrac{2}{3}$, the only possible cases for $(q_1,\ldots, q_5)$ are $(2,2,2,2,2)$ and $(2,2,2,2,3)$.

0.2cm

[**(1-B-1)**]{} $D=K_X+\dfrac{1}{2}(P_1+P_2+\ldots+P_5);\quad (2,2,5;10).$

[**(1-B-2)**]{} $D=K_X+\dfrac{1}{2}(P_1+P_2+P_3+P_4)+\dfrac{2}{3}P_5;
\quad (2,2,3;8).$ 0.2cm Henceforce we assume $r=4$ and express $D$ by $(q_1,q_2,q_3,q_4)$ and [**we always assume**]{} $q_1\le q_2\le q_3\le q_4$. In this case, $a=2$ and $3\le b\le c$. Hence $\deg D\le \dfrac{1}{2}$. Since $4.\dfrac{2}{3}-2>\dfrac{1}{2}$, $q_1=2$ and $q_4\ge 3$.

Let $s$ be the number of $q_i>2$ ($1\le s\le 3$). Then since $\deg [3D] = -6 + 8 -s$, $\dim R_3=0,1,2$ when $s=1,2,3$, respectively.

If $s=3$, $\dim R_2+\dim R_3=3$ and we must have $(a,b,c;h)=(2,3,3;9)$. 0.2cm [**(1-B-3)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{2}{3}(P_2+P_3+P_4);\quad (2,3,3;9).$ 0.2cm If $s=2$, $a=2,b=3$ and $c\ge 4$. Hence $\deg D = \dfrac{1}{6}+\dfrac{1}{c}
\le \dfrac{5}{12}$. Also, since $-2+(\dfrac{1}{2}+\dfrac{1}{2}
+\dfrac{2}{3}+\dfrac{3}{4})=\dfrac{5}{12}$, we have $2$ types.

0.2cm [**(1-B-4)**]{} $D=
K_X+\dfrac{1}{2}(P_1+P_2)+\dfrac{2}{3}(P_3+P_4);\quad (2,3,6;12).$

[**(1-B-5)**]{} $D=
K_X+\dfrac{1}{2}(P_1+P_2)+\dfrac{2}{3}P_3+\dfrac{3}{4}P_4;
\quad (2,3,4;10).$ 0.2cm Now we trat the case $(2,2,2,q)$, $q\ge 3$. In this case, $R_3=0$ and $\dim R_4=1$ or $2$ according to $q=3$ or $q\ge 4$. In the latter case, $\dim R_5=0$ or $1$ according to $q=4$ or $q\ge 5$. Hence, if $q\ge 5$, we have already $3$ generators of $R$. 0.2cm [**(1-B-6)**]{} $D=
K_X+\dfrac{1}{2}(P_1+P_2+P_3)+\dfrac{4}{5}P_4
;\quad (2,4,5;12).$

[**(1-B-7)**]{} $D=
K_X+\dfrac{1}{2}(P_1+P_2+P_3)+\dfrac{3}{4}P_4
;\quad (2,4,7;14).$

[**(1-B-8)**]{} $D=
K_X+\dfrac{1}{2}(P_1+P_2+P_3)+\dfrac{2}{3}P_4
;\quad (2,6,9;18).$ 0.2cm We have 8 types in this case. 0.5cm [**Case 1 - C. The case $g=0$ and $r=3$.**]{} 0.3cm We have to determine $(q_1,q_2,q_3)$. In this case, $ R_1= R_2=0$ and $\dim R_3=1$ or $0$ according to $q_1=2$ or $q_1\ge 3$.

0.2cm [**Case 1. $q_1\ge 3$.**]{}

In this case, $a=3$ and $4\le b\le c$. Hence $\deg D\le \dfrac{1}{4}$. Hence either $q_1=3$ or $q_1=q_2=q_3=4$.

0.2cm [**(1-C-1)**]{} $D=
K_X+\dfrac{3}{4}(P_1+P_2+P_3);\quad (3,4,4;12).$ 0.2cm [**Henceforce we assume $q_1=3$.** ]{}

$R_4\ne 0$ if and only if $q_2\ge 4$. In this case, $a=3,b=4$ and $c\ge 5$. Hence $\deg D\le \dfrac{13}{60}
=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}-2$. Hence we have only 2 possibilities; 0.2cm [**(1-C-2)**]{} $D=
K_X+\dfrac{2}{3}P_1+\dfrac{3}{4}P_2+\dfrac{4}{5}P_3;\quad (3,4,5;13).$

[**(1-C-3)**]{} $D=
K_X+\dfrac{2}{3}P_1+\dfrac{3}{4}(P_2+P_3); \quad (3,4,8;16).$ 0.2cm

Next, assume $q_1=q_2=3$. Hence $\deg D= \dfrac{q_3-1}{q_3}-\dfrac{2}{3}$. On the other hand, since $R_4=0$, $a=3$, $b\ge 5$ and $c\ge 6$ and $\deg D \le \dfrac{1}{6}$. This implies $q_3\le 6$.

0.2cm [**(1-C-4)**]{} $D=
K_X+\dfrac{2}{3}(P_1+P_2)+\dfrac{5}{6}P_3;\quad (3,5,6;15).$

[**(1-C-5)**]{} $D=
K_X+\dfrac{2}{3}(P_1+P_2)+\dfrac{4}{5}P_3;\quad (3,5,9;18).$

[**(1-C-6)**]{} $D=
K_X+\dfrac{2}{3}(P_1+P_2)+\dfrac{3}{4}P_3;\quad (3,8,12;24).$ 0.2cm

This completes the case $q_1=3$.

0.2cm

[**Case 2. $q_1=2$.**]{}

0.2cm In this case, $a\ge 4$ and $R_4\ne 0$ if and only if $q_2\ge 4$.

0.2cm [**First, we consider the case $q_1=2$ and $q_2=3$ ($q_3\ge 7$).** ]{}

In this case, $\deg [4D]=-1=\deg [5D]=\deg [7D], \deg [6D]=0$. Hence $a=6$ and $b\ge 8$. Hence $\deg D\le \dfrac{1}{18}=\dfrac{8}{9}-\dfrac{5}{6}$. This shows that $7\le q_3\le 9$ and actually these cases gives the hypersurfaces. 0.2cm [**(1-C-7)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{2}{3}P_2+\dfrac{6}{7}P_3;\quad (6,14,21;42).$

[**(1-C-8)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{2}{3}P_2+\dfrac{7}{8}P_3;\quad (6,8,15;30).$

[**(1-C-9)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{2}{3}P_2+\dfrac{8}{9}P_3;\quad (6,8,9;24).$

0.2cm [**Next, we consider the case $q_1=2$ and $q_2\ge 4$.** ]{}

In this case, $\deg [4D]=0$ and $a=4,b\ge 5, c\ge 6$. Hence $\deg D\le \dfrac{2}{15}=(\dfrac{1}{2}+\dfrac{4}{5}+\dfrac{5}{6})-2$. Hence $q_2\le 5$ and if $q_2=5$, the possibility is the following 2 cases.

0.2cm [**(1-C-10)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{4}{5}P_2+\dfrac{5}{6}P_3;\quad (4,5,6;16).$

[**(1-C-11)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{4}{5}(P_2+P_3);\quad (4,5,10;20).$

0.2cm [**The remaining case is $q_1=2,q_2=4\; (q_3\ge 5)$.** ]{}

Since $\dim R_4=1$ and $R_5=0$ and hence $a=4,b\ge 6,c\ge 7$ and $\deg D= \dfrac{q_3-1}{q_3}-\dfrac{3}{4}\le \dfrac{3}{28}$. Hence $5\le q_3\le 7$ and actually these cases give hypersurfaces.

This finishes the classification ! 0.2cm [**(1-C-12)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{3}{4}P_2+\dfrac{4}{5}P_3;\quad (4,10,15;30).$

[**(1-C-13)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{3}{4}P_2+\dfrac{5}{6}P_3;\quad (4,6,11;22).$

[**(1-C-14)**]{} $D=
K_X+\dfrac{1}{2}P_1+\dfrac{3}{4}P_2+\dfrac{6}{7}P_3;\quad (4,6,7;18).$

0.8cm

**The case $\alpha=2$**

0.8cm

We assume that $R=R(X,D)\cong k[u,v,w]/(f)$ with $$\deg(u,v,w;f)=(a,b,c;h); \quad h=a+b+c+2.$$

We always assume $(a,b,c)=1$ and $a\le b\le c$. Since $R$ is Gorenstein with $a(R)=2$, $2D$ is linearly equivalent to $K_X+\FRC(D)$. Hence we may assume that

$$D=E+\sum_{i=1}^r \dfrac{q_i-1}{2q_i}P_i,$$ where $2E \sim K_X$ and every $q_i$ is odd.

