数理解析研究所講究録 1437 巻 2005 年 136-144

136
On Weierstrass 7-semigroups
神奈川工科大学

米田 二良 (Jiryo Komeda)

Kanagawa Institute of Technology

51. Introduction.
Let be the additive semigroup of non-negative integers. A subsemigroup $H$ ofis called a numerical semigroup if the complementof $H$ in is a finite
set. For any positive integer a numerical semigroup $H$ is called an n-semigroup
if $H$ starts with , i.e., the minimum positive integer in $H$ is . For a non-singularcomplete irreducible curve over an algebraically closed field of characteristic
0 (which is called a curve in this paper) and its point we set$\mathrm{N}$$\mathrm{N}$$\mathrm{N}$$\mathrm{N}\backslash H$

$n$

$n$

$n$

$C$

$k$

$P$

$H(P)=$ {$n\in \mathrm{N}|\exists$

a rational function

$f$

on

$C$

with $(f)_{\infty}=nP$ }.

A numerical semigroup is Weierstrass if there exists a curve with its pointsuch that $H(P)=H$ . We are interested in the following problem:
$C$

$P$

Problem 1. Is every -semigroup Weierstrass ?
$n$

We have the following positive results:
Fact 2. For nevery -semigroup is Weierstrass. (For n $=2$ , classical, forn $=3$ , see [8] and for n $=4,$ 5, see [4] ,[5] respectively.)$\leqq 5$$n$

But we know the negative result as follows:
Fact 3. For any n, there exists a non- Weierstrass -sernigroup. (For n $=13$ ,[1] and for nsee, for example, [7].)$\geqq 13$$n$$\geqq 14$

Thus, we have the following problem:

Problem 4. For $6\leqq n\leqq 12$ , is every -semigroup Weierstrass or is there a
non-Weierstrass semigroup ?
$n$

$n$

In this paper we are devoted to the study of 7-semigroups. In Section 2 we determine the 7-semigroups which are the semigroups $H(P)$ of ramification points on
cyclic coverings of the projective linewith degree 7. In Section 3 we divide theWeierstrass 7-semigroups generated by 4 elements into 31 cases and investigate
whether such a 7-semigroup is of toric type in each case where a numerical semigroup is said to be of toric type if roughly speaking, the monomial curve associatedto the numerical semigroup is a specialization of some affine toric variety, because
we know that a numerical semigroup of toric type is Weierstrass ([4]).
$P$$\mathrm{P}^{1}$

137\S 2. Cyclic 7-semigroups.
An -semigroup is said to be cyclic if it is the semigroup $H(P)$ for some totallywith degree .
ramification point on a cyclic covering of the projective line
In this section we describe a necessary and sufficient condition on a 7-semigroup
to be cycle. Moreover, some non-cycle Weierstrass 7-semigroups are given. We
use the following notation: For an -semigroup $H$ we set
$n$

$P$

$\mathrm{P}^{1}$

$n$

$n$

$S(H)=\{n, s_{1,7}\ldots s_{n-1}\}$mod }. We have the following necessary condition
where $s_{i}={\rm Min}$ { $h\in H|th$on an -semigroup to be cycle if is prime.
$\equiv \mathrm{i}$

$n$

$n$

$n$

Fact 5 ([9]). Let$p$be a prime number. if $H$ is a cyclic -semigroup with$p$

$S(H)=\{p, s_{1}, \ldots, s_{p-1}\}$,then$s_{i}+s_{\mathrm{p}-i}=s_{j}+s_{p-j}$, all$\mathrm{i},j$.

We had already obtained an answer to the converse problem of the above stat.ement.
Fact 6. i) For a prime number $7\leqq 7$ , the converse of Fact 5 is true (See [9]).
For any prime number $p\geqq 11$ , the converse of Fact 5 is false (See [3]).
$\mathrm{i}\mathrm{i})$By Fact 6 i) we get the following:

Proposition 7. Let H be a 7-semigroup with
$S(H)=\{7, s_{1}, \ldots, s_{6}\}$.

