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arXiv:2105.00383v2 [math.AC] 27 May 2021
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS
OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
Abstract. Let Hbe a numerical semigroup minimally generated by an almost arithmetic
sequence. We give a complete description of the row-factorization (RF) matrices associated
with the pseudo-Frobenius elements of H. We use the information from RF-matrices to give
a characterization of the generic nature of the defining ideal. We also show that the defining
ideal of gluing of two numerical semigroups is never generic. Further, when His symmetric
and has embedding dimension 4 or 5, we prove that the defining ideal is minimally generated
by RF-relations.
1. Introduction
Numerical semigroups and semigroup rings are widely studied objects in the literature
(see [12]). Recently, A. Moscariello [6] introduced a new tool associated with the pseudo-
Frobenius elements of numerical semigroups, namely, row-factorization (RF) matrices, to
explore the properties of semigroup rings. Moscariello used this object to investigate the
type of almost symmetric semigroups of embedding dimension four and proved the conjecture
given by T. Numata in [7], which states that the type of an almost symmetric semigroup of
embedding dimension four is at most three. The connection between the RF-matrices and
numerical semigroup rings is further studied by K. Eto, J. Herzog and K.-i. Watanabe in
([3], [2], [1], [5]), where they explore the set theoretic complete intersection property, minimal
free resolution, generic property, etc for the defining ideal of certain monomial curves.
Let kbe a field and H=hm0, m1,...,mpibe a numerical semigroup minimally generated
by {m0, m1,...,mp}. For an indeterminate tover k, the semigroup ring of His a subring
of k[t], defined as k[H] = k[tm0,...,tmp].The ring k[H] can be represented as a quotient
of a polynomial ring using a canonical surjection π:k[x0, x1,...,xp]→k[H] given by
π(xi) = tmi,for all i. The kernel of this k-algebra homomorphism π, say I(H),is a prime
ideal (called toric ideal) and is the defining ideal of k[H]. The ring k[H] is called a toric ring.
If a toric ideal has a minimal generating set consisting of binomials with full support, then
it is called generic. The notion of genericity for lattice ideals, and in particular for toric
ideals, is introduced by I. Peeva and B. Sturmfels in [11]. The authors give a characteristic
independent, minimal free resolution of the toric ring when the toric ideal is generic. Generic
toric rings are further studied in [4], where the authors prove that such rings are Golod and
so the Poincar´e series of the residue field is rational. In [3], K. Eto gave necessary and
sufficient conditions for a toric ideal to be generic and as a result proved that the defining
The second author is supported by the Early Career Fellowship at IIT Gandhinagar.
The third author is the corresponding author; supported by the MATRICS research grant
MTR/2018/000420, sponsored by the SERB, Government of India.
2010 Mathematics Subject Classification: 20M14, 20M25, 13A02.
Keywords: Almost arithmetic numerical semigroup, pseudo-Frobenius elements, row-factorization matrix,
generic toric ideal.
1
2 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
ideal of almost Gorenstein monomial curves is not generic if their embedding dimension is
greater than three.
In this paper, we consider the numerical semigroups generated by an almost arithmetic
sequence. In particular, we have H=hm0, m1,...,mp, nisuch that mi=m0+id for
i= 0,...,p,d > 0 and gcd(m0, n, d) = 1.The aim of this paper is to characterize the
generic nature of the defining ideal of the semigroup Husing the information from the row-
factorization matrices. This, in the first place, requires knowledge of the pseudo-Frobenius
elements of H. For this, we refer to the computations in [8], where the authors give an explicit
description of pseudo-Frobenious elements for the numerical semigroups of this class.
We now summarize the contents of the paper. In Section 2, we recall some basic definitions,
along with the definition of RF-matrices. We also recall some results from the papers [8] and
[10], which are used in the further sections. In Sections 3 and 4, we give the computation
of RF-matrices. Section 5 contains the main result of the paper (Theorem 5.3), where we
characterize the generic nature of the defining ideal of a numerical semigroup minimally
generated by an almost arithmetic sequence. We conclude that I(H) is never generic, except
when p= 0 or under suitable conditions with p= 1.We use Eto’s result ([3, Theorem 1]) to
give this characterization. Using Eto’s result, we also prove that the defining ideal of a gluing
of two numerical semigroups is never generic. Consequently, defining ideal of the semigroup
ring defined by a complete intersection numerical semigroup of embedding dimension at least
3 is never generic.
A numerical semigroup Hof type 1 is called symmetric. It is known that k[H] is Gorenstein
if and only if His a symmetric numerical semigroup. In section 6, we identify the cases in
which a numerical semigroup H, minimally generated by an almost arithmetic sequence, is
symmetric. We compute the RF-matrices in each of these cases. The rows of RF-matrices
are known to produce binomials in the ideal I(H).These binomials are called RF-relations.
Since I(H) may not always be minimally generated by RF-relations, one may ask when is
such a scenario possible. This question was raised by Herzog and Watanabe in [5]. They
answered the question in the affirmative for the cases when the embedding dimension is 3
and when His pseudo-symmetric or almost symmetric with embedding dimension 4. In
Theorem 6.3 and Theorem 6.4, we prove that I(H) is minimally generated by RF-relations
when His a symmetric numerical semigroup, minimally generated by an almost arithmetic
sequence with embedding dimension 4 or 5.
2. Notation and Preliminaries
Let Zand Nbe the sets of integers and non-negative integers respectively. Let p > 0 and
m0, m1,...,mp∈Nsuch that gcd(m0, m1,...,mp) = 1.Then
hm0, m1,...,mpi=(p
X
i=0
aimi|ai∈N,∀i)
is called the numerical semigroup generated by m0, m1,...,mp.
Definition 2.1. Let Hbe a numerical semigroup generated by a sequence m0, m1,...,mpof
positive integers. Then the set Ap(H, m0) = {h∈H|h−m0/∈H}is called the Ap´ery set
of Hwith respect to m0.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 3
Definition 2.2. Let Hbe a numerical semigroup. An element f∈Z\His called a pseudo-
Frobenius number if f+h∈Hfor all h∈H\{0}. We will denote the set of pseudo-Frobenius
numbers of Hby PF(H).
Note that if f∈PF(H), then f+mi∈Hfor all i∈[0, p].Hence, for all i∈[0, p],there
exist ai0, ai1,...,aip ∈Nsuch that
f+mi=
p
X
j=0
aijmj.
In the above expression, aii = 0.Because if aii >0,then it would imply f∈H. We now
recall the notion of row-factorization matrix (RF-matrix) given by A. Moscariello in [6].
Definition 2.3. Let Hbe a numerical semigroup generated by m0,...,mpand f∈PF(H).
We say that A= (aij )∈Mp+1(Z)is an RF-matrix for fif for all i∈[0, p],
p
X
j=0
aijmj=f,
where aii =−1for all i∈[0, p], and aij ∈Nfor all i, j ∈[0, p]and i6=j. For f∈PF(H),
we will denote an RF-matrix of fby RF(f).
