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arXiv:1104.3600v1 [math.GR] 18 Apr 2011

GENUS OF NUMERICAL SEMIGROUPS GENERATED BY

THREE ELEMENTS

HIROKATSU NARI, TAKAHIRO NUMATA, AND KEI-ICHI WATANABE

Abstract. Let H = ?a,b,c? be a numerical semigroup generated by three ele-

ments and let R = k[H] be its semigroup ring over a field k. We assume H is not

symmetric and assume that the definig ideal of R is defined by maximal minors

?

Yβ′

Zγ′

Xα′

determined by the Frobenius number F(H) and αβγ or α′β′γ′. In particular, we

show that H is pseudo-symmetric if and only if αβγ = 1 or α′β′γ′= 1.

Also, we will give a simple algorithm to get all the pseudo-symmetric numerical

semigroups H = ?a,b,c? with give Frobenius number.

of the matrix

Xα

Yβ

Zγ

?

. Then we will show that the genus of H is

1. Introduction

Let N be the set of nonnegative integers. A numerical semigroup H is a subset of

N which is closed under addition and N\H is a finite set. We always assume 0 ∈ H.

We define F(H) := max{n | n ?∈ H},andg(H) := Card(N \ H). We call F(H)

the Frobenius number of H, and we call g(H) the genus of H. Then it is known

that 2g(H) ≥ F(H)+1. We denote by H = ?a1,a2,...,an? the numerical semigroup

generated by a1,a2,...,an. Namely, H =

semigroup admits a unique minimal system of generators.

We say that H is symmetric if F(H) is odd and for every a ∈ Z, either a ∈ H

or F(H) − a ∈ H, or equivalently, 2g(H) = F(H) + 1. We say that H is pseudo-

symmetric if F(H) is even and for every a ∈ Z,a ?= F(H)/2, either a ∈ H or

F(H) − a ∈ H, or equivalently, 2g(H) = F(H) + 2.

For a fixed field k, a variable t over k, let R = k[H] = k[th| h ∈ H] be the

semigroup ring of H.Then it is known that H semigroup is symmetric (resp.

pseudo-symmetric) if and only R = k[H] is a Gorenstein (resp. Kunz) ring (see

[BDF]). The a-invariant of the semigroup ring R ([GW]) is defined to be a(R) =

max{n | [H1

is, F(H) = a(R).

We say that an integer x is a pseudo-Frobenius number of H if x ?∈ H and x+s ∈ H

for all s ∈ H,s ?= 0. We denote by PF(H) the set of pseudo-Frobenius numbers of H.

The cardinality in PF(H) is called the type of H, denoted by t(H). Since x ∈ PF(H)

if and only if txis in the socle of H1

type of k[H]. Since F(H) ∈ PF(H), t(H) = 1 if and only if H is symmetric.

?n

i=1aiN. Moreover, every numerical

m(R)]n?= 0}. Since H1

m(R)∼= k[t,t−1]/R, a(R) = max{m | m ?∈ H}, that

m(k[H]), t(H) = r(k[H]), the Cohen-Macaulay

Date: April 20, 2011.

2000 Mathematics Subject Classification. Primary 20M15, Secondary 13F99, 13A02, 16S36.

Key words and phrases. numerical semigroup, pseudo-symmetric semigroup, Frobenius number,

genus of a semigroup.

1

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2HIROKATSU NARI, TAKAHIRO NUMATA, AND KEI-ICHI WATANABE

In this paper, we investigate numerical semigroups generated by three elements,

which is not symmetric. We put H = ?a,b,c? and always assume that H is not

symmetric.

Let We now let ϕ : S = k[X,Y,Z] → R = k[H] = k[ta,tb,tc] the k algebra

homomorphism defined by ϕ(X) = ta, ϕ(Y ) = tb, and ϕ(Z) = tcand let p = p(a,b,c)

be the kernel of ϕ. Then it is known that if H is not symmetric, then the ideal

p = Ker(ϕ) is generated by the maximal minors of the matrix

(1.1)

?Xα

Yβ′

Yβ

Zγ′

Zγ

Xα′

?

for some positive integers α,β,γ,α′,β′,γ′(cf. [He]). We want to describe g(H) by

α,β,γ,α′,β′,γ′and the main goal of this paper is the following Theorem.

