arXiv:1710.09298v1 [] 25 Oct 2017

GLUING SEMIGROUPS AND STRONGLY
INDISPENSABLE FREE RESOLUTIONS
MESUT ŞAHİN AND LEAH GOLD STELLA
Abstract. We study strong indispensability of minimal free resolutions
of semigroup rings. We focus on two operations, gluing and extending,
used in literature to produce more examples with a special property from
the existing ones. We give a naive condition to determine whether gluing of two semigroup rings has a strongly indispensable minimal free
resolution. As applications, we determine extensions of 3-generated
non-symmetric, 4-generated symmetric and pseudo symmetric numerical semigroups as well as obtain infinitely many complete intersection
semigroups of any embedding dimension, having strongly indispensable
minimal free resolutions.

1. introduction
Let N denote the set of non-negative integers and consider the affine
semigroup S generated minimally by m1 , . . . , mn ∈ Nr . Turning the additive
structure of S into a multiplicative one yields an algebra K[S], for a field
K, called the affine semigroup ring associated to S. Any polynomial ring
A = K[x1 , . . . , xn ], can be graded by S, via degS (xi ) = mi , yielding a graded
ir
map A → K[S], sending xi to tmi := t1mi1 · · · tm
r , whose kernel denoted by
IS is called the toric ideal of S. Then, K[S] is isomorphic to the coordinate
ring A/IS of the affine toric variety V (IS ).
Toric ideals generated by indispensable binomials are of special importance for some emerging problems arising from Algebraic Statistics, as they
have a unique minimal generating set. This leads to a search for criteria to
characterize indispensability, see e.g. [5, 6, 9, 12, 15, 21]. Indispensable binomials are constant multiples of those binomials that appear in every minimal
binomial generating set. Strongly indispensable binomials are those whose
constant multiples are present in every (non necessarily binomial) minimal
generating set. In the same vein, as introduced for the first time in [3, 4],
indispensable (rest. strongly indispensable) higher syzygies are those that
are present in every (resp. non-necessarily) simple minimal free resolution.
Semigroups all of whose higher syzygy modules are generated minimally by
strongly indispensable elements are said to have a strongly indispensable
minimal free resolution, SIFRE for short. The statistical models having
Date: October 26, 2017.
The first author is supported by the project 114F094 under the program 1001 of the
Scientific and Technological Research Council of Turkey.
1

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MESUT ŞAHİN AND LEAH GOLD STELLA

SIFREs or equivalently having uniquely generated higher syzygy modules
are a subclass of those having a unique Markov basis and therefore have a
better potential statistical behaviour.
It is difficult to construct examples having SIFRE. Generic lattice ideals
have SIFRE by [16, Theorem 4.2] and [3, Theorem 4.9]. Numerical semigroups having SIFREs are classified in [2, 19] for some small embedding
dimensions.
In this article, we focus on two operations, gluing and extending, used
in literature to produce more examples with a special property from the
existing one, see e.g. [22, 13, 18, 14, 8]. In section 2, we rephrase the general
method given by [3, Theorem 4.9] to check if a given semigroup has a SIFRE,
see Lemma 2.1. In section 3, we study the gluing S of S1 and S2 . We show
that a minimal graded free resolution for K[S] is obtained from that of K[S1 ]
and K[S2 ] via tensor product of three complexes, see Theorem 3.2. As a
consequence we get the Betti S-degrees, see Lemma 3.3, which is key for our
refined criterion. We then give a naive criterion to determine whether K[S]
has a SIFRE, see Theorem 3.5. This section generalizes the main results of
[9], see Remark 3.6. We conclude the section with Example 3.7 illustrating
the main advantage of our criterion over method given by [3, Theorem 4.9].
In the last section, we focus on a particular gluing also known as extension
and get an even more refined criterion special for them. It turns out that
this condition is very helpful for producing infinitely many examples having
SIFRE from a single example. As applications, we determine extensions of
3-generated non-symmetric, 4-generated symmetric and pseudo symmetric
numerical semigroups as well as obtain infinitely many complete intersection
semigroups of any embedding dimension, having SIFREs.
2. strongly indispensable minimal free resolutions
Motivated by the open questions listed at the end of [4], our main aim
here is to refine Theorem 4.9 in [3] for checking strong indispensability of
minimal free resolutions of affine semigroup rings. For a graded minimal
free A-resolution
φ

φk−1

φ2

φ1

k
F : 0 −→ Aβk −→
Aβk−1 −→ · · · −→ Aβ1 −→ Aβ0 −→K[S]−→0

of K[S], let Aβi be generated in degrees si,j ∈ S, which we call i-Betti Sβi
M
βi
A[−si,j ]. The resolution (F, φ) is called strongly
degrees, i.e. A =
j=1

indispensable if for any graded minimal resolution (G, θ), we have an
injective complex map i : (F, φ) −→ (G, θ).
The following general criterion about strong indispensability is nothing
but Theorem 4.9 in [3] stated slightly different. Denote by pd(S) the projective dimension of K[S].

GLUING SEMIGROUPS AND STRONGLY INDISPENSABLE FREE RESOLUTIONS 3

Lemma 2.1. A minimal graded free resolution of K[S] is strongly indispensable if and only if the differences between different i-Betti degrees do
not belong to S for all 1 ≤ i ≤ pd(S).
Proof. This follows from Theorem 4.9 in [3], as the differences between different i-Betti degrees do not belong to S means that i-Betti degrees are
different and minimal.

