EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 4, 2024, 2621-2650
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Binomials Arising from Buchberger Algorithm on
Polyomino Ideals

Yoshua Yonatan Hamonangan1,∗, Intan Muchtadi-Alamsyah2

1 Doctoral Program of Mathematics, Faculty of Mathematics and Natural Sciences,
Bandung Institute of Technology, Bandung, West Java, Indonesia
2 Algebra Research Group, Faculty of Mathematics and Natural Sciences,
Bandung Institute of Technology, Bandung, West Java, Indonesia

Abstract. A polyomino is a finite set of unit squares joined side by side on the Cartesian plane.
Qureshi introduced an ideal constructed from a polyomino which is called ”polyomino ideal”. In
this paper, we study the binomials arising from Buchberger Algorithm on polyomino ideals. We
also introduce socket wrench polyominoes and study the Gröbner bases of the ideal IP and some
algebraic properties of K[P].

2020 Mathematics Subject Classifications: 05B50, 05E40, 13C05

Key Words and Phrases: Buchberger Algorithm, Gröbner Bases, Polyomino, Radical Ideal

1. Introduction

A polyomino is a finite set of unit squares joined side by side on the Cartesian plane.
They are discussed in a lot of papers. Look at: [2, 3] for Combinatorics; [17–19] for its
relation to the tiling problem on the plane; [12] for the relation between polyominoes and
Dyck Words and Motzkin Word; and [41] for statistical physics.

The relation between polyominoes and commutative algebra was introduced by Qureshi,
introducing an ideal constructed from a polyomino which is called polyomino ideal [34].
The polyomino ideal is a generalization of ideals generated by the set of 2-minor of a
matrix. Generally, the ideal of t-minors is a central topic in Commutative Algebra and
has some applications in algebraic statistics [33, 40]. There were a lot of research related
to the ideal generated by the set of t−minor of a matrix [22, 27].

Since it was introduced by Qureshi in 2012, many interesting question have arisen
about polyomino ideal. Here are some recent works and related results:

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5016

Email addresses: yoshua.yonatan.h@gmail.com (Y. Y. Hamonangan),
ntan@itb.ac.id (I. Muchtadi-Alamsyah)

https://www.ejpam.com 2621 Copyright: © 2024 The Author(s). (CC BY-NC 4.0)



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2622

• The primality of the polyomino ideal is studied in many articles [5, 7, 24–26, 31, 32,
35, 36, 38]. In [24, 25, 36], it is proved that K[P] is a domain if P is simple. In [31],
it is proved that if K[P] is a domain then the polyomino have no zig-zag walks. They
also conjectured that the converse direction is true. Later, it was proved in [5] and
[7] that the conjecture holds for two special classes of non-simple polyominoes, called
closed path and weakly closed paths. Still, a complete classification of polyominoes
with prime polyomino ideal is not known.

• The algebraic properties, like when K[P] is Cohen-Macaulay or Gorenstein, are
known only for some specific polyominoes. In [36] the authors show that if P is
simple then K[P] is a normal Cohen-Macaulay domain, by identifying their quotient
ring with the toric ring of a weakly chordal graph. In [6], the authors show that if P
is a closed path polyomino having no zig-zag walks then K[P] is a Cohen-Macaulay
domain. In [34], Qureshi established that the Cohen-Macaulay property holds for
convex polyominoes and characterized all stack polyominoes P for which K[P] is
Gorenstein. The Gorensteiness is also studied in [1, 8, 10, 16, 35, 37].

• Gröbner basis of polyomino ideals are studied in [6, 20, 25, 26, 32, 34].

• The König type property is studied for simple thin polyominoes in [21], for closed
path polyominoes in [13], and for grid polyominoes in [14].

• The linearly related polyominoes are studied in [15].

• The Charney-Davis conjecture for simple thin-polyominoes are studied in [29].

• The primary decomposition of polyomino ideals, like closed paths, and more in
general for polyocollections is studied in [9].

• Another challenging problem is to compute the h-polynomial of K[P] in terms of
the rook polynomial of P [8, 16, 28, 30, 35, 37].