Theorem. Let H be a numerical semigroup as above. Then

(1) if β′b > αa, then 2g(H) − (F(H) + 1) = αβγ,

(2) if β′b < αa, then 2g(H) − (F(H) + 1) = α′β′γ′.

As a direct consequence of this Theorem, we can get the characterization of

pseudo-symmetric semigroups generated by 3 elements.

Corollary.

symmetric if and only if either α = β,= γ = 1 or α′= β′= γ′= 1.

Also, we will give an algorithm to classify all pseudo-symmetric numerical semi-

group H generated by 3 elements with given Frobenius number F(H).

Let H be a numerical semigroup as above.Then H is pseudo-

2. Numerical semigroups generated by three elements

Let H = ?a,b,c? be a numerical semigroup and R = k[H]∼= k[X,Y,Z]/p be its

semigroup ring over a field k. Then it is known that the ideal p of S = k[X,Y,Z] is

generated by the maximal minors of the matrix

?Xα

Yβ′

Yβ

Zγ′

Zγ

Xα′

?

, where α, β, γ,

α′, β′, and γ′are positive integers. Since k[H]/(ta)∼= k[Y,Z]/(Yβ+β′,Yβ′Zγ,Zγ+γ′),

the defining ideal of k[H]/(ta) is generated by the maximal minors of the matrix

?0

Yβ′

Zγ′

0

and likewise for b,c, we get the equalitions

Yβ

Zγ

?

. Since a = dimkk[H]/(ta) = dimkk[Y,Z]/(Yβ+β′,Yβ′Zγ,Zγ+γ′),

(2.1.1)

a = βγ + β′γ + β′γ′,

b= γα + γ′α + γ′α′,

c= αβ + α′β + α′β′.

We put l = Zγ+γ′−Xα′Yβ, m = Xα+α′−Yβ′Zγ, and n = Yβ+β′−XαZγ′. There

are obvious relations

Xαl + Yβm + Zγn = Yβ′l + Zγ′m + Xα′n = 0.

We put p = deg(Zγ+γ′), q = deg(Xα+α′), r = deg(Yβ+β′), s = deg(Xα) + p,

t = deg(Yβ′) + p. Since pdS(R) = 2, we get a free resolution of R

0 → S(−s) ⊕ S(−t) → S(−p) ⊕ S(−q) ⊕ S(−r) → S → R → 0.

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GENUS OF NUMERICAL SEMIGROUPS GENERATED BY THREE ELEMENTS3

Taking HomS(∗,KS) = HomS(∗,S(−x)), we get

0 → S(−x) → S(p − x) ⊕ S(q − x) ⊕ S(r − x) → S(s − x) ⊕ S(t − x) → KR→ 0,

where x = a + b + c and KR=Ext2

Since KR is generated by the elements of degree −PF(H), from this exact se-

quence, we have that PF(H) = {s − x,t − x}. We put f = s − x and f′= t − x.

By the above argument, we obtain the following results.

S(R,KS).

Proposition 2.1. If H = ?a,b,c? is not symmetric, then

(1) (α + α′)a = β′b + γc and α + α′= min{n | an ∈ ?b,c?},

(2) (β + β′)b = αa + γ′c and β + β′= min{n | bn ∈ ?a,c?},

(3) (γ + γ′)c = α′a + βb and γ + γ′= min{n | cn ∈ ?a,b?}.

Proposition 2.2. If H = ?a,b,c? is not symmetric, then PF(H) = {f,f′} where

(1) f = αa + (γ + γ′)c − (a + b + c),

(2) f′= β′b + (γ + γ′)c − (a + b + c).

