Tel Aviv University

The Raymond and Beverly Sackler Faculty of Exact Sciences

School of Mathematical Sciences

Applications of the minimal

transversal method in numerical

semigroups

A thesis submitted for the degree ”Doctor in Philosophy”

by

Eli Leher

Prepared under the supervision of Prof. Shmuel Rosset

Submitted to the Senate of Tel Aviv University

July 2007

This work was prepared under the Supervision of Prof. Shmuel Rosset.

I thank him deeply for the many conversations we had during the years.

I have learnt a lot.

Moreover, it was fun.

To the people I love, I hope they know who they are.

To my mom, whom I hardly knew, yet I miss a lot.

And to the future.

I thank Lior Bary-Sorocker for all his encouragement, help and support.

Contents

1 Introduction 11.1 The coin exchange problem . . . . . . . . . . . . . . . . . . . 11.2 History of the problem . . . . . . . . . . . . . . . . . . . . . . 41.3 About this work . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 The minimal transversal 82.1 The minimal transversal . . . . . . . . . . . . . . . . . . . . . 82.2 Example: Smooth sequences . . . . . . . . . . . . . . . . . . . 10

2.2.1 Example: The products semigroup . . . . . . . . . . . 132.2.2 Example 2: The binomial semigroup . . . . . . . . . . 142.2.3 Example 3: ”geometric progression” . . . . . . . . . . . 14

2.3 Example: Arithmetic progressions . . . . . . . . . . . . . . . . 142.4 Example: Delorme’s construction . . . . . . . . . . . . . . . . 19

2.4.1 Minimal Generation . . . . . . . . . . . . . . . . . . . 22

3 Bounds 243.1 Bounding the Frobenius number and the genus of a numerical

semigroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 Upper bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2.1 The Schur-Brauer bound . . . . . . . . . . . . . . . . 253.2.2 The bound of Nijenhuis and Wilf . . . . . . . . . . . . 27

3.3 Lower bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Presentation of the minimal transversal 324.1 Canonical domains . . . . . . . . . . . . . . . . . . . . . . . . 32

4.1.1 The presentation . . . . . . . . . . . . . . . . . . . . . 324.1.2 Bounds for C . . . . . . . . . . . . . . . . . . . . . . . 344.1.3 The method . . . . . . . . . . . . . . . . . . . . . . . . 37

i

4.2 Semigroups with small generating set . . . . . . . . . . . . . . 384.2.1 Two generators . . . . . . . . . . . . . . . . . . . . . . 384.2.2 Three generators . . . . . . . . . . . . . . . . . . . . . 40

4.2.3 Four generators . . . . . . . . . . . . . . . . . . . . . . 43

4.2.4 Smooth sequences . . . . . . . . . . . . . . . . . . . . . 45

5 Symmetric semigroups 475.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.2 3-generated symmetric semigroups . . . . . . . . . . . . . . . . 48

6 The minimal transversal in affine semigroups 516.1 Numerical Affine semigroups . . . . . . . . . . . . . . . . . . . 51

6.1.1 The rank of the semigroup . . . . . . . . . . . . . . . . 516.1.2 Dimension and codimension . . . . . . . . . . . . . . . 526.1.3 The Hilbert series of the semigroup . . . . . . . . . . . 52

6.2 Semigroups of codimension 1 . . . . . . . . . . . . . . . . . . . 536.2.1 First Proof . . . . . . . . . . . . . . . . . . . . . . . . 556.2.2 Second proof . . . . . . . . . . . . . . . . . . . . . . . 59

A The semigroup polynomial 61A.1 The polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 61

A.1.1 Basic facts . . . . . . . . . . . . . . . . . . . . . . . . . 61A.1.2 The case of two generators . . . . . . . . . . . . . . . . 63

A.2 Cyclotomic polynomials . . . . . . . . . . . . . . . . . . . . . 67A.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 67A.2.2 General results . . . . . . . . . . . . . . . . . . . . . . 69

A.3 On the cyclotomic polynomials of order pqr. . . . . . . . . . . 72A.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 72A.3.2 The result . . . . . . . . . . . . . . . . . . . . . . . . . 72

B The Frobenius number revisited 75B.1 Two-generated semigroups . . . . . . . . . . . . . . . . . . . . 75B.2 Three-generated semigroups . . . . . . . . . . . . . . . . . . . 76

ii

Chapter 1

Introduction

1.1 The coin exchange problem

Let N = {0, 1, 2, ...} be the set of natural numbers. A number m ∈ N issaid to be a natural combination of the elements a1, ..., an ∈ N if thereare numbers m1, ..., mn ∈ N such that m =

∑ni=1 miai. Let A = {a1, ..., an}

be a set of naturals and S = S(A) = S(a1, ..., an) be the set of all naturalcombinations of the ai’s. Then S is a semigroup (that is it is closed underaddition). We call S the semigroup generated by A, and the elementsai ∈ A the generators of S. The semigroup S is numerical if its complementS∗ = N\S is finite. Hence S is numerical if and only if there is a numberN ∈ N such that every s > N belongs to S.

Proposition 1 The semigroup S = S(a1, ..., an) is numerical if and only ifa1, ..., an are relatively prime.

Proof: Suppose first that a1, ..., an are relatively prime. We work by induc-tion on n.(i) If n = 1, the result is clear (since a1 = 1 and S = N).(ii) Assume the result holds for n generators. Let S = S(a1, ..., an, an+1)be a semigroup generated by n + 1 elements. Let d = gcd{a1, ..., an} andbi = ai/d, i = 1, ..., n. Let S2 = S(b1, ..., bn). Since b1, ..., bn are relativelyprime, it follows from the induction hypothesis that there is a number N suchthat for every s > N, s ∈ S2. We claim that every number t > (d−1)an+Ndbelongs to S.Indeed since a1, ..., an+1 are relatively prime, so are d and an+1. Hence the

1

set {kan+1 | k = 0, ..., d− 1} contains an element of every residue class mod-ulo d. Let t be a natural number. Then there is a unique 0 ≤ j < d suchthat t = jan+1 + md for some m ∈ Z moreover if t > (d − 1)an + Nd, wehave md = t − jan+1 > Nd and m > N . Thus m ∈ S2 and md ∈ S (ifm =

∑ni=1 mibi then md =

∑ni=1 miai). Since jan+1 ∈ S too, it follows that

t = jan+1 + md ∈ S, which completes the proof (of the ”if” part).On the other direction, we notice that if gcd{a1, ..., an} = d then d dividesevery element of S. Hence if d > 1 then N\S is infinite. ¤

Let A = {a1, ..., an} be a set of relatively prime positive integers and S =S(A). Hence S is numerical. Let S∗ = N\S. The Frobenius number of Sis the number F (S) = max S∗, the maximal natural number which does notbelong to S. The terminology is in honor of Ferdinand Georg Frobenius, fa-ther of representation theory, who used to raise the question of finding F (S)(or at least a good bound for it) repeatedly in his lectures [8]. The numberd(S) = F (S) + 1 is called in the literature the conductor of S.

The genus of S is the number g(S) of elements in S∗. For every naturalnumber k = 0, ..., F (S) at least one of the numbers k, F (S) − k does notbelong to S (had both of them belonged to S so would their sum F (S), acontradiction). Thus

g(S) ≥ F (S) + 1

2.

We say that S is symmetric if equality occurs. In particular if S is sym-metric, then F (S) is odd.

Remark: It is also customary to speak of the Frobenius number and genusof A, and denote them by F (A) = F (a1, ..., an) and g(A) = g(a1, ..., an) re-spectively. We use this notation occasionally.

The Hilbert series of the numerical semigroup S is the formal power series

HS(x) =∑s∈S

xs ∈ C[[x]].

As a function over the complex field, H is analytic in the (open) unit disc.Moreover, let H∗(z) =

∑s∈S∗ zs. Then H(z)+H∗(z) =

∑∞k=0 zk = 1

1−z. Thus

H = 11−z

−H∗(z) where H∗ is a polynomial. Hence H has a continuation to

2

a meromorphic function with one simple pole at z = 1.

The diophantine Frobenius problem for numerical semigroups is to find theirFrobenius number, genus and Hilbert series in terms of the generators a1, ..., an.

The Frobenius problem arises in many questions from various areas of math-ematics. We give two examples:

(a) To a numerical semigroup S one can associate a complete Noetherianlocal ring as follows. Let k be a field and let k[[S]] be the ring of formalpower series

∑s∈S ast

s with coefficients in k. This ring is the completion ofthe semigroup ring kS in the I-adic topology where I is the ideal generatedby S in kS. It is easy to see that the integral closure of k[[S]] is the full ringof power series k[[t]] and S is the set of values the valuation of k[[t]] achieveson k[[S]]. This example is the background to Kunz’s classic result relatingsymmetric numerical semigroups to 1-dimensional Gorenstein rings [19]:

Theorem 2 Let R be a one dimensional analytically irreducible noetherianlocal ring. Let R be its integral closure in the quotient field K and ν : K → Zbe the corresponding valuation. Assume that R and R have the same residueclass field. Then R is Gorenstein if and only if the value semigroup ν(R) issymmetric.

(b) An affine space curve C with the parametric equations xi = tai , i = 1, ..., nwhere a1, ..., an are relatively prime positive integers is a global ideal-theoreticcomplete intersection if and only if the semigroup S = S(a1, ..., an) is com-plete intersection (in the sense of [17]). It is known [12] that all completeintersection numerical semigroups are symmetric, however not all symmetricsemigroups are complete intersections.

Nevertheless the Frobenius problem is also interesting on its own merit. Someof its charm follows from the fact that it is a problem that non mathemati-cians can understand too (as is its solution in the case of 2 generators). Infact, it is known in the literature also as the coin exchange problem,where the problem is to determine the largest bill which can not be ex-changed to coins of values a1, ..., an (under the joyful assumption we haveunlimited amount of coins of each kind). In this connection we mention thatcoin exchange questions are the first problems in Polya and Szego’s classic[24].

3

1.2 History of the problem

The Frobenius number of two relatively prime positive integers is given by

F (a, b) = ab− a− b.

The original source for this formula is unknown, though it seems a commonmistake to attribute it to James Joseph Sylvester (who never published it).However, in a problem he presented in the Educational times [30], Sylvesterasked to prove (in modern terms) that

g(a, b) =1

2(a− 1)(b− 1)

(or, equivalently, that every 2-generated numerical semigroup is symmetric),which indicates he probably knew about the formula for the Frobenius num-ber as well. An answer to Sylvester’s problem was given by Curran-Sharp[31] later that year.

Although Frobenius urged his students to come up with a solution to thegeneral case, no progress in the subject was noted for many years. Only in1935, about half a century after Sylvester’s result I.Schur, in his last lecturein Berlin, gave an upper bound for F (S) in the general case. This boundwas published and improved by Brauer [6].

The Frobenius number for 3-generated numerical semigroups was found in1962 by Brauer and Shockly [8]. Their solution was not a polynomial in thegenerators and it involved magnitudes which could not be expressed by thegenerators (namely the minimal positive multiples of one generator whichbelongs to the semigroup generated by the other two). Later on, more solu-tions to this case were found by using different methods (including continuedfractions [28]). However all of these methods needed to use ”outside” terms,which could not be written explicitly by the generators. The connectionbetween some of the formulae given for the Frobenius number in this caseis still unexplained (especially compare [8] and [15]). The Hilbert series of3-generated semigroups was given recently by Denham [13].

During the years some more instances of the general problem were solved.Froberg [16] used the relations of Bresinsky [10] to write the Frobenius num-ber of 4-generated symmetric semigroups. The case of complete intersections

4

([17]) was characterized by Delorme [12], who also gave a recursive formulafor the Frobenius number of these semigroups and proved that they are sym-metric. The Frobenius number of semigroups generated by an arithmeticprogression has also been found ([6, 27, 26, 18, 14]). However, the generalproblem remained unsolved.

In 1990 F.Curtis [11] explained why this problem turned out to be so hard,proving that there is no non zero polynomial P (x1, ..., xn, y) such thatP (a1, ..., an, F (a1, ..., an)) = 0 for every relatively prime positive integersa1, ..., an.

In recent years there is a lot of research on efficient algorithms for find-ing the Frobenius number and on a multidimensional version of the problem(numerical affine semigroups).

For a detailed account of the classical Frobenius problem, one may consultthe thorough book of Ramirez Alfonsin [25].

1.3 About this work

In this work we suggest a unifying approach to the problems involving nu-merical semigroups, namely using the minimal transversal of the semigroupwith respect to one of its elements. Let S be a numerical semigroup andb ∈ N. The minimal transversal of S modulo b is the set of all the mini-mal elements of every residue class modulo b in S. These elements are welldefined since S is numerical.

The first to use these numbers is apparently Roger Apery, who gave a sym-metry criterion for the semigroup in terms of the structure of its minimaltransversal (with respect to any element of S). In fact, the sets Ap(b) ={s ∈ S | s− b /∈ S} are sometimes referred to as Apery sets in the literature.Although these sets coincide with the minimal transversals when b ∈ S, wechose to keep the term ”minimal transversal” (which we used before we wereaware of Apery’s work), since it seems more suggestive.

Brauer and Shockley [8] and Selmer [27] used the minimal transversal withrespect to an element ai in the generating set to give formulae for the Frobe-

5

nius number and genus (respectively) of the semigroup. However it seemsthat these formulae were hardly ever used.

In this work we use the minimal transversal to solve many questions involv-ing numerical semigroups. We observe that the formulae mentioned aboveapply for every element in the semigroup (and not merely for a member ofthe minimal generating set) and give a new formula for the Hilbert series ofthe semigroup (Theorem 5). Then we solve the Frobenius problem for somesemigroups. We also use the minimal transversal to give general upper andlower bounds for the Frobenius number and the genus, characterize symmet-ric semigroups generated by 3 elements and calculate the Hilbert series ofnumerical affine semigroups of codimension one.

In chapter 2 we explain how to solve the Frobenius problem by using theminimal transversal and demonstrate this method in some examples (includ-ing the cases of complete intersection semigroups and semigroups generatedby a generalized arithmetic sequence).

In chapter 3 we expand the method to give upper and lower bounds forthe Frobenius number and the genus of numerical semigroups. We give sim-ple proofs of the classic bound of Brauer and Schur [6] and for the upperbound of Nijenhuis and Wilf [23]. We also present new lower bounds for thegenus and the Frobenius number of the semigroup.

In chapter 4 we describe a general method of solving the Frobenius prob-lem using a presentation of the minimal transversal of the semigroup withrespect to one of its elements. We demonstrate the method on semigroupsgenerated by 2 and 3 elements, and explain the difficulties arising in imple-menting it in the case of more generators by discussing a concrete exampleof 4-generated semigroup.

In chapter 5 we use Apery’s symmetry criterion [1] to reprove Herzog’s the-orem: Every symmetric 3-generated semigroup is a complete intersection.

In chapter 6 we generalize the concept of minimal transversals to numer-ical affine subsemigroups of Nm. Then we use it to calculate the Hilbertseries of affine semigroups of codimension 1.

6

In the appendices we discuss some questions that arose in our work butare not related directly to the minimal transversal.

In appendix A we define a (seemingly new) invariant of the semigroup whichis closely related to the Hilbert series, namely the power series PS(x) = (1−x)HS(x). It is easy to see that PS is a polynomial of degree d(S) = F (S) + 1hence we call it the semigroup polynomial of S. As it turns out, Forevery two relatively prime positive integers a, b, the semigroup polynomial ofS(a, b) is a product of cyclotomic polynomials. We use this to prove someresults about general cyclotomic polynomials. Amongst which we mentionthe structure of the cyclotomic polynomial of order pq where p, q are twodifferent primes and a curious result on the coefficients of the cyclotomicpolynomial of order pqr where p, q, r are three different primes.

We conclude with another method of finding the Frobenius number whichdoes not rely on the minimal transversal (Appendix B). This new methodgives the classic result in the case of 2 generators and a new formula in thecase of 3 generators.

7

Chapter 2

The minimal transversal

2.1 The minimal transversal

Let S be a numerical semigroup. Let b > 1 be a fixed integer. For ev-ery residue k = 0, ..., b − 1 define the coset of k modulo b in S as the setCk = Ck,b = {s ∈ S | s = k (mod b)}. Thus S is the disjoint union of thecosets Ck for k = 0, 1, ..., b − 1. We note that for every residue k, the cosetCk is not empty (since S contains every natural number bigger than F (S)).

The minimal element tk in the coset Ck is called the minimal representa-tive of the residue k modulo b in S.

Definition 3 Let S be a numerical semigroup. Let b > 1 be a fixed integer.The minimal transversal of S modulo b is the set T = T (b) of all theminimal representatives modulo b in S.

Thus a natural number t belongs to T if and only if t ∈ S and for every s ∈ Ssuch that s < t we have s 6= t (mod b). In particular T contains exactly belements.

The main importance of the minimal transversal is when b itself belongs tothe semigroup. This follows from the next simple yet important observation:

Observation 4 Let S be a numerical semigroup, b ∈ S an element of S andT = T (b) the minimal transversal of S modulo b.

8

Every integer n ∈ Z has a unique presentation as the sum n = t + kb witht ∈ T and k ∈ Z. Moreover n ∈ S if and only if in this presentation k ≥ 0.

Thus if b ∈ S, the coset Ck contains exactly the numbers of the form tk + jbfor j ≥ 0.

Proof: There is one and only one such presentation since T contains aunique representative of every residue class modulo b. If in this presentationk ≥ 0 then n ∈ S (since t, b both belong to the semigroup). If k < 0, we haven < t and n = t (mod b) hence n does not belong to S (since t is minimal inits coset). ¤

Now we are ready to explain how to solve the Frobenius problem in terms ofthe minimal transversal.

Theorem 5 Let S be a numerical semigroup and let b be an element of S.Let T = T (b) be the minimal transversal of S modulo b. Then(a)

F (S) = max T − b

(b)

g(S) =1

b

∑t∈T

t− b− 1

2

(c)

HS(x) =1

1− xb

∑t∈T

xt.

Proof: (a) Let m = max T . Then m − b /∈ S (since m is minimal inits coset). Now let n be a natural number such that n > m − b. Let rbe the minimal representative of the coset of n. We claim that r ≤ n (sinceotherwise r ≥ n+b > m, contradicting the maximality of m). Thus n = r+lbfor some l ≥ 0 and n ∈ S which completes the proof.(b) Let k be a fixed residue modulo b. A number n which equals k modulo bbelongs to S if and only if n ≥ tk (where tk is the minimal representative ofthe class). Hence the cardinality of the set of natural numbers which equal kmodulo b and are not in S is (tk− k)/b. Summing up over all of the residueswe get that the genus of S is

g(S) =b−1∑

k=0

tk − k

b=

1

b

(b−1∑

k=0

tk −b−1∑

k=0

k

)=

1

b

∑t∈T

t− b− 1

2.