When S is symmetric, it is sufficient to check the condition above for the
first half of the indices.
Lemma 2.2. If S is symmetric, then K[S] has a SIFRE if and only if the
differences between different i-Betti degrees do not belong to S, for 1 ≤ i ≤
⌊pd(S)/2⌋.
Proof. The proof of [2, Lemma 21] extends from numerical semigroups to
arbitrary affine semigroups, as it uses the symmetry in the minimal graded
free resolution of K[S], which is true for any graded Gorenstein K-algebra
by Stanley’s second proof of Theorem 4.1. in [20].

3. Gluing Strongly Indispensable Resolutions
Let S1 = N{a1 , . . . , am } and S2 = N{b1 , . . . , bn } be two affine semigroups. If there is an α ∈ S1 ∩S2 such that ZS1 ∩ZS2 = Zα then S = S1 +S2
is said to be the gluing of S1 and S2 . When
α = u1 a1 + · · · + um am = v1 b1 + · · · + vn bn ,
the binomial fα = xu1 1 · · · xumm − y1v1 · · · ynvn has S-degree α and the toric ideal
IS = IS1 + IS2 + hfα i ⊂ R = K[x1 , . . . , xm , y1 , . . . , yn ], see [17]. Let
φp

φ2

φ1

F : 0 → Fp → · · · → F1 → F0 →0
be a minimal S1 graded minimal free resolution of IS1 with H0 (F) = R/IS1 ,
Φq

Φ

Φ

G : 0 → Gq → · · · →2 G1 →1 G0 →0
be a minimal S2 graded minimal free resolution of IS2 with H0 (G) = R/IS2 .
Our aim is to compute a minimal S graded minimal free resolution of IS
using the complexes F and G. Since IS = IS1 + IS2 + hfα i the idea is to
fα

tensor these complexes and the complex Cfα : 0 → R −→ R → 0. This shall
work if fα is a non zerodivisor on R/(IS1 + IS2 ) which we address first.
Lemma 3.1. The gluing binomial fα is a non zerodivisor on R/(IS1 + IS2 ).
Proof. For
let fα = xu −yv = xu1 1 · · · xumm −y1v1 · · · ynvn .
Pnotationalz convenience,
Let g = z,w cz,w x yw ∈ R with gfα ∈ IS1 + IS2 . As IS1 + IS2 is gener′
′
ated by binomials of P
the form xz − xz and yw − yw , these appear in the
z+u yw − xz yw+v ). In other words, each
expansion of gfα =
z,w cz,w (x
′
′
′
′
monomial xz+u yw has a match of type xz +u yw or xz yw +v such that
′
′
′
′
′
′
xz+u yw −xz +u yw or xz+u yw −xz yw +v is divisible by xz −xz or yw −yw .

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MESUT ŞAHİN AND LEAH GOLD STELLA

In the first case, this is possible only if z = z′ or w = w′ . If z = z′ , then
′
′
′
′
xz+u yw − xz +u yw = xz+u (yw − yw ) = xu (xz yw − xz yw ). This means
′
′
that the term xz yw of g has a match xz yw such that xz yw − xz yw is
′
divisible by yw − yw . One can prove similarly that this happens for the
other cases. Hence, terms in g may be rearranged so that it is an algebraic
′
′
combination of binomials xz − xz and yw − yw , that is g ∈ IS1 + IS2 . 
We are now ready to prove the following key result.
Theorem 3.2. Let S be the gluing of S1 and S2 . If F is a minimal S1 graded
minimal free resolution of IS1 and G is a minimal S2 graded minimal free
resolution of IS2 , then Cfα ⊗ F ⊗ G is an S graded minimal free resolution
of IS .
Proof. Recall that the tensor product of F and G is a complex
δp+q

δ

δ

2
1
F ⊗ G : 0 −→ Fp ⊗ Gq −→ · · · −→
F1 ⊗ G0 ⊕ F0 ⊗ G1 −→
F0 ⊗ G0 −→0

with terms (F ⊗ G)i = ⊕p+q=i Fp ⊗ Gq and maps given by
X
X
δi (
ap ⊗ bq ) =
φp (ap ) ⊗ bq + (−1)p ap ⊗ Φq (bq ).
p+q=i

p+q=i

It is well known that Hi (F ⊗ G) = Hi (F ⊗ R/IS2 ) where F ⊗ R/IS2 is the
complex
∆p

∆

∆

0 → Fp ⊗ R/IS2 → · · · →2 F1 ⊗ R/IS2 →1 F0 ⊗ R/IS2 →0,
where ∆i (ai ⊗ b) = φi (ap ) ⊗ b. It is easy to see that
Ker(∆i ) = [Ker(φi ) ⊗ R/IS2 ] ∪

ri−1
M
−1
IS2 )
[(φi (
j=1

⊗ R/IS2 ],

where ri−1 = rank(Fi−1 ), and Im(∆i ) = Im(φi ) ⊗ R/IS2 . Since, Im(φi )
involves the variables xj only and IS2 involves the variables yk only, it follows
ri−1
that φ−1
i (⊕j=1 IS2 ) = {0}. Thus, Hi (F ⊗ G) = Hi (F ⊗ R/IS2 ) = 0, for all
i > 0. Since H0 (F ⊗ G) = R/IS1 ⊗ R/IS2 ∼
= R/(IS1 + IS2 ), it follows that
F ⊗ G is an S graded minimal free resolution of R/(IS1 + IS2 ).
Let f = fα for notational convenience. As before, Hi (Cf ⊗ F ⊗ G) =
Hi (Cf ⊗ R/(IS1 + IS2 )), where Cf ⊗ R/(IS1 + IS2 ) is
f ⊗1

0 → R ⊗ R/(IS1 + IS2 ) −→ R ⊗ R/(IS1 + IS2 ) → 0.
Note that H1 (Cf ⊗ R/(IS1 + IS2 )) = ((IS1 + IS2 ) : f ) ⊗ R/(IS1 + IS2 ) = {0}
as f is a non zerodivisor on R/(IS1 + IS2 ). Since H0 (Cf ⊗ R/(IS1 + IS2 )) ∼
=
R/(IS1 + IS2 + hf i), it follows that Cf ⊗ F ⊗ G gives an S graded minimal
free resolution of IS .