An important class of ideals other than the prime ideal is the radical ideal. Radical
ideal plays an important role in Algebraic Geometry, for example the Strong Nullstellen-
satz Theorem [11]. Qureshi gave an example of a non-simple polyomino with non-prime
polyomino ideal [36] that is radical.

The radicality of an ideal can be studied from the Gröbner bases of the ideal. If we
can define a monomial order such that every element in the Gröbner bases has square-free
initial monomial then the ideal is radical [23, Problem 1.8(b)]. The Gröbner bases of an
ideal can be computed by using Buchberger Algorithm [23, Section 1.3].

In this paper, we study some elements arising from Buchberger Algorithm to polyomino
ideals. In the second section, polyominoes and some terminologies related to our study will
be defined. In the third section, we will perform the Buchberger Algorithm in polyomino
ideals. In the fourth sections, we will apply the results from previous sections to a class
of polyominoes that we call socket wrench polyominoes. We prove that for the socket
wrench polyomino P, the ideal IP has square-free quadratic Gröbner bases for a suitable
monomial order. We also study some algebraic properties of the K-algebra K[P] = S/IP .



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2623

2. Preliminaries

In this section, we will recall the definitions and terminologies about polyomino and
polyomino ideal from [5] and [34]. Consider the set Z2 and define the partial order:
(i, j) ≤ (k, ℓ) if and only if i ≤ k and j ≤ ℓ.

(i) Let a, b ∈ Z2 with a ≤ b. The set [a, b] =
{
c ∈ Z2 | a ≤ c ≤ b

}
is called an interval.

(ii) Let a = (i, j) and b = (k, ℓ). The elements a and b are called the diagonal corners
of the interval [a, b], and the elements (i, ℓ) and (k, j) are called the antidiagonal
corners of the interval [a, b]. Particularly, the elements (i, j) and (k, ℓ) are called left
lower corner and the right upper corner, respectively, of the interval [a, b]. Similarly,
the elements (i, ℓ) and (k, j) are called the left upper corner and the right lower
corner, respectively, of the interval [a, b].

(iii) If b = a+ (1, 1) then the interval [a, b] is called a cell.

(iv) The edges of a cell [a, a + (1, 1)] are the intervals: [a, a + (1, 0)], [a, a + (0, 1)],
[a+ (0, 1), a+ (1, 1)], and [a+ (1, 0), a+ (1, 1)].

(v) Let P be a finite collection of cells in Z2. The collection of all vertices of P, denoted
by V (P) is the union of all corners from each cells in P.

(vi) Let a = (i, j), b = (k, ℓ) ∈ Z2. The vertices a and b are called in horizontal position
if j = ℓ and in vertical position if i = k.

(vii) Let P be a finite collection of cells in Z2. Let C and D be two cells in P. The cells
C and D are called connected if there exists a sequence of cells C = C1, . . . , Cm = D
in P such that Ci ∩ Ci+1 is an edge of Ci for all i = 1, 2, . . . ,m− 1.

(viii) A finite collection of cells P in Z2 is called a polyomino if any two cells in P are
connected.

(ix) A walk from cell C to cell D in Z2 is a sequence of cells C : C = C1, . . . , Cm = D in
Z2 such that Ci∩Ci+1 is an edge of Ci and Ci+1 for all i = 1, 2, . . . ,m−1. If Ci ̸= Cj

for all i ̸= j then C is called a path. A polyomino P is called simple if for any two
cells C and D not belonging to P, there exist a path C : C = C1, . . . , Cm = D such
that Ci /∈ P for all i = 1, . . . ,m.

(x) Let P be a polyomino and (i, j), (k, ℓ) ∈ V (P) such that i < k and j < ℓ. The
interval [(i, j), (k, ℓ)] is called an inner interval of P if any cell [(r, s), (r + 1, s+ 1)]
is an element in P for all i ≤ r ≤ k − 1 and j ≤ s ≤ ℓ− 1.