Assume that$s_{1}+s_{6}=s_{2}+s_{5}=s_{3}+s_{4}$.

Then $H$ is cyclic, hence Weierstrass.
, $b_{m}>$ denotes the semigroup gener,,
For any positive integers ,ated by , . . . ’ . We give examples of cyclic 7-semigroups.
$b_{0}$$b_{0}$$\ldots$$b_{m},$$<b_{0}$$\ldots$$b_{m}$

Example 8. (1) Let $H=<7,8$ , 10, $12>$ . Then $S(H)=\{7, 8, 10, 12, 16, 18, 20\}$ .
Since $8+20$ $=16+12=10+18$ , $H$ is cyclic, hence Weierstrass.
(2) Let $H=<7,15,16,17$ , 25, 26, $27>$ . Then $S(H)=\{7,15,16,17,25,26, 27\}$ .
Since $15+27=16+26=17+25,$ $H$ is cyclic, hence Weierstrass.

We also have non-cyclic Weierstrass 7-semigroups.

138

Fact 9. For integers

$g$

and

$s$

with $7\leqq g\leqq s\leqq 12$ , let$H_{s,g}$

be a 7-semigroup with$\mathrm{N}_{0}\backslash H_{s,g}=\{1, \ldots, 6,8 +s-g, \ldots, s+1\}$.

Then we have the following:of degree 3 with non-ramification point $P\in C$i) There exists a covering
is a Weierstrass 7-semigroup (See [2]).
such that $H(P)=H_{s},il$ . Hence,is non-cyclic. For example, $H_{11,9}=$
If $(s, g)\neq(9_{7}9),$ $(12, 9)$ , $(12, 12)$ , then$<7,8,9,13$ , $19>$ and $H_{12,10}=<7,8$ , 9, 19, $20>$ are non-cyclic Weierstrass 7semigroups.
$Carrow \mathrm{P}^{1}$$H_{s,g}$$H_{s,g}$$\mathrm{i}\mathrm{i})$Fact 10. Let $H$ be the 7-semigroup $<7,9,10,11,12$ , $13>$ . Then there is a cyclic
andcovering of an elliptic curve of degree 8 with only two ramification points, which are totally ramified, such that $H(P_{1})=H(P_{2})=H$ . Hence$<7,9,10,11,12$ , $13>$ is a non-cyclic Weierstrass 7-semigroup (See [6]).
$P_{1}$$P_{2}$

Q3. 7-semigroups of toric type.
For a numerical semigroup $H$ we denote by $M(H)$ the minimal set of generators for$H$ . In this section we are interested in 7-semigroups $H$ with $M(H)=$, a2, }{
which satisfy the following condition:$7,$ $a_{1}$$a_{3}$

Definition 11. Let $H$ be a numerical semigroup with $\# M(H)$ $=m+1$ . The
semigroup $H$ is said to be of toric type if: a positive integer,: a saturated subsemigroup of generated by , . . . ,which generates
group
a
as
and
$(j=1, \ldots, l +m)$ : monomials insuch that$\exists l$

$]$

$S$

$\mathbb{Z}^{l}$$b_{1}$$b_{l+m}$$\mathbb{Z}^{l}$$k[X_{0}, X_{1_{7}}\ldots, X_{m}]$$\exists g_{j}’ \mathrm{s}$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k_{\lfloor}^{\acute{1}}H]$$\llcornerarrow$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[S][X_{0}, X_{1}, \ldots, X_{m}]$$\square$$\downarrow$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k$$\llcorner+$(0)–$\downarrow$$8\mathrm{p}\mathrm{e}\mathrm{c}k[Y_{1}, \ldots, Y_{l+m}]$the origin

where the right vertical map is induced by the -algebra homomorphism$k$$\eta_{S}$which sends