Denote by Bi (S) the set of i−Betti S-degrees of a minimal free resolution
of K[S] for 1 ≤ i ≤ pd(S) and let Bi (S) = {0} otherwise.

GLUING SEMIGROUPS AND STRONGLY INDISPENSABLE FREE RESOLUTIONS 5

Lemma 3.3. Let S be the gluing of S1 and S2 . Then,

 

[
[
Bp (S1 ) + Bq (S2 ) + {α} .
Bp (S1 ) + Bq (S2 ) ∪ 
Bi (S) = 
p+q=i−1

p+q=i

Proof. By Theorem 3.2, Cfα ⊗ F ⊗ G is an S graded minimal free resolution
of IS . Hence, the proof follows from the following
M
M
R ⊗ Fp ⊗ Gq +
R(−α) ⊗ Fp ⊗ Gq ,
(Cfα ⊗ F ⊗ G)i =
p+q=i

p+q=i−1

since S degrees of elements in Fp ⊗Gq constitute the set Bp (S1 )+Bq (S2 ).



We use the following simple observation in the proof of our main result.
Lemma 3.4. Let S be the gluing of S1 and S2 . Fix j ∈ {1, 2}, and b, b′ ∈ Sj .
Then, b − b′ ∈ Sj ⇐⇒ b − b′ ∈ S.
Proof. WLOG, assume that j = 1. As S1 ⊂ S, b − b′ ∈ S1 ⇒ b − b′ ∈ S.
For the converse, take b, b′ ∈ S1 with b − b′ ∈ S. Then, b − b′ = s1 + s2 , for
some s1 ∈ S1 and s2 ∈ S2 , since S = S1 + S2 . So, s2 = b − b′ − s1 ∈ ZS1 .
Since ZS1 ∩ ZS2 = Zα and s2 ∈ S2 , we have s2 = kα for a positive integer
k. Hence, b − b′ = s1 + kα ∈ S1 .

We are now ready to prove our main result which gives a practical method
to produce infinitely many affine semigroups having SIFRE.
Theorem 3.5. Let bi,j denotes an element of Bi (Sj ) for i = 1, . . . , pd(Sj ),
j = 1, 2. Then, IS has SIFRE ⇐⇒ IS1 and IS2 have SIFRE and the
following hold
(1) ∓(α + bi−1,j − bi,j ) ∈
/ Sj ,
(2) ∓(bp,1 + bq,2 − b′r,1 − b′s,2 ) ∈
/ S, for p − r ≥ 2, where p + q = i = r + s,
′
′
/ S for p − r ≥ 2, where p + q = i =
(3) ∓(bp,1 + bq,2 − br,1 − bs,2 − α) ∈
r + s + 1.
Proof. Let us prove necessity first. By Lemma 2.1, the differences between
the elements in Bi (S) do not belong to S. Let bi,j , b′i,j ∈ Bi (Sj ), for j = 1, 2.
By Lemma 3.3, we have Bi (Sj ) ⊂ Bi (S), and thus bi,j −b′i,j ∈
/ S. This implies
bi,j − b′i,j ∈
/ Sj , by Lemma 3.4, which means that IS1 and IS2 have SIFRE by
the virtue of Lemma 2.1. As the elements in the conditions (1)-(3) are the
differences of some elements in Bi (S), they do not belong to S. So, (2)-(3)
hold. Lemma 3.4 implies (1) now.
Let us prove sufficiency now. If b, b′ ∈ Bi (S), then there are three possibilities due to Lemma 3.3:
(i) b, b′ ∈ Bp (S1 ) + Bq (S2 ), for p + q = i,
(ii) b, b′ ∈ Bp (S1 ) + Bq (S2 ) + α, for p + q = i − 1,
(iii) b ∈ Bp (S1 ) + Bq (S2 ), b′ ∈ Br (S1 ) + Bs (S2 ) + α, for p + q = i = r + s + 1.
Case (i): Let b = bp,1 + bq,2 and b′ = b′r,1 + b′s,2 with p + q = i = r + s.
Suppose now that b − b′ ∈ S. Then, b − b′ = bp,1 + bq,2 − b′r,1 − b′s,2 = s1 + s2 ,