Observe that for a given f∈PF(H),RF(f) may not be not unique. A factorization
of f+migives the (i+ 1)th row of an RF-matrix for f. Thus, if f+midoes not have
unique factorization in H, then the matrix RF(f) is not unique. Nevertheless, RF(f) will
be the notation for one of the possible RF-matrices of f. If His symmetric, i.e. type(H) =
|PF(H)|= 1,then RF(H) denotes an RF-matrix of the only pseudo-Frobenius number of
H.
Assume that the sequence m0, m1,...,mpis a strictly increasing arithmetic sequence of
positive integers and nis a positive integer such that gcd(m0,...,mp, n) = 1.Also, as-
sume that {m0,...,mp, n}is a minimal generating set for the numerical semigroup H. The
following result was proved by D.P. Patil and B. Singh in [10].
Lemma 2.4 ([10, Lemma 3.1, 3.2]).Let dbe a positive integer such that mi=m0+id for all
0≤i≤p. Let nbe an arbitrary positive integer with gcd(m0, d, n) = 1.Let H′=Pp
i=0 Nmi
and H=H′+Nn. For t∈N,let qt∈Z, rt∈[1, p]and gt∈H′defined by t=qtp+rtand
gt=qtmp+mrt.
(1) Let u= min{t∈N|gt/∈Ap(H, m0)}and v= min{b≥1|bn ∈H′}.Then there
exist unique integers z∈[0, u −1], w ∈[0, v −1], λ ≥1, and µ≥0such that
(i) gu=λm0+wn;
(ii) vn =µm0+gz;
(iii) gu−z+ (v−w)n=νm0.Moreover, ν=λ+µ+ǫwhere ǫ= 1 or 0according as
ru−z< ruor ru−z≥ru.
(2) gs+gt=ǫm0+gs+twith ǫ= 1 or 0according as rs+rt≤por rs+rt> p.
In addition to the above notation, set r=ru, r′=ru−z, q =qu, q′=qu−z, and W=
[u−z, u −1] ×[v−w, v −1].Note that if w= 0 or z= 0, then Wis empty.
Lemma 2.5 ([8, Lemma 2.4]).With the above notation,
(1) u≥p+ 1,q≥1and q′∈[0, q].
4 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
(2) rz=(r−r′if r > r′,
p+r−r′if r≤r′.
(3) if q′=qand z6= 0,then r′< r and r≥2.
(4) if µ= 0,then z > p and q′< q.
(5) if λ= 1,then w6= 0.
(6) if λ= 1 and µ= 0,then r′< r and r≥2.
In this article, we aim to describe the structure of RF-matrices for pseudo-Frobenius
numbers of a numerical semigroup Hdefined by an almost arithmetic sequence, using the
above results. We discuss the cases W=∅and W6=∅separately. We use the above results
throughout the paper without repeated reference.
3. The case W=∅
Throughout this section we will assume that W=∅and H=hm0, m1,...,mp, ni; where
mi=m0+id for i∈[1, p] and gcd(m0, n, d) = 1.
Let γk=g(q−1)p+r+k−1+ (v−1)n−m0,for k∈[1, p].In this case, using [8, Proposition
3.3], it follows that the set of pseudo-Frobenius numbers is
PF(H) = (γk|k∈[1, p]if r= 1,
γk|k∈[p−r+ 2, p]if r≥2.
We find the factorizations of γk+mjfor j∈[0, p], and γk+nin H. From these factoriza-
tions, we form an RF-matrix for γk.
Proposition 3.1. Let r= 1 and λ > 1.Then for every element γkof PF(H),k∈[1, p],
there exists an RF-matrix of the following type
RF(γk) =
−1 0 0 ··· ··· ··· 0q−1v−1
0−1 0 ··· ··· ··· 0q−1v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q−1v−1
λ−2 0 0 · · · −1 0 ··· 0w+v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
λ−2 0 0 ··· ··· ··· 0−1w+v−1
ap+2,1ap+2,2ap+2,3··· ··· ··· ··· ap+2,p+1 −1
+
R(0)
R(1)
.
.
.
R(p−k)
R(p−k+1)
.
.
.
R(p)
R(n)
,
where
(1) R(0) = (a(0),...,a(k),...,a(n))such that a(k)= 1 and all other a(j)’s equal to zero.
(2) For all j∈[1, p −k], R(j)= (a(0),...,a(k+j),...,a(n))such that a(k+j)= 1 and all
other a(j)’s equal to zero.
(3) For all j∈[p−k+1, p], R(j)= (a(0) ,...,a(k+j−p−1),...,a(n))such that a(k+j−p−1) = 1
and all other a(j)’s equal to zero.
(4) If µ≥1,then (ap+2,1, ap+2,2,...,ap+2,p+1,−1) = (µ−1,0,...,0, q +qz−1,−1)
and R(n)= (a(0),...,a(k),...,a(rz),...,a(n))such that a(k)=a(rz)= 1 and all
other a(j)’s equal to zero. Otherwise, when µ= 0,(ap+2,1, ap+2,2,...,ap+2,p+1,−1) =
(λ−2,0,...,0, qz−1,−1) and R(n)= (a(0),...,a(k−1) ,...,a(rz),...,a(n))such that
a(k−1) =a(rz)= 1 and all other a(j)’s equal to zero.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 5
Proof. Since r= 1, PF(H) = {γk|k∈[1, p]}where γk=g(q−1)p+k+ (v−1)n−m0.
(1) γk+m0=g(q−1)p+k+ (v−1)n= (q−1)mp+mk+ (v−1)n.
(2) If j∈[1, p −k],then
γk+mj=g(q−1)p+k+ (v−1)n−m0+mj=g(q−1)p+k+gj+ (v−1)n−m0
=m0+g(q−1)p+k+j+ (v−1)n−m0
= (q−1)mp+mk+j+ (v−1)n.
(3) If j=p−k+ 1,then
γk+mj=g(q−1)p+k+ (v−1)n−m0+mp−k+1
= (q−1)mp+mk+ (v−1)n−m0+mp−k+1
= (q−1)mp+ (v−1)n−m0+mp+m1
=gu+ (v−1)n−m0= (λ−1)m0+ (w+v−1)n.
If j > p −k+ 1 and λ > 1,then
γk+mj=g(q−1)p+k+ (v−1)n−m0+mj= (q−1)mp+mk+ (v−1)n−m0+mj
=qmp+ (v−1)n−m0+mk+j−p
=gu+ (v−1)n−2m0+mk+j−p−1
= (λ−2)m0+mk+j−p−1+ (w+v−1)n.
(4) If µ > 0, then
γk+n=g(q−1)p+k+vn −m0=g(q−1)p+k+ (µ−1)m0+gz
= (q−1)mp+mk+ (µ−1)m0+qzmp+mrz
= (µ−1)m0+mk+mrz+ (q+qz−1)mp.