(xi) Let P be a polyomino. The interval [(i, j), (k, j)] with i < k is called in a horizontal
edge interval of P if the interval [(ℓ, j), (ℓ + 1, j)] are edges of cells of P for all
ℓ = i, . . . , k − 1. If [(i− 1, j), (i, j)] and [(k, j), (k, j)] are not edges af cells if P then
the interval [(i, j), (k, j)] is called a maximal horizontal edge interval of P. We define
the vertical edge interval and the maximal vertical edge interval similarly.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2624

(xii) Let P be a polyomino and K be a field. Define the polynomial ring S over K with
variables xij for all (i, j) ∈ V (P). Each inner interval [(i, j), (k, ℓ)] in P is associated
to xijxkℓ − xiℓxkj ∈ S, that is called the inner 2-minor of P. The set of all inner
2-minors of P is denoted by S2.

(xiii) Let P be a polyomino. The ideal IP ⊆ S generated by S2 is called the polyomino
ideal of P and K[P] = S/IP the coordinate ring of P.

(xiv) Let J ⊆ S be a binomial ideal and f = f+ − f− be a binomial in J . The binomial
f is called redundant if it can be expressed as a linear combination of binomials in
J od lower degree. The binomal f is called irredundant if it is not redundant. We
also denote by V +

f the set of vertices v such that xv divides f+ and by V −
f the set

of vertices v such that xv divides f−.

3. Buchberger Algorithm in Polyomino Ideal

Let P be a polyomino. We define an ordering in the set V (P) in the following way:
(i, j) <P (k, ℓ) if and only if

• j < ℓ or

• ȷ = ℓ and i < k.

By this ordering, all the vertices in a polyomino with n vertices can be labelled with
positive integer 1, 2, . . . , n from left to right, starting from the vertices with the lowest
ordinate to the vertices with the highest ordinate. Below is an example of such labelling.

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23

Figure 1: Labelling the Set V (P).

By this labelling, the polynomial ring associated to the polyomino ideal is S = K[x1, . . . , xn].
We use the lexicographic monomial order with x1 > x2 > · · · > xn. For the sake of sim-
plicity, the elements xa ∈ R will be written with a. We define the degree of monomials
xa11 xa22 . . . xann with

∑n
i=1 ai. We also define the interval determined by {a, b} as the in-

terval with diagonal corners {a, b} or antidiagonal corners {a, b}. Now, we are ready to
perform the Buchberger Algorithm. Since S2 consists of inner 2-minors then the polyno-
mial obtained by the Buchberger Algorithm in each step is again a binomial consisting of



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2625

two monomials of the same degree. The degree of this binomial is defined as the degree of
both monomials.

3.1. Binomials of Degree Three

The result in this subsection can also be derived from [34, Theorem 4.1] and [32,
Proposition 3.2]. We start the Buchberger Algorithm by computing the S-polynomial
S(F,G) for every F,G ∈ S2. The S-polynomial S(F,G) is defined by

S(F,G) =
lcm(in<(F ), in<(G))

cF · in<(F )
− lcm(in<(F ), in<(G))

cG · in<(G)

where in<(F ) (resp. in<(F )) denotes the initial monomial of F (resp. G) with respect to
< and cF (resp. cG) denotes the coefficient of in<(F ) (resp. in<(G)) in F (resp. G).

• If the initial monomial of F and G are relatively prime then S(F,G) is reduced to
zero.

• If the greatest common divisor of their initial monomials is a monomial of degree
two then F = G and S(F,G) = 0.

• If the greatest common divisor of their initial monomials is a monomial of degree
one then S(F,G) is a binomial of degree three.

We will compute S(F,G) in the last possibility and find the condition for the S-
polynomial to be not reduced to zero. Consider the case when F and G have common
factor in their non-initial monomials (reader may see [6, Remark 1] for more general result).
So, let F = ab− pq and G = ac− pr with initial monomials ab and ac, respectively. Note
that S(F,G) = p(br − cq) and the interval determined by {b, r} is an inner interval. We
conclude that S(F,G) is reduced to zero.

Now we assume that F and G have no common factor in their non-initial monomial.
Let F and G be the inner 2-minors associated to inner intervals [a, b] and [c, d], respectively.
Without losing of generality, assume that a ≤P c. Write F = ab − pq and G = cd − rs
with p <P q and r <P s. We conclude that a <P d. We consider three cases.