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MESUT ŞAHİN AND LEAH GOLD STELLA

for some s1 ∈ S1 and s2 ∈ S2 . Thus, bp,1 − b′r,1 − s1 = s2 − bq,2 + b′s,2 = kα
being an element of ZS1 ∩ ZS2 = Zα. By the condition (2), we need only to
check the difference for p = r and p = r + 1.
When p = r, it follows that bp,1 − b′p,1 = s1 + kα ∈ S1 if k ≥ 0, and that
bq,2 − b′q,2 = s2 + (−k)α ∈ S2 if k < 0, contradicting to hypothesis by Lemma
2.1.
When p = r + 1, it follows that br+1,1 − b′r,1 − s1 = s2 − bq,2 + b′q+1,2 = kα.
Since the resolution of IS2 is S2 -graded, there is s′2 ∈ S2 such that b′q+1,2 =
bq,2 + s′2 . So, kα = s2 + s′2 ∈ S2 . Since S2 ∩ (−S2 ) = {0}, we have k > 0.
But then, br+1,1 − b′r,1 − α = s1 + (k − 1)α ∈ S1 , contradicts to the condition
(1).
Case (ii): follows from Case (i).
Case (iii): Let b = bp,1 +bq,2 and b′ = b′r,1 +b′s,2 +α with p+q = i = r+s+1.
Suppose that b−b′ ∈ S. Then, b−b′ = bp,1 +bq,2 −b′r,1 −b′s,2 −α = s1 +s2 , for
some s1 ∈ S1 and s2 ∈ S2 . It follows that bp,1 −b′r,1 −s1 = s2 −bq,2 +b′s,2 +α =
kα, for some k ∈ Z. By the condition (3), we need only to check the
difference for p = r and p = r + 1.
When p = r, we have br,1 − b′r,1 = s1 + kα ∈ S1 if k > 0, and bs+1,2 −
′
bs,2 − α = s2 + (−k)α ∈ S2 if k ≤ 0, which give rise to contradictions.
When p = r + 1, we have br+1,1 − b′r,1 − α = s1 + (k − 1)α ∈ S1 if k > 0,
and bs,2 − b′s,2 = s2 + (−k)α ∈ S2 if k ≤ 0, which give rise to contradictions.
One proves that b′ − b ∈
/ S, similarly.

Remark 3.6. This section generalizes the main results of [9] as we briefly
explain now. Our Lemma 3.3 specializes to B1 (S) = B1 (S1 ) ∪ B1 (S2 ) ∪ {α},
which is exactly [9, Theorem 10]. Furthermore the condition (1) in our
Theorem 3.5, specializes to ∓(α − b1,j ) ∈
/ Sj , since b0,j = 0, for j = 1, 2.
This is exactly the condition in [9, Theorem 12] by the virtue of Lemma 3.4.
The following illustrates the advantage of our criterion over the general
method given by [3, Theorem 4.9].
Example 3.7. S1 = hq · 31, q · 37, q · 41i, S2 = hp · 4, p · 5i has SIFRE. But
it is difficult to determine p and q for which S = S1 + S2 has SIFRE using
[3, Theorem 4.9] since it requires checking if 68 elements does not lie in S
for every choice of p and q, and gaps of S depend on p and q. On the other
hand, the conditions (2) − (3) of Theorem 3.5 hold automatically and it is
sufficient to check if 15 elements do not belong to one of h31, 37, 41i and
h4, 5i. Note that this time everything is independent of p and q. One can
use gaps of h31, 37, 41i and h4, 5i to see only q = 109 or q = 150 yield (1)
to holds. When q = 109 taking p = 9 and for q = 150 assuming p = 13
produces gluings with SIFREs. Similarly one can see using our criterion
that h6, 7, 10i or h8, 9, 11i does not give rise to a gluing with SIFRE.

GLUING SEMIGROUPS AND STRONGLY INDISPENSABLE FREE RESOLUTIONS 7

4. Extending Strongly Indispensable Resolutions
We determine some semigroups having SIFREs in this section. We focus
on a particular case of gluing where the second semigroup is generated by
a single element. These semigroups are known as extensions in literature.
Given an affine semigroup S generated minimally by m1 , . . . , mn , recall
that an extension of S is an affine semigroup denoted by E and generated
minimally by ℓm1 , . . . , ℓmn and m, where ℓ is a positive integer coprime
to a component of m = u1 m1 + · · · + un mn for some non-negative integers
u1 , . . . , un . Note that E is the gluing of S1 = ℓS and S2 = N{m}, with
α = ℓm.
Theorem 4.1. K[E] has a SIFRE if and only if K[S] has a SIFRE and
∓(m + b′ − b) ∈
/ S, for all b ∈ Bi (S), b′ ∈ Bi−1 (S), and 1 ≤ i ≤ pd(S) + 1.
Proof. K[E] has a SIFRE ⇐⇒ e − e′ ∈
/ E for all e, e′ ∈ Bi (E) by Lemma
2.1. It follows from Lemma 3.3 that Bi (E) = ℓBi (S) ∪ ℓ[Bi−1 (S) + m], and
so we have three possibilities if e, e′ ∈ Bi (E):
(1) e, e′ ∈ ℓBi (S),
(2) e, e′ ∈ ℓ[Bi−1 (S) + m],
(3) e ∈ ℓBi (S) and e′ ∈ ℓ[Bi−1 (S) + m].
In the first two cases e − e′ = ℓ(b − b′ ) ∈
/ E ⇐⇒ b − b′ ∈
/ S, by Lemma
3.4, which is equivalent to K[S] having a SIFRE by Lemma 2.1. In the last
one, ∓(e − e′ ) = ∓ℓ(b − b′ − m) ∈
/ E ⇐⇒ ∓(b − b′ − m) ∈
/ S, which
completes the proof.