If µ= 0,then z > p. This implies that w= 0, qz−p=qz−1 and rz−p=rz. Thus
γk+n=g(q−1)p+k+vn −m0=g(q−1)p+k−m0+gz
=g(q−1)p+k−m0+gp+gz−p
=gqp+k−m0+gz−p
=qmp+mk−m0+gz−p
=gu+mk−1−2m0+gz−p
= (λ−2)m0+mk−1+ (qz−1)mp+mrz.
6 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
Proposition 3.2. Let r= 1 and λ= 1.Then for every element γkof PF(H), k ∈[1, p],
there exists an RF-matrix of the following type
RF(γk) =
−1 0 0 ··· ··· ··· 0q−1v−1
0−1 0 ··· ··· ··· 0q−1v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q−1v−1
µ−1 0 0 · · · −1 0 ··· 0w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
µ−1 0 0 ··· ··· ··· 0−1w−1
µ−1 0 0 ··· ··· ··· ··· q−1−1
+
R(0)
R(1)
.
.
.
R(p−k)
R(p−k+1)
.
.
.
R(p)
R(n)
,
where
(1) R(0) = (a(0),...,a(k),...,a(n))such that a(k)= 1 and all other a(j)’s equal to zero.
(2) For all j∈[1, p −k],R(j)= (a(0),...,a(k+j),...,a(n))such that a(k+j)= 1 and all
other a(j)’s equal to zero.
(3) For all j∈[p−k+ 1, p],R(j)= (a(0),...,a(k+j−p−1),...,a(n))having a(k+j−p−1) = 1
and all other a(j)’s equal to zero.
(4) R(n)= (a(0),...,a(k),...,a(n))such that a(k)= 1 and all other a(j)’s equal to zero.
Proof. For (1) and (2), see the proof of Proposition 3.1.
(3) Since λ= 1 and r= 1, using lemma 2.5 we get w, µ 6= 0. Also, W=∅implies that
z= 0 and hence ν=µ+ 1.Put ℓ=k+j−p. Since k+j≥p+ 1, we get ℓ∈[1, p] and
γk+mj=g(q−1)p+k+ (v−1)n−m0+mj=g(u+l−1) + (v−1)n−m0
=g(u+l−1) +νm0−gu+ (w−1)n−m0
= (ν−2)m0+mℓ−1+ (w−1)n
= (µ−1)m0+mk+j−p−1+ (w−1)n.
(4) Since z= 0,we have gz= 0 and therefore,
γk+n=g(q−1)p+k+vn −m0=g(q−1)p+k+ (µ−1)m0
= (q−1)mp+mk+ (µ−1)m0.
Corollary 3.3. Suppose p= 1.Then PF(H) = {γ1}and there exists an RF-matrix of γ1of
either of the following type
−1u−1v−1
λ−1−1w+v−1
µ−1u+z−1−1
,
−1u−1v−1
λ−1−1w+v−1
λ−1z−1−1
,or
−1u−1v−1
µ−1w−1
µ−1u−1−1
.
Proof. Since p= 1,we get rt= 1, qt=t−1 for all tand PF(H) = {γ1}.Therefore, we have
k= 1.Let λ > 1.Using Proposition 3.1, we have
γ1+m0=gq+ (v−1)n= (u−1)m1+ (v−1)n,
γ1+m1= (λ−1)m0+ (w+v−1)n.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 7
If µ > 0, then
γ1+n= (µ−1)m0+ (q+qz+ 1)m1= (µ−1)m0+ (u+z−1)m1.
If µ= 0,then γ1+n= (λ−1)m0+ (qz−1+ 1)m1= (λ−1)m0+ (z−1)m1.
Let λ= 1.Using Proposition 3.2, we get
γ1+m0= (u−1)m1+ (v−1)n,
γ1+m1=µm0+ (w−1)n,
γ1+n= (µ−1)m0+qm1.
Thus, we have the above possible three matrices.
Proposition 3.4. Let r≥2.Then for every element γkof PF(H),k∈[p−r+ 2, p],there
exists an RF-matrix of the following type
RF(γk) =
−1 0 0 ··· ··· ··· 0q v −1
0−1 0 ··· ··· ··· 0q v −1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q v −1
λ−1 0 0 · · · −1 0 ··· 0w+v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
λ−1 0 0 ··· ··· ··· 0−1w+v−1
ap+2,1ap+2,2ap+2,3··· ··· ··· ··· ap+2,p+1 −1
+
R(0)
R(1)
.
.
.
R(e)
R(e+1)
.
.
.
R(p)
R(n)
,
where e∈[1, p −1] such that r+k+e= 2p+ 1 and
(1) R(0) = (a(0),...,a(p−e),...,a(n))such that a(p−e)= 1 and all other a(j)’s equal to zero.
(2) For all j∈[1, e],R(j)= (a(0),...,a(p−e+j),...,a(n))such that a(p−e+j)= 1 and all
other a(j)’s equal to zero.
(3) For all j∈[e+ 1, p],R(j)= (a(0),...,a(k+j−p−1),...,a(n))such that a(k+j−p−1) = 1
and all other a(j)’s equal to zero.
(4) If µ≥1then (ap+2,1, ap+2,2,...,ap+2,p+1,−1) = (µ−1,0,...,0, q +qz,−1) and
R(n)= (a(0),...,a(p−e),...,a(rz),...,a(n))such that a(p−e)=a(rz)= 1 and all other
a(j)’s equal to zero. Otherwise, when µ= 0 then (ap+2,1, ap+2,2,...,ap+2,p+1,−1) =
(λ−1,0,...,0, qz−1,−1) and R(n)= (a(0),...,a(k−1) ,...,a(rz),...,a(n))such that
a(k−1) =a(rz)= 1 and all other a(j)’s equal to zero.
Proof. Since r≥2, PF(H) = {γk|k∈[p−r+ 2, p]}.Therefore,
(1) γk+m0=g(q−1)p+r+k−1+ (v−1)n=qmp+mp−e+ (v−1)n.
(2) If j∈[1, e], then r+k+j−p−1≤pand
γk+mj=g(q−1)p+r+k−1+ (v−1)n−m0+mj=m0+qmp+mp−e+j+ (v−1)n−m0
=qmp+mp−e+j+ (v−1)n.
8 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
(3) If j∈[e+ 1, p],then r+k+j−p−1> p and
γk+mj=g(q−1)p+r+k−1+ (v−1)n−m0+mj=gu−p+j+k−1−(v−1)n−m0
=gu−m0+gj+k−p−1−(v−1)n
= (λ−1)m0+wn +mk+j−p−1+ (v−1)n
= (λ−1)m0+mk+j−p−1+ (w+v−1)n.
(4) If µ≥1,then
γk+n=g(q−1)p+r+k−1+ (v−1)n−m0+n=qmp+mp−e+ (µ−1)m0+gz
=qmp+mp−e+ (µ−1)m0+qzmp+mrz
= (µ−1)m0+mp−e+mrz+ (q+qz)mp.