We divide the cases according to (A) $a=b=1$, (B) $a=1,b\ge 2$, and (C) $a>1$. 0.3cm [**Case 2 - A.**]{} When $a=b=1$, we have the following $4$ types. 0.2cm [**(2-A-1)**]{} $g=6, D=E$ with $2E\sim K_X;   
\quad (1,1,1;5).$ 0.2cm [**(2-A-2)**]{} $g=4, D=E$ with $2E\sim K_X;    
\quad (1,1,2;6).$ 0.2cm [**(2-A-3)**]{} $g=3, D=E+\frac{1}{3}P$ with $2E\sim K_X;   
\quad (1,1,3;7).$ 0.2cm [**(2-A-4)**]{} $g=3, D=E$ with $2E\sim K_X;   
\quad (1,1,4;8).$ 0.3cm [**Case 2 - B.**]{} $a=1$ and $b\ge 2$ 0.2cm If $b=2$, we have $g=\dim R_2=2$ and we have the following types. 0.2cm [**(2-B-1)**]{} $g=2, D=E+\dfrac{1}{3}P$ with $2E\sim K_X;  
\quad (1,2,3;8).$ 0.2cm [**(2-B-2)**]{} $g=2, D=E$ with $2E\sim K_X;    
\quad (1,2,5;10).$ 0.2cm If $b\ge 3$, then $g=1$. Since $2E\sim 0$ and $R_1\ne 0$, we have $E=0$ and $\deg [3D]=r$. Hence $r\le 3$ and $\deg D
\le \dfrac{9}{1\cdot 3\cdot 3}=1.$ If $r\ge 2$, then $a=1$ and $b=3$. We have the following cases. 0.2cm [**(2-B-3)**]{} $g=1, D=\dfrac{1}{3}(P_1+P_2+P_3);  
\quad (1,3,3;9).$ 0.2cm [**(2-B-4)**]{} $g=1, D=\dfrac{1}{3}P_1+\dfrac{2}{5}P_2;  
\quad (1,3,5;11).$ 0.2cm [**(2-B-5)**]{} $g=1, D=\dfrac{1}{3}(P_1+P_2);  
\quad (1,3,6;12).$ 0.2cm If $r=1$, then $b\ge 5$ and $\deg D \le \dfrac{15}{1\cdot 5\cdot 7}=\dfrac{3}{7}$. We have the following cases. 0.2cm [**(2-B-6)**]{} $g=1, D=\dfrac{3}{7}P;  
\quad (1,5,7;15).$ 0.2cm [**(2-B-7)**]{} $g=1, D=\dfrac{2}{5}P;  
\quad (1,5,8;16).$ 0.2cm [**(2-B-8)**]{} $g=1, D=\dfrac{1}{3}P;  
\quad (1,6,9;18).$ 0.2cm This finishes the case $a=1,b\ge 2$. 0.2cm [**Case 2 - C.**]{} $a\ge 2$. 0.2cm This is equivalent to say that $R_1=H^0(X,\OO_X(D))=0$. If this is the case, we have $$\deg D \le \dfrac{9}{2\cdot 2\cdot 3}=\dfrac{3}{4}<1.$$

Since $\deg D\ge g-1$, $g=0$ or $1$ in this case. 0.2cm First, assume $g=1$. Then $2E\sim 0$ and $\deg E=0$. Since $\deg [3D]
\ge 1$, $R_3\ne 0$. Hence $a=2$ and $b=3$ in this case. We have the following cases. 0.2cm [**(2-C-1)**]{} $g=1, D=E+\dfrac{1}{3}P$ with $E\ne 0, 2E\sim 0
  ;\quad (2,3,7;14).$ 0.2cm [**(2-C-2)**]{} $g=1, D=E+\dfrac{2}{5}P$ with $E\ne 0, 2E\sim 0  ;\quad (2,3,5;12).$ 0.2cm Next, assume $g=0$. Then since $2E\sim K_X$, $\deg E=-1$ and $\deg [3D]=-3+r\ge 0$, which implies $a=3$. Since $\dim R_3\le 2$, $r=3$ or $4$.

If $\dim R_3=2$ and $r=4$, we have only one case. 0.2cm [**(2-C-3)**]{} $g=0, D=E+\dfrac{1}{3}(P_1+\ldots +P_4)$ with $\deg E=-1 ;\quad (3,3,4;12).$ 0.2cm If $r=3$, then $a=3, b\ge 5$ and $\deg D\le 
\dfrac{15}{3\cdot 5\cdot 5}=\dfrac{1}{5}$. We have the following cases. 0.2cm [**(2-C-4)**]{} $g=0, D=E+\dfrac{2}{5}(P_1+P_2 +P_3)$ with $\deg E=-1 ;\quad (3,5,5;15).$ 0.2cm [**(2-C-5)**]{} $g=0, D=E+\dfrac{1}{3}P_1+\dfrac{2}{5}P_2 +
\dfrac{3}{7}P_3$ with $\deg E=-1 ;\quad (3,5,7;17).$ 0.2cm [**(2-C-6)**]{} $g=0, D=E+\dfrac{1}{3}P_1+\dfrac{2}{5}(P_2 +
P_3)$ with $\deg E=-1 ;\quad (3,5,10;20).$ 0.2cm [**(2-C-7)**]{} $g=0, D=E+\dfrac{1}{3}(P_1+P_2) +\dfrac{4}{9}P_3$ with $\deg E=-1 ;\quad (3,7,9;21).$ 0.2cm [**(2-C-8)**]{} $g=0, D=E+\dfrac{1}{3}(P_1+P_2) +\dfrac{3}{7}P_3$ with $\deg E=-1 ;\quad (3,7,12;24).$ 0.2cm [**(2-C-9)**]{} $g=0, D=E+\dfrac{1}{3}(P_1+P_2) +\dfrac{2}{5}P_3$ with $\deg E=-1 ;\quad (3,10,15;30).$ 0.2cm This finishes the case $a(R)=2$.

0.8cm

**The case $\alpha=3$**

0.8cm

We assume that $R=R(X,D)\cong k[u,v,w]/(f)$ with $$\deg(u,v,w;f)=(a,b,c;h); \quad h=a+b+c+3.$$

Since $R$ is Gorenstein with $a(R)=3$, $3D$ is linearly equivalent to $K_X+\FRC(D)$. By \[Fund\] (1), we may assume that

$$\leqno{(3.1.1)}\quad 
D=E+\sum_{i=1}^r{}_{q_i \equiv 1 (\mod 3)} \dfrac{q_i-1}{3q_i}P_i+
\sum_{i=1}^s {}_{q_i \equiv 2 (\mod 3)}\dfrac{2q_i-1}{3q_i}Q_i.$$

By \[Fund\] (1), we have $$\leqno{(3.1.2)} \quad 
3E + \sum_{i=1}^s Q_i \sim K_X; \quad \deg E=\dfrac{2g-2-s}{3}.$$

Since $\dfrac{q_i-1}{3q_i}\ge \dfrac{1}{4}$ if $q_i \equiv 1 (\mod 3)$, $q_i>1$ and $\dfrac{2q_i-1}{3q_i}\ge \dfrac{1}{2}$ if $q_i \equiv 2 (\mod 3)$, we have $$\leqno{(3.1.3)} \quad 
\deg D\ge \deg E+\dfrac{r}{4}+\dfrac{s}{2}=\dfrac{2g-2}{3}+
\dfrac{r}{4}+\dfrac{s}{6}.$$

0.3cm We divide the cases according to (A) $a=b=1$, (B) $a=1,b\ge 2$, (C) $a\ge 2$. 0.5cm [**Case 3 - A. $a=b=1$.**]{} 0.5cm In this case, the type of $R$ is the form $(1,1,c; c+5)$. By Lemma \[bd-of-c\], $c\le 5$ and $c\ne 3$. So we have for cases $c=1,2,4,5$. We can calculate the genus $g$ by $g=\dim R_3$. 0.2cm[**(3-A-1)**]{} $g=10$, $D=E$ with $\deg E=6$, $3E\sim K_X$ ;$(1,1,1;4).$ $X$ is not hyperelliptic. 0.2cm[**(3-A-2)**]{} $g=6,D=E+\dfrac{1}{2}Q$ with $\deg E =3, 3E+Q\sim K_X$; $(1,1,2;7)$.