9

(c) Similarly, the sum∑

xs over powers s ∈ S which equal k modulo b is:

∞∑j=0

xtk+jb = xtk

∞∑j=0

(xb)j = xtk/(1− xb)

and summing over all the residues yields:

HS(x) =1

1− xb

∑t∈T

xt.

¤

In practice, the summations in the previous proposition can be calculatedmany times by using the sum of geometric progression. In the next sectionswe demonstrate the method on semigroups generated by smooth sequences(see definition below) and by (generalized) arithmetic progression. We alsouse it to solve the problem to semigroups which are complete intersectionsusing their structure theorem of Delorme [12]. We remark that this methodcan be applied on some other instances of the problem as well (for examplewhen the generators form a broken arithmetic progression).

2.2 Example: Smooth sequences

Let (a1, a2, ..., an) be a sequence of relatively prime positive integers. For ev-ery k = 1, ..., n , let dk = gcd(a1, ..., ak). Since a1, ..., an are relatively prime,dn = 1. Let ck = dk−1/dk where k = 2, ..., n. We note, for further use thatcn = dn−1 and

∏ki=2 ci =

∏ki=2

di−1

di= d1

dk= a1

dkin particular

∏ni=2 ci = a1

1= a1.

Let Sk be the semigroup generated by a1, ..., ak, and S = Sn. We say thatthe sequence (a1, a2, ..., an) is smooth if ckak ∈ Sk−1 for every k = 2, ..., n.This condition seems to play a special role in the theory of numerical semi-groups. In [6] and [7], Brauer and Seelbinder showed that F (a1, ..., an) ≤∑n

i=2 ciai −∑n

i=1 ai with equality if and only if (a1, ..., an) is smooth. Thiswas reproved in a different way in [23]. In [17] Herzog proved that the semi-groups generated by smooth sequences are complete intersections (ci) andasked whether all numerical semigroups that are ci are such. That this is not

10

the case was shown by Delorme [12]. However Delorme proved that smoothsequences are the construction stones of complete intersections numericalsemigroups.

We remark that if dk = dk+1, the smoothness condition implies that 1 ·ak+1 ∈Sk. Thus ak+1 is a natural combination of a1, ..., ak and can be omitted fromthe sequence without changing neither its smoothness nor the generated semi-group S.

Proposition 6 Let s be an element of S, then s can be written as a naturalcombination

∑ni=1 kiai with 0 ≤ ki < ci for i = 2, ..., n.

Proof: Since s ∈ S, we have s =∑n

i=1 kiai for some k1, ..., kn ∈ N. Assumethat this presentation does not fulfill the condition. Let j ≥ 2 be the max-imal index for which kj ≥ cj. Write kj = bcj + r where 0 ≤ r < cj. Sincecjaj ∈ Sj−1, we can write bcjaj as the combination

∑j−1i=1 liai. Hence s can

also be written as∑j−1

i=1 (ki + li)ai + raj +∑n

i=j+1 kiai. In this new presen-tation, either the coefficients fulfill the condition of the proposition, or themaximal index j′ for which the coefficient violates the condition is strictlysmaller than j, in which case we can proceed in the same manner, until weget a presentation in which all of the coefficients fulfill the condition. ¤

Let T = T (a1) be the minimal transversal of S with respect to a1. Ourmain observation is the following:

Theorem 7 Every element of T has exactly one natural combination∑n

i=2 kiai

with 0 ≤ ki < ci.

Proof: Every element s ∈ S has a presentation of the form s =∑n

i=1 kiai

with 0 ≤ ki < ci for i = 2, ..., n. If k1 > 0 then s /∈ T . Thus every elementt ∈ T has a presentation of the form

∑ni=2 kiai with 0 ≤ ki < ci.

Now, the number of these combinations is∏n

i=2 ci = a1. Since there are a1

elements in T and every one of them has at least one such combination wededuce that each combination yields a different representative. ¤

Remark: This result can be extended to claim that every element in s ∈ Shas a unique presentation s =

∑ni=1 kiai with 0 ≤ ki < ci for i = 2, ..., n.

11

Corollary 8 Let S = S(a1, ..., an) where (a1, ..., an) is smooth then:(a)

F (S) =n∑

i=2

ciai −n∑

i=1

ai.

(b) S is symmetric.(c)

H(S) =

∏ni=2 (1− xciai)∏ni=1(1− xai)

.

Proof: (a) The Frobenius number of S is:

F (S) = max T − a1 =n∑

i=2

(ci − 1)ai − a1 =n∑

i=2

ciai −n∑

i=1

ai.

(b) We start by summing up the elements in T :

∑t∈T

t =

c2−1∑

k2=0

c3−1∑

k3=0

...

cn−1∑

kn=0

(k2a2 + k3a3 + ... + knan) =

= C2

c2−1∑

k2=0

k2a2 + C3

c3−1∑

k3=0

k3a3 + ... + Cn

cn−1∑

kn=0

knan

where Ci =∏

j 6=i cj =∏n

j=2 cj/ci = a1

ci.

Thus∑

t∈T t =∑n

i=212Cici(ci − 1)ai = 1

2a1

∑ni=2 (ci − 1)ai.

Now the genus of S is g(S) = 1a1

∑t∈T t− a1−1

2. Thus

g(S) =1

2(

n∑i=2

(ci − 1)ai− (a1− 1)) =1

2(

n∑i=2

ciai−n∑

i=1

ai + 1) =1

2(F (S) + 1).

and the semigroup is symmetric.

The Hilbert series of S is H(x) = 11−xa1

∑t∈T xt where T be the minimal

transversal of S modulo P .

12

Now∑t∈T

xt =

c2−1∑

k2=0

c3−1∑

k3=0

...

cn−1∑

kn=0

xk2a2+k3a3+...+knqn =

= (

c2−1∑

k2=0

xk2a2) · (c3−1∑

k3=0

xk3a3) · ... · (cn−1∑

kn=0

xknan) =

=n∏

i=2

1− xciai

1− xai

Dividing by (1− xa1) we get the result. ¤

2.2.1 Example: The products semigroup

Let p1, ..., pn be pairwise relatively prime positive integers. Let P =∏n

i=1 pi

be their product, and let Pi = P/pi (thus Pj is the product of all thepi’s except pj). Let ai = qiPi where q1 = 1 and qi is not divisible by pi.Thus a1, a2, ..., an are relatively prime. Moreover, the sequence (a1, ..., an) issmooth. Indeed dk =

∏ni=k+1 pi (for k = 1, ..., n) , ck = pk (for k = 2, ..., n)

and ckak = qkP = qkp1 · a1 ∈ Sk−1 (in fact ckak ∈ S1 for every k ≥ 2).Let S = S(a1, a2, ..., an). Then we have:

F (S) = (n∑

i=2

qi)P −n∑

i=1

ai

and

H(S) =

∏ni=2 (1− xqiP )∏ni=1(1− xai)

.

13

2.2.2 Example 2: The binomial semigroup

Another family of smooth sequences is (am, bam−1, ..., bm−1a, bm), where a, bare relatively prime integers greater than 1. Indeed, let ak = am−kbk, k =0, ...,m then it is easy to verify that the sequence (ak) is smooth (with ck = afor k = 1, ..., m and ckak = bak−1 ∈ S(a0, ..., ak−1) ).Using the formulae from the previous section we achieve:

F (a0, ..., am) =m∑

k=1

am+1−kbk −m∑

k=0

am−kbk.

and the Hilbert series of the semigroup S(a0, ..., am) is

HS(x) =m∏

k=1

(1− xam+1−kbk

)/m∏

k=0

(1− xam−kbk

).

2.2.3 Example 3: ”geometric progression”

We end this section with another example of smooth sequences, namely thesewhich are generated by a somewhat generalized ”geometric progression”.

Let a0 = qk, a1 = s1qk + tqj−1, a2 = s2q

k + tqj−2, ..., aj = sjqk + t, where

t > 0, j ≤ k, t and q are relatively prime and si−1 ≤ qsi for i = 2, ..., j.

The sequence (ai)ni=0 is smooth. To show this, we note that di = gcd{a0, ..., ai} =

qj−i for i ≥ 1. Thus ci = di−1/di = q for i ≥ 2 and ciai = ai−1 +(siq−si−1)a0

(here we use the non negativity of the expression siq − si−1).Finally, d0 = qk and c1 = qk+1−j. It is easy to verify that c1a1 is a multipleof a0.

2.3 Example: Arithmetic progressions

In this section we solve the Frobenius problem of Semigroups generated by ageneralized arithmetic progression. By this we mean a sequence (a0, a1, ..., an)

14

in which a0 = a and ak = ha + kd for k = 1, ..., n, where a, d are relativelyprime positive integers and h ≥ 0. The Frobenius number in this case wasdetermined by Selmer [27] who generalized the results of Brauer [6] (whoproved it for d = 1, h = 1) and Roberts [26] (for h = 1 and d > 0). Estradaand Lopez found the criteria for symmetry of these sequences [14] generaliz-ing a result of Juan [18]. Here we use the minimal transversal to write thegenus and the Hilbert series of these semigroups as well.

We find the minimal transversal of S modulo a. Then we apply Theorem 5(since a ∈ S).

Lemma 9 Every minimal representative of S can be written in the form san

or san + ar for some 1 ≤ r < n.

Proof: Suppose that p =∑n

i=0 kiai is the minimal representative of somecoset of S modulo a. Since p is minimal in its coset, p =

∑ni=1 kiai (that is

k0 = 0).From all the representations of p with maximal kn, choose one with minimal∑n−1

i=1 ki . We claim that in this presentation∑n−1

i=1 ki ≤ 1. Indeed, supposethat there are two non zero coefficients ku, kv where both u, v < n, (or onecoefficient ku, u < n such that ku > 1, that we treat as the case v = u).Thus p = au + av +

∑n−1i=1 liai + knan, where li are natural numbers. Let

w = u + v. If w ≤ n then p − ha = aw +∑n−1

i=1 liai + knan ∈ S whicheither contradicts the minimality of p (if h ≥ 1) or the minimality of thesum of the coefficients (if h = 0 and w < n) or the maximality of kn (ifh = 0 and w = n) and in any way it is impossible. If w > n then p can bewritten as aw−n +

∑n−1i=1 liai + (kn + 1)an (notice w < 2n) and again we get

a contradiction to the maximality of kn. Hence the result. ¤

Theorem 10 Let a− 1 = sn + r, 0 ≤ r < n. Let T = T (a) be the minimaltransversal modulo a of S. Then the numbers in T are:kan, k = 0, ..., s ,kan + al, k = 0, ..., s− 1, l = 1, ..., n− 1 andsan + al, l = 1, ..., r.

Proof: By the previous lemma, Every element in T can be written in theform kan or kan +ar for some 0 < r < n. The numbers listed in the theoremare the first minimal numbers in S of these forms. Moreover kan = knd (mod

15

a) and kan + ar = (kn + r)d (mod a). Hence these numbers are of differentresidue classes modulo a. Since there are a numbers in our list, it containsexactly the minimal transversal of S modulo a. ¤

Corollary 11 If n divides a− 1, say a− 1 = sn, then:(a) F (S) = san − a = s(ha + nd)− a = a(sh + d− 1)− d.(b) g(S) = 1

2(a− 1)(sh + h + d− 1).

(c) HS(x) = 11−xa · (1 + xa1 (1−xnd)(1−xsan)

(1−xd)(1−xan )).

Proof: If a− 1 = sn then the minimal transversal of S modulo a is:

T = {0} ∪ {kan + a` | k = 0, ..., s− 1 , ` = 1, ..., n}.Thus:(a) is immediate because max T = san.

(b) Recall that g(S) = 1a

∑t∈T t− a−1

2. Now:

∑t∈T

t =s−1∑

k=0

n∑

l=1

(kan + al) =

=s−1∑

k=0

(nkan + nha + dn(n + 1)/2) =

= ns(s− 1)an/2 + nsha + nsd(n + 1)/2 =

=ns

2[(s− 1)an + 2ha + d(n + 1)].

Recalling that ns = a− 1 we get

∑t∈T

t =a− 1

2((s− 1)an + 2ha + d(n + 1)).

Now (s− 1)an = (s− 1)(ha + nd) = sha− ha + (a− 1)d− nd. Thus

∑t∈T

t =a− 1

2((s + 1)ha + ad) =

a(a− 1)(sh + h + d)

2.

Hence the result.

16

The semigroup is symmetric if F (S)+ 1 = 2g(S). Substituting the results ofparts (a) and (b), yields the condition h(a− s− 1) = 0. Since a− 1 = sn weget hs(n− 1) = 0. If s = 0 then a = 1 and the semigroup is N. In the othercases the semigroup is generated by two elements. Indeed, for n − 1 = 0this is clear and if h = 0, then S = S(a, d, 2d, ..., nd) which amounts to thecase in which S is generated by a and d. It is well known that 2-generatednumerical semigroups are indeed symmetric [30].

(c) As usual,

HS(x) =1

1− xa

∑t∈T

xt.

Notice that for a fixed 0 ≤ k < s the elements xkan+a` , ` = 1, ..., n form ageometric sequence (with quotient xd) whose sum is

f(k) = xkan+a11− xnd

1− xd.

Now the elements f(k), k = 0, ..., s− 1 form also a geometric sequence (withquotient xan) whose sum is:

xa1 · 1− xnd

1− xd· 1− xsan

1− xan.

To this we have to add x0 = 1. Hence the result. ¤

Now we turn to the case where a− 1 = sn + r with 0 < r < n.

Corollary 12 Suppose that a− 1 = sn + r and 0 < r < n. Then:(a) F (S) = san + ar − a = (sh + h− 1 + d)a− d.(b) g(S) = 1

2((a− 1 + r)h(s + 1) + (d− 1)(a− 1)).

(c) HS(x) = 11−xa · (1 + xa1( (1−xnd)(1−xsan)

(1−xd)(1−xan )+ xsan 1−xrd

1−xd ).

Proof: In this case then the minimal transversal of S modulo a is:

T = {0} ∪ {kan + al | k = 0, ..., s− 1 , l = 1, ..., n} ∪ {san + al | l = 1, ..., r}.

(a) is immediate because max T = san + ar.

17

(b) again g(S) = 1a

∑t∈T t − a−1

2. Now we saw that the sum of the mid-

dle set in T is:

s−1∑

k=0

n∑

l=1

(kan + al) =1

2ns[(s− 1)an + 2ha + d(n + 1)].

To sum up all the elements in T , we have to add the right set:

r∑

l=1

(san + al) = rsan + rha +1

2r(r + 1)d.

Thus the sum of all the elements of T is:

1

2(ns2an − nsan + 2nsha + n2sd + nsd + 2rsan + 2rha + r(r + 1)d) =

=1

2(san(ns+r)+rsan−nsha−n2sd+2nsha+n2sd+nsd+2rha+r2d+rd) =

(Remembering that sn + r = a− 1)

=1

2(san(a− 1) + rsan + nsha + dns + 2rha + r2d + rd) =

=1

2(sha2 +snda−sha−snd+rsha+rsnd+nsha+dns+2rha+r2d+rd) =

=1

2(sha2 + snda− sha + rsha + rsnd + ha(ns + r) + rha + r2d + rd) =

=1

2(sha2 + snda− sha + rsha + rd(sn + r) + ha2 − ha + rha + rd) =

=1

2(sha2 + snda− sha + rsha + rda− rd + ha2 − ha + rha + rd) =

=a

2(sha + snd− sh + rsh + rd + ha− h + rh) =

=a

2(sh(a− 1 + r) + d(sn + r) + h(a− 1 + r) =

=a

2((a− 1 + r)h(s + 1) + d(a− 1))

Hence the result.

18

The semigroup is symmetric if F (S) + 1 = 2g(S). Substituting the resultsof parts (a) and (b), yields the condition h(r − 1)(s + 1) = 0. Of course,s + 1 6= 0. If h = 0, then, again, the semigroup is generated by a and d(and is symmetric as every numerical semigroup generated by two elements).Thus the only really interesting case is when r = 1 that is a− 1 = sn + 1 ora = 2(mod n).For example: the semigroup S = S(5, 8, 11, 14) in which a = 5, h = 1, n = 3is symmetric. Indeed, writing a−1 = sn+r yields s = 1, r = 1. Substitutingthis in our formulae yields F (S) = 17, g(S) = 9 which can be checked directly.

(c) Again,∑

t∈T xt = 1 +∑s−1

k=0

∑n−1l=1 xkan+al +

∑rl=1 xsan+al . We have al-

ready calculated the sum in the middle. As for the right sum it, it is easy tosee that it is equal to xsan+a1

∑r−1l=0 xld = xsan+a1 · 1−xrd

1−xd . Hence the result. ¤

2.4 Example: Delorme’s construction

Let A = {a1, ..., an}, B = {b1, ..., bm} be two sets of relatively prime positiveintegers. Let S1 = S(A), S2 = S(B) be the numerical semigroups generatedby A and B respectively. Let a ∈ S1, b ∈ S2 be two relatively prime positiveintegers and c = ab. Finally, let S = bS1 + aS2 = {bs1 + as2 | si ∈ Si}. ThenS is generated by ba1, ..., ban, ab1, ..., abm. Since its generators are relativelyprime, S is a numerical semigroup.

The importance of semigroups which are constructed this way was shownby Delorme who proved in [12] that a non trivial numerical semigroup Sis a complete intersection ([17]) if and only if S is constructed in this wayfrom numerical semigroups S1, S2 which are themselves complete intersec-tions. Thus we will refer to this construction as the Delorme’s construction.

Let a1, ..., an, an+1 be relatively prime positive integers such that the se-quence (ai)

n+1i=1 is smooth (cf. Section 2.2). Let d = gcd(a1, ..., an). Hence

dan+1 ∈ S(a1, ..., an). For i = 1, ..., n let bi = ai/d. Thus the numbers bi arerelatively prime and S1 = S(b1, ..., bn) is numerical. It is easy to check thatS = S(a1, ..., an+1) = dS1 + an+1N where an+1 ∈ S1 (and of course d ∈ N).Since the sequence b1, ..., bn is also smooth, we get that S is obtained by

19

successive applications of Delorme’s construction with N factors.