If µ= 0,then z > p. This implies that w= 0, qz−p=qz−1 and rz−p=rz. Thus
γk+n=g(q−1)p+r+k−1+ (v−1)n−m0+n=gu−p+k−1+vn −m0
=gu+gz−p+gk−1−m0
= (λ−1)m0+mk−1+gz−p
= (λ−1)m0+mk−1+ (qz−1)mp+mrz.
Example 3.5. Let H=hm0= 14, m1= 17, m2= 20, m3= 23, m4= 26, n = 21i.Then
H′=h14,17,20,23,26iand p= 4.The Ap´ery set of Hwith respect to 14 is
Ap(H, 14) = {0,17,20,21,23,26,38,41,43,44,46,47,64,67}.
The smallest isuch that gi/∈Ap(H, 14) is g7(= 49).Hence, u= 7, q = 1,and r= 3.
Since 21 /∈H′but 42 ∈H′,we have v= 2.Also, 49 = 21 + 2 ×14,42 = 3 ×14,implies
that w= 1,z= 0, λ = 2, µ = 3, r′= 3, q′= 1,and ν= 5.Since r≥2, we get
PF(H) = {γ3, γ4}={50,53}.Using Proposition 3.4,
γ3+m0=m1+m4+n, γ3+m1=m2+m4+n, γ3+m2=m3+m4+n,
γ3+m3= 2m4+n, γ3+m4=m0+m2+ 2n, γ3+n= 2m0+m1+m4.
We can therefore write an RF-matrix for γ3= 50 from these expressions,
RF(γ3) =
−1 1 0 0 1 1
0−1 1 0 1 1
0 0 −1 1 1 1
000−1 2 1
1 0 1 0 −1 2
21001−1
.
Similarly,
γ4+m0=m2+m4+n, γ4+m1=m3+m4+n, γ4+m2= 2m4+n,
γ4+m3=m0+m2+ 2n, γ4+m4=m0+m3+ 2n, γ4+n= 2m0+m2+m4.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 9
The RF-matrix for γ4= 53 from the above expressions is as follows
RF(γ4) =
−1 0 1 0 1 1
0−1 0 1 1 1
0 0 −1 0 2 1
101−1 0 2
1 0 0 1 −1 2
20101−1
.
Corollary 3.6. Let r= 2. Then H is symmetric and there exists an RF(H)of the following
type
RF(H) =
−1 1 0 0 ··· ··· 0q v −1
0−1 1 0 ··· ··· 0q v −1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 0 0 · · · −1 1 q v −1
0 0 0 0 ··· ··· −1q+ 1 v−1
λ−1 0 0 0 ··· ··· 1−1w+v−1
ap+2,1ap+2,2ap+2,3··· ··· ··· ··· ap+2,p+1 −1
,
where (ap+2,1,...,ap+2,p+1 ,−1) = (µ−1,1,0,...,0, q +qz,−1) +(a(0),...,a(rz),...,a(n))such
that a(rz)= 1 and all other a(j)’s equal to zero when µ > 0. Otherwise, when µ= 0, we have
(ap+2,1,...,ap+2,p+1,−1) = (λ−1,0,...,0,1, qz−1,−1) + (a(0),...,a(rz),...,a(n))
such that a(rz)= 1 and all other a(j)’s equal to zero.
Proof. Since r= 2,we get k=p. Therefore, PF(H) = {γp}. Hence, His symmetric.
Since k=pand r= 2 then r+k+j−p−1> p only for j=p.
Therefore, we have
γp+m0=gqp+1 + (v−1)n=qmp+m1+ (v−1)n,
γp+mj=mj+1 +qmp+ (v−1)n, for all j∈[1, p −1],
γp+mp= (λ−1)m0+mp−1+ (w+v−1)n.
If µ > 1,then γp+n= (µ−1)m0+m1+mrz+ (q+qz)mp,and
if µ= 0,then γp+n= (λ−1)m0+mp−1+mrz+ (qz−1)mp. Hence the result follows.
Example 3.7. Let H=hm0= 10, m1= 19, m2= 28, m3= 37, n = 35i.Then, p= 3, u = 5,
v= 2, w = 1, z = 0, λ = 3, µ = 7, ν = 10, r = 2, q = 1.Since z= 0 and r= 2, therefore
W=∅and PF(H) = {γ3}={81}.
RF(H) =
−1 1 0 1 1
0−1 1 1 1
0 0 −1 2 1
201−1 2
6 1 0 1 −1
.
10 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
4. The case W6=∅
Throughout this section we assume W6=∅and H=hm0, m1,...,mp, ni, where mi=
m0+id for i∈[1, p] and gcd(m0, n, d) = 1.In this case, there are two types of pseudo-
Frobenius numbers. We denote them by PF1(H) and PF2(H).
•Let αi=g(q′−1)p+i+ (v−1)n−m0.Then PF1(H)⊂ {αi;i∈I}, where
I=
pif r′= 1 and q′= 0,
[1, p] if r′= 1 and q′>0,
[p+ 1, p +r′−1] if r′≥2.
•Let βj=g(q−1)p+r+j−1+ (v−w−1)n−m0for j∈[1, p].Then,
PF2(H)⊂(βj|j∈[1, p]if r= 1,
βj|j∈[p−r+ 2, p]if r≥2.
For the complete description of PF1(H) and PF2(H), see [8, Proposition 4.6, 4.7] and [8,
Proposition 5.6, 5.7, 5.8, 5.9].
4.1. Structure of RF-matrices for pseudo-Frobenius numbers in PF1(H).
Proposition 4.1. Let r′= 1, µ > 0and q′= 0.Then PF1(H) = {αp}and there exists an
RF(αp)of the following type
RF(αp) =
−1 0 0 0 ··· ··· 0 0 v−1
ν−1−1 0 0 ··· ··· 0 0 w−1
ν−2 1 −1 0 ··· ··· 0 0 w−1
ν−2 0 1 −1··· ··· 0 0 w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
ν−2 0 0 0 ··· ··· 1−1w−1
ap+2,1ap+2,2ap+2,3··· ··· ··· ··· ap+2,p+1 −1
,
where (ap+2,1,...,ap+2,p+1,−1) = (µ−1,0,··· ,0, qz,−1) + (a(0) ,...,a(rz),...,a(n))such
that a(rz)= 1 and all other a(i)’s equal to zero.
Proof. Since r′= 1 and q′= 0, we get PF1(H) = {αp}.Also, u−z=q′p+r′= 1.
(i) αp+m0=g0+ (v−1)n= (v−1)n.
(ii) For j∈[1, p],
αp+mj= (v−1)n−m0+mj= (v−w)n+ (w−1)n−m0+mj
=νm0−g(u−z)+ (w−1)n−m0+mj
= (ν−2)m0+mj−1+ (w−1)n.