4.1. Symmetric affine semigroups. As another application, we obtain
infinitely many complete intersection semigroup rings with SIFRE. When E
is symmetric (or equivalently S is symmetric), it is sufficient to check the
condition above for the first half of the indices.
Corollary 4.2. If E is symmetric, then K[E] has a SIFRE if and only if
K[S] has a SIFRE and ∓(m + b′ − b) ∈
/ S, for b ∈ Bi (S), b′ ∈ Bi−1 (S), and
1 ≤ i ≤ ⌊pd(E)/2⌋.
Proof. The proof mimics the proof of Theorem 4.1 applying this time Lemma
2.2 instead of Lemma 2.1.

Let n > 1 and ei denote the canonical basis of Nn , for i = 1, . . . , n.
Let u1 , . . . , un be some positive integers and S be the semigroup generated
minimally by u1 e1 , . . . , un en . It is clear that IS = (0) and thus K[S] =
K[x1 , . . . , xn ] has a SIFRE.
Fix a = (−u1 , u2 , . . . , un ), a0 = (0, u2 , . . . , un ) ∈ S and consider the
extensions of S defined recursively as follows:
• E1 = 2S + N{a1 }, where a1 = 2a0 − a = (u1 , u2 , . . . , un ) ∈ S, and
• Ej = 2Ej−1 + N{aj }, where aj = aj−1 + 2aj−2 ∈ Ej−1 , for j ≥ 2.
Proposition 4.3. With the notations above, we have

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MESUT ŞAHİN AND LEAH GOLD STELLA

(1) aj − 2aj−1 = (−1)j a, for all j ≥ 1,
(2) aj + b′ − b = ua, for some u ∈ Z − {0} and for all b′ ∈ Bi−1 (Ej−1 ),
b ∈ Bi (Ej−1 ), where j ≥ 2 and 1 ≤ i ≤ ⌊j/2⌋,
(3) K[Ej ] has a SIFRE for all j ≥ 1.
Proof. We use induction on j in all items.
(1) The claim follows from the definition of a1 = 2a0 − a when j = 1.
Assuming that the claim is true for j = p − 1, we have ap − 2ap−1 =
−(ap−1 − 2ap−2 ) = −(−1)p−1 a = (−1)p a, since ap = ap−1 + 2ap−2 ,
for all p ≥ 2.
(2) When j = 2 and 1 ≤ i ≤ ⌊j/2⌋ = 1, we have a2 + b′ − b = a2 − 2a1 =
a, by (1), for all b′ ∈ B0 (E1 ) = {0}, b ∈ B1 (E1 ) = {2a1 }, as IE1
is a principal ideal generated by y 2 − xa11 · · · xann of E1 -degree 2a1 .
Assume now that the claim is true for all indices 3 ≤ j ≤ p − 1. We
need to study ap + b′ − b, for all b′ ∈ Bi−1 (Ep−1 ), b ∈ Bi (Ep−1 ),
where p ≥ 4 and 1 ≤ i ≤ ⌊p/2⌋. There are four cases to consider
since by Lemma 3.3, Bi (Ep−1 ) = 2Bi (Ep−2 ) ∪ 2[Bi−1 (Ep−2 ) + ap−1 ]:
Case (i): b′ = 2c′ , c′ ∈ Bi−1 (Ep−2 ) and b = 2c, c ∈ Bi (Ep−2 ). In
this case, ap + b′ − b = ap − 2ap−1 + 2(ap−1 + c′ − c) = (−1)p a + 2ua,
for some u ∈ Z − {0}, by induction hypothesis.
Case (ii): b′ = 2(c′ + ap−1 ), c′ ∈ Bi−2 (Ep−2 ) and b = 2(c + ap−1 ),
c ∈ Bi−1 (Ep−2 ). In this case, ap + b′ − b = ap − 2ap−1 + 2(ap−1 + c′ −
c) = (−1)p a + 2ua, for some u ∈ Z − {0}, by induction hypothesis.
Case (iii): b′ = 2(c′ + ap−1 ), c′ ∈ Bi−2 (Ep−2 ) and b = 2c, c ∈
Bi (Ep−2 ). Taking d ∈ Bi−1 (Ep−2 ) in this case, we have ap + b′ − b =
ap −2ap−1 +2(ap−1 +c′ −d)+2(ap−1 +d−c) = (−1)p a+2u1 a+2u2 a,
for some u1 , u2 ∈ Z − {0}, by induction hypothesis.
Case (iv): b′ = 2c′ , c′ ∈ Bi−1 (Ep−2 ) and b = 2(c + ap−1 ), c ∈
Bi−1 (Ep−2 ). So, ap + b′ − b = ap − 2ap−1 + 2(c′ − c) = (−1)p a +
2(c′ − c). Note that the proof will be complete if we show that
c′ − c = va, for some v ∈ Z. Let us prove this by verifying the
claim that c′ − c = va, for some v ∈ Z, and for all c, c′ ∈ Bi−1 (Eq ),
1 ≤ i ≤ ⌊p/2⌋, using induction on 1 ≤ q ≤ p − 2. For q = 1, the
claim is trivial with v = 0 as i = 1 and B0 (E1 ) = {0}. Assume
now that it is true for q = r − 1, and consider c, c′ ∈ Bi−1 (Er ). By
Lemma 3.3, we have three possibilities as as before and in two of
them c′ − c = 2(d′ − d), for either d′ , d ∈ Bi−1 (Er−1 ) or d′ , d ∈
Bi−2 (Er−1 ). So, we are done by induction hypothesis on q. In the
third one, c′ − c = 2(ap + d′ − d) = 2ua, by induction hypothesis on
p, where d ∈ Bi−1 (Er−1 ) or d′ ∈ Bi−2 (Er−1 ).
(3) As IE1 is a principal ideal, the projective dimension of K[E1 ] is 1 and
B0 (E1 ) = {0} and B1 (E1 ) = {2a1 }. So, when j = 1, there is nothing
to check in Corollary 4.2 as K[S] has a SIFRE and ⌊j/2⌋ = 0. So,
K[E1 ] has a SIFRE. Assume that the claim is true for j = p − 1, so
K[Ep−1 ] has a SIFRE for all p ≥ 2. We first note that the projective