Delorme found the Frobenius number of S (in terms of F (S1), F (S2)) andproved that complete intersections semigroups are symmetric. Here we usethe minimal transversal to compute recursively the Frobenius number andgenus of every semigroup which is built this way (not necessarily completeintersection). In addition we compute the Hilbert series of the Delorme’sconstruction given the Hilbert series of its components.

Theorem 13 Let T1 be the minimal transversal of S1 relative to a and T2

the minimal transversal of S2 relative to b. Then the minimal transversal ofS relative to c is bT1 + aT2.

Proof: For r = 0, ..., a − 1 let kr be the minimal representative of theresidue r modulo a in S1. It is easy to see that bkr, r = 0, ..., a − 1 are theminimal representatives modulo a of bS1. Similarly if lq, q = 0, ..., b− 1 arethe minimal representatives of the residue classes modulo b in S2 then alq areminimal representatives of the residue classes modulo b of aS2.

The set T = {bkr + alq | r = 0, ..., a − 1, q = 0, ..., b − 1} contains ex-actly one element of each residue modulo c by the Chinese Remainder The-orem. Note that T ⊂ S. Moreover we claim that the numbers in T areall minimal representatives modulo c of their classes. Indeed every m ∈ Scan be written as a combination m = bm1 + am2 where mi ∈ Si. Thusm = b(kr + ta)+a(lq +sb) = bkr +alq +(t+s)c for some 0 ≤ r < a, 0 ≤ q < band natural t, s. Hence m is a minimal representative modulo c only ift = s = 0 that is m ∈ T . Since T contains every minimal representativemodulo c and |T | = c it follows that T is the minimal transversal modulo c.¤

Theorem 14 Using the notation above, let S = bS1 + aS2.(a) Let Fi be the Frobenius number of Si then:

F (S) = bF1 + aF2 + c.

(b) Let gi be the genus of Si then:

g(S) = bg1 + ag2 +1

2(a− 1)(b− 1).

20

In particular S is symmetric if and only if S1 and S2 are.

(c) Let Hi be the Hilbert series of Si then:

HS(x) = (1− xc)H1(xb)H2(x

a).

Proof: Let T1 be the minimal transversal of S1 modulo a, T2 be the minimaltransversal of S2 modulo b and T be the minimal transversal of S moduloc = ab. Recall that T = bT1 + aT2. To avoid repeating formulae, we alsodenote a by α1 and b by α2 for b.

(a) As for the Frobenius number, recall that Fi = max Ti − αi. ThusF (S) = max T − c = b max T1 + a max T2 − c = b(F1 + a) + a(F2 + b)− c =bF1 + aF2 + c.

(b) Let hi =∑

t∈Tit and h =

∑t∈T t. Noticing that T1 consists of a ele-

ments and T2 consists of b elements we get: h =∑

s∈T1

∑r∈T2

(bs + ar) =b∑

s∈T1bs + a

∑r∈T2

ar = b2h1 + a2h2.

Now, by the formula of the genus hi = αigi + αi(αi−1)2

for i = 1, 2. Henceh = b2a(g1 + a−1

2) + a2b(g2 + b−1

2).

Finally, g(S) = 1ch− c−1

2= bg1 + ag2 + 1

2(b(a− 1) + a(b− 1)− ab + 1).

Now b(a − 1) + a(b − 1) − ab + 1 = ab − b − a + 1 = (a − 1)(b − 1), henceg(S) = bg1 + ag2 + 1

2(a− 1)(b− 1).

As for the symmetry of the semigroup, 2g = 2bg1 + 2ag2 + c − a − b + 1 ≥b(F1 + 1) + a(F2 + 1) + c− a− b + 1 = bF1 + aF2 + c + 1 = F (S) + 1.Equality holds if and only if 2gi = Fi + 1 for i = 1, 2. Hence S is symmetricif and only if both S1 and S2 are. It follows easily that complete intersectionnumerical semigroups are symmetric.

(c) Finally, as to the Hilbert series of S. Denote Hi(x) = HSi(x). No-

tice that∑

t∈Tixt = (1 − xαi)Hi(x), i = 1, 2. Again working with T (the

minimal transversal of S modulo c) we get that

HS(x) =1

1− xc

∑t∈T

xt =1

1− xc

∑t1∈T1

∑t2∈T2

xbt1+at2 =1

1− xc(∑t1∈T1

xbt1)(∑t2∈T2

xat2) =

21

1

1− xc(1− xab)H1(x

b)(1− xba)H2(xa) =

(1− xc)H1(xb)H2(x

a).

This completes the proof. ¤

Example 1:Let a, b be relatively prime positive integers. Then S(a, b) = bN + aN isclearly constructed this way. Since F (N) = −1 and g(N) = 0 we get thatF (a, b) = bF (N)+aF (N)+ab = ab−a−b and g(S) = 1

2(a−1)(b−1). Hence

the semigroup is symmetric and we have proved Sylvester’s result. Finally,the Hilbert series of N is

∑∞k=0 xk = 1

1−x. Hence the Hilbert series of S(a, b)

is1− xab

(1− xa)(1− xb).

Example 2:Let a1, a2 and b1, b2 be two pairs of relatively prime integers. Let a ∈S(a1, a2) and b ∈ S(b1, b2) be two relatively prime integers. Then S =S(ba1, ba2, ab1, ab2) is a complete intersection and F (S) = bF (S(a1, a2)) +aF (S(b1, b2)) + ab = b(a1a2 − a1 − a2) + a(b1b2 − b1 − b2) + ab.For specific example we take the numbers from Delorme’s paper. Let a =10, b = 15, c = 14, d = 21. Here b = 5, a1 = 2, a2 = 3 and a = 7, b1 = 2, b2 = 3and indeed a ∈ S(a1, a2) and b ∈ S(b1, b2). Then

F (10, 15, 14, 21) = 5 · 1 + 7 · 1 + 35 = 47.

The Hilbert series of S(2, 3) is 1−x6

(1−x2)(1−x3)(Example 1 above). It follows that

the Hilbert series for S(10, 15, 14, 21) is

HS(x) = (1− x35) · 1− x30

(1− x10)(1− x15)· 1− x42

(1− x14)(1− x21).

2.4.1 Minimal Generation

Let again S1 = S(a1, ..., an), S2 = S(b1, ..., bm) be numerical semigroups withboth generating sets minimal, i.e. no proper subset generates its semigroup.Let a ∈ S1 and b ∈ S2 be relatively prime.

22

Claim 15 If a /∈ {a1, ..., an} and b /∈ {b1, ..., bm} then

C = {ba1, ..., ban, ab1, ..., abm}

is a minimal generating set for the semigroup S = bS1 + aS2.

Proof: It is clear that C generates S. We show that it is minimal. Indeedif this is not the case, then one of these numbers can be written as a naturalexpression of the rest. Say

bak =∑

i6=k

αibai +∑

j

βjabj,

where αi, βj ∈ N. Dividing by b we get

ak =∑

i 6=k

αiai + a

∑j βjbj

b,

where∑

j βjbj

bis an integer since b is relatively prime to a. But the only

presentation of ak as a natural combination of elements from S(a1, ..., an) isak = ak (by the minimality of {a1, ..., an}). Thus all αi must be zero, a = ak

and∑

j βjbj

b= 1. However this possibility was excluded. ¤

23

Chapter 3

Bounds

3.1 Bounding the Frobenius number and the

genus of a numerical semigroup

In this chapter we use ”bounds” for the minimal transversal to give lower andupper bounds for the Frobenius number and genus of a numerical semigroup.

Let x = (xi), y = (yi) be two finite sequences of real numbers of the samelength. We say that y majors x and write x ≤ y if xi ≤ yi for every i.

Theorem 16 Let S be a numerical semigroup. Let s ∈ S be an elementof S. Let T be the minimal transversal of S modulo s and let t = (ti) bea sequence of the elements of T in some order (thus t is a sequence of selements).Let r = (ri),u = (ui) be two other sequences of real numbers of length s suchthat r ≤ t ≤ u. Then:(a)

max {ri} − s ≤ F (S) ≤ max {ui} − s

(b)

1

s

s∑i=1

ri − s− 1

2≤ g(S) ≤ 1

s

s∑i=1

ui − s− 1

2.

Proof: The theorem follows easily from Theorem 5 and the following facts:(a) F (S) = max {ti} − s and max {ri} ≤ max {ti} ≤ max {ui}.

24

(b) g(S) = 1s

∑ti − s−1

2and ri ≤ ti ≤ ui, for every i = 1, ..., s. ¤

Remark From the proof we get that if u 6= t then g(S) < 1s

∑ui − s−1

2

(with strict inequality). However, we may have F (S) = max {ui} − s evenif u 6= t (because their maximal element may still be the same). Analogousresults hold for r.

In particular we get the following: For a positive integer s, a set L ⊆ Nis called a transversal of s if for every residue k = 0, ..., s− 1 there is exactlyone element `k ∈ L such that `k ≡ k( mod s). We call `k the representativeof the residue k in L.

Let s be an element of S. If U ⊆ S is a transversal of s such that all of itselements belong to the semigroup S, then U majors the minimal transversalT (that is there are orderings t = (ti),u = (ui) of T, U respectively such thatti ≤ ui for every i). Indeed, let tk, uk be the representatives of the residue kin T, U respectively. Then clearly tk ≤ uk for every k = 0, ..., s−1. Similarly,if R is a transversal in which the representatives belong either to T or toN\S then T majors R. Hence we get:

Corollary 17 The result of the previous theorem apply whenever we take uto be a transversal of s which is composed of elements of S and r a transversalwhich is composed of elements in T ∪ (N\S).

3.2 Upper bounds

3.2.1 The Schur-Brauer bound

In [6] Brauer gave a bound for the Frobenius number of S which is due toSchur. Here we give a new proof for this bound and use this proof to boundthe genus of S too. We start with the following lemma.

Lemma 18 Let a1 < a2 < ... < an be relatively prime positive integers. LetS = S(a1, ..., an) be the numerical semigroup they generate. Let T be theminimal transversal of a1 in S. Let to < t1 < ...ta1−1 be the sequence ofelements of T arranged by size. Then tk ≤ kan for every k = 0, ..., a1 − 1.

25

Proof: By induction on k. First, t0 = 0 = 0an. Now, assume that ti ≤ ian

for every i < k. We show that tk < kan.Since 0 < tk ∈ S, it is a natural combination of the ai’s. Hence we canwrite tk = aj + m for some 1 ≤ j ≤ n and m ∈ S. We claim that m ∈ T .Indeed, m = tr + la1 for some 1 ≤ r ≤ n and l ≥ 0. Thus tk = aj + tr + la1.Since tk is minimal in its coset and aj + tr ∈ S we must have l = 0 andm = tr where r ≤ k − 1 since m < tk. Thus, using the induction hypothesistk = aj + tr ≤ an + ran ≤ an + (k − 1)an = kan as we wanted to prove. ¤

Theorem 19 Let S = S(a1, ..., an) be a numerical semigroup where a1 <a2 < ... < an. Then:(a)

F (S) ≤ a1an − a1 − an

(b)

g(S) ≤ (an − 1)(a1 − 1)

2

where both equalities hold if and only if n = 2 (that is S is generated by twoelements).

Proof: The theorem is a simple corollary of the preceding lemma andTheorem 16. We use the sequence u = (ui) = (ian) where i = 0, ..., a1 − 1.Then:(a) F (S) ≤ max {ui} − a1 = (a1 − 1)an − a1.(b)

g(S) ≤ 1

a1

∑ui − a1 − 1

2=

1

a1

a1−1∑j=0

jan − a1 − 1

2=

1

a1

· a1(a1 − 1)an

2− a1 − 1

2=

(a1 − 1)(an − 1)

2.

By the remark after Theorem 16, the equality for the genus holds if and onlyif u is the minimal transversal of a1. This happens if and only if a2, ..., an−1

all belong to S(a1, an). (Indeed, take any aj. The minimal element of itscoset is of the form man for some 0 ≤ m < a1. Hence aj = man + la1 forsome l ≥ 0 as claimed). Thus the bound for the genus is tight if and only ifS is generated by 2 elements. But g(S) ≥ (F (S) + 1)/2. Hence equality inpart (a) implies equality in part (b). On the other direction, it is well knownthat for 2-generated semigroups both equalities hold ([30]). ¤

26

3.2.2 The bound of Nijenhuis and Wilf

Let (a1, a2, ..., an) be a sequence of relatively prime positive integers. For ev-ery k = 1, ..., n , let dk = gcd(a1, ..., ak). Since a1, ..., an are relatively prime,dn = 1. Let ck = dk−1/dk where k = 2, ..., n. We note, for further use thatcn = dn−1 and

∏ki=2 ci =

∏ki=2

di−1

di= d1

dk= a1

dkin particular

∏ni=2 ci = a1

1= a1.

Let Sk = S(a1, ..., ak). Recall that the sequence (a1, a2, ..., an) is smoothif ckak ∈ Sk−1 for every k = 2, ..., n (Section 2.2).

In [6] Brauer proved that F (S) ≤ ∑ni=2 ciai −

∑ni=1 ai and in [7] it was

proved that equality holds if and only if the sequence (ai)ni=1 is smooth. In

[23] this was reproved by showing that g(S) ≤ 12(∑n

i=2 ciai −∑n

i=1 ai + 1)with equality, again, only for smooth sequences of generators.

Here we give a simple proof of these results.

Theorem 20 In the notation above

(a)

F (S) ≤n∑

i=2

ciai −n∑

i=1

ai

(b)

g(S) ≤ 1

2(

n∑i=2

ciai −n∑

i=1

ai + 1).

in both cases equality holds if and only if the sequence of generators is smooth.

Proof: We claim that the set U = {∑ni=2 kiai | 0 ≤ ki < ci} is a transversal

of S with respect to a1. We will show this by noticing that every choice of ki

yields a number of different residue modulo a1 (and in particular a differentnumber). Hence U contains

∏ni=2 ci = a1 elements, no two of those have the

same residue modulo a1. Thus it is indeed a transversal.

Assume that∑n

i=2 kiai =∑n

i=2 liai (mod a1) where 0 ≤ ki, li < ci. Weclaim that ki = li for every i = 2, ..., n. Indeed, if this is not the case thenthere is an index i such that ki 6= li. Let j be the maximal index of thissort. Since dj−1|a1 we have

∑ni=2 kiai =

∑ni=2 liai (mod dj−1). Now dj−1|ai

27

for every i = 1, ..., j − 1, hence∑n

i=j kiai =∑n

i=j liai (mod dj−1). By ourassumption ki = li for every i > j hence kjaj = ljaj (mod dj−1). If dj = dj−1,cj = 1 and kj = lj = 0. Thus we have dj < dj−1. Dividing the last equalityby dj we get kj

aj

dj= lj

aj

dj(mod cj) where

aj

djis relatively prime to cj. Hence

kj = lj (mod cj). But 0 ≤ kj, lj < cj hence kj = lj in contradiction to ourassumption. Hence ki = li for every i = 2, ..., n and U is indeed a transversalmodulo a1.

Let T be the minimal transversal of S modulo a1. Since U is composedof elements from S, U majors T . Thus we have:(a) F (S) ≤ max U − a1 =

∑ni=2 (ci − 1)ai − a1 =

∑ni=2 ciai −

∑ni=1 ai.

(b) g(S) ≤ 1a1

∑u∈U u− a1−1

2. Now

∑u∈U

u =

c2−1∑

k2=0

c3−1∑

k3=0

...

cn−1∑

kn=0

(k2a2 + k3a3 + ... + knan) =

= C2

c2−1∑

k2=0

k2a2 + C3

c3−1∑

k3=0

k3a3 + ... + Cn

cn−1∑

kn=0

knan

where Ci =∏

j 6=i cj =∏n

j=2 cj/ci = a1

ci

Thus∑

u∈U u =∑n

i=2 [12Cici(ci − 1)ai] = 1

2a1

∑ni=2 (ci − 1)ai and the genus

of S satisfies

g(S) ≤ 1

a1

∑u∈U

u−a1 − 1

2=

1

2(

n∑i=2

(ci − 1)ai−(a1−1)) =1

2(

n∑i=2

ciai−n∑

i=1

ai+1).

Finally, we claim that U is the minimal transversal if and only if the sequence(a1, ..., an) is smooth. Indeed, if (a1, ..., an) is smooth, then every element ofS can be presented uniquely as a combination k1a1 +

∑ni=2 kiai with k1 ≥ 0

and 0 ≤ ki < ci. (Every s ∈ S can be written as a combination∑n

i=1 liai

with li ≥ 0. If lj ≥ cj we can decrease it by cj without changing the totalsum s by increasing only coefficients with smaller index, hence we get thewanted result). In particular if t ∈ S is a minimal representative modulo a1,we must have k1 = 0 and t =

∑ni=2 kiai for some 0 ≤ ki < ci. Hence t ∈ U

and U is indeed the minimal transversal.

In the other direction, suppose that U is the minimal transversal. Let

28

2 ≤ j ≤ n be some index. The number cjaj is divisible by dj−1 (sincecj = dj−1/dj and dj|aj). Now every element in U of the form

∑j−1i=2 kiai where

0 ≤ ki < ci, i = 2, ..., j − 1 is also divisible by dj−1. Notice that each choiceof ki yields a different residue modulo a1 and there are

∏j−1i=2 ci = a1/dj−1

such choices. Now there are a1/dj−1 residues modulo a1 which are divisibleby dj−1. Since cjaj is of one of these residues, we get that the minimal repre-sentative of cjaj is of the form

∑j−1i=2 kiai and cjaj ∈ Sj−1. Thus the sequence

(a1, ..., an) is smooth.

The bound for the genus is strict if and only if U is the minimal transver-sal modulo a1 thus if and only if (ai) is smooth. Also if this holds, thebound for the Frobenius number is tight too. Finally, if the bound forthe Frobenius number is tight, then so is the bound for the genus (sinceg(S) ≥ (F (S) + 1)/2) hence a1, ..., an is smooth. ¤

3.3 Lower bounds

We conclude this chapter with a new lower bound for the genus. We startwith a general combinatorial lemma.

Lemma 21∑k

i=0

(m + i− 1

m− 1

)=

(m + k

m

)

Proof: The number of ways to throw i identical balls into m cells is(m + i− 1

m− 1

). Hence the sum on the left side is the number of ways to

throw at most k identical balls into m cells. But this is equivalent to throw-ing exactly k identical balls into m+1 cells and ignoring the last cell. Hencethe result. ¤

Let A = {a, a1, ..., an} be a set of n + 1 relatively prime positive integersand assume that a1 < ai for every i > 1 (we do not care whether a1 < a ora1 > a). Let S = S(A) be the numerical semigroup generated by A. LetT = {ti | i = 1, ..., a} be the minimal transversal of S modulo a. We assume0 = t1 < t2 < t3 < ... < ta.