(iii) αp+n= (v−1)n−m0+n= (µ−1)m0+gz= (µ−1)m0+qzmp+mrz.
Hence, the result follows.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 11
Example 4.2. Let H=hm0= 11, m1= 13, m2= 15, m3= 17, m4= 19, n = 21i.Then,
p= 4, u = 5, v = 3, w = 1, z = 4, λ = 1, µ = 4, ν = 5, r = 1, r′= 1, q = 1, q′= 0.Since
r′= 1, q′= 0, we get PF1(H) = {α4}={31}and
RF(31) =
−1 0 0 0 0 2
4−1 0 0 0 0
3 1 −1 0 0 0
301−1 0 0
3001−1 0
30001−1
.
Proposition 4.3. Let r′= 1, µ > 0and q′>0.Then for each αi∈PF1(H),i∈[1, p], there
exists an RF(αi)of the following type
RF(αi) =
−1 0 0 ··· ··· ··· 0q′−1v−1
0−1 0 ··· ··· ··· 0q′−1v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q′−1v−1
ν−2 0 0 · · · −1 0 ··· 0w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
ν−2 0 0 ··· ··· ··· 0−1w−1
µ−1 0 0 ··· ··· ··· ··· q′+qz−1−1
+
R(0)
R(1)
.
.
.
R(p−i)
R(p−i+1)
.
.
.
R(p)
R(n)
,
where
(1) R(0) = (a(0),...,a(i),...,a(n))such that a(i)= 1 and all other a(j)’s equal to zero.
(2) For all j∈[1, p −i],R(j)= (a(0),...,a(i+j),...,a(n))such that a(i+j)= 1 and all
other a(j)’s equal to zero.
(3) For all j∈[p−i+ 1, p],R(j)= (a(0),...,a(i+j−p−1),...,a(n))such that a(i+j−p−1) = 1
and all other a(j)’s equal to zero.
(4) R(n)= (a(0),...,a(i),...,a(rz),...,a(n))such that a(rz), a(i)= 1 and all other a(j)’s
equal to zero.
Proof. Since µ > 0 and q′>0,we have PF1(H) = {αi|i∈[1, p]}.
(1) αi+m0=g(q′−1)p+i+ (v−1)n= (q′−1)mp+mi+ (v−1)n.
(2) If j∈[1, p −i],then i+j≤pand
αi+mj=g(q′−1)p+i+ (v−1)n−m0+mj= (q′−1)mp+mi+ (v−1)n−m0+mj
= (q′−1)mp+mi+j+ (v−1)n.
(3) If j∈[p−i+ 1, p],then i+j > p and
αi+mj=g(q′−1)p+i+ (v−1)n−m0+mj=g(q′−1)p+i+j+ (v−1)n−m0
=gu−z−r′−p+i+j+ (v−w)n+ (w−1)n−m0
=gi+j−p−1+ (ν−2)m0+ (w−1)n
= (ν−2)m0+mi+j−p−1+ (w−1)n.
12 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
(4) Now,
αi+n=g(q′−1)p+i+vn −m0=g(q′−1)p+i+µm0+gz−m0
= (q′+qz−1)mp+ (µ−1)m0+mi+mrz.
Note that if p= 1 and PF1(H)6=∅,then PF1(H) = {α1}.
Corollary 4.4. Suppose p= 1 and PF1(H)6=∅.Then there exists an RF-matrix of α1of
the type
−1u−z−1v−1
ν−1−1w−1
µ−1u−1−1
,or
−1 0 v−1
ν−1−1w−1
µ−1z−1
.
Proof. Since p= 1,we get rt= 1, qt=t−1 for all t. Now use Proposition 4.1 and 4.3 to get
desired RF-matrices accordingly as q′>0 or q′= 0.
Proposition 4.5. Let r′= 1, µ = 0, q′>0and r≥2. Then for every element of PF1(H) =
{αi|i∈[1, p −r+ 1]}, there exists an RF(αi)of the following type
RF(αi) =
−1 0 0 ··· ··· ··· 0q′−1v−1
0−1 0 ··· ··· ··· 0q′−1v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q′−1v−1
ν−2 0 0 · · · −1 0 ··· 0w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
ν−2 0 0 ··· ··· ··· 0−1w−1
0 0 0 ··· ··· ··· 0q−1−1
+
R(0)
R(1)
.
.
.
R(p−i)
R(p−i+1)
.
.
.
R(p)
R(n)
,
where
(1) R(0) = (a(0),...,a(i),...,a(n))such that a(i)= 1 and all other a(j)’s equal to zero.
(2) For all j∈[1, p −i],R(j)= (a(0),...,a(i+j),...,a(n))such that a(i+j)= 1 and all
other a(j)’s equal to zero.
(3) R(n)= (a(0),...,a(r+i−1),...,a(n))such that a(r+i−1) = 1 and all other a(j)’s equal to
zero.
Proof. For (1),(2) and (3), see the proof of Proposition 4.3.
(4) Since r≥2 and r′= 1,we get rz=r−1.Also, we have i∈[1, p −r+ 1] and hence
i+rz≤p. Therefore,
αi+n=g(q′−1)p+i+vn −m0=g(q′−1)p+i+gz−m0
=gu−z−p+i−1+gz−m0
=gu−p+i−1
=g(q−1)p+r+i−1= (q−1)mp+mr+i−1.
Note that PF(H) = ∅in the following cases and so we proceed to the case r′≥2.
(1) r′= 1, µ = 0 and q′= 0,(2) r′= 1, µ = 0, q′>0,and r= 1.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 13
Proposition 4.6. Let r′≥2. Then for each αi∈PF1(H),i∈[p+ 1, p +r′−1], there exists
an RF(αi)of the following type
RF(αi) =
−1 0 0 ··· ··· ··· 0q′v−1
0−1 0 ··· ··· ··· 0q′v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q′v−1
ν−1 0 0 · · · −1 0 ··· 0w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
ν−1 0 0 ··· ··· ··· 0−1w−1
ap+2,1ap+2,20··· ··· ··· ··· ap+2,p+1 −1
+
R(0)
R(1)
.
.
.
R(2p−i)
R(2p−i+1)
.
.
.
R(p)
R(n)
,
where θi∈[1, r′−1] such that i=p+θi.
(1) R(0) = (a(0),...,a(θi),...,a(n))such that a(θi)= 1 and all other a(j)’s equal to zero.
(2) For all j∈[1,2p−i],R(j)= (a(0),...,a(θi+j),...,a(n))such that a(θi+j)= 1 and all
other a(j)’s equal to zero.
(3) For all j∈[2p−i+ 1, p],R(j)= (a(0),...,a(θi+j−r′),...,a(n))such that a(θi+j−r′)= 1
and all other a(j)’s equal to zero.