Since $g=6$ and $\deg D=\dfrac{7}{2}=\dfrac{10}{3}+\dfrac{1}{6}$, this is the only possible equation for $D$. 0.2cm[**(3-A-3)**]{} $g=4,D=E+\dfrac{1}{4}P$ with $\deg E=2, 3E\sim K_X$;$(1,1,4;9).$ 0.2cm[**(3-A-4)**]{} $g=4,D=E$ with $\deg E=2$, $3E\sim K_X;\quad (1,1,5;10).$ 0.5cm [**Case 3 - B. $a=1,b\ge 2$.**]{} 0.5cm In this case, since $R_1=H^0(X,O_X(E))\ne 0$, we may assume $E\ge 0$. Since $g=\dim R_3$, we have $g\le 3$. Also, $g\ge 2$ if and only if $b\le 3$ and $g=1$ otherwise.

If $g =3$, then $c\le 3$ and we have the following cases. 0.2cm[**(3-B-1)**]{} $g=3,D=\dfrac{1}{2}\sum_{i=1}^4Q_i$ with $\sum_{i=1}^4Q_i\sim K_X$;$(1,2,2;8).$ 0.2cm[**(3-B-2)**]{} $g=3,D=E+\dfrac{1}{2}Q$ with $\deg E=1$, $3E+Q\sim K_X;\quad (1,2,3;9).$ 0.2cm If $g=2$, then $b\le 3$ and $c\ge 4$. Aso, by (3.1.2), $E=0, s=2$ and $Q_1+Q_2\sim K_X$. Hence $\dim R_2 = h^0(K_X)=2$. Thus we have $b=2$.

Since $E=0$, $\deg [4D] = r + 2s \ge 4$ and $c=4$ if and only if $r>0$. 0.2cm[**(3-B-3)**]{} $g=2,D=\dfrac{1}{4}P+\dfrac{1}{2}(Q_1+Q_2)$ with $Q_1+Q_2\sim K_X;
\quad (1,2,4;10).$

Now, since $E=0$ and $s=2$, we have $\deg D\ge 1$. On the other hand, since $a=1, b=2$, $deg D= \dfrac{c+6}{2c}$ and $\deg D\ge 1$ if and only if $c\le 6$. Hence we have the following cases. 0.2cm[**(3-B-4)**]{} $g=2,D=\dfrac{3}{5}Q_1+\dfrac{1}{2}Q_2;$ with $Q_1+Q_2
\sim K_X;\quad (1,2,5;11).$ 0.2cm[**(3-B-5)**]{} $g=2,D=\dfrac{1}{2}(Q_1+Q_2)$ with $Q_1+Q_2\sim K_X;\quad (1,2,6;12).$ 0.2cm In the remaining cases, $a=1$ and $b\ge 4$. Then we have $g=1$ and by (3.1.2) we have $E=0$ and $s=0$. Also, since $\deg [4D] =r$, $b=4$ if and only if $r\ge 2$. If this is the case, we have the following $3$ cases. 0.2cm[**(3-B-6)**]{} $g=1,D=\dfrac{1}{4}(P_1+P_2+P_3)
  ;\quad (1,4,4;12).$ 0.2cm[**(3-B-7)**]{} $g=1,D=\dfrac{1}{4}P_1+\dfrac{2}{7}P_2
  ;\quad (1,4,7;15).$ 0.2cm[**(3-B-8)**]{} $g=1,D=\dfrac{1}{4}(P_1+P_2)
  ;\quad (1,4,8;16).$

If $g=1, s=0$ and $r=1$, $b\ge 7$ and $c\ge 10$. Hence $\deg D
 \le \dfrac{21}{70}=\dfrac{3}{10}$. We have the following $3$ types. 0.2cm[**(3-B-9)**]{} $g=1,D=\dfrac{1}{4}P 
  ;\quad (1,8,12;24).$ 0.2cm[**(3-B-10)**]{} $g=1,D=\dfrac{2}{7}P 
  ;\quad (1,7,11;22).$ 0.2cm[**(3-B-11)**]{} $g=1,D=\dfrac{3}{10}P
  ;\quad (1,7,10;21).$ 0.5cm [**Case 3 - C. $a\ge 2$.**]{} 0.5cm In this case, $\deg D \le \dfrac{10}{2\cdot 2\cdot 3}<1$.

Since $g=\dim R_3$, $g\le 2$. But since the hypersurfaces of type $(3,3,c; c+9)$ cannot be normal, $g\le 1$.

If $g=1$, by (3.1.2), either $\deg E=0 =s$ or $\deg E= -1$ and $s=3$. In the first case, since $3E \sim 0, R_1=0, E\not\sim 0$ and $R_2 =0$. Hence $\deg D \le  
\dfrac{15}{3\cdot 4\cdot 5}=\dfrac{1}{4}$ and we have the following case. 0.2cm[**(3-C-1)**]{} $g=1, D=E+\dfrac{1}{4}P$, $3E\sim 0,E\not\sim 0 
  ;\quad (3,4,5;15).$ 0.2cm In the latter case, since $\deg [2D] \ge 1$, we have $a=2, b=3$. Since $\deg D \le \dfrac{12}{2\cdot 3\cdot 4}= \dfrac{1}{2}$, the following case is the only possiblity. 0.2cm[**(3-C-2)**]{} $g=1, D=E+\dfrac{1}{2}(P_1+P_2+P_3)$, $3E+P_1+P_2+P_3\sim 0 
 ;\quad (2,3,4;12).$ 0.2cm Now, until the end of the case $\alpha =2$, we assume $g=0$. By (3.1.2), we have $\deg E = - \dfrac{s+2}{3}$ and $\deg [2D] = s + 2\deg E = 
\dfrac{s-4}{3}$.

If $s=7$, then $a=b=2$ and $\deg D \ge \dfrac{1}{2}$, which implies $c\le 7$ and we have the following $2$ cases.