3 to: $k[Y_{1}, \ldots, Y_{l+m}]arrow k[S][X_{0}, X_{1}, \ldots, X_{m}]$$T^{b_{j}}-g_{j}$, that is to say,$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[H]$$\mathrm{c}arrow$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[X_{0}, X_{1}, \ldots, X_{m}]$$\square$$\downarrow$$\downarrow$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[S]$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[Y_{1}, \ldots, Y_{\mathrm{J}+m}]$$\mathrm{L}arrow$

where the horizontal maps are the embeddings through the generators and theright vertical map is induced by the -algebra morphism from $k[Y_{1}, \ldots, Y_{l+m}]$ to$k[X_{0}, X_{1}, \ldots, X_{m}]$ sending
to .
$k$

$Y_{j}$$g_{j}$

139
We explain how to find a subsemigroup

of

$S$

$\mathbb{Z}^{l}$as in Definition 11 below.

Remark 12. Let $H$ be a numerical semigroup with $M(H)=\{a_{0},, a_{1}, \ldots, a_{m}\}$ .
, , i.e., a set ofi) Determine a generating system of relations am ong , ,.
generators for the ideal of the monomial curve, , i.e., a basis
, ,
among
of
relationssystemDetermine a fundamentalof the relation -module among , , . . . ’ .
from the fundamental system In thisWe construct a subsemigroup of
as a group naturally.
and generates
case, is generated by $l+m$ elements
in $k[X_{0}, \ldots, X_{m}]$monomials
toforMoreover, we associate the generatorssuch that we have the fiber products in Definition 11.
The remaining problem is whether the semigroup is saturated or not. We
note that is saturated if and only if the semigroup ring $k[S]$ is normal, i.e.,
is an affine toric variety. If is saturated, the numerical semigroup $H$
become of toric type.
$a_{0}$$a_{1}$$a_{m}$$\ldots$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[H]$$a_{0}$$\mathrm{i}\mathrm{i})$$\mathbb{Z}$$a_{0}$$\ldots$$a_{m}$$a_{m}$$a_{1}$$S$$\mathrm{i}\mathrm{i}\mathrm{i})$$a_{1}$$\mathbb{Z}^{l}$$\mathbb{Z}^{l}$$S$$b_{j}’ \mathrm{s}$$S$$b_{j}’ \mathrm{s}$$g_{\mathrm{j}}’ \mathrm{s}$$S$$\mathrm{i}\mathrm{v})$

$S$

$S$

$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[S]$From now on we treat only 7-semigroups generated by 4 elements.
, $M(H)=$
Lemma 13. Let $H$ be 1-semigroup generated by 4 elements,
and. Renumbering ,satisfies one of the following:(I) $a_{1}+a_{2}+a_{3}\equiv 0(7)$ ,(I) $a_{1}+a_{2}$ $\equiv 0(7)$ ,(III) $2a_{1}+a_{2}\equiv 0(7)$ and $2a_{2}+a_{3}\equiv 0(7)$ .
$\mathrm{i}.e.$$a$$\{7, a_{1}, a_{2}, \text{\^{a}} \}$$\mathrm{a}\mathrm{i}$$a_{2}$$a_{3}$$ii$

We give the construction of a saturated subsemigroupin (I) and some cases of (II).

$S$

of

$\mathbb{Z}^{l}$as in Definition 11

Case (I) $a_{1}+a_{2}+a_{3}\equiv 0(7)$ . A fundamental system of relations consists of$\frac{a_{1}+a_{2}+a_{3}}{7}a_{0}=a_{1}+a_{2}+a_{3},2a_{1}=\frac{2a_{1}-a_{2}}{7}a_{0}+a_{2},2a_{2}=\frac{2a_{2}-a_{3}}{7}a_{0}+a_{3}$.