(4) If µ > 0, then (ap+2,1, ap+2,2,...,ap+2,p+1,−1) = (µ−1,0,...,0, q′+qz,−1) and R(n)=
(a(0),...,a(θi),...,a(rz),...,a(n))such that a(rz), a(θi)= 1 and all other a(j)’s equal to
zero. Otherwise, if µ= 0 then (ap+2,1, ap+2,2,...,ap+2,p+1,−1) = (0,...,0, q′+qz,−1)
and R(n)= (a(0), . . . , a(θi+rz),...,a(n))such that a(θi+rz)= 1 and all other a(j)’s equal
to zero.
Proof. Since r′≥2, we have PF1(H)⊆ {αi|i∈[p+ 1, p +r′−1]}.
(1) αi+m0=g(q′−1)p+i+ (v−1)n=gq′p+θi+ (v−1)n=q′mp+mθi+ (v−1)n.
(2) If j∈[1,2p−i], then i+j≤2pand hence θi+j≤p. Therefore,
αi+mj=g(q′−1)p+i+ (v−1)n−m0+mj=gq′p+θi+ (v−1)n−m0+gj
=gq′p+θi+j+ (v−1)n
=q′mp+mθi+j+ (v−1)n.
(3) If j∈[2p−i+ 1, p], then i+j > 2p. Therefore,
αi+mj=gq′p−p+i+ (v−1)n−m0+mj=gu−z−r′−p+i+gj+ (v−w)n+ (w−1)n−m0
=gu−z+gi+j−p−r′+ (v−w)n+ (w−1)n−m0
= (ν−1)m0+mi+j−p−r′+ (w−1)n.
(4) Assume µ > 0. Then
αi+n=g(q′−1)p+i+vn −m0=g(q′−1)p+i+ (µ−1)m0+gz
=gq′p+θi+ (µ−1)m0+gz
= (µ−1)m0+mrz+mθi+ (qz+q′)mp.
14 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
If µ= 0,then vn =gz. Therefore,
αi+n=g(q′−1)p+i+vn −m0=gq′p+θi+gz−m0
=q′mp+qzmp+mθi+mrz−m0.
Since we have rz+θi≤p, this implies αi+n= (q′+qz)mp+mθi+rz.
4.2. Structure of RF-matrices for pseudo-Frobenius numbers in PF2(H).
Proposition 4.7. Let r= 1 and λ > 1. Then for each βj∈PF2(H),j∈[1, p], there exists
an RF(βj)of the following type
RF(βj) =
−1 0 0 ··· ··· ··· 0q−1v−w−1
0−1 0 ··· ··· ··· 0q−1v−w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q−1v−w−1
λ−2 0 0 · · · −1 0 ··· 0v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
λ−2 0 0 ··· ··· ··· 0−1v−1
ν−ǫ0··· ··· ··· ··· 0qz+j−p−1−1
+
R(0)
R(1)
.
.
.
R(p−j)
R(p−j+1)
.
.
.
R(p)
R(n)
,
where ǫ= 2 or 1according as r′+rz−p+j−1≤por r′+rz−p+j−1> p and
(1) R(0) = (a(0),...,a(j),...,a(n))such that a(j)= 1 and all other a(k)’s equal to zero.
(2) For all k∈[1, p −j],R(k)= (a(0),...,a(j+k),...,a(n))such that a(j+k)= 1 and all
other a(k)’s equal to zero.
(3) For all k∈[p−j+1, p],R(k)= (a(0),...,a(k+j−p−1),...,a(n))such that a(k+j−p−1) = 1
and all other a(k)’s equal to zero.
(4) R(n)= (a(0),...,a(rz+j−p−1),...,a(n))such that a(rz+j−p−1)= 1 and all other a(k)’s
equal to zero.
Proof. Since r= 1, βj=g(q−1)p+j+ (v−w−1)n−m0.
(1) βj+m0= (q−1)mp+mj+ (v−w−1)n.
(2) If k∈[1, p −j],then k+j≤p. Therefore,
βj+mk=g(q−1)p+j+ (v−w−1)n−m0+mk= (q−1)mp+mj+k+ (v−w−1)n.
(3) If k∈[p−j+ 1, p],then k+j≥p+ 1.Given λ > 1,if k+j > p + 1 then
βj+mk=g(q−1)p+j+ (v−w−1)n−m0+mk=gu+k+j−p−1+ (v−1)n+λm0−gu−m0
= (λ−2)m0+gj+k−p−1+ (v−1)n
= (λ−2)m0+mj+k−p−1+ (v−1)n.
If k+j=p+ 1,then βj+mk= (λ−1)m0+ (v−1)n.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 15
(4) We consider two subcases z≥pand z < p.
(a) Assume z≥p.
βj+n=g(q−1)p+j+ (v−w−1)n−m0+n=g(q−1)p+j+ (ν−1)m0−gu−z
=gu−z+z−p+j−1+ (ν−1)m0−gu−z.
Since z≥p, we have z−p+j−1≥0.Therefore,
βj+n=gz−p+j−1−ǫm0+ (ν−1)m0= (ν−1)m0−ǫm0+mrz+j−p−1+qz−p+j−1mp,
with ǫ= 1 or 0 according as r′+rz−p+j−1≤por r′+rz−p+j−1> p.
(b) If z < p, then z=rz=p+r−r′=p+1−r′.Also in this case PF2(H) = {βj|j∈[r′, p]}
(see [8, Proposition 5.7]). Since j≥r′=p−z+ 1,we get z−p+j−1≥0.Hence we are
done by subcase(a).
Proposition 4.8. Let r= 1 and λ= 1. Then for each βj∈PF2(H),j∈[1, p], there exists
an RF(βj)of the following type
RF(βj) =
−1 0 0 ··· ··· ··· 0q−1v−w−1
0−1 0 ··· ··· ··· 0q−1v−w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q−1v−w−1
λ−1 0 0 · · · −1 0 ··· 0v−1
ν−2 0 0 ··· ··· ··· 0−1w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
ν−2 0 0 ··· ··· ··· 0−1w−1
ν−ǫ0··· ··· ··· ··· 0qz+j−p−1−1
+
R(0)
R(1)
.
.
.
R(p−j)
0
R(p−j+2)
.
.
.
R(p)
R(n)
,
where ǫ= 2 or 1according as r′+rz−p+j−1≤por r′+rz−p+j−1> p and
(1) R(0) = (a(0),...,a(j),...,a(n))such that a(j)= 1 and all other a(k)’s equal to zero.
(2) For all k∈[1, p −j],R(k)= (a(0),...,a(j+k),...,a(n))such that a(j+k)= 1 and all
other a(k)’s equal to zero.
(3) For all k∈[p−j+2, p],R(k)= (a(0),...,a(k+j−p−2),...,a(n))such that a(k+j−p−2) = 1
and all other a(k)’s equal to zero.
(4) R(n)= (a(0),...,a(rz+j−p−1),...,a(n))such that a(rz+j−p−1)= 1 and all other a(k)’s
equal to zero.