0.2cm[**(3-C-3)**]{} $g=0, D=E+\dfrac{1}{2}(P_1+\ldots +P_6)+\dfrac{3}{5}P_7, \deg E=-3
  ;\quad (2,2,5;12).$ 0.2cm[**(3-C-4)**]{} $g=0, D=E+\dfrac{1}{2}(P_1+\ldots +P_6+P_7), \deg E=-3
  ;\quad (2,2,7;14).$ 0.2cm If $s=4$ and $\deg E = -2$, we have $a=2$ and since $\deg [4D] = r \le 1$, $b=4$ if and only if $r=1$, otherwise, $b\ge 5$. If $r=1$, then $\deg D \ge \dfrac{1}{4}$ and $a=2,b=4$ and $c\le 9$ and we have the following $3$ cases. 0.2cm[**(3-C-5)**]{} $g=0, D=E+\dfrac{1}{4}P+\dfrac{3}{5}Q_1+\dfrac{1}{2}
(Q_2+Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,4,5;14).$ 0.2cm[**(3-C-6)**]{} $g=0, D=E+\dfrac{2}{7}P+\dfrac{1}{2}
(Q_1+\ldots+Q_4)$, $\deg E=-2
  ;\quad (2,4,7;16).$ 0.2cm[**(3-C-7)**]{} $g=0, D=E+\dfrac{1}{4}P+\dfrac{1}{2}
(Q_1+\ldots+Q_4)$, $\deg E=-2
  ;\quad (2,4,9;18).$ 0.2cm If $s=4$ and $r=0$, then $b\ge 5$ and we have $\dfrac{1}{10}\le \deg D \le 
\dfrac{15}{2\cdot 5\cdot 5}$, we have the following $6$ cases. 0.2cm[**(3-C-8)**]{} $g=0, D=E+\dfrac{1}{2}Q_1+\dfrac{3}{5}(Q_2+Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,5,5;15).$ 0.2cm[**(3-C-9)**]{} $g=0, D=E+\dfrac{5}{8}Q_1+\dfrac{3}{5}Q_2+\dfrac{1}{2}
(Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,5,8;18).$ 0.2cm[**(3-C-10)**]{} $g=0, D=E+\dfrac{3}{5}(Q_1+Q_2)+\dfrac{1}{2}
(Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,5,10;20).$ 0.2cm[**(3-C-11)**]{} $g=0, D=E+\dfrac{7}{11}Q_1+\dfrac{1}{2}
(Q_2+Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,8,11;24).$ 0.2cm[**(3-C-12)**]{} $g=0, D=E+\dfrac{5}{8}Q_1+\dfrac{1}{2}
(Q_2+Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,8,13;26).$ 0.2cm[**(3-C-13)**]{} $g=0, D=E+\dfrac{3}{5}Q_1+\dfrac{1}{2}
(Q_2+Q_3+Q_4)$, $\deg E=-2
  ;\quad (2,10,15;30).$ 0.2cm If $s=1$, since $\deg (K_X+\FRC(D))>0$, we must have $r\ge 2$. On the other hand, since $R_2=R_3=0$ and $\deg [4D] = r-2\le 1$, $r\le 3$. Hence $a\ge 4$ and $\deg D\le \dfrac{16}{4\cdot 4\cdot 5}=\dfrac{1}{5}$. If $r\ge 3$, then $\deg D\ge \dfrac{1}{4}$. Hence $r=2$, $a=4,b\ge 5$. We have the following $7$ cases. 0.2cm[**(3-C-14)**]{} $g=0, D=E+\dfrac{3}{5}Q+\dfrac{2}{7}P_1+\dfrac{1}{4}P_2$, $\deg E=-1
  ;\quad (4,5,7;19).$ 0.2cm[**(3-C-15)**]{} $g=0, D=E+\dfrac{5}{8}Q+\dfrac{1}{4}(P_1+P_2)$, $\deg E=-1
  ;\quad (4,5,8;20).$ 0.2cm[**(3-C-16)**]{} $g=0, D=E+\dfrac{3}{5}Q+\dfrac{1}{4}(P_1+P_2)$, $\deg E=-1
  ;\quad (4,5,12;24).$ 0.2cm[**(3-C-17)**]{} $g=0, D=E+\dfrac{1}{2}Q_1+\dfrac{2}{7}P_1+\dfrac{3}{10}P_2$, $\deg E=-1  ;\quad (4,7,10;24).$ 0.2cm[**(3-C-18)**]{} $g=0, D=E+\dfrac{1}{2}Q_1+\dfrac{2}{7}(P_1+P_2)$, $\deg E=-1  ;\quad (4,7,14;28).$ 0.2cm[**(3-C-19)**]{} $g=0, D=E+\dfrac{1}{2}Q_1+\dfrac{1}{4}P_1+\dfrac{3}{10}P_2$, $\deg E=-1  ;\quad (4,10.17;34).$ 0.2cm[**(3-C-20)**]{} $g=0, D=E+\dfrac{1}{2}Q_1+\dfrac{1}{4}P_1+\dfrac{2}{7}P_2$, $\deg E=-1  ;\quad (4,14,21;42).$ 0.2cm This finishes the case $a(R)=3$.

0.8cm

**The case $\alpha=4$**

0.8cm

We assume that $R=R(X,D)\cong k[u,v,w]/(f)$ with $$\deg(u,v,w;f)=(a,b,c;h); \quad h=a+b+c+4.$$

Since $4D \sim K_X+\FRC(D)$, we may assume that

$$\leqno{(4.1.1)}\quad 
D=E+\sum_{i=1}^r{}_{q_i \equiv 1 (\mod 4)} \dfrac{q_i-1}{4q_i}P_i+
\sum_{i=1}^s {}_{q_i \equiv 3 (\mod 4)}\dfrac{3q_i-1}{4q_i}Q_i,$$ where $E$ is an integral divisor on $X$.

Since $4D\sim K_X+\FRC(D)$, we have $$\leqno{(4.1.2)} \quad 
4E + 2 \sum_{i=1}^s Q_i \sim K_X; \quad \deg E=\dfrac{g-1-s}{2}.$$

Since $\dfrac{q_i-1}{4q_i}\ge \dfrac{1}{5}$ if $q_i \equiv 1 (\mod 4)$, $q_i>1$ and $\dfrac{3q_i-1}{4q_i}\ge \dfrac{2}{3}$ if $q_i \equiv 3 (\mod 4)$, we have $$\leqno{(4.1.3)}\quad 
\deg D\ge \deg E+\dfrac{r}{5}+\dfrac{2s}{3}=\dfrac{g-1}{2}+
\dfrac{r}{5}+\dfrac{s}{6}.$$

0.3cm We divide the cases according to (A) $a=b=1$, (B) $a=1,b\ge 2$, (C) $a\ge 2$.

0.5cm [**Case A. $a=b=1$.**]{} 0.5cm In this case, the type of $R$ is the form $(1,1,c; c+6)$. By \[bd-of-c\], $c\le 6$ and we have the following cases. 0.2cm[**(4-A-1)**]{} $g=15$, $D=E$ with $\deg E=6$, $3E\sim K_X$ ;$(1,1,1;4).$ $X$ is not hyperelliptic. 0.2cm[**(4-A-2)**]{} $g=9,D=E$ with $\deg E=4, 4E\sim K_X$; $(1,1,2;8)$.

0.2cm[**(4-A-3)**]{} $g=7,D=E$ with $\deg D=3, 3E\sim K_X ;\quad (1,1,3;9).$ 0.2cm[**(4-A-4)**]{} $g=5,D=E+\dfrac{1}{5}$ with $\deg E=2$, $4E\sim K_X;
\quad (1,1,4;10).$ 0.2cm[**(4-A-5)**]{} $g=5$, hyperelliptic, $D=E$ with $\deg D=2$, $4D\sim K_X;
\quad (1,1,6;12).$ 0.5cm [**Case B. $a=1,b\ge 2$.**]{} 0.5cm In this case, since $R_1=H^0(X,O_X(E))\ne 0$, we may assume $E\ge 0$. Since $g=\dim R_4$, $g\ge 2$ if and only if $b\le 4$ and $g=1$ otherwise. Checking the cases of type $(1,2,c; c+7)$ and $(1,3,c;c+8)$, we have the following types. 0.2cm[**(4-B-1)**]{} $g=4,D=E+\dfrac{2}{3}(Q$ with $4E+2Q\sim K_X$; $(1,2,3;10)$. 0.2cm[**(4-B-2)**]{} $g=3,D=E+\dfrac{1}{5}Q$ with $4E\sim K_X$; $(1,2,5;12)$. 0.2cm[**(4-B-3)**]{} $g=3$, hyperelliptic; $D=E$ with $4E\sim K_X$; $(1,2,7;14)$. 0.2cm[**(4-B-4)**]{} $g=3,D=E$ with $4E\sim K_X$; $(1,3,4;12)$. 0.2cm[**(4-B-5)**]{} $g=2,D=\dfrac{1}{5}P+\dfrac{2}{3}Q$ with $2Q\sim K_X$; $(1,3,5;13)$. 0.2cm[**(4-B-6)**]{} $g=2,D=\dfrac{5}{7}Q$ with $2Q\sim K_X$; $(1,3,7;15)$. 0.2cm[**(4-B-7)**]{} $g=2,D=\dfrac{2}{3}Q$ with $2Q\sim K_X$; $(1,3,8;16)$. 0.2cm These finishes the case $a=1$ and $2\le b\le 3$. We can also check that the type $(1,4,c;c+9)$ can not give a normal ring. Hence we may assume $b\ge 5$ and hence $g=\dim R_4=1$. Also, we have $\deg D \le \dfrac{15}{5\cdot 5}=\dfrac{3}{5}$. Hence $E=0$ and $s=0$. Also, $b=5$ if and only if $r\ge 2$. Since $\dim R_5\le 3$, we have also $r\le 3$ and if $r=3$, then $\deg D\ge 
\dfrac{3}{5}$. 0.2cm[**(4-B-8)**]{} $g=1,D=\dfrac{1}{5}(P_1+P_2+P_3); \quad (1,5,5;15)$. 0.2cm If $r=2$, $b=5$ and $c\ge 9$. Hence $\deg D \le \dfrac{16}{45}=\dfrac{1}{5}+
\dfrac{2}{9}$. 0.2cm[**(4-B-9)**]{} $g=1,D=\dfrac{1}{5}P_1+\dfrac{2}{9}P_2; \quad (1,5,9;19)$. 0.2cm[**(4-B-10)**]{} $g=1,D=\dfrac{1}{5}(P_1+P_2); \quad (1,5,10;20)$. 0.2cm If $r=1$, $b\ge 9$ and $c\ge 13$. Hence $\deg D \le \dfrac{27}{9\cdot 13}=
\dfrac{3}{13}$. Thus we have only the following possibilities. 0.2cm[**(4-B-11)**]{} $g=1,D=\dfrac{1}{5}P; \quad (1,10,15;30)$. 0.2cm[**(4-B-12)**]{} $g=1,D=\dfrac{2}{9}P; \quad (1,9,14;28)$. 0.2cm[**(4-B-13)**]{} $g=1,D=\dfrac{3}{13}P; \quad (1,9,13;27)$. 0.2cmThis finishes the case $R_1\ne 0$. 0.5cm [**Case C. $a\ge 2$.**]{} 0.5cm Since $\deg D\le \dfrac{11}{2\cdot 2\cdot 3}<1$, we have $g\le 2$ by (4.1.3). But if $\dim R_4\ge 2$, $2$ among $a,b,c$ should be even and $R$ will not be normal. Hence $g\le 1$.