(i) If a = c then S(F,G) = brs− dpq.

Consider the location of vertex d, there are 4 cases to discuss (see the figure below).

a p

q b

r

s d

a pr

q b

ds

a p r

q b

s d

a p

q b

d

r

s

Figure 2: Case a = c.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2626

For the first three cases, the interval determined by {d, q} is an inner interval. Since
s, d are in horizontal position and s, q are in vertical position, then there is a vertex
y such that the inner interval determined by {q, d} is the inner interval determined
by {s, y}. By checking all possible configurations (see the figure below), we see that
the inner interval determined by {b, r} is the inner interval determined by {p, y}.

a p

q b

r

s d

a pr

q b

ds

a p r

q b

s d

y y
y

Figure 3: The first three cases.

Note that
S(F,G) = brs− dpq = (br − py)s+ p(ys− dq).

Therefore, S(F,G) is reduced to zero.

a p r

q b

d
s

y

Figure 4: The fourth case.

For the fourth case, the interval determined by {d, p} is an inner interval (see the
figure above). Similarly, there is a vertex y such that the inner interval determined
by {p, d} is the inner interval determine by {r, y}. Note that

S(F,G) = brs− dpq = (bs− qy)r + q(ry − dp).

Therefore, S(F,G) is reduced to zero.

(ii) If a ̸= c and b = d then S(F,G) = ars− cpq

a p

q b

y r

s

c

Figure 5: Case a ̸= c and b = d.

Note that there exists a vertex y such that q, c are the antidiagonal corners of the
inner interval determined by {q, c}. Since

S(F,G) = (ar − py)s+ p(ys− cq)

we conclude that S(F,G) is reduced to zero.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2627

a p

q
b

r

s d

y

z

Figure 6: Case a ̸= c and b ̸= d.

(iii) If a ̸= c and b ̸= d then b = c and S(F,G) = ars− dpq.

Note that ars is the initial monomial of S(F,G) and this monomial can only be
divided by initial monomials ar or as from an inner 2-minor. This is the only case
when [a, r] or [a, s] is an inner interval. If [a, r] is an inner interval then

S(F,G) = (ar − qy)s+ q(ys− pd)

with y is the antidiagonal corner other than q from the inner interval [a, r]. If [a, s]
is an inner interval then

S(F,G) = (as− pz)r + p(zr − qd)

with z is the antidiagonal corner other than p from the inner interval [a, s].

From the observation above, we conclude the following theorem.

Theorem 1. Let F and G be the inner 2-minors associated to inner interval [a, b] and
[c, d], respectively, with a ≤P c. The binomial S(F,G) is not reduced to zero by all inner
2-minors if and only if

• b = c and

• the interval determined by {a, r} for all r that is an antidiagonal corner of [c, d] is
not an inner interval.

Definition 1. Let P be a polyomino. Define S3 to be the set of all binomials a1a3a5 −
a2a4a6 such that

• ai, ai+1 are in vertical position for i = 1, 3, 5

• ai, ai+1 are in horizontal position for i = 2, 4, 6 (with a7 = a1)

• a1 <P a2 <P a3 <P a4 and a1 <P a6 <P a5 <P a4

• both intervals determined by {a1, a3} and {a1, a5} are not inner intervals

• the interval determined by {a1, b} and {b, a4} are inner interval with b is the inter-
section of the segments a2a3 and a5a6.

Note that there are no elements in S3 whose initial monomial is divisible by the initial
monomial of an inner 2-minor. Therefore, we conclude that S3 is the set of all binomials
of degree three arising from Buchberger Algorithm in the polyomino ideal with respect to
the given monomial order <P .



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2628

a1

a2 a3

a4a5

a6

b

: every cell is contained in the polyomino

: some cells are not in the polyomino

Figure 7: Binomials in S3.