Proof. Same as 4.7 except when j+k≥p+ 2.
If j+k=p+ 2,then βj+mk= (ν−1)m0+ (w−1)n.
If j+k > p +2,then βj+mk= (ν−2)m0+gj+k−p−2+ (w−1)n. Hence the result follows.
Note that if p= 1,then PF2(H) = {β1}.
Corollary 4.9. Suppose p= 1. Then there exists an RF-matrix of β1of the following type
−1u−1v−w−1
λ−1−1v−1
ν−1z−1−1
.
Proof. Use Proposition 4.7 and 4.8.
16 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
Let Hbe a numerical semigroup generated by a sequence m0,...,me. Let kbe a field.
The semigroup ring k[H] of His a k-subalgebra of the polynomial ring k[t].In other words,
k[H] = k[tm0, tm1,...,tme].Set R=k[x0,...,xe] and define a map π:R→k[H] given by
π(xi) = tmifor all i= 0,...,e. Set deg xi=mifor all i= 0,...,e. Observe that Ris a
graded ring and that πis a degree preserving surjective k-algebra homomorphism. We say
kernel of πis defining ideal of k[H],denoted by I(H).
For a vector b= (b1, . . . , bn)∈Zn, we let b+denote the vector whose i-th entry is biif
bi>0, and is zero otherwise, and we let b−=b+−b. Then b=b+−b−with b+, b−∈Nn.
For a pseudo-Frobenius element f, the rows of RF(f) produce binomials in I(H).
Definition 4.10. For some pseudo-Frobenius element fof H, let δ1,...,δndenote the row
vectors of RF(f).Set δ(ij)=δj−δi, for all 1≤i < j ≤n. Then φij =xδ+
(ij)−xδ−
(ij)∈I(H)
for all i < j. We call φij an RF(f)-relation. We call a binomial relation φ∈I(H)an
RF-relation if it is an RF(f)-relation for some f∈PF(H).
Proposition 4.11. If p= 1 and µ > 0,then I(H)is minimally generated by RF(β1)-
relations.
Proof. Since p= 1 and µ > 0,using [8, Proposition 4.6, 5.6] it follows that PF1(H) = {α1}
and PF2(H) = {β1}.Therefore, we get PF(H) = {α1, β1}and His not symmetric. Also, as
p= 1,we have qt=t−1 for all t. In [9, Theorem 4.5], it is proved that in the above setup,
I(H) = hxν
0−xq′+1
1xv−w
2, xq+1
1−xλ
0xw
2, xv
2−xµ
0xq−q′
1i.
In order to prove the result, it is sufficient to show that these generators arise from either
RF(α1) or RF(β1)-relations. Consider the matrix
RF(β1) :=
−1u−1v−w−1
λ−1−1v−1
ν−1z−1−1
.
If δ1, δ2, δ3represent the row vectors of RF(β1), then we have the following RF(β1)-relations:
δ12 =xu
1−xλ
0xw
2, δ13 =xν
0−xu−z
1xv−w
2and δ23 =xv
2−xµ
0xz
1.Hence the result.
Proposition 4.12. Let r≥2. Then for each βj∈PF2(H),j∈[p−r+ 2, p], there exists
an RF(βj)of the following type
RF(βj) =
−1 0 0 ··· ··· ··· 0q v −w−1
0−1 0 ··· ··· ··· 0q v −w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 · · · −1 0 ··· 0q v −w−1
λ−1 0 0 · · · −1 0 ··· 0v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
λ−1 0 0 ··· ··· ··· 0−1v−1
ν−ǫ0··· ··· ··· ··· 0qz+j−p−1−1
+
R(0)
R(1)
.
.
.
R(e)
R(e+1)
.
.
.
R(p)
R(n)
,
where e∈[1, p −1] such that r+j+e= 2p+ 1 and ǫ= 2 or 1according as r′+rz−p+j−1≤p
or r′+rz−p+j−1> p, and
(1) R(0) = (a(0),...,a(p−e),...,a(n))such that a(p−e)= 1 and all other a(k)’s equal to zero.
(2) For all k∈[1, e],R(k)= (a(0),...,a(p−e+k),...,a(n))such that a(p−e+k)= 1 and all
other a(k)’s equal to zero.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 17
(3) For all k∈[e+ 1, p],R(k)= (a(0),...,a(j+k−p−1),...,a(n))such that a(j+k−p−1) = 1
and all other a(k)’s equal to zero.
(4) R(n)= (a(0),...,a(rz+j−p−1),...,a(n))such that a(rz+j−p−1)= 1 and all other a(k)’s
equal to zero.
Proof. Use similar arguments as in Proposition 3.4 and Proposition 4.7.
5. Generic toric ideals
Let R=k[x0,...,xn] be a polynomial ring over the field k. A binomial ideal is said to
be generic if it has a minimal generating set consisting binomials of full support (i.e each
variable should appear in each binomial of the generating set). This concept was introduced
by Peeva and Sturmfels in [11]. In this section, we give a characterization of the generic
nature of toric ideals when the semigroup is minimally generated by an almost arithmetic
sequence. We use the following result of Eto for the same, which gives a necessary and
sufficient condition for a toric ideal to be generic.
Theorem 5.1 ([3, Theorem 1]).Let Hbe a numerical semigroup. Then I(H)is generic if
and only if for each f∈PF(H),RF(f) = (ai,j )is unique and ai,j 6=ai′,j if i6=i′.
Example 5.2. Let H=h5,6,7i.Then PF(H) = {8,9}and
RF(8) =
−1 1 1
0−1 2
3 0 −1
,RF(9) =
−1 0 2
3−1 0
2 1 −1
,
where RF(8) and RF(9) are unique RF-matrices. Hence by Theorem 5.1,I(H)is generic.
On the other hand, I(H) = (x2
1−x0x2, x4
0−x1x2
2, x3
0x1−x3
2)and hence is generic.
Theorem 5.3. Let Hbe a numerical semigroup minimally generated by an almost arithmetic
sequence, i.e H=hm0, . . . , mp, ni, where mi=m0+id for i∈[1, p]and gcd(m0, n, d) = 1.
(1) If p= 0,then I(H)is generic.
(2) If p= 1,and if W6=∅, µ > 0,then I(H)is generic. Otherwise, it is never generic.
(3) If p > 1, then I(H)is not generic.
Proof. (1) It is obvious.
(2) If p= 1, W 6=∅,and µ > 0,then by Proposition 4.11, we conclude that I(H) is generic.
If W6=∅and µ= 0,then λ=νand by Corollary 4.9 and Theorem 5.1 we get I(H) is not
generic.
If W=∅,then either w= 0 or z= 0.Using Corollary 3.3 and Theorem 5.1, we conclude
that I(H) is not generic.