If $g=1$, then $s$ is even and $\deg E=-\dfrac{s}{2}$ by (4.1.2). Also by (4.1.3), $s\le 4$ and we have always $\deg [2D] =0$. Hence $\deg D \le \dfrac{12}{2\cdot 3\cdot 3}=\dfrac{2}{3}$. If $s=4$, $\deg D \ge \dfrac{2}{3}$ and we have equality. 0.2cm[**(4-C-1)**]{} $g=1,D=E+\dfrac{2}{3}(P_1+P_2+P_3+P_4),
\deg E =-2, -2E\sim P_1+P_2+P_3+P_4; \quad 
(2,3,3;12)$. 0.2cm If $g=1, \deg E=-1$ and $s=2$, $\deg D\ge \dfrac{1}{3}$ by (5.3.1). On the other hand, if $R_2=0$, then $\deg D \le \dfrac{16}{3\cdot 4\cdot 5}$ and we have a contradiction. Thus we must have $\dim R_2=1$, that is, $2E +P_1+P_2\sim 0$. If $r\ge 1$, then $\deg D \ge \dfrac{2}{3}+\dfrac{1}{5}$ and on the other hand, $\deg D \le \dfrac{16}{3\cdot 4\cdot 5}$ and we have a contradiction. If $r=0$, the only possibility is the following; 0.2cm[**(4-C-2)**]{} $g=1, D=E+\dfrac{2}{3}P_1+\dfrac{5}{7}P_2,
\deg E =-1, -2E\sim P_1+P_2; \quad (2,3,7;16)$. 0.2cm If $g=1, s=0$ and $\deg E=0$, then $\deg D\le \dfrac{16}{2\cdot 5\cdot 5}
<\dfrac{1}{3}$ and we must have $r=1$. Since we must have $\dim R_4=1$, the following case is the only possible one. 0.2cm[**(4-C-3)**]{} $g=1, D=E+\dfrac{2}{9}P, E\ne 0, 2E\sim 0; 
\quad (2,5,9;20)$. 0.2cm Now, we assume $g=0$. Since $R_4=0$, $a\ge 3$ and in (4.1.1) we have $\deg  E= \dfrac{-s-1}{2}$ by (4.1.2). Since $\deg [3D]= \dfrac{s-3}{2}\le 1$, we get $s\le 5$. If $s=5$, we have the following. 0.2cm[**(4-C-4)**]{} $g=0, D=E+\dfrac{2}{3}(Q_1+\ldots Q_5), \deg E=-3, 
\quad (3,3,5;15)$. 0.2cm Next, we assume $s=3, \deg E=-2.$ In this case, $a=3, b\ge 5, \dim R_5= r$ and $\dim R_6=r+1$. Hence if $r>0, r=1, (a,b,c;h)=(3,5,6;18)$ and $\deg D = \dfrac{1}{5}$. 0.2cm[**(4-C-5)**]{} $g=0, D=E+\dfrac{2}{3}(Q_1+Q_2+ Q_3)+\dfrac{1}{5}P, 
\deg E=-2, \quad (3,5,6;18)$. 0.2cm If $r=0$, $\deg [7D] = -2 +t$, where $t$ is the number of $q_i's$ with $q_i\ge 7$. We have the following cases; 0.2cm[**(4-C-6)**]{} $g=0, D=E+\dfrac{5}{7}(Q_1+Q_2+ Q_3), 
\deg E=-2, \quad (3,7,7;21)$. 0.2cm[**(4-C-7)**]{} $g=0, D=E+\dfrac{2}{3}Q_1+\dfrac{5}{7}(Q_2+ Q_3), 
\deg E=-2, \quad (3,7,14;28)$. 0.2cm[**(4-C-8)**]{} $g=0, D=E+\dfrac{2}{3}Q_1+\dfrac{5}{7}Q_2+\dfrac{8}{11}Q_3,
\deg E=-2, \quad (3,7,11;25)$. 0.2cm[**(4-C-9)**]{} $g=0, D=E+\dfrac{2}{3}(Q_1+Q_2)+\dfrac{11}{15}Q_3, 
\deg E=-2, \quad (3,11,15;33)$. 0.2cm[**(4-C-10)**]{} $g=0, D=E+\dfrac{2}{3}(Q_1+Q_2)+\dfrac{8}{11}Q_3, 
\deg E=-2, \quad (3,11,18;36)$. 0.2cm[**(4-C-11)**]{} $g=0, D=E+\dfrac{2}{3}(Q_1+Q_2)+\dfrac{5}{7}Q_3, 
\deg E=-2, \quad (3,14,21;42)$. 0.2cm Then we treat the case $s=1,\deg E=-1$. Since $s+r\ge 3$, $r\ge 2$ and we have $a=5, b=6$ and the only possible case is; 0.2cm[**(4-C-12)**]{} $g=0, D=E+\dfrac{2}{3}Q+\dfrac{1}{5}(P_1+P_2), 
\deg E=-1, \quad (5,6,15;30)$. 0.2 cm This finishes the classification of the case with $a(R)=4$.

0.8cm

**The case $\alpha=5$**

0.8cm

We assume that $R=R(X,D)\cong k[u,v,w]/(f)$ with $$\deg(u,v,w;f)=(a,b,c;h); \quad h=a+b+c+5.$$

Since $5D$ is linearly equivalent to $K_X+\FRC(D)$, we may assume that

$$\leqno{(5.1.1)}\quad \begin{array}{rcl}
D& = & E+\sum_{i=1}^r{}_{q_i \equiv 1 (\mod 5)} \dfrac{q_i-1}{5q_i}P_i+
\sum_{i=1}^s {}_{q_i \equiv 3 (\mod 5)}\dfrac{2q_j-1}{5q_j}P'_j\\
& & +\sum_{k=1}^t {}_{q_k \equiv 2 (\mod 5)}\dfrac{3q_k-1}{5q_k}Q_k+
\sum_{l=1}^u {}_{q_l \equiv 4 (\mod 5)}\dfrac{4q_l-1}{5q_l}Q'_l,
\end{array}$$ where $E$ is an integral divisor on $X$.

Since $5D\sim $ $K_X+\FRC(D)$, we have $$\leqno{(5.1.2)} \quad \begin{array}{cl} &
5E +  \sum_{j=1}^s P'_j+2\sum_{k=1}^t Q_k+3\sum_{l=1}^t Q'_l
 \sim K_X;\\
& \deg E=\dfrac{2g-2-s-2t-3u}{5}\ge \dfrac{2g-2}{5}+\dfrac{r}{6}+
\dfrac{2s}{15}+\dfrac{t}{10}+\dfrac{3u}{20}.
\end{array}$$ 0.3cm We divide the cases according to (A) $a=b=1$, (B) $a=1,b\ge 2$, (C) $a\ge 2$.