Remark 2.3. Formulas related to our results in this section can be found in [RG1],

[RG].

3. Main results

The following is the key lemma to prove our main theorem.

Lemma 3.1. Let H = ?a,b,c? be as in the previous section. We assume that

β′b > αa, or equivalently, f′> f. Then

(1) for p,q,r ∈ N, f′− f + pa + qb + rc ?∈ H if and only if p < α,q < β and

r < γ.

(2) Card{h ∈ H | f′− f + h ?∈ H} = αβγ.

(3) Card[[(f − H) ∩ N] \ (f′− H)] = αβγ.

Proof. Since f′− f + αa = bγ,f′− f + βb = γ′c,f′− f + γc = α′a ∈ H, f′− f +

pa + qb + rc ∈ H if p ≥ α or q ≥ β or r ≥ γ. Conversely, assume p < α,q < β and

r < γ and f′− f + pa + qb + rc = ua + vb + wc ∈ H for some u,v,w ∈ N. Then

we have (β′+ q − v)b = (α − p + u)a + (v − r)c. If v ≥ r, then this contradicts

Proposition 2.1 (2). If r > v, we have (α−p+u)a = (β′+q −v)b+(r −v)c. Then

by Proposition 2.1 (1), we must have p − u ≥ α′and again we have a contradiction

since r −v < γ. This finishes the proof of (1) and (2) is a direct consequence of (1).

To show (3), it suffices to note that for h ∈ H, f − h ?∈ f′− H if and only if

f′− (f − h) ?∈ H.

Thus we have Card[(f−H)\(f′−H)] = Card{h ∈ H | f′−f+h ?∈ H} = αβγ.

?

Theorem 3.2. Let H = ?a,b,c? be a numerical semigroup. Then

(1) if β′b > αa, then 2g(H) − (F(H) + 1) = αβγ,

(2) if β′b < αa, then 2g(H) − (F(H) + 1) = α′β′γ′.

Proof. We may assume β′b > αa. Then by Proposition 2.2, F(H) = f′. Since

N \ H = ((f′− H) ∩ N) ∪ ((f − H) ∩ N), we get

g(H) = Card[(f′− H) ∩ N] + Card[[(f − H) ∩ N] \ (f′− H)]

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4 HIROKATSU NARI, TAKAHIRO NUMATA, AND KEI-ICHI WATANABE

hence by Lemma 3.1,

g(H) = (F(H) + 1 − g(H)) + αβγ.

?

As a corollary, we find a characterization of pseudo-symmetric numerical semi-

groups generated by 3 elements.

Corollary 3.3. H is pseudo symmetric if and only if

(1) if β′b > αa, then α = β = γ = 1 and

(2) if β′b < αa, then α′= β′= γ′= 1.

Proof. We may assume that β′b > αa. By Theorem 3.2, 2g(H)−(F(H)+1) = αβγ.

Since H is pseudo-symmetric if and only if 2g(H) = F(H) + 2, we obtain that

αβγ = 1, or equivalently, α = β = γ = 1.

?

4. The structure of a pseudo-symmetric numerical semigroup

generated by three elements

In this section, we assume that H = ?a,b,c? is a pseudo-symmetric numerical

semigroup. Our purpose is to classify, for any fixed fixed even integer f, all the

pseudo-symmetric numerical semigroups H = ?a,b,c? with F(H) = f. For example,

it is shown in Exercise 10.8 of [RG] that there is no pseudo-symmetric numerical

semigroup H = ?a,b,c? with F(H) = 12. Actually, we can give many examples of

such even integer f for which there does not exist numerical semigroup H = ?a,b,c?

with F(H) = f. (It is shown in [RGG] that every even integer is the Frobenius

number of some numerical semigroup generated by at most 4 elements.)

As is mentioned before, p = p(a,b,c) of k[X,Y,Z] is generated by the maximal

minors of the matrix as in (1.1) and by Corollary 3.3, we can always assume that

α′= β′= γ′= 1. Recall that in this case we have by (2.1.1),

(4.1.1)a = βγ + γ + 1,b = γα + α + 1,c = αβ + β + 1.