Lemma 22 Let `p be the minimal number such that

(n + `p

n

)≥ p. Let

rp = `pa1. Then rp ≤ tp.

29

Proof: Every element of T is of the form∑n

i=1 kiai for some ki ∈ N. Writetq =

∑ni=1 kqiai and call the tuple vq = (kq1, ..., kqn) the corresponding tuple

of tq. Note that tq may have more than one presentation of the form∑n

i=1 kiai

in which case we choose one for its corresponding tuple.

The number of natural tuples (k1, ..., kn) with∑n

i=1 ki = j is

(n + j − 1

n− 1

)

(throwing j identical balls into n cells) hence the number of tuples (k1, ..., kn)with

∑ni=1 ki < `p is

∑`p−1j=0

(n + j − 1

n− 1

)=

(n + `p − 1

n

)(by Lemma 21) and this number is

strictly smaller than p (by the definition of `p). Since the correspondencebetween the elements of T and their corresponding tuples is injective, theremust be an index q ≤ p such that vq = (kq1, kq2, ..., kqn) with

∑ni=1 kqi ≥ `p.

Hence tp ≥ tq =∑n

i=1 kqiai ≥∑n

i=1 kqia1 ≥ `pa1. ¤Theorem 23 In the notation above(a)

F (S) ≥ `aa1 − a

(b)

g(S) ≥ a1

a[n

(n + `a − 1

n + 1

)+ (a−

(n + `a − 1

n

))`a]− a− 1

2

Proof: (a) This follows easily from F (S) = max T − a = ta − a ≥ ra − a.(b) The heart of the proof is that we ”know” the sequence (`p). Since(

n + `p

n

)=

∑`p

j=0

(n + j − 1

n− 1

)we get that the sequence `p is composed

(according to order) of

(n + j − 1

n− 1

)elements which equal j. Thus `1 = 0.

Then the next n elements equal 1 and they are followed by 12n(n+1) elements

which equal 2 and so forth. Our goal is to compute∑a

p=1 `p.

Let L =∑`a−1

j=0

(n + j − 1

n− 1

)=

(n + `a − 1

n

). Since L < a (by the defini-

tion of `a), we have∑a

p=1 `p =∑L

p=1 `p+∑a

p=L+1 `p =∑`a−1

j=0 [j

(n + j − 1

n− 1

)]+

(a− L)`a. Now

`a−1∑j=0

j

(n + j − 1

n− 1

)=

`a−1∑j=1

j

(n + j − 1

n− 1

)=

`a−1∑j=1

n

(n + j − 1

n

)=

30

`a−2∑

k=0

n

(n + k

n

)= n

(n + `a − 1

n + 1

).

where the last equality follows again from Lemma 21 with m = n + 1.

Hence∑a

p=1 rp = a1(n

(n + `a − 1

n + 1

)+ (a − L)`a). The result now fol-

lows from g(S) ≥ 1a

∑ap=1 rp − a−1

2. ¤

Example: Let A = {α, β} be a set of two elements (hence n = 1). Choose

a = α then a1 = β. The minimal index `a such that

(1 + `a

1

)= `a +1 ≥ α

is `a = α− 1.Thus we have F (S) ≥ (α − 1)β − α (Indeed it is well known that equalityholds) and

g(S) ≥ βα[

(α− 1

2

)+ (α−

(α− 1

1

))(α− 1)]− α−1

2= (α−1)(β−1)

2which is

tight too. Notice that in this case we could have satisfied with the ”trivial”bound for the genus g(S) ≥ (M + 1)/2 where M is the lower bound for theFrobenius number. However, in general, our formula improves this ”trivial”bound.

Example: Let A = {20, 54, 81, 105, 187} where n = 4. Direct computa-tion gives F (A) = 298 and g(A) = 177. If we choose a = 20 we have a1 = 54and `a = 3. Hence we get the estimates F (A) ≥ 142 and g(A) ≥ 96. Wecould have chosen a = 54 (in which case a1 = 20 and `a = 4) but this choiceleads to much poorer estimates.

Example: Let A = {30, 4074, 6790, 10185} where n = 3. Let a = 30.Thus a1 = 4074 and `a = 4. Direct computation gives F (A) = 40, 061 andg(A) = 20, 031. Our formulae yield F (A) ≥ 16, 266 and g(A) ≥ 7455.

We remark that there are cases in which the formulae yield quite poor lowerbounds (for example if A forms an arithmetic progression, where every ele-ment of T has many corresponding tuples). We believe that by taking theseoverlaps into consideration one will be able to achieve more accurate bounds.

31

Chapter 4

Presentation of the minimaltransversal

4.1 Canonical domains

We present a general solution to the Frobenius problem of numerical semi-groups using a presentation of the minimal transversal. Although this methodis mostly theoretical, it can be used effectively for semigroups generated by2 or 3 elements as well as in some other cases.

4.1.1 The presentation

Let Fn = Nn be the free commutative semigroup on n generators. Lete1, ..., en be the unit vectors of Fn. We use the same letter for a tuple andits components (thus x = (x1, ..., xn), y = (y1, ..., yn) etc.).

We define a partial ordering on Fn in the following way: x < y if xi ≤ yi forevery i = 1, ..., n, with at least one strict inequality. Alternatively, x ≥ y ifand only if x− y ∈ Fn.

The support of a tuple x is the set supp(x) of indices i for which xi > 0. Ify < x then supp(y) ⊆ supp(x).

The lexicographic order on Fn is defined as follows: For every elementx ∈ Fn let the k’th beginning of x be the tuple x(k) = (x1, ..., xk). We say

32

that x <L y if there is a number 1 ≤ k ≤ n such that x(k) < y(k) (as tuplesin Nk). Thus x <L y if there is an index 1 ≤ k ≤ n such that x1 = y1,x2 = y2, ... , xk−1 = yk−1 and xk < yk.

We note that <L forms a well ordering on Fn which preserves addition (thatis if x1 <L y1 and x2 <L y2 then x1 + x2 <L y1 + y2). We also note that ifx < y then x <L y. However x <L y implies only that either x < y or xand y are incomparable by <.

Let S be a numerical semigroup and A = {a0, a1, ..., an} be a generatingset of S. Let T = T (a0) be the minimal transversal of S modulo a0. Noticethat A may be any generating set of S, hence a0 can be any element of thesemigroup. However, usually we choose A to be minimal (that is none of theai’s is a natural combination of the others).

Let f : Fn → S be the semigroup homomorphism defined by f(x) =∑ni=1 xiai. Let S1 = Imf . We claim that T ⊂ S1. Indeed, every element

t ∈ S can be written as a combination∑n

i=0 kiai. If k0 6= 0 then t = s + k0a0

where s =∑n

k=1 kiai ∈ S and t is not minimal in its class. Hence t ∈ T iffor every natural combination t =

∑ni=0 kiai we have k0 = 0. In particular

t ∈ S1. Finally since T is finite and S1 is not, T ⊂ S1 (with a strict inclusion).

If f(x) = s, we say that x is a source of s. Notice that an element s ∈ S1

may have more than one source. Our goal is to describe a set C ⊂ Fn whichis mapped bijectively by f onto T . The most important property of C willbe that is downwards closed with respect to <. That is if x ∈ C and y < xthen y ∈ C too.

The presentation f defines a natural equivalence relation on Fn: two tu-ples u,v ∈ Fn are equivalent (written u ∼ v) if f(u) = f(v). This relationis actually a congruence: If u1 ∼ v1 and u2 ∼ v2 then u1 + u2 ∼ v1 + v2.

Definition 24 An element x ∈ Fn is canonic if(a) f(x) ∈ T and(b) x ≥L y for every y ∼ x in Fn.

Saying it differently, an element x ∈ Fn is canonic if and only if x is the max-imal element (with respect to <L) in f−1(t) for some t ∈ T . (This maximum

33

is well defined since f−1(s) is finite for every s ∈ S). If x = (x1, ..., xn) is thecanonical source of some t ∈ T we write t :=

∑ni=1 xiai and call the sum

on the right the canonical presentation of t.

Let C = C(T ) be the set of all canonical sources of T . Then, by defini-tion, f maps C bijectively onto T . In addition we claim the following:

Lemma 25 The set C is downwards closed with respect to <.

Proof: Let x ∈ C and y < x. We want to show that y ∈ C too. First weshow that f(y) ∈ T . Indeed f(y) = t + ka0 for some t ∈ T and k ≥ 0. Sincex > y we have x− y ∈ Fn. Thus f(x) = f(y)+f(x− y) = (t+f(x− y))+ka0 where t + f(x− y) ∈ S. But f(x) ∈ T hence k = 0. Thus f(y) = t ∈ Ttoo.We show that y is the canonical source of t. Indeed, suppose there was anelement z >L y in Fn such that f(z) = f(y). Let w = z + (x − y). Thenw ∈ Fn (since x− y ∈ Fn), w >L y + (x − y) = x and f(w) = f(x) (sincez ∼ y), in contradiction to the (maximality part of the) canonicity of x.Thus y is maximal in f−1(t) and y ∈ C. ¤

Definition 26 The set C is called the canonical domain of a0.

4.1.2 Bounds for C

Next, we describe the set C in terms of its (strict) bounds in Fn.

Definition 27 An element v ∈ Fn is a bound for C if v /∈ C(T ) but forevery u < v , u ∈ C(T ) (thus v is minimal in Fn\C with respect to <). Theset of all bounds of C in Fn is denoted by B = B(C).

Let a ∈ N be a natural number. We write a + S for the set {a + s | s ∈ S}.An element s ∈ S belongs to aj + S for some 0 ≤ j ≤ n if and only if thereis a natural combination s =

∑ni=0 kiai with kj > 0. In particular if in the

tuple v = (v0, ..., vn) we have vj > 0, then f(v) ∈ aj + S (however f(v) maybelong to aj +S even if vj = 0 since it may have other natural combinations).Finally, an element s ∈ S belongs to T = T (a0) if and only if s /∈ a0 + S.

The following is almost obvious:

34

Lemma 28 Let v ∈ Fn be an element with support I. If there is an indexj < min I such that f(v) ∈ aj + S then v /∈ C .

Proof: If f(v) ∈ a0 +S then clearly f(v) /∈ T . Otherwise f(v) = aj + s forsome 0 < j < min I and s =

∑ni=1 kiai. Let u = (u1, ..., un) where ui = ki

for i 6= j and uj = kj + 1. Then u is another source in of f(v) in Fn andu >L v. Thus again v is not canonic. ¤

Lemma 29 Let v = (v1, ..., vn) ∈ B be a bound for C and let j be theminimal index such that vj 6= 0. Then there is a tuple u = (u0, ..., un) ∈ Nn+1

such that:(a) ui > 0 for some i < j.(b) For every i = 1, ..., n either vi = 0 or ui = 0.(c)

∑ni=1 viai =

∑ni=0 uiai.

Proof: Since v is a bound, v does not belong to C(T ). Hence eitherf(v) /∈ T or f(v) = t for some t ∈ T but v is not the canonical source of tin Fn. We divide into cases:If f(v) /∈ T then f(v) = t + u0a0 for some t ∈ T and u0 > 0. Writingt =

∑ni=1 uiai yields

∑ni=1 viai =

∑ni=0 uiai. Thus conditions (a) and (c)

hold. If there was an index 0 < k such that both vk > 0 and uk > 0 we hadv − ek ∈ Fn and f(v − ek) =

∑i6=k viai+(vk−1)ak =

∑i6=k uiai+(uk−1)ak ∈

a0 + S. Thus f(v − ek) /∈ T and v − ek /∈ C contradicting the minimality ofv in Fn\C(T ) (with respect to <), since v − ek < v. Hence part (b).If on the other hand f(v) = t belongs to T (a0), then every source off(v) belongs to Fn. Since v is not canonic, there is another source uof t in Fn such that u >L v (hence condition (c)). If there is an indexk ∈ supp(v) ∩ supp(u) then v − ek <L u− ek both belong to Fn andf(v − ek) = f(u− ek) = t − ak. Hence v − ek is not the canonical sourceof t − ak, and v − ek /∈ C in contradiction again to the minimality of v inFn\C(T ) (with respect to <). Hence the supports are disjoint. Condition(a) now follows from u >L v. ¤

We conclude this discussion with some bounds for C that always exist:

Lemma 30 For every k = 1, ..., n there is exactly one bound of the formbk = mkek. In this bound mk is the minimal positive integer such thatmkak = ai + S for some i < k. In particular, B is finite.

35

Proof: By Lemma 28 mkek does not belong to C. Hence we need to showthat for every ` < m we have `ek ∈ C. Indeed in every natural combination`ak =

∑ni=0 uiai we must have u0 = u1 = uk−1 = 0 (if ui > 0 for some

i < k then `ak ∈ ai + S contradicting the minimality of mk) and uk ≤ `with equality if and only if ui = 0 for every i > k (because all the ui’s arenon negative). Hence `ak ∈ T and `ek is its canonical source. Thus `ek ∈ Cwhich completes the proof. ¤Lemma 31 There is exactly one bound b = (b1, ..., bn) such that f(b) is amultiple of a0. In this case b is the maximal element with respect to <L

in f−1(m0a0) where m0 is the minimal positive integer such that m0a0 ∈S(a1, ..., an).

Proof: Let m0 be the least positive integer such that m0a0 ∈ S(a1, ..., an)and let b be its maximal source in Fn. Then b /∈ C (since m0a0 /∈ T ). In orderto prove that b is a bound we need to show that every u =

∑ni=1 uiei < b

belongs to C. We do this by reductio ad absurdum. Suppose that u /∈ C,then there are two possibilities: f(u) /∈ T or f(u) ∈ T but u is not thecanonical source of f(u).If f(u) /∈ T , then f(u) = t + ka0 for some t ∈ T and 0 < k < m0 (noticethe strict inequality in k < m0 since f(u) < f(b) = m0a0). Let t be thecanonical source of t in C. Since b > u we have r = t + (b− u) ∈ Fn.But f(r) = t + m0a0 − (ka0 + t) = (m0 − k)a0 where 0 < m0 − k < m0

in contradiction to the minimality of m0 (since we got that (m0 − k)a0 =f(r) =

∑ni=1 riai ∈ S(a1, ..., an)). Hence f(u) ∈ T . As for the maximality

of u, had there been v >L u such that f(v) = f(u) we would have anelement w = v + (b− u) such that w >L u + (b− u) = b and f(w) = f(b)contradicting the <L-maximality of b in f−1(m0a0). Hence u is <L-maximaltoo and u ∈ C.Finally we prove the uniqueness of b. Let z ∈ Fn be a bound of C suchthat f(z) = `a0 (notice ` ≥ m0). We will prove that z = b. Let z− b =∑n

i=1 (zi − bi)ei. Let I ⊆ {1, ..., n} be the set of indices i such that bi > zi

and J ⊆ {1, ..., n} be the set of indices j such that zj > bj. Then I and J aremutually disjoint. Let u =

∑i∈I (bi − zi)ei and w =

∑j∈J (zj − bj)ej. Then

u,w ∈ Fn, u < b and w < z. In particular u,w both belong to C (since band z are bounds). Now w−u = z−b hence f(w) = f(u)+(`−m0)a0. Sincef(w) ∈ T we deduce that `−m0 = 0 hence f(w) = f(u). Now u,w ∈ C areboth canonic and have the same image. Hence u = w. Since supp(u) = Iand supp(w) = J are disjoint sets, it follows that u = w = 0 and z = b. ¤

36

Corollary 32 There is at most one bound v ∈ B such that supp(v) ={1, ..., n}.

Proof: For every bound v ∈ B such that supp(v) = {1, ..., n} we have∑ni=1 viai = u0a0 (by Lemma 29). By Lemma 31, there is at most one such

bound. Moreover, this bound, if it exists, equals b (but it may happen thatsupp(b) ⊂ {1, ..., n} in which case there is no such bound). ¤

4.1.3 The method

We now explain how to find the Frobenius number of the semigroup S =S(a0, ..., an).Let T be the minimal transversal of S modulo a0 and C = C(T ) be the cor-responding canonical domain. Let f : Fn → S be the map f(x) =

∑ni=1 xiai.

step 1: collecting the bounds

We start by finding B, the set of bounds of C. Let v ∈ Fn be an tuple in Fn

and let I = supp(v). Then v may be a bound for C only if f(v) =∑

i∈I viai

can be written as a natural combination∑

j /∈I kjaj with kj > 0 for somej < min I (by Lemma 29). Also no two bounds are <-comparable (sinceevery bound is minimal in Fn\C). We remark that the bounds bi, i = 1, ..., nlimit the magnitude of the components of v.

step 2: Finding maximal elements in C

Let v ∈ B be a bound of C. If x ∈ Fn satisfies x ≥ v then clearly x /∈ C(since C is downwards closed). Thus for every bound v = (v1, ..., vn) ∈ B wehave the following condition:

For every x = (x1, ..., xn) ∈ C we have

(x1 < v1) or (x2 < v2) or ... or (xn < vn).

We call this condition the condition induced by the bound v and denote itby α(v). If vi = 0 then the condition (xi < 0) is always false, thus indices ithat do not belong to supp(v) can be omitted from α(v).

37

We note that every x /∈ C satisfies x ≤ v for some v ∈ B (since B isthe set of all minimal elements in Fn\C), hence every x /∈ C violates at leastone of the conditions α(v).

Hence x ∈ C if and only if x satisfies the conditions α(v) for every v ∈ B.Thus an element s ∈ S is a minimal representative modulo a0 if and only ifit can be written as a combination s =

∑ni=1 xiai where the coefficients tuple

x = (x1, ..., xn) ∈ Fn fulfills the conditions α(v) for every v ∈ B.

step 3: Finding the maximal element of T

In order to find F (S) we need only to find m = max T . Clearly, if m is thecanonical source of m then m is maximal in C with respect to <. Hence forevery k = 1, ..., n we have m+ek ≥ vk for some bound vk ∈ B. Let M be theset of the maximal elements in C with respect to <. Then m = max(ImM).

When the structure of C is simple, one can also compute the genus of thesemigroup and its Hilbert series. However, the main problem of the methodis that C can become rather complicated. If this is the case, this task as wellas finding B and M become quite difficult.

4.2 Semigroups with small generating set

4.2.1 Two generators

The Frobenius number of numerical semigroups generated by two elementsis known at least from the 19th century. Sylvester and Curran Sharp [31]proved that such semigroups are symmetric. During the years more proofsfor these statements were found (For some of them see [25]).