0.5cm [**Case A. $a=b=1$.**]{} 0.5cm In this case, the type of $R$ is the form $(1,1,c; c+7)$. By \[bd-of-c\], $c\le 7$ and we have the following cases. 0.2cm[**(5-A-1)**]{} $g=21$, $D=E$ with $\deg E=8$, $5E\sim K_X$ ;$(1,1,1;8).$ 0.2cm[**(5-A-2)**]{} $g=12,D=E+\dfrac{1}{2}P$ with $\deg E=4, 5E+2P\sim K_X$; $(1,1,2;9).$ 0.2cm[**(5-A-3)**]{} $g=9,D=E+\dfrac{1}{3}P$ with $\deg E=3, 5E+P\sim K_X$; $(1,1,3;10).$ 0.2cm[**(5-A-4)**]{} $g=6,D=E+\dfrac{1}{6}P$ with $\deg E=2, 5E\sim K_X$; $(1,1,6;13).$ 0.2cm[**(5-A-5)**]{} $g=6,D=E$ with $\deg E=2, 5E\sim K_X$; $(1,1,7;14).$

0.5cm [**Case B. $a=1,b\ge 2$.**]{} 0.5cm First, we list the cases $a=1$ and $b=2,3$. 0.2cm[**(5-B-1)**]{} $g=6$, $D=\dfrac{1}{2}(P_1+\ldots +P_5)$ with $2(P_1+\ldots +P_5)\sim K_X$;$(1,2,2;10).$ 0.2cm[**(5-B-2)**]{} $g=5,D=E+\dfrac{1}{2}P$ with $\deg E=4, 5E+2P\sim K_X$; $(1,2,3;11).$ 0.2cm[**(5-B-3)**]{} $g=4$, $D=\dfrac{1}{2}(P_1+P_2 +P_3)$ with $2(P_1+P_2 +P_3)\sim K_X$ ;$(1,2,4;12).$ 0.2cm[**(5-B-4)**]{} $g=3,D=\dfrac{1}{2}P_1+\dfrac{1}{3}(P_2 +P_3)$ with $2P_1+P_2 +P_3\sim K_X$; $(1,2,6;14).$ 0.2cm[**(5-B-5)**]{} $g=3$, $D=\dfrac{1}{2}P_1+\dfrac{4}{7}P_2$ with $2(P_1+P_2)\sim K_X$ ;$(1,2,7;15).$ 0.2cm[**(5-B-6)**]{} $g=3,D=\dfrac{1}{2}(P_1+P_2)$ with $2(P_1+P_2 )\sim K_X$; $(1,2,8;16).$ 0.2cm[**(5-B-7)**]{} $g=3$, $D=\dfrac{1}{3}(P_1+\ldots +P_4)$ with $(P_1+\ldots +P_4)\sim K_X$;$(1,3,3;12).$ 0.2cm[**(5-B-8)**]{} $g=3,D=\dfrac{1}{3}P_1+\dfrac{3}{4}P_2$ with $P_1+3P_2\sim K_X$; $(1,3,4;13).$ 0.2cm[**(5-B-9)**]{} $g=2$, $D=\dfrac{1}{3}(P_1+P_2)+\dfrac{1}{6}P_3$ with $P_1+P_2\sim K_X$;$(1,3,6;15).$ 0.2cm[**(5-B-10)**]{} $g=2,D=\dfrac{1}{3}(P_1+P_2)$ with $P_1+P_2\sim K_X$; $(1,3,9;18).$ 0.2cm If $a=1$ and $b\ge 4$, then $\deg D<1$ and $g\le 2$. If $g=2=\dim R_5$, then $b\le 5$. 0.2cm[**(5-B-11)**]{} $g=2,D=\dfrac{1}{2}P_1+\dfrac{1}{6}P_2$ with $2P_1\sim K_X$; $(1,4,6;16).$ 0.2cm[**(5-B-12)**]{} $g=2,D=\dfrac{1}{2}P$ with $2P\sim K_X$; $(1,4,10;20).$ 0.2cm In the case $a=1$ and $b\ge 6$, then $D\ge 0, \deg D\le \dfrac{1}{2}$ and $g=1$. 0.2cm[**(5-B-13)**]{} $g=1,D=\dfrac{1}{6}(P_1+P_2+P_3)$; $(1,6,6;18).$ 0.2cm[**(5-B-14)**]{} $g=1,D=\dfrac{1}{6}(P_1+P_2)$;$(1,6,12;24).$ 0.2cm If $a=1$ and $b\ge 7$, then $\deg D< \dfrac{1}{3}$. Hence only possibility is of the form $D=\dfrac{q-1}{5q}$ with $q\equiv 1$ (mod $5$). 0.2cm[**(5-B-15)**]{} $g=1,D=\dfrac{2}{11}$; $(1,11,17;34).$ 0.2cm[**(5-B-16)**]{} $g=1,D=\dfrac{3}{16}P$; $(1,11,16;33).$ 0.2cm[**(5-B-17)**]{} $g=1,D=\dfrac{1}{6}P$; $(1,12,18;36).$ 0.2cm This finishes the case $a=1$. 0.5cm [**Case C. $a\ge 2$.**]{} 0.5cm In this case, $\deg D \le \dfrac{12}{2\cdot 2\cdot 3}=1$. Since $g=\dim R_5$, $g\le 2$ and $g=2$ only in the following cases. 0.2cm[**(5-C-1)**]{} $g=2,D=E+\dfrac{1}{2}(P_1+P_2)$, with $E\ne 0, 2E+P_1+P_2\sim K_X$; $(2,2,3;12).$ 0.2cm[**(5-C-2)**]{} $g=2,D=\dfrac{1}{6}P$; $(2,3,5;15).$ 0.2cm If $g=1$, $a=2$ then we have the following cases. 0.2cm[**(5-C-3)**]{} $g=1,D=E+\dfrac{1}{2}P_1+\dfrac{1}{2}P_2+
\dfrac{1}{3}P_3+E$, with $\deg E=-1, 2E+P_1+P_2\sim E+P_3\sim 0$; $(2,3,7;17).$ 0.2cm[**(5-C-4)**]{} $g=1,D=\dfrac{3}{8}P+E$ with $2E\sim 0, 5E+P\sim 0; \quad (2,3,8;18).$ 0.2cm[**(5-C-5)**]{} $g=1,D=\dfrac{1}{3}P+E$ with $2E\sim 0, 5E+P\sim 0; \quad (2,3,10;20).$ 0.2cm If $g=1$ and $a\ge 3$, we have $5E\sim 0$ since $\dim R_5=1$ and by (6.1.2), $s=t=u=0$. Since $E\ne 0$, $R_n=0$ for $1\le n\le 4$ and we have $a=5,b\ge 6$. Hence $\deg D\le \dfrac{22}{5\cdot 6\cdot 6}
<\dfrac{1}{6}$. On the other hand, since $r>0$, we should have $\deg E\ge \dfrac{1}{6}$. Hence there is no case with $g=1,a\ge 3$. 0.2cm Now we treat the case $g=0$. We divide our discussion into the following cases 0.2cm

\(i) $R_2\ne 0$, (ii) $R_2=0$ and $R_3\ne 0$, (iii) $R_2=R_3=0$.