The following is the key for our goal.

Theorem 4.1. Let H = ?a,b,c? be a pseudo-symmetric numerical semigroup and as-

sume that p(a,b,c) is generated by the maximal minors of the matrix

?

Xα

Y

Yβ

Z

Zγ

X

?

.

Then we have

αβγ =F(H)

2

+ 1.

Proof. From our hypothesis and Corollary 3.3, we have f′< f. Thus by Proposition

2.2, F(H) = f = αa + (γ + 1)c − (a + b + c) = 2αβγ − 2.

?

Now, given a positive even integer f, we can list all possibilities of the set {α,β,γ}

by prime factorization of

2

+ 1.

F(H)

Remark 4.2. Let σ be a permutation of {α,β,γ}. Then it is easy to see that if σ

is an even permutation, then the set {a,b,c} obtained by {σ(α),σ(β),σ(γ)} as in

(4.1.1) is the same and hence the semigroup H = ?a,b,c? does not change.

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GENUS OF NUMERICAL SEMIGROUPS GENERATED BY THREE ELEMENTS5

But if σ is an odd permutation, then the set {a,b,c} does change. So, from the

factorization ofF(H)

2

+ 1, we get 2 different semigroups in general.

Example 4.3. For example, let us classify all pseudo-symmetric semigroup H =

?a,b,c? with F(H) = f = 18. Since we have αβγ = f/2 + 1 = 10 by Theorem 4.1,

we have {α,β,γ} = {10,1,1} or {5,2,1}. But if we put {α,β,γ} = {10,1,1} in any

order to (4.1.1), a,b,c are all multiple of 3 and we don’t get a numerical semigroup.

Thus we get 2 semigroups with F(H) = 18; if (α,β,γ) = (5,2,1) we get H =

?4,11,13? and if (α,β,γ) = (5,1,2), then we get H = ?5,16,7?.

If f is an even integer not divisible by 12, then there is a pseudo-symmetric

semigroup H = ?a,b,c? with F(H) = f by [RGG].

Proposition 4.4. [RGG] Let H = ?a,b,c? be a numerical semigroup and F(H) = f.

Then

(1) If f is an even integer not divisible by 3, then

?

3,f

2+ 3,f + 3

?

is a pseudo-symmetric numerical semigroup with Frobenius number f. We

put (α,β,γ) = (f/2 + 1,1,1).

(2) If f is a multiple of 6 and not a multiple of 12, then if we put (α,β,γ) =

((f + 2)/4,2,1), we get

H =

?

4,f

2+ 2,f

2+ 4

?

,

which is pseudo-symmetric with F(H) = f.

If f is divisible by 12, there are many cases such that there does not exist pseudo-

symmetric semigroup H = ?a,b,c? with F(H) = f.

Proposition 4.5. We suppose 12 | f. If there exists a pseudo-symmetric numerical

semigroup H = ?a,b,c? with F(H) = f, then f/2+1 has a prime factor of the form

3k + 2 (k ≥ 1).

Proof. Otherwise, since α,β,γ are divisors of f/2 + 1, we get α ≡ β ≡ γ ≡ 1 (mod

3). Then by (4.1.1) we see that a,b,c are divisible by 3 and H = ?a,b,c? is not a

numerical semigroup.

?

Example 4.6. Let f be an integer divisible by 12.

(1) By Proposition 4.5, there is no pseudo-symmetric semigroup H = ?a,b,c?

with F(H) = 12,24,36,60,72,84,96,120,132,144,156,180,192.

(2) On the other hand, there exists pseudo-symmetric semigroups H = ?a,b,c?

with F(H) = 48,168.Actually, H = ?7,11,31? is the unique pseudo-

symmetric semigroup generated by 3 elements, with F(H) = 48 and H =

?19,11,103? is the unique pseudo-symmetric semigroup generated by 3 ele-

ments with F(H) = 168