For exampte, the relation2$a_{3}= \frac{2a_{3}-a_{1}}{7}a_{0}+a_{1}$is derived from the addition of the three relations. The determinant of the matrix
consisting of the coefficients of the three relations is$|\begin{array}{l}(a_{1}+a_{2}+a_{3})/7-12-1-1-(2a_{1}-a_{2})/7-(2a_{2}-a_{3})/702\end{array}|$

$=$

a3.

satisfying the above conA numerical semigroup $H$ with $M(H)$dition is said to be 1-neat Under the above condition we get a saturated subsemias in Definition 11 from the fundamental systemgroup of$=\{a_{0}, a_{1}, a_{2}, a_{3}\}$$S$$\mathbb{Z}^{6}$

140.
Case (II-1) $a_{1}+$ a2 $\equiv 0(7)$ andCase (II-l-i) $2a_{2}<a_{1}+2a_{3}$ and $2a_{3}<3a_{2}$ . A generating system for relations
consists of$2a_{1}\equiv a_{3}(7)$$\frac{a_{1}+a_{2}}{7}a_{0}=a_{1}+a_{2},2a_{1}=\frac{2a_{1}-a_{3}}{7}a_{0}+a_{3},3a_{2}=\frac{3a_{2}-2a_{3}}{7}a_{0}+2\mathrm{a}3$$3a_{3}= \frac{3a_{3}-a_{2}}{7}a_{0}+a_{2},$$\frac{a_{2}+a_{3}-a_{1}}{7}a_{0}+a_{1}=a_{2}+a_{3}$$\frac{a_{1}+2a_{3}-2a_{2}}{7}a_{0}+2a_{2}=a_{1}+2a_{3}$$\mathrm{i}.\mathrm{e}.$,.

, the kernel of$\varphi_{H}$: $k[X_{0}, X_{1}, X_{2}, X_{3}]$$X_{i}$$arrow$$k[t^{a_{0}}, t^{a_{1}}, t^{a_{2}}, t^{a_{3}}]$$\vdash+$$t^{a_{i}}$is generated by$X_{0^{7}}-X_{1}X_{2}[perp] a\pm\underline{a}\mathrm{a}$$X_{3}-X_{0}^{7}X_{2}\underline{3a}_{\mathrm{A}^{-}R}a$,,$X_{1}^{2}-X_{0}X_{3}\underline{2}a\infty_{7}^{-\Delta}a$$X_{0}^{\mathrm{m}_{7}}X_{1}-X_{2}X_{3}a+a-a$

A fundamental system of relations is the follow,$X_{2}^{3}-X_{0}^{\ovalbox{\tt\small REJECT}_{7}\mathrm{f}\mathrm{i}}X_{3}3a-\underline{2a}$,,$X_{0}^{\mathrm{B}}X_{2}^{2}-X_{1}X_{3}^{2}\circ+2a_{7}-2a$$\mathrm{i}\mathrm{n}\mathrm{g}$.

:$\frac{a_{1}+a_{2}}{7}a_{0}=a_{1}+a_{2},2a_{1}=\frac{2a_{1}-a_{3}}{7}a_{0}+a_{3},3a_{2}=\frac{3a_{2}-2a_{3}}{7}a_{0}+3\mathrm{a}3$.

For example, the addition of the first and second relations$\frac{a_{1}+a_{2}}{7}a_{0}+2a_{1}=(a_{1}+a_{2})+(\frac{2a_{1}-a_{3}}{7}a_{0}+$$a_{3})$

induces the fifth relation. To get a subsemigroup of we divide this case into
three cases again.
Case (II-l-i-A) $a_{1}+2a_{2}>3a_{3}$ . We divide the coefficients in the fundamental
system of relations into the following:$S$$\mathbb{Z}^{l}$$(\alpha_{0}’+\alpha_{0}’+\alpha_{0}’)a_{0}=\alpha_{01}a_{1}+\mathrm{a}02\mathrm{a}22\alpha_{01}a_{1}=(\alpha_{0}’+\alpha_{0}’)a_{0}+$$(2\alpha_{02}+\alpha_{2}’)a_{2}$We associate elements of$\alpha_{0}’a_{0}$$\vdasharrow \mathrm{b}_{1}=\mathrm{e}_{1}$,$\mathbb{Z}^{5}$$=(\alpha_{0}’+\alpha_{0}’)a_{0}+\alpha_{23}a_{3}$.