GLUING SEMIGROUPS AND STRONGLY INDISPENSABLE FREE RESOLUTIONS 9

dimension of K[Ep ] is p, by Theorem 3.2. So, we need to verify that
ap + b′ − b ∈
/ Ep−1 , for all b′ ∈ Bi−1 (Ep−1 ), b ∈ Bi (Ep−1 ), where
p ≥ 2 and 1 ≤ i ≤ ⌊p/2⌋, which is true by (2) as Ep−1 ⊂ Nn and
ua ∈
/ Nn , for u ∈ Z − {0}.
So, K[Ej ] has a SIFRE for j ≥ 1.

4.2. Numerical semigroups. In this section, we characterize extensions
of some numerical semigroups having SIFRE. As we classified extensions of
3-generated symmetric numerical semigroups in [2, Theorem 25], we start
with 3-generated non-symmetric numerical semigroups here. It is known
that they have SIFREs, see [2, Example 20]. As a first application, we
determine their extensions which have SIFREs, using the following results.
Theorem 4.4. [11, Proposition 3.2] Let αp be the smallest positive integer
such that αp mp = αpq mq + αpr mr , for some αpq , αpr ∈ N, where {p, q, r} =
{1, 2, 3}. Then S = hm1 , m2 , m3 i is 3-generated not symmetric if and only
if αpq > 0 for all p, q, and αqp + αrp = αp , for all {p, q, r} = {1, 2, 3}. Then
K[S] = A/(f1 , f2 , f3 ), where
f1 = xα1 1 − xα2 12 xα3 13 , f2 = xα2 2 − xα1 21 xα3 23 , f3 = xα3 3 − xα1 31 xα2 32 .
Theorem 4.5. [7, Lemma 2.5] If S is a 3-generated semigroup which is not
symmetric then K[S] has a minimal graded free A-resolution
φ2

φ1

0 −→ A2 −→ A3 −→ A−→K[S] −→ 0,
 α23

x3
xα2 32
where φ1 = (f1 f2 f3 ), and φ2 =  xα1 31 xα3 13  .
xα2 12 xα1 21
Remark 4.6. As the resolution above is graded, B1 (S) = {d1 , d2 , d3 }, where
dp = αp mp = αpq mq + αpr mr , for all p, q, r ∈ {1, 2, 3}. Since the entries in
φ1 φ2 = 0 are S-homogeneous, we have B2 (S) = {b1 , b2 }, where
b1 = α23 m3 + d1 = α31 m1 + d2 = α12 m2 + d3 ,
b2 = α32 m2 + d1 = α13 m3 + d2 = α21 m1 + d3 .
Theorem 4.7. Let S = hm1 , m2 , m3 i be a non-symmetric numerical semigroup and E be an extension of S, where m = u1 m1 + u2 m2 + u3 m3 .
Then, K[E] has a SIFRE if and only if 0 < up < min{αqp , αrp } for all
{p, q, r} = {1, 2, 3}. In particular, S does not have an extension with a
SIFRE if and only if αpq = 1 for some p, q ∈ {1, 2, 3}.
Proof. If K[E] has a SIFRE, then m + di − bj ∈
/ S and di − m ∈
/ S, for all
i, j ∈ {1, 2, 3} by Theorem 4.1. By Remark 4.6, there are i, j ∈ {1, 2, 3} such
that m + di − bj = (up − αqp )mp + uq mq + ur mr . When, up ≥ αqp for some
p, q ∈ {1, 2, 3}, m + di − bj ∈ S, a contradiction. Thus, up < αqp for all p, q ∈
{1, 2, 3}. If ui = 0, for some i, then di −m = (αip −up )mp +(αiq −uq )mq ∈ S,
a contradiction.

10

MESUT ŞAHİN AND LEAH GOLD STELLA

Conversely, assume that 0 < up < min{αqp , αrp } for all {p, q, r} =
{1, 2, 3}. We claim m − dp = up mp + (uq − αpq )mq + (ur − αpr )mr ∈
/
S. If not, m − dp = vp mp + vq mq + vr mr , for some vi ∈ N, and so
(up − vp )mp = (uq + vq − αpq )mq + (ur + vr − αpr )mr > 0 which contradicts
to αp being the smallest positive integer with this property. Next, we prove
dp −m = (αp −up )mp −uq mq −ur mr ∈
/ S. If not, dp −m = vp mp +vq mq +vr mr ,
for some vi ∈ N, and so (αp − up − vp )mp = (uq + vq )mq + (ur + vr )mr > 0,
which contradicts to αp being the smallest positive integer with this property.
By [2, Corollary 11], P F (S) = {b1 −N, b2 −N }, where N = m1 +m2 +m3 ,
are the pseudo-Frobeneous elements of S. In particular, bi − N ∈
/ S. This
implies that bi − m − dj ∈
/ S, since bi − N = (bi − m − dj ) + (αp + up −
1)mp + (uq − 1)mq + (ur − 1)mr . Finally, by Remark 4.6, for all i, j there
are p, q, r such that m + di − bj = (up + αip )mp + (uq − αpq )mq + (ur −
αpr )mr . If m + di − bj = vp mp + vq mq + vr mr , for some vi ∈ N, then
(up + αip − vp )mp = (vq + αpq − uq )mq + (vr + αpr − ur )mr > 0, which implies
that up + αip − vp ≥ αp , as αp is the smallest with this property. But this
contradicts to the assumption up < min{αqp , αrp }.