3.2. Binomials of Degree Four

Now, we have the set S2 ∪ S3. We want to compute S(F,G) for F ∈ S3, G ∈ S2 or
F,G ∈ S3. For the case F ∈ S3 and G ∈ S2, note that if F = a1a3a5 − a2a4a5 ∈ S3 then
the pairs (a1, a3), (a1, a5), and (a3, a5) are not pair of diagonal corners of an inner interval,
thus we only need to consider the case when the greatest common divisor of their initial
monomials has degree one. For the case both F,G ∈ S3, we will consider the case when
the greatest common divisor of their initial monomials is a monomial of degree two.

3.2.1. The Case F ∈ S3 and G ∈ S2

Let F = abc− def and G = pq − rs, with |{a, b, c} ∩ {p, q}| = 1 as illustrated below.

a

b

c

d

e

f p r

s q

Figure 8: The Binomials F and G.

Recall that the intervals determined by {a, b} and {c, d} are not inner intervals. First,
we observe the case if F and G also have common monomial divisor on their non-initial
monomials. We show that S(F,G) can be reduced to zero. From the structure of F and
G, the possible cases are:

(i) a = p and d = s;

(ii) a = p and f = r;

(iii) b = p and e = s;

(iv) b = q and d = s;



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2629

(v) c = p and e = r;

(vi) c = q and r = f .

The first two cases can occur simultaneously. In that case, we notice that q is the
intersection of the segments cf and bd. Notice that S(F,G) = df(bc − qe) is reduced to
zero since bc− qe ∈ S2. Now, we examine all the cases separately.

(i) If a = p and d = s then S(F,G) = d(bcr − efq).

• If a < r < f then consider the interval determined by {r, c}. If it is an inner
interval then bcr − efq is reduced to zero by similar argument in section 3.1
(case (iii)). If it is not an inner interval, since the interval determined by {r, b}
is not an inner interval, then bcr − efq ∈ S3 and it is reduced to zero.

a

b

c

d

e

fr

q

Figure 9: Case a = p, d = s and a < r < f .

• If a < f < r, let z be the intersections of segments cf and bd, then

bcr − efq = (−e)(qf − zr) + (−r)(ez − bc)

is reduced to zero.

a

b

c

d

e

f r

qz

Figure 10: Case a = p, d = s and a < f < r.

(ii) The case a = p and f = r is similar with the first case.

(iii) If b = p and e = s then S(F,G) = e(acr − dfq). Note the the interval determined
by {a, r} and the interval determined by {a, c} are not inner intervals. Therefore
acr − dfq ∈ S3 and S(F,G) is reduced to zero.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2630

(iv) If b = q and d = s then S(F,G) = d(acr − pef). Since the interval determined by
{a, b} is not inner interval then a < p < d and the interval determined by {a, r} is
not an inner interval. Therefore acr − pef ∈ S3 and S(F,G) is reduced to zero.

(v) The case c = p and e = r is similar with the case (iii).

(vi) The case c = q and r = f is similar with the case (iv).

Now, we can assume that F and G have no common monomial divisor in their non-
initial monomials. We consider every possibility of {a, b, c} ∩ {p, q}.

(i) If q = c then S(F,G) = abrs − pdef . Note that r, c, f are in vertical position.
Consider the vertex r.

If b <P r then the interval determined by {p, e} is an inner interval with s as one
of the antidiagonal corner. Let y be the other antidiagonal corner. It is clear that
y, b, e are in vertical position and b <P y <P e. Therefore,

S(F,G) = (abr − dfy)s+ df(ys− pe).

The binomial abr−dfy is reduced to zero by S2 or an element of S3. Hence, S(F,G)
is reduced to zero.

a

b

c

d

e

f

s

p
r

y

a

b

c

d

es

p r
y

f

Figure 11: Case q = c and b <P r.

a

b

c

d

e

f

s

p

y

r

a

b

c

d

e

f

s

p

y

r

Figure 12: Case q = c and r <P b.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2631

If r <P b then the interval determined by {p, f} is an inner interval. Let y be the
vertex such that the interval determined by {y, r} is the inner interval determined
by {p, f}. It is clear that a, y, f are in horizontal position and a <P y <P f since
the interval determined by {a, c} is not an inner interval. Therefore

S(F,G) = (abs− dey)r + de(yr − pf)

and is reduced to zero of the previous arguments.

(ii) If q = b then S(F,G) = acrs − pdef . Let u be the intersection of the segments cf
and db. Consider the vertex s.