to the components of the above system as follows:$\alpha_{0}’’a_{0}\}arrow \mathrm{b}_{2}=\mathrm{e}_{2}$$\alpha_{2}’a_{2}\vdash+$a $13a3$ ,$\mathrm{b}_{5}=\mathrm{e}_{5}$,,$\alpha_{0}’’’a_{0}\}arrow$$\mathrm{b}_{3}=\mathrm{e}_{3}$$\mathrm{a}02\mathrm{a}2\vdash\Rightarrow \mathrm{b}_{6}=(1, 1, 1,$,$\alpha_{01}a_{1}\vdasharrow \mathrm{b}_{4}=\mathrm{e}_{4_{7}}$$-1, 0)$ ,141$\alpha_{13}a_{3}\vdasharrow \mathrm{b}_{7}=$$(-1, -1_{7}0,2, 0)$ , $\alpha_{23}a_{3}\vdash\Rightarrow \mathrm{b}_{8}=(1, 2,1, -2,1)$.

denotes the vector whose i-th component is 1 and j-th component is 0
wheregenerated by , . . . , . We can show
. Let
ifbe the subsemigroup ofthat$\mathrm{e}_{i}$$j\neq \mathrm{i}$$\mathbb{Z}^{5}$$S$$\mathrm{b}_{8}$$\mathrm{b}_{1}$$\sum_{i=1}^{8}\mathbb{R}_{+}\mathrm{b}_{i}\cap \mathbb{Z}^{5}\subseteq S$

where$\mathbb{R}_{+}$

denotes the set of non-negative real numbers. Hence,$S$

is saturated.

Case (II-l-i-B) $a_{1}+2a_{2}<3a_{3}$ . We divide the coefficients in the fundamental
system of relations into the following:,$(\alpha_{0}’+\alpha_{10}+\alpha_{20})a_{0}=\alpha_{01}a_{1}+\alpha_{02}a_{2},2\alpha_{01}a_{1}=\alpha_{10}a_{0}+\alpha_{13}a_{3}$$(2\alpha_{02}+\alpha_{2}’)a_{2}=\alpha_{20}a_{0}+\alpha_{23}a_{3}$We associate elements of$\alpha_{0}’a_{0}\vdasharrow \mathrm{b}_{1}=\mathrm{e}_{1}$,$\mathbb{Z}^{5}$$\alpha_{13}a_{3}-arrow$

Let

$S$

to the components of the above system as follows:$\alpha_{10}a_{0}\vdash\Rightarrow \mathrm{b}_{2}=\mathrm{e}_{2}$$\alpha_{2}’a_{2}\vdash+$.

,$\alpha_{20}a_{0}-\succ$$\mathrm{b}_{3}=\mathrm{e}_{3}$,$\alpha_{01}a_{1}-\neq \mathrm{b}_{4}=\mathrm{e}_{4}$,,$\mathrm{b}_{5}=\mathrm{e}_{5},\alpha_{02}a_{2}\vdash-\succ \mathrm{b}_{6}=(1,1,1, -1,0)$$\mathrm{b}_{7}=(\mathrm{O}, -1,0,2,0)$be the subsemigroup of$\mathbb{Z}^{\mathrm{S}}$, $\alpha_{23}a_{3}\vdash+\mathrm{b}_{8}=(2,2,1, -2,1)$ .

generated by$\mathrm{b}_{1}$,$\ldots$,$\mathrm{b}_{8}$. Then$S$

is saturated.