Example 4.8. Take S = h7, 9, 10i. Then K[S] = A/(f1 , f2 , f3 ), where
f1 = x41 − x22 x3 ,

f2 = x32 − x1 x23 ,

f3 = x33 − x31 x2 .

Since α13 = 1, no extension of S will have a SIFRE. On the other hand,
the following semigroups will lead to infinitely many families of extensions
having SIFRE. For S = h31, 37, 41i, Macaulay 2 computes the following
generators
f1 = x91 − x22 x53 ,

f2 = x52 − x21 x33 ,

f3 = x83 − x71 x32 .

So, u1 = 1, u2 = 1 and 1 ≤ u3 ≤ 2 give m = 109 and m = 150. Hence,
E = h31ℓ, 37ℓ, 41ℓ, 109i and E = h31ℓ, 37ℓ, 41ℓ, 109i have SIFREs, for any
ℓ, with gcd(ℓ, 109) = 1 and with gcd(ℓ, 150) = 1, respectively. Similarly,
S = h67, 91, 93i leads to 6 and S = h71, 93, 121i leads to 14 different infinite
families having SIFREs.
Now, we study extensions of a symmetric 4-generated not complete intersection numerical semigroup using the following.
Theorem 4.9. [1, Theorem 5, Theorem 3] The semigroup S is 4-generated
symmetric, not complete intersection, if and only if there are integers αi and
αij , such that 0 < αij < αi , for all i, j,
α1 = α21 + α31 , α2 = α32 + α42 , α3 = α13 + α43 , α4 = α14 + α24 and
m1 = α2 α3 α14 + α32 α13 α24 ,
m3 = α1 α4 α32 + α14 α42 α31 ,

m2 = α3 α4 α21 + α31 α43 α24 ,
m4 = α1 α2 α43 + α42 α21 α13 .

GLUING SEMIGROUPS AND STRONGLY INDISPENSABLE FREE RESOLUTIONS 11

Then, K[S] = A/(f1 , f2 , f3 , f4 , f5 ), where
f1 = xα1 1 − xα3 13 xα4 14 ,
f4 = xα4 4 − xα2 42 xα3 43 ,

f2 = xα2 2 − xα1 21 xα4 24 ,
f3 = xα3 3 − xα1 31 xα2 32 ,
f5 = xα3 43 xα1 21 − xα2 32 xα4 14 .

If S = hm1 , m2 , m3 , m4 i is a symmetric not complete intersection numerical semigroup then K[S] has a SIFRE by [2, Theorem 27]. Let E be an
extension of S with m = u1 m1 + u2 m2 + u3 m3 + u4 m4 . Then, we have the
following result.
Theorem 4.10. K[E] has a SIFRE if and only if up < min{αqp , αrp } for
all p, q, r ∈ {1, 2, 3, 4} and at most one up = 0 such that
u1
u2
u3
u4

=0
=0
=0
=0

=⇒
=⇒
=⇒
=⇒

α32 − α42
α43 − α13
α31 − α21
α21 − α31

< u2
< u3
< u1
< u1

or
or
or
or

α13 − α43
α24 − α14
α14 − α24
α42 − α32

< u3 ,
< u4 ,
< u4 ,
< u2 .

Proof. We use [2, Corollary 13] and Theorem 4.9 for all the relations involving ai and di , where di = degS (fi ), and ai is the S-degree of a first
syzygy, for i = 1, . . . , 5. We first prove the necessity of these conditions.
Assume K[E] has a SIFRE. Then by Corollary 4.2, m + dj − ak ∈
/ S
and di − m ∈
/ S, Given αpi there X
are j, k with dj − ak = −αpi mi . So,
uq mq + (ui − αpi )mi ∈ S. Thereif ui ≥ αpi , then m + dj − ak =
q6=i