Let a0 = a, a1 = b be two relatively prime positive integers and S = S(a, b).Let f : F1 → S be the map f(x) = xb. Let T = T (a) be the minimaltransversal of S modulo a and C be the canonical domain of a.

In this case b1 = m1e1 where m1 is the minimal positive number such thatm1b = ua. Thus m1 = a (since a and b are relatively prime). By Corollary

38

32, there are no other bounds in B (in particular b = b1).

Thus C = {xe1 | x < a} and T = {xb | x = 0, ..., a − 1}. Hencemax T = (a − 1)b. Recalling that F (S) = max T − a we get the classicresult

F (S) = ab− a− b.

Genus of two generators

Recall that

g(S) =1

a

∑t∈T

t− a− 1

2.

Now,∑a−1

t=0 tb = 12(a−1)ab. Hence we get g(S) = (a−1)b

2− a−1

2= 1

2(a−1)(b−

1) = 12(F (S) + 1). Thus

Corollary 33 Every numerical semigroup which is generated by two ele-ments is symmetric.

Hilbert series of two generators

The Hilbert series of S is

HS(x) =1

1− xa

∑t∈T

xt.

Now,∑a−1

t=0 xtb = 1−xab

1−xb . Hence we get

HS(x) =1− xab

(1− xa)(1− xb).

39

4.2.2 Three generators

The problem for numerical semigroups generated by 3 elements was appar-ently first solved by Brower and Shockly [8]. During the years more solutionsof different forms were found by different methods (See for example [28], [16]and [15]). Denham [13] found the Hilbert series for these semigroups usingthe free resolution over the polynomial ring. Herzog [17] showed that thesesemigroups can be presented using two or three relators. Our method givesnew elementary proofs for these results.

Let a0 = a, a1 = b, a2 = c be three relatively prime positive integers. LetS = S(a, b, c). Let f : F2 → S be the map f(x, y) = xb + yc. Let T = T (a)be the minimal transversal of S modulo a and C = C(T ) be the canonicaldomain of a0.

We start with the bounds bi for i = 1, 2.

b1 = se1 where s is the minimal positive number such that sa1 ∈ a0 + S.By Lemma 29 we get that s is the minimal integer for which the equationsb = s1a + s2c has a solution in natural numbers with s1 > 0.

Similarly, b2 = te2 where t is the minimal integer such that ta2 ∈ ai + S forsome i < 2 (or, equivalently, the equation tc = t1a + t2c has a solution innatural numbers not both of them zero).

So far we have found two bounds for C namely (s, 0) and (0, t). Hence if(x, y) ∈ C(T ) then x < s and y < t. Equivalently, every element t ∈ T has anatural combination t = xb + yc with x < s and y < t.

We move on to the bound b. let r be the minimal positive integers suchthat ra ∈ S(b, c). Let (p, q) be the maximal source of ra in F2 with respectto the lexicographic order. Then b = (p, q) is also a bound for C and every(x, y) ∈ C satisfies (x < p) or (y < q). By Corollary 32 there are no otherbounds for C.

Since (p, q) is a bound for C(T ) it is not bigger than (s, 0), (0, t). Hencethere are two cases:

40

The case of two bounds

First there is the possibility that b = b1 or b = b2.

Here T = {xb+ yc | 0 ≤ x < s, 0 ≤ y < t}. Thus max T = (s− 1)b+(t− 1)cand

F (S) = max T − a = sb + tc− a− b− c.

genus

Now

z =∑τ∈T

τ =s−1∑

k=0

t−1∑

l=0

kb + lc = t

s−1∑

k=0

kb + s

t−1∑

l=0

lc =(s− 1)s

2tb +

(t− 1)t

2sc =

st

2(sb + tc− b− c).

and the genus is:

g(S) =z

a− a− 1

2.

Every choice of 0 ≤ k < s, 0 ≤ l < t yields a different representative kb+ lc ∈T . Since T contains exactly a elements, it follows that st = a and

g(S) =1

2(sb + tc− a− b− c + 1).

Since F (S) = sb + tc− a− b− c we see that S is symmetric.

Hilbert series

Finally,∑τ∈T

xτ =s−1∑

k=0

xkb ·t−1∑

l=0

xlc =1− xsb

1− xb· 1− xtc

1− xc.

Hence

HS(x) =(1− xsb)(1− xtc)

(1− xa)(1− xb)(1− xc).

41

The case of three bounds

If b = (p, q) 6= bi for i = 1, 2 we must have 0 < p < s and 0 < q < t(since the three bounds are not < comparable). In this case every element(x, y) ∈ C(T ) satisfies (x < s) and (y < t) and [(x < p) or (y < q)].Hence T = {xb + yc | (x < s and y < q) or (x < p and y < t)}. Thusmax T = max {(s− 1)b + (q − 1)c, (p− 1)b + (t− 1)c} and

F (S) = max{pb + tc, sb + qc} − a− b− c.

genus

The sum of the elements in T is:

z =s−1∑

l=0

q−1∑m=0

(lb + mc) +

p−1∑

l=0

t−1∑m=0

(lb + mc)−p−1∑

l=0

q−1∑m=0

(lb + mc) =

1

2sq[(s− 1)b + (q− 1)c] +

1

2pt[(p− 1)b + (t− 1)c]− 1

2qp[b(p− 1) + c(q− 1)] =

1

2[sq(sb + qc) + pt(pb + tc)− qp(bp + cq)− (sq + pt− qp)(b + c)].

Since the number of elements in T is sq + pt − qp = a and ra = bp + cq weget that the genus is:

g(S) =z

a− a− 1

2=

1

2(1

a[sq(sb + qc) + pt(pb + tc)]− a− b− c + 1− rqp).

Hilbert series

Finally,

∑τ∈T

xτ =s−1∑

k=0

q−1∑

l=0

xkb+lc +

p−1∑

k=0

s−1∑

l=0

xkb+lc −p−1∑

k=0

q−1∑

l=0

xkb+lc

=(1− xsb)(1− xqc) + (1− xpb)(1− xtc)− (1− xpb)(1− xqc)

(1− xb)(1− xc).

42

arranging the nominator and remembering that bp + qc = ra we get:

HS(x) =1− xra − xsb − xtc + xsb+qc + xpb+tc

(1− xa)(1− xb)(1− xc).

Remark: It is easy to verify that the three relations sb = s1a + s2c, tc =t1a + t2b and ra = pb + qc generate (in the sense of Herzog [17]) the kernelof the natural presentation h : F3 → S defined by h(x, y, z) = xa + yb + zc.Hence one can reprove Herzog’s theorem: ker h is generated by at most 3relators.

4.2.3 Four generators

The case of semigroups generated by 4 (or more) elements is still an openproblem. In [16] Froberg used Bresinsky’s relations [10] to calculate theFrobenius number of 4-generated symmetric semigroups. Here we use ourmethod to theoretically solve this case and discuss the difficulties arising inpractice.

Let a0 = a, a1 = b, a2 = c, a3 = d be four relatively prime positive inte-gers and S = S(a, b, c, d). Let f : F3 → S be the natural map f(x, y, z) =xb + yc + zd. Let T = T (a) be the minimal transversal of S modulo a andC = C(T ) be the canonical domain of a0.

Again we start with the bounds bi = miei, i = 1, 2, 3 where mi is theminimal positive integer such that miai ∈ aj + S for some j < i.

Next we move to the bound b. Let m0 be the minimal positive integer suchthat m0a ∈ S(b, c, d). Then the bound b is the maximal source (p1, p2, p3) ofm0a in F3 (with respect to <L).

By Corollary 32 there is no other bound v with support {1, 2, 3}. Howeverthere may be various bounds of the form riei+qjej for (i, j) = (2, 3), (3, 4), (2, 4)which occur if riai + qjaj ∈ ak + S for some k < i. We need to take onlyminimal bounds of these forms

43

Remark: It may happen that for a certain pair (i, j) the only bounds ofthere form riei + qjej are miei,mjej which we already got. On the otherhand, it may happen that we have more than one bound on the same sup-port. In [9] it is shown that the number of such bounds can become arbitrarylarge (although it is bounded by a polynomial in the generators)

Now for every bound v we get a condition α(v). Thus every element t ∈ Tcan be written (uniquely) as a combination t = x2b + x3c + x4d in which(x2, x3, x4) fulfills all of the conditions α(v). Thus(a) xi < mi.(b) (xi < ri) or (xj < qj) For every bound of the form riei + qjej where(i, j) = (2, 3), (3, 4), (2, 4). (we can include here the bounds miei where ei-ther ri or qj vanish).(c) x1 < p1 or x2 < p2 or x3 < p3 (Recall that b = (p1, p2, p3)).

In order to find the Frobenius number of S we need to find max T . Thecandidates for this job are the numbers xb + yc + zd ∈ T which correspondto <-maximal elements in C (thus (x, y, z) ∈ C but (x + 1, y, z), (x, y + 1, z)and (x, y, z + 1) all do not belong to C).

Example: Let a0 = a = 31, a1 = b = 49, a2 = c = 40, a3 = d = 15 befour relatively prime integers. We seek the minimal transversal of S(a, b, c, d)modulo a.

First we seek the minimal numbers mi (i = 1, 2, 3) such that miai ∈ aj + Sfor some j < i. It is easy to verify that m2 = 3,m3 = 2 and m4 = 8 (indeed3b = 2a++c+3d, 2c = a+b and 8d = a+b+c. One also needs to check thatthese are minimal). The corresponding bounds are (3, 0, 0), (0, 2, 0), (0, 0, 8).Thus in every (x, y, z) ∈ C(T ) we have x < 3, y < 2 and z < 8.

Next we look for the minimal positive integers m0 such that m0a ∈ S(b, c, d).We find that m0 = 4 and 4a = b+5d where (1, 0, 5) is the <L-maximal sourceof 4a in F3. Thus (1, 0, 5) ∈ B and for every (x, y, z) ∈ C(T ) we have (x < 1or z < 5).

Finally we look for the bounds of the form riei+qjej. The only new equationwe get is 2b + c = 3a + 3d which corresponds to the bound (2, 1, 0) and tothe condition: every (x, y, z) ∈ C(T ) satisfies x < 2 or y < 1.

44

To sum up, the bounds of C are: {(3, 0, 0), (0, 2, 0), (0, 0, 8), (2, 1, 0), (1, 0, 5)}.The maximal elements in C with respect to < are {(2, 0, 4), (1, 1, 4), (0, 1, 7)}.Their images are 2 · 49 + 4 · 15 = 158, 1 · 49 + 1 · 40 + 4 · 15 = 149 and1 · 40 + 7 · 15 = 145. Hence max T = 158 and the Frobenius number of S is158− 31 = 127 as a direct computation indeed shows.

The genus and the Hilbert series of S can also be calculated using this knowl-edge. For example we calculate the genus. Every element x = (x, y, z) ∈ Csatisfies (x ≤ (2, 0, 4) or x ≤ (1, 1, 4) or x ≤ (0, 1, 7)). Thus C is the disjointunion of C1, C2 and C3 where C1 = {x | x ≤ (2, 0, 4)}, C2 = {(x, 1, z) | x ≤1, z ≤ 4} and C3 = {(0, y, z) | y = 0, 1, z = 5, 6, 7}.

Now∑

t∈C1t =

∑2i=0

∑4j=0 (ib + jd) = 5

∑2i=0 ib+3

∑4j=0 jd = 15b+30d. In

a similar manner,∑

t∈C2t = 5b + 10c + 20d and

∑t∈C3

t = 3c + 36d. Hence∑t∈C t = 20b + 13c + 86d = 2790.

Finally g(S) = 1a

∑t∈T t− 1

2(a− 1) = 75 which can be confirmed directly.

4.2.4 Smooth sequences

We conclude this chapter by using the method in case the semigroup is gen-erated by a smooth sequence (cf. Section 2.2).

Let (a0, a1, ..., an) be a sequence of relatively prime positive integers. Letdk = gcd{a0, ..., ak}, k = 0, ..., n and ck = dk−1/dk, k = 1, ..., n. LetSk = S(a0, ..., ak), k = 0, ..., n. Recall that the sequence (ai) is smooth iffor every k = 1, ..., n we have ckak ∈ Sk−1.

Let S = Sn be the numerical semigroup generated by the ai’s. Let f : Fn → Sbe the map f(x) =

∑ni=1 xiai. Let T = T (a0) be the minimal transversal of

S modulo a0 and C be the canonical domain of a0.

We look for the bounds of C. As usual we start with the bounds bk = mkek

where mk is the minimal positive number such that mkak ∈ aj + S forsome j < k. Since ckak ∈ Sk−1 we have mk ≤ ck and in particular everyx = (x1, ..., xn) ∈ C fulfills (∗) xk < ck for every k = 1, ..., n.

45

But the number of elements in Fn which fulfill (∗) is∏n

k=1 ck = d0/dn =a0/1 = a0. Since there are exactly a0 elements in C we deduce that bk = ckek

for every k = 1, ..., n and these are the only bounds of C. Hence T ={∑n

i=1 xiai | 0 ≤ xi < ci}.

46

Chapter 5

Symmetric semigroups

5.1 Preliminaries

A numerical semigroup S is said to be symmetric if for every k = 0, ..., F (S)either k ∈ S or F (S)− k ∈ S. (It is clear that at least one of these numbersdo not belong to S, since their sum F (S) does not, but it may happen thatboth of them do not). Hence S is symmetric if and only if g(S) = 1

2(F (S)+1).

In this chapter we use a test for symmetry due to Apery [1] which is basedonly on the minimal transversal of S (with respect to any element of thesemigroup). Then we give a new proof of Herzog’s theorem about symmetricsemigroups generated by 3 elements [17].

Proposition 34 Let S be a numerical semigroup. Let b ∈ S be an elementof S and T be the minimal transversal of S modulo b. Let m = max T .Then S is symmetric if and only if for every minimal representative t ∈ T ,m− t ∈ T too.

Proof: Let F = F (S) be the Frobenius number of S, then F = m − b(Theorem 5). Suppose first that S is symmetric. If t ∈ T then t− b /∈ S (bydefinition). Now by symmetry m−t = F−(t−b) ∈ S and m−t−b = F−t /∈ Shence m− t ∈ T as claimed.On the other direction, suppose that there is a minimal representative t ∈ Tsuch that m− t /∈ T . We will show that both t− b and F − (t− b) = m− tdo not belong to S, hence S is not symmetric.It is clear that t − b /∈ S since t is minimal in its class. By our choice of t

47

we also know that m − t /∈ T . Hence it suffices to show that m − t /∈ S\T .This is so because for every element a ∈ (S\T ) we have a − b ∈ S, but(m− t)− b = F − t can not belong to S since t belongs to S but F does not.¤

Example: Let S = S(a, b). The minimal transversal of S modulo a isT = {kb | k = 0, ..., a − 1} (cf. Section 4.2.1). Thus m = max T = (b − 1)aand for every t = kb ∈ T , m− t = la for some 0 ≤ l < a thus m− t ∈ T tooand S is symmetric.

5.2 3-generated symmetric semigroups

Let S = S(a, b, c) where a, b, c are relatively prime positive integers. We give anew elementary proof to Herzog’s characterization of 3-generated symmetricnumerical semigroups [17].

Theorem 35 (Herzog) S(a, b, c) is symmetric if and only if it is a completeintersection.

Actually, we prove that S(a, b, c) is symmetric if and only if there is an ar-rangement (ai)

3i=1 of the generators a, b, c (in some order) that is smooth

(hence, in particular, S is a complete intersection).

Recall that for a sequence (ai)ni=1 of relatively prime positive integers we

denote di = gcd{a1, ..., ak} (k = 1, ..., n), ck = dk−1/dk (k = 2, ..., n) andSk = S(a1, ..., ak). The sequence (ai) is said to be smooth if ckak ∈ Sk−1 forevery k = 2, ..., n.

Notice that c2a2 always belongs to S(a1) since d1 = a1, d2 = gcd{a1, a2}and c2a2 = d1

d2a2 = a1

a2

d2where a2

d2is actually a natural number. Thus say-

ing that a sequence (a1, a2, a3) of three relatively prime positive integers issmooth amounts to c3a3 ∈ S(a1, a2) where c3 = d2/d3 = d2/1 = gcd(a1, a2).The basic idea lies in the following lemma:

Lemma 36 Let α, β, γ be three relatively prime positive integers. Let d =gcd(α, β). Let r be the minimal positive integer such that rα ∈ S(β, γ). Thenif rα is a multiple of β then dγ ∈ S(α, β).

48

Proof: Since rα = pβ and r is minimal, we have r = β/d and p = α/d.Since p = α/d and β/d are relatively prime it follows that for every naturalnumber k there is a unique integer 0 ≤ j < p such that j(β/d) = k (mod p).Multiplying by d we get that there is a unique solution 0 ≤ j < p for jβ = dk(mod α) for every natural k. In particular there is a solution 0 ≤ j < p suchthat jβ = dγ (mod α). Now (p − j)β + dγ = pβ = 0 (mod α). If jβ > dγthen (p − j)β + dγ < pβ = rα and (p − j)β + dγ = mα for some some0 < m < r (m > 0 since p > j) which contradicts the minimality of r. Hencedγ ≥ jβ. Thus dγ = jβ + sα for some s ≥ 0 and indeed dγ ∈ S(α, β). ¤

We see that in order to prove that (a1, a2, a3) is smooth, it is sufficient toshow that the minimal positive multiple of a1 which belongs to S(a2, a3) isactually a multiple of a2.

Proof: (of the theorem) Since every complete intersection is symmetric [12]and every semigroup generated by a smooth sequence (a1, ..., an) is a com-plete intersection [17], in order to complete the proof it suffices to show thatevery symmetric 3-generated semigroup is generated by a smooth sequence.

Let T be the minimal transversal of S modulo a. We define the usual pa-rameters (cf. Section 4.2.2):(a) Let s be the minimal positive number such that the equation sb = s1a+s2chas a solution in natural numbers with s1 > 0.(b) Let t be the minimal positive number such that the equation tc = t1a+t2bhas a solution in natural numbers.(c) Finally, let r be the minimal positive integers such that ra ∈ S(b, c). Let(p, q) be the pair with maximal p such that ra = pb + qc.

We have seen that there are two cases:

(a) (p = s and q = 0) or (p = 0 and q = t).

In this case the minimal transversal of S modulo a is the set T = {xb + yc |0 ≤ x < s, 0 ≤ y < t}. Thus m = max T = (s − 1)b + (t − 1)c andfor every u = xb + yc ∈ T (where 0 ≤ x < s and 0 ≤ y < t), we havem − u = (s − 1 − x)b + (t − 1 − y)c ∈ T too (since 0 ≤ s − 1 − x < s and0 ≤ t− 1− y < t). Hence S is symmetric. Also, if the left condition occursthen ra = sb and (a, b, c) is smooth and if the right condition holds then

49

ra = tc and (a, c, b) is smooth.