(3) In the Propositions 3.1,3.2,3.4,4.1,4.3,4.5,4.6,4.7,4.8, and 4.12, we see that for each
f∈PF(H),there exists an RF-matrix RF(f) = (ai,j ) such that ai,p+2 =ai′,p+2 for some
i6=i′.Hence by Theorem 5.1,I(H) is not generic.
We now prove that the defining ideal of a gluing of two numerical semigroups is never
generic. As a consequence, it is easy to see that a complete intersection toric ideal is not
generic. We recall the definition of gluing first.
18 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
Definition 5.4 ([12]).Let H1and H2be two numerical semigroups minimally generated
by {m0,...,me1}and {m′
0,...,m′
e2}respectively. Let d1∈H2\ {m′
0,...,m′
e2}and d2∈
H1\ {m0,...,me1}be such that gcd(d1, d2) = 1.We say that
H=hd1m0,...,d1me1, d2m′
0,...,d2m′
e2i
is a gluing of H1and H2.
Note that if H1=H2=N, then the embedding dimension of His two and hence I(H) is
a generic ideal. Therefore, we assume that either H16=Nor H26=N.
Theorem 5.5. Let H1=hm0,...,me1iand H2=hm′
0,...,m′
e2ibe two numerical semi-
groups such that e2≥1and Hbe the gluing of H1and H2.Then I(H)is never generic.
Proof. Let H=d1H1+d2H2,where d1∈H2\ {m′
0,...,m′
e2}and d2∈H1\ {m0,...,me1}
such that gcd(d1, d2) = 1.We will show that there exists an RF-matrix for a f∈PF(H) such
that aij =ai′jfor some i6=i′.Let f1∈PF(H1) and f2∈PF(H2).Set g=d1f1+d2f2+d1d2.
Then g∈PF(H), see [13, Theorem 2.4]. Since f1∈PF(H1),we get f1+d2∈H1. Therefore,
there exist a0,...,ae1∈Z≥0such that f1+d2=Pe1
i=0 aimi. Similarly, for f2+d1∈H2,
there exist a′
0,...,a′
e2∈Z≥0such that f2+d1=Pe2
i=0 a′
im′
i.Let RF(f1) be an RF-matrix of
f1and RF(f2) be an RF-matrix of f2. Then,
RF(f1)
a′
0··· a′
e2
.
.
..
.
..
.
.
a′
0··· a′
e2
a0··· ae1
.
.
..
.
..
.
.
a0··· ae1
RF(f2)
defines an RF-matrix for g. Since a1,j =a2,j =... =ae1+1,j for each j∈[e1+ 2, e1+e2+ 2],
and ae1+2,j =ae1+3,j =... =ae1+e2+2,j for each j∈[1, e1+ 1].Then by Theorem 5.1, we
conclude that I(H) is not generic.
Corollary 5.6. Let Hbe a complete intersection numerical semigroup with embedding di-
mension at least 3. Then I(H)is not generic.
Proof. By [12, Theorem 9.10], we have that a numerical semigroup (other than N) is a
complete intersection if and only if it is a gluing of two complete intersection numerical
semigroups. Now using Theorem 5.5, we are done.
6. RF-relations when type(H)=1
In this section, we consider those numerical semigroups generated by an almost arith-
metic sequence which are symmetric. In particular, let H=hm0, m1,...,mp, ni, where
mi=m0+id for i∈[1, p] and gcd(m0, n, d) = 1 such that type(H) = |PF(H)|= 1.Using
the language from the previous sections, we list the different cases in which type(H) = 1
and also give a possible RF-matrix of the Frobenius number in each case. The aim of this
section is to prove that I(H) has a minimal generating set consisting of RF-relations in the
above setup if the embedding dimension is either 4 or 5.
ON ROW-FACTORIZATION MATRICES AND GENERIC IDEALS 19
Let Hbe a numerical semigroups generated by an almost arithmetic sequence. Then
type(H) = 1 in the following cases:
Case 1: Let W=∅.In this case, if r= 1, then His of type 1 only if p= 1 and
PF(H) = {γ1}. If r≥2,then His of type 1 only if r= 2 and PF(H) = {γp}.
Case 2: Let W6=∅.If p= 1,then His of type 1 only if µ= 0 and PF(H) = {β1}.
Suppose p≥2.The only cases in which His of type 1 are the following:
(i) r= 1, r′= 2, λ = 1, z < p, and PF(H) = {αp+1}.
(ii) r= 2, r′= 1, µ = 0, q′= 0, and PF(H) = {βp}.
(iii) r= 2, r′= 2, µ = 0, and PF(H) = {βp}.
We now consider each case separately and list an RF-matrix of the Frobenius number in
each case.
Case 1: Let W=∅.
Let p= 1, r = 1 and PF(H) = {γ1}. Then we have choices for the RF(γ1)-matrix as given
in Corollary 3.3.
Let r= 2 and p≥2.Then there exists an RF-matrix of γpas given in Corollary 3.6.
Case 2: Let W6=∅.
Suppose p= 1 and µ= 0.Then there exists an RF-matrix of β1as given in Corollary 4.9.
Let p≥2.We have the following three sub-cases:
Sub-case (i): Let r= 1, r′= 2, λ= 1 and z < p and PF(H) = {αp+1}.As a consequence,
rz=p−1,implying that z=p−1 and qz= 0. Also, λ= 1 and r′> r implies µ6= 0.Using
Proposition 4.6, there exists an RF-matrix of αp+1 of the following type
−1 1 0 0 ··· 0 0 q′v−1
0−1 1 0 ··· 0 0 q′v−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 0 ··· ··· 0−1q′+ 1 v−1
ν−1 0 0 ··· ··· 0 1 −1w−1
µ−1 1 0 ··· ··· 0 1 q′−1
.
Sub-case (ii): Let r= 2, r′= 1, µ= 0, q′= 0 and PF(H) = {βp}.Then rz= 1 and
hence rz−1=pand qz−1=qz−1.Since q′= 0,we get u−z=r′= 1 which implies
z=u−1 = qp + 1 and hence qz=q. Also r′+rz−1=p+ 1 > p implies ǫ= 1 and as µ= 0
and r′< r, we get ν=λ+ 1.By Proposition 4.12, there exists an RF-matrix of βpof the
following type
20 OM PRAKASH BHARDWAJ, KRITI GOEL, AND INDRANATH SENGUPTA
−1 1 0 0 ··· 0 0 q v −w−1
0−1 1 0 ··· 0 0 q v −w−1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
0 0 0 ··· ··· 0−1q+ 1 v−w−1
λ−1 0 0 ··· ··· 0 1 −1v−1
λ0 0 ··· ··· 0 0 q−1
.
Sub-case (iii): Let r= 2, r′= 2, µ = 0 and PF(H) = {βp}.Then rz=pand hence
rz−1=p−1 and qz−1=qz.Also, r′+rz−1=p+ 1 > p implies ǫ= 1 and as µ= 0 and r=r′,
we get ν=λ. By Proposition 4.12, there exists an RF-matrix of βpof the following type
−1 1 0 0