. We get a
Case (II-l-i-C) $a_{1}+2a_{2}=3a_{3}$ . In the Case (II-l-i-A) Jet
generated by 7 elements. Then is saturated.
subsemigroup of$\alpha_{0}’=0$

$S$

$S$

$\mathbb{Z}^{4}$

But our method does not work well in the following case.
, $2a_{2}+a_{3}\equiv 0,2a_{1}\leqq a_{2}+a_{3},2a_{2}>3a_{1}$ . We have theCase (III-2-i)
following generating system of relations$2\mathrm{a}\mathrm{x}+a_{\mathit{2}}$$\equiv 0$,(1)$4a_{1}= \frac{4a_{1}-a_{3}}{7}a_{0}+a_{3_{7}}$(2)$\frac{2a_{1}+a_{2}}{7}a_{0}=2a_{1}+a_{2}$$2a_{2}= \frac{2a_{2}-3a_{1}}{7}a_{0}+3a_{1}$$2a_{3}= \frac{2a_{3}-a_{1}}{7}a_{0}+a_{1}$

(3)

,

(4)

,

$\frac{a_{2}+a_{3}-2a_{1}}{7}a_{0}+2a_{1}=a_{2}+a_{3}$$\frac{a_{1}+a_{3}-a_{2}}{7}a_{0}+a_{2}=a_{1}+a_{3}$

.

,

(5)
(6)

142
The three equations (1), (2) and (6) in the generating system of relations form afundamental system. In fact,(1) $+(2)=(5),{}^{t}(1)+{}^{t}(2)+(6)=(3)$ an $\mathrm{d}{}^{t}(1)+{}^{t}(2)+{}^{t}(6)=(4)$ .

We divide the coefficients in the fundamental system of relations into the following:$(\alpha_{10}+\alpha_{20}+\alpha_{0}’)a_{0}=\alpha_{01}a_{1}+\alpha_{2}’a_{2}$,$\alpha_{0}’a_{0}+\alpha_{2}’a_{2}=\alpha_{1}’a_{1}+\alpha_{13}a_{3}$We associate elements of

aiOao }

$arrow \mathrm{b}_{1}=\mathrm{e}_{1}$,$\mathbb{Z}^{5}$$\alpha_{20}a_{0}$Let.

to the components of the above system as follows:}$arrow \mathrm{b}_{2}=\mathrm{e}_{2}$$\alpha_{1}’a_{1}\mathrm{k}arrow \mathrm{b}_{5}=\mathrm{e}_{5}$$\alpha_{31}a_{1}\vdash+$,,$\alpha_{0}’a_{0}-+$$\alpha_{2}’a_{2}-\succ \mathrm{b}_{6}=$$\mathrm{b}_{7}=(2,1,2, -2, -2)$,$\mathrm{b}_{3}=\mathrm{e}_{3}$,$\alpha_{01}a_{1}-+\mathrm{b}_{4}=\mathrm{e}_{4}$$(1, 1, 1_{7}-1,0)$$\mathbb{Z}^{5}$$\mathrm{b}_{8}$$\mathrm{b}_{1}$$\ldots$,,$\mathrm{a}13\mathrm{a}3\vdash\Rightarrow \mathrm{b}_{8}=(1,1, 2, -1, -1)$

be the subsemigroup of generated by ,, . ThenIn fact,
2 $(1, 1, 1,$ $-1,$ $-1)$ = $(2, 2, 2, -2, -2)=\mathrm{b}_{2}+$$S$,$(\alpha_{01}+\alpha_{1}’+\alpha_{31})a_{1}=\alpha_{10}a_{0}+\alpha_{13}a_{3}$$S$$\mathrm{b}_{7}\in S$.

is not saturated.
,but$(1, 1, 1,$ $-1,$ $-1)$

Hence,$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}k[S]$\not\in S and $(1, 1, 1, -1, -1)\in \mathbb{Z}^{5}$ .

is not a toric variety.

To check whether a 7-semigroup generated by 4 elements is of toric type we divide
them into the 31 cases in the following table. But this problem is still open in thelast three cases. The right-hand side of column in the table means the dimensionof the affine toric variety which is constructed from a numerical semigroup of given
type in our way

143

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Department of Mathematics
Kanagawa Institute of Technology
Atsugi, Kanagawa, 243-0292, Japan-mail: komeda@gen.kanagawa-it.ac.jp$\mathrm{e}$