We move to the second case:

(b) 0 < p < s and 0 < q < t.

Suppose that S is symmetric. We show that there is an arrangement ofits generators that is smooth. HereT = {xb + yc | (0 ≤ x < s and 0 ≤ y < q) or (0 ≤ x < p and 0 ≤ y < t)}and m = max T = max{(s− 1)b + (q − 1)c, (p− 1)b + (t− 1)c}.First assume the latter. Since p < s and 0 < q, pb ∈ T . Because S is sym-metric, m − pb = (t − 1)c − b also belongs to T . Therefore it is of the formkb+lc. Hence we get that (t−1−l)c = (k+1)b Now 0 < t−1−l < t (t−1−lis positive because so is (k + 1)b). But this contradicts the minimality of tas the least positive integer such that tc ∈ S(a, b).Hence m = (s− 1)b + (q− 1)c. Since q < t and 0 < p, it follows that qc ∈ T .By symmetry m − qc = (s − 1)b − c ∈ T too. Write (s − 1)b − c = kb + lcfor some k, l ∈ N. Then (s − 1 − k)b = (l + 1)c where 0 < s − 1 − k < s(notice (l +1)c is positive). but s was the minimal positive integer such thatsb = s1a+s2c with s1 > 0. Hence the minimal multiple of b which belongs toS(a, c) is actually a multiple of c (since it does not exceed s− 1− k). Hence(b, c, a) is smooth and S is again a complete intersection. ¤

50

Chapter 6

The minimal transversal inaffine semigroups

6.1 Numerical Affine semigroups

A numerical affine semigroup (nas) is a finitely generated subsemigroupof Fm = Nm, the free commutative semigroups on m generators. Thus S isnumerical affine if there are tuples a1, ..., an ∈ Nm such that S = {∑n

i=1 kiai |ki ∈ N, i = 1, ..., n}. If a1, ..., an generate S we write S = S(a1, ..., an).

We start with some terminology. Basically, we use the same definitions asHerzog [17] with minor adaptations. We use the same notation as in Chapter4. In particular we use the same letter for a tuple and its components (thusif v ∈ Fm we implicity assume that v = (v1, ..., vm) unless stated otherwise).

6.1.1 The rank of the semigroup

We claim that numerical affine semigroups have a canonic minimal generat-ing set (that is a generating set which is contained in every other generatingset), namely the set of all irreducible elements in S.

We say that an element 0 6= s ∈ S is irreducible if for every a, b ∈ Ssuch that a + b = s either a = 0 or b = 0. Let K(S) be the set of irreducibleelements of S. It is clear that for every generating set G of S, K(S) ⊆ G(an irreducible element is in the semigroup generated by G if and only if itbelongs to G). We claim that if S is a nas then K(S) generates S. To show

51

this we use a partial order < on Fm. Recall that u ≤ v if ui ≤ vi for everycomponent i = 1, ..., m. Since the number of elements smaller than everygiven v in this order is finite, we have that every set has a minimal element.

Now we can show that K(S) generates S. This follows immediately from:

Proposition 37 Let S be a subsemigroup of Nm then every element of S isa finite sum of irreducibles.

Remark: We use the convention that 0 is an empty sum of irreducibles.

Proof: By induction. The proposition holds for 0 by the remark above.Let s ∈ S. Suppose that every t < s is the sum of irreducibles. If s isirreducible then there is nothing to prove. Else, s can be written as a sums1 +s2 where both s1 and s2 are strictly smaller than s. Thus (by induction)s1 and s2 are both finite sums of irreducibles. Hence so is s. ¤

We call K(S) the minimal generating set of the nas S. The rank of S is theminimal number of elements needed to generate S. Thus rank(S) = |K(S)|.

6.1.2 Dimension and codimension

Let S = S(a1, ..., an) be a numerical affine subsemigroup of Nm. Let U(S) =spanQ{a1, ..., an} be the subspace of Qm which is spanned by the ai’s. Thedimension of S is dim S = dim U(S). The codimension of S is the numbercodimS = rank S − dim S. Since K(S) generates S, it also spans U over Q.Thus rank S ≥ dim S and the codimension of S is non negative (and it is 0if and only if K(S) is linearly independent, hence if and only if S is free).

6.1.3 The Hilbert series of the semigroup

Let s = (s1, ..., sm) be a tuple in Nm. Let x = (x1, ..., xm). We write xs forxs1

1 xs22 ...xsm

m . Let S be a numerical affine subsemigroup of Nm. The Hilbertseries of S is the formal power series H(x) = HS(x) =

∑s∈S xs.

Example Let S = S(a1, ..., an) be a semigroup of dimension dim S =

52

rankS = n. Then the vectors a1, ..., an are linearly independent over Qand every tuple s ∈ S has a unique combination s =

∑ni=1 kiai for some

k1, ..., kn ∈ N. Thus

H(x) =∑s∈S

xs =∑

(k1,...,kn)∈Nn

xk1a1+...+knan =n∏

i=1

(∞∑

ki=0

xkiai) =n∏

i=1

1

1− xai.

6.2 Semigroups of codimension 1

Let S ⊆ Nm be a numerical affine semigroup. Let K(S) = {a1, ..., an} beits minimal generating set, and U = U(S). Suppose that dim S = dim U =rankS − 1 = n− 1.

Denote Vn = Qn. We say that a tuple in Qn is natural (resp. integral) ifall of its components are natural numbers (resp. integers).

Let f be the linear transformation f : Vn → Vm which assigns to eachtuple x = (x1, ..., xn) ∈ V the vector f(x) =

∑ki=1 xiai. Then Imf = U and

rankf = dim (Imf) = n − 1. By the dimension theorem for linear transfor-mations we have dim(ker f) = 1.

For a tuple α = (α1, ..., αn) we let |α| = (|α1|, ..., |αn|). Let | ker f | = {|α| |α ∈ ker f}. Let B = | ker f | ∩Nn. Let β > 0 be a minimal non-zero elementin B (with respect to <). Notice that there is such a β since ker f obviouslycontains integral tuples other than 0 (take any tuple and multiply it by thecommon denominator of its components). Since ker f is one dimensional itfollows that β is determined uniquely (as the only tuple in B which compo-nents are relatively prime) and every other tuple in B is of the form kβ forsome k ∈ N. Let α ∈ ker f be a tuple such that β = |α| (there are two suchtuples ±α from which we choose one). Then every integral tuple in ker f isan integral multiple of α.

Let I be the set of indices i such that αi > 0 and let J be its comple-mentary set in {1, ..., n} (that is the set of indices j such that αj ≤ 0). I, Jare both not empty (since α 6= 0, f(α) = 0 and ai > 0 for every i). By

53

changing wings we write the dependency∑n

i=1 αiai = 0 in the form:

∑i∈I

λiai =∑j∈J

λjaj

where λi = αi for i ∈ I and λi = −αi for i ∈ J . Hence λi ≥ 0 for everyi = 1, ..., n. Let w =

∑i∈I λiai =

∑j∈J λjaj. Since λi > 0 for every i ∈ I,

we have w > 0.

We remark that the sets AI = {ai | i ∈ I} and AJ = {ai | i ∈ J} areboth linearly independent over Q. Let UI = spanQAI, UJ = spanQAJ andW = UI ∩ UJ . Since UI + UJ = U we have, by the dimension theorem forsubspaces, dim W = dim UI + dim UJ − dim U = 1. Hence {w} is a basis forW .

Theorem 38 Let w =∑

i∈I λiai =∑

i∈J λiai. Then

H(x) =1− xw

∏ni=1 (1− xai)

.

Example Every numerical subsemigroup of N is of dimension 1. Thus S isof codimension 1 if it is generated by two relatively prime positive integersa1 = a, a2 = b. The defining relation in this case is b ·a1 = a ·a2 and w = ab.Thus

H(S) =1− xab

(1− xa)(1− xb).

Example in N3 Let S = S(a1, a2, a3, a4) where a1 = (1, 1, 0), a2 = (1, 3, 2),a3 = (1, 0, 1) and a4 = (2, 1, 5). It is easy to check that none of the ai’s is anatural combination of the others. Thus rankS = 4. Since spanQ{a1, ..., a4} =Q3 we have dim S = 3 and codimS = 1. The defining relation of S is2a1 + a4 = a2 + 3a3. and w = 2a1 + a4 = a2 + 3a3 = (4, 3, 5). Thus ifx = (x, y, z) we have

H(S) =1− xw

(1− xa1)(1− xa2)(1− xa3)(1− xa4)=

1− x4y3z5

(1− xy)(1− xy3z2)(1− xz)(1− x2yz5).

54

Now we prove the theorem. We give two proofs: the first uses the minimaltransversal of S with respect to w and the second uses a ”trick” which worksin this case.

6.2.1 First Proof

We calculate the Hilbert series of S by using a multidimensional analogue ofthe minimal transversal with respect to a tuple b ∈ Nm.

We define an equivalence relation on S as follows: s1 ∼ s2 if and only ifs2 − s1 ∈ Zm is an integral multiple of b. It is easy to check that ∼ is indeedan equivalence relation and in fact, a congruence. If s1, s2, t1, t2 ∈ S are fourelements such that s1 ∼ s2 and t1 ∼ t2 then s1 + t1 ∼ s2 + t2.

Since the equivalence class of s ∈ S contains only elements of the forms + kb, it is clear that every equivalence class is totally ordered by < (recallthat s ≤ t if and only if si ≤ ti for every component i = 1, ..., m). More-over < forms a well ordering on every such class (since there are no infinitedescending chains in S). For every equivalence class we define the minimalrepresentative of it to be the minimal element r with respect to <. Thus Ifr is the minimal representative of its class, then every other element in it isof the form r + kb for some k > 0 (we could also use this property to definethe minimal representative r). Let T be the set of all minimal representa-tive modulo b (notice T may be infinite). As usual, we call T the minimaltransversal of S modulo b.

If b ∈ S and t ∈ T then the class of t is exactly {t + kb | k ∈ N}. Thus wehave:

Theorem 39 Let S ⊆ Nm be a numerical affine semigroup. Let b ∈ S bean element of S and T be the minimal transversal of S modulo b. Then theHilbert series of S is

H(x) =1

1− xb

∑t∈T

xt.

55

Proof: Let H(x) =∑

s∈S xs be the Hilbert series of S. Let C(t) be theequivalence class of some element t ∈ T , then C(t) = {t+kb | k ∈ N}. Thus

∑

s∈C(t)

xs = xt

∞∑

k=0

xkb =xt

1− xb.

Summing over the classes we get the result. ¤

We now return to the case of codimension 1. Recall that the defining re-lation of S is

w =∑i∈I

λiai =∑j∈J

λjaj

where I, J form a partition of {1, ..., n} to non empty disjoint sets, and allthe coefficients λi are non negative. We compute the minimal transversal ofS modulo w. Let T be this transversal. We claim that t ∈ T is the minimalrepresentative of its class if and only if there is a tuple τ = (τ1, ..., τn) ∈ Nn

such that t = f(τ) =∑n

i=1 τiai which fulfills the following condition:(∗) There are indices i ∈ I, j ∈ J such that τk < λk for k = i, j.

First we show that for every t =∑n

i=1 τiai ∈ T the tuple τ = (τ1, ..., τn) ∈ Nn

satisfies (∗). Indeed if τi ≥ λi for every i ∈ I then t =∑n

i=1 τiai =∑i∈I (τi − λi)ai +

∑j∈J τjaj + w where all the coefficients of the ai’s are

positive and t is not minimal in its class. The proof for J is analogous.

Next we show that if τ is a tuple in Fn which fulfill (∗) then f(τ) is a minimalrepresentative modulo w. Indeed, f(τ) = r+kw for some r =

∑ni=1 ρiai ∈ T

and k ≥ 0. Notice that the tuple ρ = (ρ1, ..., ρn) also fulfills (∗) by theprevious paragraph. From f(τ) = f(ρ) + w we get

∑i∈I (τi − ρi)ai =∑

j∈J (ρj − τj)aj+kw. Hence u =∑

i∈I (τi − ρi)ai ∈ UI∩UJ (since w ∈ UJ).Thus u = mw for some integer m. By unique representation (on the basisAI of UI) we get that τi − ρi = mλi for every i ∈ I. But τ fulfills (∗), hencethere is an index j ∈ I such that τj < λj. Thus τj − ρj < λj and we get thatm ≤ 0. On the other hand, ρ fulfills (∗) too. Hence there is an index l ∈ Isuch that ρl < λl. Thus τl − ρl > −λl and m ≥ 0. Thus m = 0 and τi = ρi

for every i ∈ I. Similarly (since w ∈ UI) one shows that τj = ρj for everyj ∈ J . Hence τ = ρ and f(τ) = r is a minimal representative.

Finally we show that every choice of τ which fulfill (∗) yields a different

56

element t ∈ T . Let τ = (τ1, ..., τn), ρ = (ρ1, ..., ρn) be two tuples in Fn whichfulfill (∗). Let t = f(τ) and r = f(ρ). Suppose that t = r. We want to showthat τ = ρ.Let σ = (σ1, ..., σn) = τ − ρ be a tuple in Zn. By our assumption f(σ) =f(τ − ρ) = 0. Hence σ ∈ ker f . This means that σ = `α for some integer `.But σ satisfies (∗) (because τ does) thus there are indices i ∈ I, j ∈ J suchthat σi < λi = αi and σj < λj = −αj. Hence −1 < ` < 1. Thus ` = 0, σ = 0and indeed τ = ρ.

Let C ⊂ Fn be the set of tuples in Fn which fulfill (∗). We have just seen thatf induces a bijective correspondence between C and the minimal transversalT .

Let e1, ..., en be the unit vectors of Vn. Let CI be the set of all tuplesτ =

∑i∈I τiei such that τi < λi for some index i ∈ I and CJ be the set

of all tuples τ =∑

j∈J τjej such that τj < λj for some index j ∈ J . ThenC = CI ⊕ CJ .

Now we can calculate the sum G(x) =∑

t∈T xt. First we notice that

G(x) =∑τ∈C

x∑n

i=1 τiai = (∑τ∈CI

x∑

i∈I τiai) · (∑τ∈CJ

x∑

j∈J τjaj) .

DefineGI(x) =

∑τ∈CI

x∑

i∈I τiai , GJ(x) =∑τ∈CJ

x∑

j∈J τjaj .

ThenG(x) = GI(x)GJ(x).

We proceed to computing GI . In order to do this we partition the ele-ments in CI into disjoint sets according to the minimal index i such thatτi < λi. Suppose that I = {i(1), ..., i(k)} where i(1) < i(2) < ...i(k). Forevery ` = 1, ..., k let D` be the set of tuples τ ∈ CI where i(`) is the minimalindex i for which τi < λi. Thus τ ∈ D` if and only if for every j = 1, ..., k wehave vj ≤ τi(j) < uj where(a) vj = λi(j) and uj = ∞ for j = 1, ..., `− 1(b) v` = 0 and u` = λi(`) and

57

(c) vj = 0 and uj = ∞ for j = ` + 1, ..., k

Now

s(D`) =∑τ∈D`

x∑

i∈I τiai =k∏

j=1

x∑

vj≤rj<ujrjai(j) =

(l−1∏j=1

x∑∞

r=λi(j)rai(j)) · (x

∑λi(`)−1

r=0 rai(`)) · (k∏

j=`+1

x∑∞

r=0 rai(j)) =

(l−1∏j=1

xλi(j)ai(j)

1− xai(j)) · (1− xλi(`)ai(`)

1− xai(`)) · (

k∏

j=l+1

1

1− xai(j)) =

xλi(1)ai(1)xλi(2)ai(2) ...xλi(`−1)ai(`−1)(1− xλi(`)ai(`))∏i∈I (1− xai)

.

Since the denominators of all of these sums are all the same, and the numera-tors form a telescopic sum we get that the nominator of GI(x) =

∑k`=1 s(D`)

is1− x

∑i∈I λiai = 1− xw.

Hence

GI(x) =1− xw

∏i∈I (1− xai)

.

Similarly

GJ(x) =1− xw

∏i∈J (1− xai)

.

Finally H(x) = 11−xw G(x) = 1

1−xw GI(x)GJ(x). Hence

H(x) =1− xw

∏ni=1 (1− xai)

which is what we wanted to prove.

58

6.2.2 Second proof

We use the same notation: Let Fk = Nk and Vk = Qk. Let f : Vn → Vm be theusual map f(τ) =

∑ni=1 τiai hence Imf = U(S). Let h be the restriction of f

to Fn then Imh = S. Let α, I, J be as before. Thus α =∑

i∈I λiei−∑

i∈J λjej

where ek is the kth unit vector of Vn, λi ≥ 0 for every i = 1, ..., n and everyintegral tuple in Fn is of the form kα for some integer k.

Let UI = spanQ{ai | i ∈ I}, UJ = spanQ{ai | i ∈ J} and W = UI ∩ UJ .Recall that W is a one dimensional vector space over Q and w =

∑i∈I λiai =∑

i∈J λiai spans W .

For every v ∈ Fm, let h−1(v) = {τ ∈ Fn | h(τ) = v} the set of all sources ofv. In particular, v ∈ S if and only if h−1(v) 6= ∅.

We claim that every element s ∈ S has a source σ ∈ h−1(s) such that σi < λi

for at least one index i ∈ I. Let j ∈ I be any index in I such that λj > 0(in fact by our definition of I, this happens for every i ∈ I). Let σ ∈ h−1(s)be a tuple with minimal jth component σj (if there are several such tuples,simply choose one of them). If we had σi ≥ λi for every i ∈ I then ρ = σ−αwould belong to Fn too (for i ∈ I we have ρi = σi − αi = σi − λi ∈ N by ourassumption, whereas for j ∈ J we have ρj = σj − αj = σj + λj ∈ N). Buth(ρ) = s and ρj = σj − λj < σj contradicting our choice of σ. Hence theremust be an index i ∈ I such that σi < λi.

We claim further that for every s ∈ S there is only one source σ ∈ h−1(s)such that σi < λi for at least one index i ∈ I . Let σ, τ ∈ f−1(s) be twonatural sources of s. Assume that σj < λj for some j ∈ I. Then σ − τ isan integral tuple in ker f . Hence σ − τ = kα for some integer k. Since τ isnatural, we have kαj = σj − τj ≤ σj < αj hence k ≤ 0. Now, τ = σ − kα. Ifk = 0 then τ = σ. If k < 0 then τ ≥ −kα = |k|α. In particular τi ≥ λi forevery i ∈ I. Hence the claim.