0.2cm (i) If $\dim R_2=2$, we have the following cases. 0.2cm[**(5-C-6)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_8)+\dfrac{4}{7}P_9$ with $\deg E=-4; \quad (2,2,7;16).$ 0.2cm[**(5-C-7)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_9)$ with $\deg E=-4; \quad (2,2,9;18).$ 0.2cm If $a=2,b=4$, we have the following cases. 0.2cm[**(5-C-8)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_4)+\dfrac{4}{7}P_5+\dfrac{3}{4}P_6$ with $\deg E=-3; \quad (2,4,7;18).$ 0.2cm[**(5-C-9)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_5)+\dfrac{7}{9}P_6$ with $\deg E=-3; \quad (2,4,9;20).$ 0.2cm[**(5-C-10)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_5)+\dfrac{3}{4}P_6$ with $\deg E=-3; \quad (2,4,11;22).$ 0.2cm If $a=2$ and $b\ge 6$, from $\deg[2D]=\deg[4D]=0$ and $\deg[3D]=-2$ we have $\deg E=-2, t=4, s=u=0$. $b=6$ if and only if $r>0$. 0.2cm[**(5-C-11)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_3)+\dfrac{4}{7}P_4+\dfrac{1}{6}P_5$ with $\deg E=-2; \quad (2,6,7;20).$ 0.2cm[**(5-C-12)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_4)+\dfrac{2}{11}P_5$ with $\deg E=-2; \quad (2,6,11;24).$ 0.2cm[**(5-C-13)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+\ldots +P_4)+\dfrac{1}{6}P_5$ with $\deg E=-2; \quad (2,6,13;26).$ 0.2cm If $a=2$ and $b\ge 7$, then we have $\deg E=-2, t=4, r=s=u=0$. 0.2cm[**(5-C-14)**]{} $g=0,D=E+\dfrac{1}{2}P_1+\dfrac{4}{7}
(P_2+P_3 +P_4)$ with $\deg E=-2; \quad (2,7,7;21).$ 0.2cm[**(5-C-15)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+P_2)+\dfrac{4}{7}(P_3 +P_4)$ with $\deg E=-2; \quad (2,7,14;28).$ 0.2cm[**(5-C-16)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+P_2)+\dfrac{4}{7}(P_3 +P_4)$ with $\deg E=-2; \quad (2,7,14;28).$ 0.2cm[**(5-C-17)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+P_2+P_3)+\dfrac{10}{17}P_4$ with $\deg E=-2; \quad (2,12,17;36).$ 0.2cm[**(5-C-18)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+P_2+P_3)+\dfrac{7}{12}P_4$ with $\deg E=-2; \quad (2,12,19;38).$ 0.2cm[**(5-C-19)**]{} $g=0,D=E+\dfrac{1}{2}
(P_1+P_2+P_3)+\dfrac{4}{7}P_4$ with $\deg E=-2; \quad (2,14,21;42).$ 0.2cm This finishes the case $g=0$ and $a=2$. 0.2cm [**(ii) The case $g=0$ and $a=3$.**]{} 0.2cm If $a=3$ and $b=3$ or $4$, we have the following cases. 0.2cm[**(5-C-20)**]{} $g=0,D=E+\dfrac{1}{3}
(P_1+\ldots+P_5)+\dfrac{3}{4}P_6$ with $\deg E=-2; \quad (3,3,4;15).$ 0.2cm[**(5-C-21)**]{} $g=0,D=E+\dfrac{3}{4}
(P_1+\ldots+P_4)+\dfrac{1}{3}P_5$ with $\deg E=-3; \quad (3,4,4;16).$ 0.2cm[**(5-C-22)**]{} $g=0,D=E+\dfrac{1}{3}
P_1+\dfrac{3}{4}(P_2+P_3)+\dfrac{3}{8}P_4$ with $\deg E=-2; \quad (3,4,8;20).$ 0.2cm[**(5-C-23)**]{} $g=0,D=E+\dfrac{1}{3}
(P_1+P_2)+\dfrac{3}{4}P_3+\dfrac{7}{9}P_4$ with $\deg E=-2; \quad (3,4,9;21).$ 0.2cm[**(5-C-24)**]{} $g=0,D=E+\dfrac{1}{3}
(P_1+P_2)+\dfrac{3}{4}(P_3+P_4)$ with $\deg E=-2; \quad (3,4,12;24).$ 0.2cm If $a=3$ and $b\ge 6$, then since $\deg[D]=\deg[4D]=-1$ by Lemma 0.3 and $\deg[3D]=0$, we have $\deg E=-1, s=3, t=u=0$. Also, in this case $r>0$ if and only if $b=6$. In the latter case, since $\deg D\le \dfrac{21}{3\cdot 6\cdot 7}=\dfrac{1}{6}$, $D=E+\dfrac{1}{3}(P_1+P_2+P_3)+\dfrac{1}{6}P_4$ with $\deg E=-1$. 0.2cm[**(5-C-25)**]{} $g=0,
D=E+\dfrac{1}{3}(P_1+P_2+P_3)+\dfrac{1}{6}P_4$ with $\deg E=-1; \quad (3,6,7;21).$ 0.2cm If $r=0$, we have the following cases. 0.2cm[**(5-C-26)**]{} $g=0,
D=E+\dfrac{3}{8}(P_1+P_2+P_3)$ with $\deg E=-1; \quad (3,8,8;24).$ 0.2cm[**(5-C-27)**]{} $g=0,
D=E+\dfrac{1}{3}P_1+\dfrac{3}{8}P_2+\dfrac{5}{13}P_3$ with $\deg E=-1; \quad (3,8,13;29).$ 0.2cm[**(5-C-28)**]{} $g=0,
D=E+\dfrac{1}{3}P_1+\dfrac{3}{8}(P_2+P_3)$ with $\deg E=-1; \quad (3,8,16;32).$ 0.2cm[**(5-C-29)**]{} $g=0,
D=E+\dfrac{1}{3}(P_1+P_2)+\dfrac{7}{18}P_3$ with $\deg E=-1; \quad (3,13,18;39).$ 0.2cm[**(5-C-30)**]{} $g=0,
D=E+\dfrac{1}{3}(P_1+P_2)+\dfrac{5}{13}P_3$ with $\deg E=-1; \quad (3,13,21;42).$ 0.2cm[**(5-C-31)**]{} $g=0,
D=E+\dfrac{1}{3}(P_1+P_2)+\dfrac{3}{8}P_3$ with $\deg E=-1; \quad (3,16,24;48).$ 0.2cm This finishes the case $g=0$ and $a=3$. 0.2cm [**(iii) $g=0$, $R_2=R_3=0$.**]{} 0.2cm First we treat the case $R_4\ne 0$. It is easy to see that hypersurfaces of type $(4,4,c;c+13)$ can not be normal. So, if $a=4$, then $n\ge 6$ and we have $\deg E=-2, s+u=2, t+u=3$. If $a=4$ and $b=6$, we have the following types. 0.2cm[**(5-C-32)**]{} $g=0,
D=E+\dfrac{1}{2}(P_1+P_2)+\dfrac{1}{3}P_3+\dfrac{7}{9}P_4$ with $\deg E=-2; \quad (4,6,9;24).$ 0.2cm[**(5-C-33)**]{} $g=0,
D=E+\dfrac{1}{2}(P_1+P_2)+\dfrac{1}{3}P_3+\dfrac{3}{4}P_4$ with $\deg E=-2; \quad (4,6,15;30).$ 0.2cm If $a=4$ and $b>6$, then $\deg [6D]<0$. Since $5D \sim K_X+\FRC(D)$, this is only possible if the support of $\FRC(D)$ consists of $3$ points. This implies $r=s=0, t=1,u=2$. 0.2cm [**(5-C-34)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{7}{9}P_2+\dfrac{11}{14}P_3$ with $\deg E=-2; \quad (4,9,14;32).$ 0.2cm[**(5-C-35)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{7}{9}(P_2+P_3)$ with $\deg E=-2; \quad (4,9,18;36).$ 0.2cm[**(5-C-36)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{3}{4}P_2+\dfrac{7}{9}P_3$ with $\deg E=-2;
 \quad (4,18,27;54).$ 0.2cm[**(5-C-37)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{3}{4}P_2+\dfrac{11}{14}P_3$ with $\deg E=-2; \quad (4,14,23;46).$ 0.2cm[**(5-C-38)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{3}{4}P_2+\dfrac{15}{19}P_3$ with $\deg E=-2; \quad (4,14,19;42).$ 0.2cm This finishes the case $a=4$. If $a>4$, then $a\ge 6$ and since $\deg[nD]=-1$ by Lemma 1.4, we have $\deg E=-1$. Since $6D =(K_X+\FRC(D)) +D$, $\deg [6D] = -3+r+s+t+u\ge 0$. Hence $a=6$ and it is easy to see type $(6,6,c; c+17)$ does not occur. Hence we have either $r=2,s=t=0,u=1$ or $r=s=t=1, u=0$. In the former case, $\deg D\ge 2\dfrac{1}{6}+\dfrac{3}{4}
-1=\dfrac{1}{12}$. On the other hand, since $b\ge 7$ and $c\ge 8$, $\deg D\le \dfrac{26}{6\cdot 7\cdot 8}<\dfrac{1}{12}$. So, this case does not occur and we have $r=s=t=1,u=0$. We have the following cases. 0.2cm[**(5-C-39)**]{} $g=0,
D=E+\dfrac{1}{3}P_1+\dfrac{1}{6}P_2+\dfrac{4}{7}P_3$ with $\deg E=-1; \quad (6,7,9;27).$ 0.2cm[**(5-C-40)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{3}{8}P_2+\dfrac{2}{11}P_3$ with $\deg E=-1; \quad (6,8,11;30).$ 0.2cm[**(5-C-41)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{1}{6}P_2+\dfrac{5}{13}P_3$ with $\deg E=-1; \quad (6,8,13;32).$ 0.2cm[**(5-C-42)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{1}{3}P_2+\dfrac{2}{11}P_3$ with $\deg E=-1; \quad (6,22,33;66).$ 0.2cm[**(5-C-43)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{1}{3}P_2+\dfrac{3}{16}P_3$ with $\deg E=-1; \quad (6,16,27;54).$ 0.2cm[**(5-C-44)**]{} $g=0,
D=E+\dfrac{1}{2}P_1+\dfrac{1}{3}P_2+\dfrac{4}{21}P_3$ with $\deg E=-1; \quad (6,16,21;48).$ 0.2cm This finishes the ccase $a(R)=5$.