fore, up < min{αqp , αrp } for all {p, q, r} = {1, 2, 3, 4}. If up = uq = 0,
then di − m = (αjr − ur )mr + (αts − us )ms − up mp − uq mq ∈ S. So, at
most one up = 0. If α32 − α42 ≥ u2 and α13 − α43 ≥ u3 when u1 = 0,
then a4 − d4 − m = (α32 − α42 − u2 )m2 + (α13 − α43 − u3 )m3 + (α14 −
u4 )m4 ∈ S. The others are shown similarly. Next, we prove sufficiency. Assume up < min{αqp , αrp } forPall {p, q, r} = {1, 2, 3,
P4} and at most one
up = 0. Then, di − m =
vj mj implies di =
(uj + vj )mj . Since
fP
is
indispensable,
there
are
only
two
monomials
with
S-degree di . So,
i
(uj + vj )mj must be αi mi or αpq mq + αrs ms . In any P
case, at least
two uj = 0, contradiction. So, m − di ∈
/ S. If m − di =
vj mj , then
d := (ui − vi )mi + (uj − vj )mj = (uq − αpq − vq )mq + (us − αrs − vs )ms > 0.
Since ui − vi < αi and uj − vj < αj , we get ui > vi and uj > vj by
the minimality of αi and αj . But then d <S ds = αpi mi + αqj mj , which
contradicts to the minimality of ds . So, m − di ∈
/ S. For, i 6=
P j and
(i, j) ∈
/ {(1, 3), (2, 4)}, we have
a
−d
=
α
m
.
So,
if
a
−d
−m
=
vj m j ,
i
j
pq
q
i
j
P
then (αpq − uq − vq )mq = j6=q (uj + vj )mj ≥ 0. Since αpq − uq − vq < αq ,
P
all ui = 0 for i 6= q. So, ai − dj − m ∈
/ S. If a1 − d3 − m =
vj mj , then
(α2 − u2 − v2 )m2 = (α13 + u3 + v3 )m3 + (u1 + v1 )m1 + (u4 + v4 )m4 > 0.
By the minimality of α2 , u2 = v2 = 0 in which case we have a third monomial xu1 1 +v1 x3α13 +u3 +v3 xu4 4 +v4 of S-degree d2 , a contradiction to the indispensability of f2 . Similarly, a2 − d4 − m = α1 m1 − α42 m2 − m ∈
/ S.
To prove that ai − di − m ∈
/ S, it suffices to see m + 2di − N ∈ S,

12

MESUT ŞAHİN AND LEAH GOLD STELLA

as the Frobeneous number of S, which is the biggest integer not in S, is
ai + dP
i − N = (ai − di − m) + (m + 2di − N ) by [2, Corollary 14], where
N = j mj . If all ui > 0 then m − N ∈ S, so m + 2di − N ∈ S. If up = 0 for
some p ∈ {1, 2, 3, 4}, then m+2di −N = (di +m−mj −mq −mr )+(di −mp ) ∈
S, except for (i, p) ∈ {(4, 1), (1, 2), (2, 3), (3, 4)}. When u1 = 0, we have
a4 − d4 − m = (αP
32 − α42 − u2 )m2 + (α13 − α43 − u3 )m3 + (α14 − u4 )m4 .
If a4 − d4 − m =
vj mj , then either α32 − α42 ≥ u2 or α13 − α43 ≥ u3 as
0 < α14 − u4 < α4 . If α32 − α42 ≥ u2 then α13 − α43 < u3 and thus we have
(α32 − α42 − u2 − v2 )m2 + (α14 − u4 − v4 )m4 = v1 m1 + (α43 − α13 + u3 + v3 )m3 .
This gives a binomial in IS of degree d with d <S d5 , a contradiction to
the minimality of d5 . If α13 − α43 ≥ u3 , then we get similarly a binomial of
S-degree less than d1 . The other cases can be done the same way. Finally,
we prove that mP+ dj − ai ∈
/ S. For i 6= j and (i, j) ∈
/ {(1, 3), (2,P
4)}, we have
m + dj − ai = p6=q up mp + (uq − αrq )mq . If m + dj − ai =
vp mp ∈ S,
then (ur − vr )mr + (us − vs )ms = (vp − up )mp + (αrq − uq + vq )mq , as
the other cases contradicts to the minimality of some αt . But this gives a
binomial in IS of S-degree less than dt = αpr mr + αqs ms , a contradiction.
P
Now, d1 − a1 = α31 m1 − α43 m3 − α24 m4 . If m + d1 − a1 =
vj mj , then
(u2 − v2 )m2 + (α31 + u1 − v1 )m1 = (α43 − u3 + v3 )m3 + (α24 − u4 + v4 )m4 > 0.
If one term of the left hand side is negative then we get a contradiction to
the minimality of α1 and α2 . So, this gives a binomial in IS of S-degree d.
Only d3 may be less than d, so α32 ≤ u2 − v2 ≤ u2 , a contradiction. The
rest is similar and we are done.

Remark 4.11. Using the formulas in Theorem 4.9, one can now produce
infinitely many symmetric not complete intersection S = hm1 , m2 , m3 , m4 i
having SIFRE.
We classified 4-generated pseudo symmetric semigroups having SIFRE in
[19]. The next result reveals that none of their extensions have a SIFRE.
Theorem 4.12. Let S be a 4-generated pseudo symmetric semigroup and
E be one of its extensions. Then K[E] does not have a SIFRE.
Proof. By the proof of [19, Theorem 2.5], we have B1 (S) = {d1 , . . . , d6 },
B2 (S) = {b1 , . . . , b6 } and B3 (S) = {c1 , c2 }, where
c1 = b1 + m4 = b2 + m1 = b4 + m3 = b5 + m2 .
Thus, if m = u1 m1 + · · · + u4 m4 with uj > 0, then there is some i such that
m + bi − c1 = m − mj ∈ S. The result now follows from Theorem 4.1.

Acknowledgements
The author would like to thank Anargyros Katsabekis for his suggestions
on the preliminary version of the paper. All the examples are computed by
using the computer algebra system Macaulay 2, see [10].

GLUING SEMIGROUPS AND STRONGLY INDISPENSABLE FREE RESOLUTIONS 13

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Department of Mathematics, Hacettepe University, Beytepe, 06800, Ankara,
Turkey
E-mail address: mesut.sahin@hacettepe.edu.tr
Department of Mathematics, Cleveland State University, Cleveland, OH,
USA
E-mail address: l.gold33@csuohio.edu