Definition 40 Let s ∈ S be an element of the semigroup and let σ ∈ f−1Sbe the unique tuple such that σi < λi for some i ∈ I. Then we call σ thecanonical source of s.

We have actually proved some more:

59

Proposition 41 Let λ =∑

i∈I λiei where ei is the ith unit vector in Fn. Letσ be the canonical source of s ∈ S then for every other natural source τ 6= σof s we have τ ≥ λ.

Proof: We have seen above that τ = σ + |k|α where |k| ≥ 1. The resultfollows from αi = λi for every i ∈ I ¤

Define the formal power series G(x) =∑

v∈Fmcvx

v where cv = |h−1(v)|is the number of natural sources of v. Every such source corresponds to adifferent natural solution of the equation

∑ni=1 xiai = v. Thus

G(x) =k∏

i=1

(∞∑

j=0

xjai) =k∏

i=1

1

1− xai.

The Hilbert series of S is H(x) =∑

s∈S xs =∑

v∈Fnbvx

v, where bv = 1 ifv ∈ S and bv = 0 otherwise.

Our main observation is that (∗) G(x) = xwG(x) + H(x). Assuming forthe moment this holds, we have

H(x) = (1− xw)G(x) =1− xw

∏ni=1 (1− xai)

.

Which concludes the proof.

To prove (∗) it suffices to show (by comparing coefficients) that cv = cv−w +bv. If v /∈ S this is trivial (since v −w /∈ S and cv = cv−w = bv = 0).Hence assume v ∈ S. Then bv = 1. Let σ be the canonical source of v. Letλ =

∑i∈I λiei be a source of w. The map ν : Fn → Fn defined by ν(x) = x+λ

maps h−1(v −w) (which may be empty) bijectively onto h−1(v)\{σ}. Thisis so because every natural source τ 6= σ of v satisfies τ > λ (Proposition 41).Hence, taking σ into account, the number of elements in h−1(v) is cv−w + 1.Since bv = 1 we have cv = cv−w + bv as claimed.

60

Appendix A

The semigroup polynomial

A.1 The polynomial

Let S be a numerical semigroup. Recall that the Hilbert series of S is theformal power series H(x) =

∑s∈S xs. In this chapter we define a (seemingly

new) invariant of the semigroup which is closely related to the Hilbert series,namely the power series P (x) = (1 − x)H(x). We show that P is in fact apolynomial end explore its properties. Then we use this polynomial to provesome propositions on cyclotomic polynomials.

A.1.1 Basic facts

Proposition 42 Let P (x) = (1−x)H(x). Then P is a polynomial of degreed = F (S) + 1 and P (0) = P (1) = 1 (over every field).

Proof: Let H∗(x) =∑

s/∈S xs. Then H∗ is a polynomial of degree F (S),H∗(0) = 0 (since the powers s /∈ S are all positive) and H(x) + H∗(x) =∑∞

n=0 xn = 11−x

. Thus H(x) = 11−x

− H∗(x) and P (x) = (1 − x)H(x) =1 − (1 − x)H∗(x) is indeed a polynomial of degree F (S) + 1. In addition,P (0) = P (1) = 1 as claimed. ¤

Definition 43 Let S be a numerical semigroup. The semigroup polyno-mial of S is the polynomial P (x) = PS(x) = (1− x)HS(x).

61

Example: Let S be the semigroup generated by 3, 5. Then the Frobeniusnumber is 7 and S = {0, 3, 5, 6, 8, 9, 10, ...}. Thus

HS(x) = 1 + x3 + x5 + x6 +x8

1− x=

1− x + x3 − x4 + x5 − x7 + x8

1− x

andP (x) = 1− x + x3 − x4 + x5 − x7 + x8.

Let H(x) =∑∞

k=0 akxk and P (x) =

∑∞k=0 bkx

k. The connection betweenthe coefficients of the two series is given by the following simple observation:

Observation 44 We have(a) b0 = a0 = 1 and bk = ak − ak−1 for every k > 0.(b) ak =

∑ki=0 bi for every k ∈ N.

(c) The non-zero coefficients of P are alternately 1 and (−1).

Proof: To prove (a), compare coefficients in P (x) = (1−x)H(x) (note thata0 = 1 since 0 ∈ S). Part (b) is a simple consequence of (a). Finally part (c)follows from (a) since the elements ak alternate between 1 and 0 (startingand ending with 1) ¤

Remark We note that ak = 1 if and only if k ∈ S. Hence by part (a)we have bk = 1 if and only if k ∈ S and k − 1 /∈ S. Similarly bk = −1 ifk /∈ S but k − 1 ∈ S. Finally bk = 0 if and only if k, k − 1 both belong to Sor both do not.

Definition 45 A polynomial Q(x) =∑f

k=0 ckxk of degree f is said to be

symmetric if ck = cf−k (for every k = 0, ..., f).

Recall that a numerical semigroup S is symmetric if for every number0 ≤ k ≤ F = F (S), either k or F − k belongs to the semigroup. In theHilbert series HS(x) =

∑∞k=0 akx

k this translates to ak = 0 ⇔ aF−k = 1 forevery 0 ≤ k ≤ F . Equivalently aF−k = 1− ak for every k = 0, ..., F .

Proposition 46 A numerical semigroup S is symmetric if and only if PS issymmetric.

62

Proof: Let F = F (S) be the Frobenius number of the semigroup S andd = F + 1 be its conductor. Let H(x) =

∑∞n=1 anx

n be the Hilbert series of

S and P (x) =∑d

n=0 bnxn be the semigroup polynomial.Suppose first that S is symmetric. For k = 1, ..., F we have

bd−k = bF−(k−1) =

(0 < F − (k − 1), Observation 44)

aF−(k−1) − aF−k =

(by symmetry of S)

(1− ak−1)− (1− ak) = ak − ak−1 = bk.

In addition b0 = bd = 1 (b0 = P (0) = 1 by Proposition 42, and bd = ad−aF =1− 0 by definition). Hence P is symmetric.Suppose now that P is symmetric. Hence bi = bd−i for every i = 0, ..., d.For every k = 0, ..., F we have ak + aF−k =

∑ki=0 bi +

∑F−ki=0 bi =

∑ki=0 bi +∑d−k−1

i=0 bi. But, by the symmetry of the polynomial,∑k

i=0 bi =∑k

i=0 bd−i =∑di=d−k bi. Hence ak + aF−k =

∑di=d−k bi +

∑d−k−1i=0 bi =

∑di=0 bi = P (1) = 1

(P (1) is the sum of the coefficients of P ). Hence aF−k + ak = 1 for everyk = 0, ..., F and the semigroup is symmetric. ¤

A.1.2 The case of two generators

Let a, b be two relatively prime positive integers and let S = S(a, b). InSection 4.2.1 we have shown that the Hilbert series of S is

H(x) =(1− xab)

(1− xa)(1− xb)

hence the semigroup polynomial is

P (x) =(1− x)(1− xab)

(1− xa)(1− xb).

In this section we find the coefficients of P .

We begin with a simple lemma, which yields another proof for the classicresult on the Frobenius number of two elements.

63

Lemma 47 Let a, b be two relatively prime positive integers. Then everynatural number t can be written uniquely as an integral combination t =ma + nb (where m,n ∈ Z) with 0 ≤ m < b. Moreover t ∈ S(a, b) if and onlyif in this combination n ≥ 0 too. In this case the number of different naturalsolutions (x, y) for xa + yb = t (with x, y ∈ N) is 1 + bn

ac.

Proof: By Euclid there are integers m,n such that ma + nb = t. Since(m + kb)a + (n− ka)b = t for every k ∈ Z, we can assume that 0 ≤ m < b.If in this case n ≥ 0 then t obviously belongs to S.The rest follows easily from the fact that if xa + yb = t with x, y ∈ Zthen there is an integer k such that x = m + kb and y = n − ka. Indeed,suppose that xa + yb = t = ma + nb. Then (x − m)a = (n − y)b. Sincea, b are relatively prime there is an integer k such that x − m = kb. Nowt = xa+yb = (m+kb)a+yb. Hence (y+ka)b = t−ma = nb and y = n−ka.Let t = ma + nb with 0 ≤ m < b. Thus all integral combinations of t areof the form (m + kb)a + (n− ka)b. For m + kb to be natural we must havek ≥ 0. In this case n− ka is natural if and only if ka ≤ n. Hence if t ∈ S wemust have n ∈ N and the number of natural combinations of t correspondsto the number of non negative k’s for which n − ka ≥ 0. Hence the result.¤

Corollary 48 Let a, b be two relatively prime integers then

F (a, b) = ab− a− b.

Proof: Write t = ma + nb where 0 ≤ m < b. Then t /∈ S if and only ifn < 0. Hence t ≤ (b − 1)a + (−1)b = ab − a − b where the number on theright still does not belong to S. ¤

Corollary 49 Let d = (a − 1)(b − 1). Then there is a unique combinationd = ra + sb with 0 ≤ r < b and v ≥ 0.

Proof: Since d > F (a, b) we have d ∈ S. Hence there are natural numbers0 ≤ r < b, s such that d = ra + sb (Lemma 47). Since d < ab we have s < a,hence (by Lemma 47) this combination is unique. ¤

We return to the semigroup polynomial of S(a, b).

64

Theorem 50 Let a, b be two relatively prime positive integers bigger than 1.Let d = (a− 1)(b− 1) be the conductor of S = S(a, b) and P (x) = PS(x) =∑d

k=0 bkxk. Suppose that d = ra + sb where 0 ≤ r < b. Then:

(a) bk = 1 if and only if k = ua + vb with 0 ≤ u ≤ r and 0 ≤ v ≤ s.(b) bk = −1 if and only if k + ab = ua + vb with r < u < b and s < v < a.(c) bk = 0 for all other k.

Remark 51 Since k + ab = ua + vb ⇔ k = ua + (v − a)b, we have that forevery integer k = ua + vb with 0 ≤ u < b exactly one of the following holds:either k = ua + vb with 0 ≤ u ≤ r and v ∈ Z or k + ab = ua + vb withr < u < b and v ∈ Z. In particular the conditions in (a) and (b) can notboth occur simultaneously.

Proof: First, for 0 = 0a + 0b the theorem claims that b0 = 1 which is true.Let H(x) =

∑∞n=1 anx

n be the Hilbert series of S(a, b). For every k > 0 wehave bk = ak−ak−1. Since S is symmetric ([30]) we have ak−1 = 1−aF−(k−1) =1− ad−k. It follows that bk = ak + ad−k− 1. The rest is mostly technical andrelies mainly on Lemma 47.(a) If k = ua + vb with 0 ≤ u ≤ r and 0 ≤ v ≤ s then k ∈ S (thus ak = 1).In addition d − k = (r − u)a + (s − v)b ∈ S too (since both coefficients arenatural). Hence ad−k = 1 and bk = 1.(b) If k = ua + vb with 0 ≤ u ≤ r and v > s then again k ∈ S and ak = 1.However, d−k = (r−u)a+(s− v)b /∈ S (since 0 ≤ r−u < b and s− v < 0).Hence ad−k = 0 and bk = 0.(c) If k = ua+vb with 0 ≤ u ≤ r and v < 0 then k /∈ S and ak = 0. Howeverd − k = (r − u)a + (s − v)b ∈ S (0 ≤ r − u < b and 0 ≤ s < s − v) hencead−k = 1 and bk = 0.(d) If k+ab = ua+vb with r < u < b and s < v < a then k = ua+(v−a)b /∈ Ssince v−a < 0. Thus ak = 0. We claim that d−k = (r−u)a+(s+a−v)b /∈ Stoo, hence ad−k = 0 and bk = −1.To show that d − k /∈ S we notice that d − k = (r − u)a + (s + a − v)b =(r + b − u)a + (s − v)b. It suffices to prove that 0 ≤ r + b − u < b (sinces− v < 0). Indeed r + (b− u) > 0 and r + b− u = b + (r − u) < b.(e) If k+ab = ua+vb with r < u < b and v ≤ s then again k = ua+(v−a)b /∈S and ak = 0. However d− k = (r + b− u)a + (s− v)b ∈ S (since r + (b− u)and s− v are both positive) hence ad−k = 1 and bk = 0 as claimed.(f) Finally, if k + ab = ua + vb with r < u < b and a ≤ v then k =ua + (v− a)b ∈ S and ak = 1. However d− k = (r− u)a + (s− v)b < 0 since

65

both coefficients are negative.(Recall that ra + sb = d < ab thus s < a). Inparticular d− k /∈ S, ad−k = 0 and bk = 0.By the remark above, we are done. ¤

Corollary 52 Let a, b be two relatively prime positive integers bigger than1, and suppose b is odd. Let d = (a− 1)(b− 1) = ra + sb and ` = d/2. Thenb` = (−1)r.

Proof: Since P (x) is symmetric and its ”visible” coefficients are 1,−1alternately, it is clear that the middle coefficient must be either 1 or −1(that is it can not be zero) since otherwise we would have two consecutive”1”’s or ” − 1”’s in a row. Since b` = a` − a`−1 (Observation 44) it fol-lows that b` = 1 if and only if a` = 1 that is if and only if ` ∈ S. But` ∈ S if and only if ` = ua + vb for some u, v ∈ N thus if and only ifd = ra + sb = 2` = (2u)a + (2v)b with both coefficients even, since thepresentation of d is unique (Corollary 49). Now d = F (S)+1 is even since Sis symmetric. Hence for r, s to be even it suffices that r is even (here is theonly place we use the assumption that b is odd). Hence the result. ¤

We conclude with another observation on the semigroup polynomial.

Theorem 53 Let P be the semigroup polynomial of S(a, b) where a, b arerelatively prime and b is odd. Then(a) If a is even then P (−1) = b.(b) If a, b are both odd then P (−1) = 1.

Proof: Recall that

P (x) =(1− x)(1− xab)

(1− xa)(1− xb)=

1− x

1− xb

b−1∑

k=0

xka.

Then over C(1) If a is odd then so is ab and

P (−1) =(1− (−1))(1− (−1)ab)

(1− (−1)a)(1− (−1)b)=

4

4= 1

(2) If a is even then

P (−1) =1− (−1)

1− (−1)b

b−1∑

k=0

(−1)ka =b−1∑

k=0

1 = b.

66

We note that since P ∈ Z[x] has integral coefficients, the results hold forevery ring with identity R, by the natural homomorphism Z→ R. ¤

A.2 Cyclotomic polynomials

A.2.1 Introduction

Let a, b be two relatively prime positive integers. Let P be the semigrouppolynomial of S(a, b). It is an interesting fact that P is the product ofcyclotomic polynomials. To show this we write

P (x) =(1− xab)(1− x)

(1− xa)(1− xb)=

(xab − 1)(x− 1)

(xa − 1)(xb − 1).

Since xk − 1 =∏d|k

Φd(x), where Φd(x) is the dth cyclotomic polynomial, we

see that P is a product of dth cyclotomic polynomials where d runs over allthe divisors of n = ab which are not divisors of a nor of b. In particular everysuch divisor is not a prime power. Moreover, If n = pq is a product of twodistinct primes then the semigroup polynomial of S(p, q) is exactly Φn whilefor other n (which are not powers of a prime) we have P (x) = Φn(x)

∏Φd(x)

where d < n for every d in the product. Hence we can prove theorems aboutgeneral cyclotomic polynomials by induction on their order.

In the previous sections we gave easy methods for computing P for everysemigroup which is generated by two relatively prime positive integers (onemethod is by using the Hilbert series of S (which is convenient when p, q arerelatively small) and another by representing (p − 1)(q − 1) = rp + sq. Inparticular we can apply these methods to compute Φpq(x).

Example Let p 6= q be primes. Then the semigroup polynomial of S(p, q) isexactly the pq-th cyclotomic polynomial:

P (x) =(1− xpq)(1− x)

(1− xp)(1− xq)= Φpq(x).

Example Let p 6= q be primes. The semigroup polynomial of S(p2, q) is

P (x) =(1− xp2q)(1− x)

(1− xp2)(1− xq)= Φpq(x)Φp2q(x).

67

Example Let p, q, r be 3 different primes. The semigroup polynomial ofS(pq, r) is

P (x) =(1− xpqr)(1− x)

(1− xpq)(1− xr)= Φpr(x)Φqr(x)Φpqr(x).

Example: Polynomials of complete intersections

We have seen (in section 2.4) that the Hilbert series of the semigroup S =bS1 + aS2 where a ∈ S1 and b ∈ S2 are two relatively prime positive integersis given by

HS(x) = (1− xab)H1(xb)H2(x

a)

where Hi is the Hilbert series of Si.

Let Pi be the semigroup polynomial of Si for i = 1, 2. Thus Hi(x) =Pi(x)/(1− x). If P is the semigroup polynomial of S we have

P (x) = (1−x)H(x) = (1−xab)(1−x) · P1(xb)

1− xb· P2(x

a)

1− xa= Pa,b(x)P1(x

b)P2(xa)

where Pa,b is the semigroup polynomial of S(a, b).

Now if Φ(x) is a cyclotomic polynomial then Φ(x`) is a product of cyclo-tomic polynomials for every positive integer `. Thus if P1, P2 are both prod-ucts of cyclotomic polynomials then so is P . We deduce by induction that thesemigroup polynomial of complete intersections are products of cyclotomicpolynomials (since every complete intersection is of the form bS1 +aS2 whereS1, S2 are themselves complete intersections).

For a specific example take S = S(10, 15, 14, 21). We saw in Section 2.4that its Hilbert series is

HS(x) = (1− x35) · 1− x30

(1− x10)(1− x15)· 1− x42

(1− x14)(1− x21).

Hence its semigroup polynomial is

PS = Φ26Φ30Φ35Φ42,

68

showing that PS may have multiple roots.

A.2.2 General results

Let p, q be two different primes. As we have seen, the semigroup polynomialof S(p, q) is P (x) = Φpq(x) where Φn is the cyclotomic polynomial of order n.

First we restate the results of the last section for the pqth cyclotomic poly-nomial. For other proofs of the structure theorm for Φpq see, for example,[20], [22] or [2]).

Theorem 54 Let p < q be two prime numbers (notice q is odd). Let Φ(x) =Φpq(x) =

∑dk=0 bkx

k be the pq-th cyclotomic polynomial where d = (p−1)(q−1) = rp + sq. Then:(a) Φ is symmetric.(b) The coefficients of Φ which do not equal zero are alternately 1,−1.(c) b0 = bd = 1.(d) Φ(1) =

∑dk=0 bk = 1.