0.8cm

**The case $\alpha=6$**

0.8cm

Since $6D$ is linearly equivalent to $K_X+\FRC(D)$, we may assume that

$$\leqno{(6.1.1)}\quad 
D=E+\sum_{i=1}^r{}_{q_i \equiv 1 (\mod 6)} \dfrac{q_i-1}{6q_i}P_i+
\sum_{i=1}^s {}_{q_i \equiv 5 (\mod 6)}\dfrac{5q_i-1}{6q_i}Q_i,$$ where $E$ is an integral divisor on $X$.

Since $6D\sim K_X+\FRC(D)$, we have $$\leqno{(6.1.2)} \quad 
6E + 4\sum_{i=1}^s Q_i \sim K_X; \quad \deg E=\dfrac{g-1-2s}{3}.$$

Since $\dfrac{q_i-1}{6q_i}\ge \dfrac{1}{7}$ if $q_i \equiv 1 (\mod 6), 
q_i>1$ and $\dfrac{5q_i-1}{6q_i}\ge \dfrac{4}{5}$ if $q_i \equiv 5
 (\mod 6)$,we have $$\leqno{(6.1.3)}\quad 
\deg D\ge \deg E+\dfrac{r}{7}+\dfrac{4s}{5}=\dfrac{g-1}{3}+
\dfrac{r}{7}+\dfrac{2s}{15}.$$

0.3cm We divide the cases according to (A) $a=b=1$, (B) $a=1,b\ge 2$, (C) $a\ge 2$.

0.5cm [**Case A. $a=b=1$.**]{} 0.5cm In this case, we have the following cases. We can calculate the genus $g$ by $g=\dim R_6$. 0.2cm[**(6-A-1)**]{} $g=28$, $D=E$ with $\deg E=9$, $6E\sim K_X$ ;$(1,1,1;9).$ $X$ is not hyperelliptic. 0.2cm[**(6-A-2)**]{} $g=16$, $D=E$ with $\deg E=5$, $6E\sim K_X$; $(1,1,2;10).$ $X$ is not hyperelliptic. 0.2cm[**(6-A-3)**]{} $g=10$, $D=E$ with $\deg E=3$, $6E\sim K_X$; $(1,1,4;12).$ 0.2cm[**(6-A-4)**]{} $g=7$, $D=E+\dfrac{1}{7}$ with $\deg E=2$, $6E\sim K_X$;$(1,1,7;15).$ 0.2cm[**(6-A-5)**]{} $g=7$, $D=E$ with $\deg E=2$, $6E\sim K_X$ ;$(1,1,8;16).$ 0.5cm [**Case B. $a=1, b\ge 2$.**]{} 0.5cm First we list the cases with $a=1,b=2,3,4$. 0.2cm[**(6-B-1)**]{} $g=7$, $D=E$ with $\deg E=2$, $6E\sim K_X$;$(1,2,3;12).$ 0.2cm[**(6-B-2)**]{} $g=4$, $D=E+\dfrac{1}{7}$ with $\deg E=1$, $6E\sim K_X$;$(1,2,7;16).$ 0.2cm[**(6-B-3)**]{} $g=4$, $X$ is hyperelliptic, $D=E$ with $\deg E=1$, $6E\sim K_X$;$(1,2,9;18).$ 0.2cm[**(6-B-4)**]{} $g=4$, $D=E$ with $\deg E=1$, $6E\sim K_X$;$(1,3,5;15).$ 0.2cm[**(6-B-5)**]{} $g=3$, $D=\dfrac{4}{5}P$ with $4P\sim K_X$;$(1,4,5;16).$ 0.2cm If $b\ge 5$, then $\deg D\le \dfrac{3}{5}$. Hence $E=0$ and $s=0$. This implies $\deg [nD] =0$ for $n\le 6$. Thus we have $g=1$ and $b\ge 7$, $\deg D\le \dfrac{3}{7}$. 0.2cm [**(6-B-6)**]{} $g=1$, $D=\dfrac{1}{7}(P_1+P_2+P_3); \quad (1,7,7;21).$ 0.2cm [**(6-B-7)**]{} $g=1$, $D=\dfrac{1}{7}(P_1+P_2+P_3); \quad (1,7,7;21).$ 0.2cm [**(6-B-6)**]{} $g=1$, $D=\dfrac{1}{7}(P_1+P_2); \quad (1,7,14;28).$ 0.2cm [**(6-B-7)**]{} $g=1$, $D=\dfrac{1}{7}P_1+\dfrac{2}{13}P_2; \quad (1,7,13;27).$ 0.2cm [**(6-B-8)**]{} $g=1$, $D=\dfrac{3}{19}P; \quad (1,13,19;39).$ 0.2cm [**(6-B-9)**]{} $g=1$, $D=\dfrac{2}{13}P; \quad (1,13,20;40).$ 0.2cm [**(6-B-10)**]{} $g=1$, $D=\dfrac{1}{7}P; \quad (1,14,21;42).$ 0.2cm This finishes the case $a=1,b\ge 2$. 0.5cm [**Case C. $a\ge 2$.**]{} 0.5cm We can easily see that $(a,b)=(2,2),(2,3),(2,4),(3,3)$ does not occur. By a similar computation, we can assert $\deg D<\dfrac{1}{3}$ and we know that $g\le 1$. 0.2cm If $g=1$, then $6E+4\sum_{i-1}^s Q_j\sim 0$. If $\deg E=-2$, then $s=3$ and $\deg D\ge \dfrac{2}{5}$, which contradicts our previous computation. Hence, if $g=1$, then $\deg E=0$ and $s=0$. This implies that $a=2,3$ or $6$ and $\deg[nD]=0$ for $1\le n\le 6$. We have the following cases. 0.2cm [**(6-C-1)**]{} $g=1$, $D=E+\dfrac{1}{7}P$ with $E\ne 0$, $2E\sim 0; \quad (2,7,15;30).$ 0.2cm [**(6-C-2)**]{} $g=1$, $D=E+\dfrac{2}{13}P$ with $E\ne 0$, $2E\sim 0; \quad (2,7,13;28).$ 0.2cm [**(6-C-3)**]{} $g=1$, $D=E+\dfrac{1}{7}P$ with $E\ne 0$, $3E\sim 0; \quad (3,7,8;24).$ 0.2cm This finishes the case $g=1$. If $g=0$, we have $$6E + 4\sum_{i-1}^s Q_j\sim K_X, \quad \deg E=\dfrac{-1-2s}{3}.$$

If $deg E=-1, s=1$, then $R_n=0$ for $n\le 6$ and this implies $\deg D\le \dfrac{28}{7\cdot 7\cdot 8}=\dfrac{1}{14}$. On the other hand, since $r+s\ge 3$, $\deg D\ge -1+\dfrac{4}{5}
+\dfrac{2}{7}=\dfrac{6}{35}$, a contradiction ! 0.2cm Hence we have $s=4$ and $\deg E=-3$. Then $\deg [4D]=0$ and $\deg [5D] =1$. Hence we are restricted to the following type. 0.2cm [**(6-C-4)**]{} $g=0$, $D=E+\dfrac{4}{5}(P_1+\ldots +P_4)$ with $\deg E=-3;  \quad (4,5,5;20).$ 0.2cm This finishes the case $a(R)=6$.

0.8cm Now, we list the table of numbers of cases with given $a(R)\le 6$. In the table, br denotes the number of brances of $D$, which is equal to the number of branches of the graph of the minimal resolution of $\Spec(R)$. In other word, br $=\deg \lceil\FRC(D)\rceil$. 0.8cm

       $a(R)$        1    2    3    4    5    6
  ----------------- ---- ---- ---- ---- ---- ----
                                             
    $g=0$, br$=3$    14   6    7    7    11   0
   $g=0$, br$\ge4$   8    1    10   2    22   1
        $g=1$        6    8    8    7    8    8
        $g=2$        2    2    3    3    6    0
        $g=3$        1    2    2    3    5    1
      $g\ge 4$       0    2    4    6    8    9
        Total        31   21   34   28   58   19

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M. Tomari, Multiplicity of filtered rings and simple K3 singularities of multiplicity two, Publ. Res. Inst. Math. Sci. **38** (2002), 693–724.

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[^1]: See Remark \[negative\] for the case $\alpha <0$

[^2]: See Remark \[negative\] for the case $\alpha \le 0$.

[^3]: A resolution whose exceptional set consists of smooth curves with normal crossings and which is minimal resolution satisfying this condition; namely, which contains no $(-1)$ curve intersecting to at most $2$ other irreducible curves.