(e) (1) bk = 1 if and only if k = up + vq with 0 ≤ u ≤ r and 0 ≤ v ≤ s.(2) bk = −1 if and only if k + pq = up + vq with r < u < q and s < v < p.(3) bk = 0 for all other k.

(f) Let l = (a− 1)(b− 1)/2. Then bl = (−1)r.(g) If p = 2 then Φ(−1) = p− 1 else Φ(−1) = 1.

Now we can turn to general cyclotomic polynomials.

Theorem 55 Let Φn be the cyclotomic polynomial of order n where n > 1has at least two different prime factors. Then(a) Φn(1) = 1.(b) Φn is symmetric.(c) The middle coefficient of Φn is odd.(d) If n is odd then Φn(−1) = 1.(e) Let n = 2tm where t > 0 and m > 1 odd. then(i) if t = 1 and m = qs where q is a prime number, Φn(−1) = q(ii) otherwise Φn(−1) = 1.

69

Proof: In most of the proofs we work on induction on the number of (notnecessarily different) prime factors of n. Where for the basis of the induction(where n is the product of two different primes) the result holds by theprevious theorem.Let n be a product of more than two primes. Since n has at least two differentprime factors, we can write it as a non trivial product n = ab where a, b < nare relatively prime positive integers. Let P be the semigroup polynomial ofS(a, b). Thus

P (x) = Φn(x)∏

Φd(x)

where for every d in the product d | n and d < n, in particular d has less (notnecessarily different) prime factors than n. Thus our induction hypothesis isvalid for every Φd.(a) Substituting x = 1 we get

P (1) = Φn(1)∏

Φd(1).

Now P (1) = 1 (Proposition 42) and Φd(1) = 1 by induction. Hence Φn(1) =1.

(b) Recall that a polynomial Q(x) =∑f

i=0 cixi is symmetric if ci = cf−i

for every i = 0, ..., f . One method of proving is to show that if Q1, Q2 ∈ k[x]are two polynomials over a field k and Q1 is symmetric, then their productQ1Q2 is symmetric if and only if Q2 is symmetric. Hence we can proceed byinduction.Here we use another method. Let Φ(x) = Φn(x) =

∑di=0 cix

i ∈ C[x] be thenth cyclotomic polynomial where d = ϕ(n) is the number of primitive nthroots of the unity. Let Ψ(x) = xdΦ( 1

x) =

∑bix

i. Then Ψ is a polynomialof degree f ≤ d whose roots in C include the primitive nth roots of theunity (because z is such a root if and only if 1/z is). Since there are d suchroots it follows that f = d and Ψ = mΦ for some scalar m ∈ C. SinceΨ(1) = 1 = Φ(1) we conclude that m = 1 hence ci = bi for every i = 0, ..., d.But bi = cd−i hence the result.

(c) This is immediate from the fact that Φn is a symmetric polynomialand that the sum of its coefficients is odd. (The sum of its coefficients isΦn(1) = 1).

70

(d) If n is odd then so are a, b and every d in the product. Hence P (−1) = 1(Theorem 53) and by induction we are done.

(e) (i) Suppose that n = 2qs where q is an odd prime and s > 0. Wework on induction on s. If s = 1, Φn is the semigroup polynomial of S(2, q)hence the result (Theorem 53).If s > 1, let P be the semigroup polynomial of S(2, qs). Then

P (x) =(1− x2qs

)(1− x)

(1− x2)(1− xqs)= Φ2qs(x)

s−1∏

k=1

Φ2qk(x).

Since P (−1) = qs (Theorem 53) and for every k = 1, ..., s− 1, Φ2qk(−1) = qby induction we have Φ2qs(−1) = q as claimed.

(ii) By induction on n. Let n = 2tm where m > 1 is odd and either t > 1 orm has at least two distinct prime factors. Let q be a prime factor of m. Writem = m′qs where q does not divide m′. Let P be the semigroup polynomialof S(2tm′, qs). Then

P (x) = Φn(x)∏

d

Φd(x)

where d < n is a divisor of 2tm′qs but does not divide 2tm′ or qs. In particular,all numbers of the form 2qk, k = 1, ..., s satisfy this condition. Substitutingx = −1 we get

P (−1) = Φn(−1)∏

d

Φd(−1).

Now, P is the semigroup polynomial of S(2tm′, qs) and since a = 2tm′ iseven P (−1) = qs. For k = 1, ..., s, we have Φ2qk(−1) = q (by part (i)). Forevery other d in the product we have Φd(−1) = 1 either by part (d) (if d isodd) or by induction (for even d). Substituting this to the equation we getqs = qsΦn(−1) and Φn(−1) = 1.This completes the proof. ¤

71

A.3 On the cyclotomic polynomials of order

pqr.

A.3.1 Introduction

Let Φ = Φn be the nth cyclotomic polynomial where n = pqr is the product ofthree different primes p < q < r. In the late sixties of the last century, SisterMarion Beiter, determined the coefficients of Φ [3], [4]. It had been provedby Bang [2] that the absolute value of these coefficients can not exceed p−1,and Beiter improved this bound to 3

4p. Furthermore, she has conjectured the

following:

Conjecture 56 (Beiter) Let n be the product of 3 primes p < q < r. Thenthe actual bound for the coefficients of Φn is p+1

2.

Although this was found to be the bound in some peculiar cases, no generalproof was given.Our first goal was to prove Beiter’s conjecture. To get a sense of the subject,we calculated a few polynomials. Then we found that for n = 17 ·29 ·41 someof the coefficients in Φn equal −10 (hence the magnitude the coefficients mayexceed p+1

2).

However, examining the cyclotomic polynomials of order pqr (for the smallprimes in our examples) revealed an interesting phenomena: the differencebetween two consecutive coefficients was always 0, 1 or −1.In this section we prove that if n is the product of 3 different primes then theabsolute value of the difference of consecutive coefficients in Φn is at most 4.We do not know if this bound is tight.

A.3.2 The result

Theorem 57 Let p < q < r be three primes. Let Φ(x) =∑d

i=0 aixi be the

cyclotomic polynomial of order n = pqr (thus d = ϕ(n) = (p− 1)(q − 1)(r −1) ). Then for every k = 1, ..., d we have

| ak − ak−1 |≤ 4

Proof: Define:

F (x) =(xpqr − 1)(xp − 1)(xq − 1)(xr − 1)

(xpq − 1)(xqr − 1)(xpr − 1)(x− 1).

72

It is easy to see that F is a rational function which have no poles, hence apolynomial of degree d = (p− 1)(q − 1)(r − 1). Furthermore the roots of Fin C are exactly the primitive nth roots of the unity. Since F is monic, it isthe cyclotomic polynomial of order n. Hence F (x) =

∑di=0 aix

i.

Let U be the power series which is defined by

U(x) =∞∑i=0

bixi =

(xp − 1)(xq − 1)(xr − 1)

(xpq − 1)(xqr − 1)(xpr − 1)(x− 1).

Since F (x) = (xpqr − 1)U(x), we have ai = −bi for every i = 0, 1, ..., pqr− 1.In particular bi = −ai for every i ≤ d.

Now let V (x) be the formal power series

V (x) =∞∑i=0

cixi =

(xp − 1)(xq − 1)(xr − 1)

(xpq − 1)(xqr − 1)(xpr − 1).

Since V (x) = (x−1)U(x) we have ci = bi−1−bi for every i > 0. In particularci = ai − ai−1 for every i = 1, ..., d. Hence we need to show that −4 ≤ ci ≤ 4for every i = 1, ..., d.

We write V (x) = R(x)S(x)T (x) where

R(x) =xp − 1

xpq − 1=

∑rix

i,

S(x) =xq − 1

xqr − 1=

∑six

i,

and

T (x) =xr − 1

xrp − 1=

∑tix

i.

Let m > 1. We claim that the coefficients of the power series

Km(x) =x− 1

xm − 1=

∞∑

k=0

kixi

satisfy ki = 1 if m | i, ki = −1 if i = 1(mod m) and ki = 0 otherwise.This follows from (xm − 1)Km(x) = x− 1 by comparison of coefficients.

73

Now R(x) = Kq(xp) hence ri = 1 if i = kpq (for some natural number k),

ri = −1 if i = kpq + p and ri = 0 otherwise. Similarly, S(x) = Kr(xq) hence

si = 1 if i = kqr (for some natural number k), si = −1 if i = kqr + q andsi = 0 otherwise and T (x) = Kp(x

r) (and ti = 1 if i = kpr (for some naturalnumber k), ti = −1 if i = kpr + r and ti = 0 otherwise).

Now cn =∑

i+j+k=n risjtk. There are eight cases in which the product risjtkdoes not vanish which we list below:

case 1: n = ipq + jqr + krp where ripqsjqrtkrp = 1case 2: n− p = ipq + jqr + krp where ripq+psjqrtkrp = −1case 3: n− q = ipq + jqr + krp where ripqsjqr+qtkrp = −1case 4: n− r = ipq + jqr + krp where ripqsjqrtkrp+r = −1case 5: n− p− q = ipq + jqr + krp where ripq+psjqr+qtkrp = 1case 6: n− q − r = ipq + jqr + krp where ripqsjqr+qtkrp+r = 1case 7: n− p− r = ipq + jqr + krp where ripq+psjqrtkrp+r = 1case 8: n− p− q − r = ipq + jqr + krp where ripq+psjqr+qtkrp+r = −1

The triplet (i, j, k) contributes to cn if and only if it fulfills (exactly) oneof the equations above. In this case its contribution is the product in theright (either 1 or −1).

But every m < pqr has at most one natural representation as the sumipq + jqr + krp (by the Chinese Reminder Theorem). Hence no two tripletsthat sum up to n fall in the same condition. Hence n has at most 8 tripletswhich contribute to the sum. Since there are four cases in which this contri-bution is 1 while in the other four it is −1 we deduce −4 ≤ cn ≤ 4 for everyn = 1, ..., d. ¤

74

Appendix B

The Frobenius numberrevisited

We conclude this work with another new method of finding the Frobe-nius number of numerical semigroups, which does not rely on the minimaltransversal of the semigroup. This method yields the well known solution for2-generated semigroups and a new formula for 3-generated semigroups. Un-fortunately, the attempt to implement the method to semigroups generatedby 4 or more generators runs into difficulties.

B.1 Two-generated semigroups

Let S = S(a, b) be a 2-generated numerical semigroup where a, b are relativelyprime positive integers. Let F be the Frobenius number of S. Since F +a ∈ Swe can write F + a = la + kb where k, l ∈ N. Since F /∈ S we must havel = 0. Hence F + a = kb for some positive integer k. For the same reasonkb /∈ a + S(a, b) (where a + S = {a + s | s ∈ S}). Hence k < a (otherwisekb = ba+(k−a)b ∈ a+S). We claim that kb /∈ a+S for every k < a. Indeedif kb = ia+jb with i, j ∈ N and i > 0, then 0 ≤ j < k and ia = (k−j)b. Sincea, b are relatively prime a divides k − j > 0 and in particular k ≥ k − j ≥ a.By the maximality of F , F + a = (a− 1)b and

F (a, b) = ab− a− b.

75

B.2 Three-generated semigroups

Let S = S(a, b, c) be a numerical semigroup generated by three relativelyprime positive integers a, b, c. Let F be the Frobenius number of S. SinceF + a ∈ S we can write F + a = ia + kb + lc. But i must be zero (orelse F would belong to S). Thus F + a = kb + lc. Moreover kb /∈ a + S.(Indeed lc /∈ a + S too, but we do not need this in the sequel). Similarly,F + b = ma + nc where ma /∈ b + S(a, c).

Thus, F = kb+ lc−a = ma+nc−b and we get (k+1)b+ lc = (m+1)a+nc.Suppose first that n ≥ l. Then (k+1)b = (m+1)a+(n−l)c. Since kb /∈ a+Swe get that k is the minimal number such that (k+1)b ∈ a+S (in fact, by itsminimality (k + 1)b ∈ a + S(a, c)) . Now l must be such that kb + lc /∈ a + S(since F = kb + lc− a /∈ S). It is clear that if kb + jc belongs to a + S thenso does kb + ic for every i > j. Hence in the representation F = kb + lc, lis the minimal number such that kb + (l + 1)c ∈ a + S by the maximality ofF (in fact kb + (l + 1)c ∈ a + S(a, b)). The case l ≥ n can be analyzed in asimilar way. Hence we get the following:

Theorem 58 Let a, b, c be three relatively prime positive integers. Let k′ bethe minimal number such that k′b ∈ a+S(a, c) and l′ be the minimal naturalnumber such that (k′− 1)b + l′c ∈ a + S(a, b). Let m′ be the minimal numbersuch that m′a ∈ b + S(b, c) and n′ be the minimal natural number such that(m′ − 1)a + n′c ∈ b′ + S(a, b). Then

F (S) = max{k′b + l′c,m′a + n′c} − a− b− c.

Proof: We claim that both (k′−1)b+(l′−1)c−a and (m′−1)a+(n′−1)c−bdo not belong to S. Since we saw in the discussion above that either F +a =kb + lc for k = k′ − 1 and l = l′ − 1 or F + b = (m − 1)a + (n − 1)c form = m′ − 1, n = n′ − 1, this leads to the conclusion (by the maximality ofF ).To see this, assume, for example, that (k′ − 1)b + (l′ − 1)c = ra + sb + tcwith r > 0. Then either s < k′ − 1 or t < l′ − 1. If t < l′ − 1 then(k′−1)b+(l′−1− t)c = ra+sb ∈ a+S(a, b) contradicting the minimality ofl′. Hence t ≥ (l′−1)c and s < k′−1. But then (k′−1−s)b = ra+(t+1−l′)c ∈S(a, c) contradicting our choice of k′. Thus (k′−1)b+(l′−1)c /∈ a+S(a, b, c).¤

76

Example Let a = 5, b = 9, c = 21. It is easy to check that k′ = 4 (4 ·9 = 3 · 5 + 21) and l = 1 (since 3 · 9 + 1 · 21 = 6 · 5 + 2 · 9). Similarlym = 6 (6 · 5 = 9 + 21) and n = 1 (5 · 5 + 21 = 4 · 9 + 2 · 5). HenceF (a, b, c) = max{4b + c, 6a + c} − a − b − c = max{57, 51} − 35 = 22 asdirect computation confirms.

77

Bibliography

[1] R.Apery, Sur les branches superlineaires des courbes algebriques, C.R.Acad. Sci. Paris 222 (1946), 1198-1200.

[2] A.S. Bang, Om Ligningen φn(x) = 0, Nyt Tidsskrift for Mat. 6 (1895),6-12.

[3] M. Beiter, Magnitude of the coefficients of the cyclotomic polynomialFpqr(x), Amer. Math. Monthly 75 (1968), 370-372.

[4] M. Beiter, Magnitude of the coefficients of the cyclotomic polynomialFpqr(x) II, Duke Math. J. 38,no.3 (1971), 591-594.

[5] D.M.Bloom, On the coefficients of the cyclotomic polynomials, Amer.Math. Monthly 75 (1968), 372-377.

[6] A.Brauer, On a problem of partitions, Amer. J. of Math. 64 (1942),299-312.

[7] A.Brauer, B.M.Seelbinder, On a problem of partition II, Amer. J. ofMath. 76 (1954), 343-346.

[8] A.Brauer, J.E.Shockley On a problem of Frobenius, J. Reine Angew.Math. 211 (1962), 215-220.

[9] H.Bresinsky, On prime ideals with generic zero xi = tni , Proc. Amer.Math. Soc. 47(2) (1975), 329-332.

[10] H.Bresinsky, Symmetric semigroups of integers generated by 4 elements,Manuscripta Math. 17 (1975), 205-219.

[11] F.Curtis, On formulas for the Frobenius number of a numerical semi-group, Math. Scand. 67 (1990), 190-192.

78

[12] C.Delorme, Sous-monoıdes d’intersection completes de N , Ann. Sci. Ec.Norm. Sup. 4 serie, t. 9, (1976), 145-154.

[13] G.Denham, Short generating functions for some semigroup algebras,Electron. J. Combin. 10 (2003).

[14] M.Estrada, A.Lopez, A note on symmetric semigroups and almost arith-metic sequences, Comm. Algebra 22(10) (1994), 3903-3905.

[15] L.Fel, Frobenius problem for semigroups S(d1, d2, d3), Technionmanuscript (2004).

[16] R.Froberg, The Frobenius number of some semigroups, Comm. Algebra22 (1994), 6021-6024.

[17] J. Herzog, Generators and relations of abelian semigroups and abeliansemigroup rings, Manuscripta Math. 3 (1970), 175-193.

[18] L.Juan, Some results about symmetric semigroups, Comm. Algebra21(10) (1993), 3637-3645.

[19] E.Kunz, The value semigroup of a one dimensional Gorenstein ring,Proc. Amer. Math. Soc. 25 (1970), 748-751.

[20] T.Y.Lam and K.H.Leung, On the cyclotomic polynomial φpq(x), Amer.Math. Monthly 103 (1996) 562-564.

[21] E.Leher, Bounds for the genus of a numerical semigroups, Inter. J. ofNumber Theory, To appear.

[22] A. Migotti, Zur Theorie der Kreisteilungsgleichung, S.-B. der Math.-Naturwiss. Classe der Kaiserlichen Akademie der Wissenschaften, Wien(2) 87 (1883) 7-14.

[23] M.Nijenhuis, H.S.Wilf, Representations of integers by linear forms innonnegative integers, J.Number Theory 4 (1972), 98-106.

[24] G.Polya, G.Szego, Problems and Theorems in Analysis I, Springer-Verlag (1972) pp.1-2.

[25] J.L.Ramirez Alfonsin, The Diophantine Frobenius Problem, Oxford Lec-ture Series in Mathematics and Its Applications 30 (2005).

79

[26] J.B.Roberts, Note on linear forms, Proc. Amer. Math. Soc. 7 (1956),465-469.

[27] E.S.Selmer, On the linear diophantine problem of Frobenius, J. ReineAngew. Math. 293/294(1) (1977), 1-17.

[28] E.S.Selmer, O.Beyer, On the linear diophantine problem of Frobenius inthree variables, J. Reine Angew. Math. 301 (1978), 161-170.

[29] Z.Skupien, A generalization of Sylvester’s and Frobenius’ problems onnumerical semigroups, Acta Arith. 4 (1993), 353-366.

[30] J.J.Sylvester, Problem 7382, Educational Times 37 (1884), 26.

[31] J.J.Sylvester, W.J.Curran Sharp, Problem 7382, Educational Times 41(1884), 171-178.

80