If u <P s then [p, e] is an inner interval with r as one of the antidiagonal corner. Let
y be the other antidiagonal corner. Therefore,

S(F,G) = (acs− dyf)r + df(yr − pe)

and is reduced to zero.

a

b

c

d

e

f

s

p

y

r

a

b

c

d

e

s

p

y

r

f

Figure 13: Case q = b and u <P s.

If s ≤P u then the interval determined by {p, d} is an inner interval.

a

b

c

d

e

f

s

p
y r

a

b

c

d

e

s

p y
r

f

Figure 14: Case q = b and s ≤P u.

Let y be a vertex such that the interval determined by {p, d} is the interval deter-
mined by {y, s}. Note that a, y, d are in vertical position and a <P y <P d since the



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2632

interval determined by {a, b} is not an inner interval. Therefore

S(F,G) = (arc− yef)s+ ef(ys− pd)

and is reduced to zero.

(iii) If q = a then S(F,G) = bcrs − pdef . Let u be the intersection of the segments cf
and bd. Note that

S(F,G) = (−rs)(ue− bc) + (−e)(pdf − usr)

and therefore is reduced to zero.

a

b

c

d

e

f

p r

s

u

Figure 15: Case q = a.

(iv) If p = a then S(F,G) = bcrs − qdef . Consider the vertex r which is in horizontal
position with a and f .

If a <P r <P f then the interval determined by {q, d} is an inner interval and it is
an interval determined by {s, y} for some vertex y. Therefore,

S(F,G) = (bcr − efy)s+ ef(ys− qd)

is reduced to zero.

a

b

c

d

e

f

q

r

s

y

a

b

c

d

e

r

qs

y

f

Figure 16: Case a <P r <P f .

If a <P f <P r then the interval determined by {q, f} is an inner interval and it is
an interval determined by {r, y} for some vertex y. Note that

S(F,G) = (bcs− dey)r + ed(ry − fq)



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2633

is reduced to zero if a <P s <P d. If a <P d <P s, note that y <P c since [a, c] is
not an inner interval. The binomial bcs− dey is not in S3. But

bcs− dey = (bc− ez)s+ (−e)(dy − sz)

with z is the intersections of the segments cf and bd. So, S(F,G) is reduced to zero.

a

b

c

d

e

r

qs y

f

z

a

b

c

d

e

f r

qs
y

Figure 17: Case a <P f <P r.

(v) If p = b then S(F,G) = acrs − qdef . Note that the vertices b, e, s are in vertical
position. Consider the vertex s.

If e <P s then there is a vertex y such that the inner interval [e, q] has s, y as the
antidiagonal corners. Therefore,

S(F,G) = (arc− dyf)s+ (−df)(qe− ys)

is reduced to zero.

a

b

c

d

e

f

r

qs

y

Figure 18: Case p = b and e <P s.

If s <P e, note that there is no inner 2-minor whose initial monomial divides the ini-
tial monomial of S(F,G), which is acrs. Hence the binomials whose initial monomial
may divide acrs are the elements of S3. From Theorem 1, the only initial monomials
of the element in S3 that may divide acrs are ars, acr, or asc. Let x, y, z be the
vertices such that ars− xdq, acr− dyf , asc− efz are the corresponding binomials,
respectively.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2634

a x

b

c

d

e

f

r

q
s

y

z

Figure 19: Case p = b and s <P e.

• The binomial ars− xdq is not in S3 since [a, b] is not an inner interval.

• The binomial acr−dyf is contained in S3 if and only if the interval determined
by {q, e} is an inner interval. If the interval is an inner interval then

S(F,G) = (acr − dyf)s+ df(ys− qe)

is reduced to zero.

• The binomial asc−efz is contained in S3 if and only if the interval determined
by {q, d} is an inner interval. If the interval is an inner interval then

S(F,G) = (asc− efz)r + ef(zr − qd).

is reduced to zero.

(vi) If p = c then S(F,G) = abrs− qdef . Note that the vertices e, c, r are in horizontal
position. Consider the vertex r.

If e <P r then there exists a vertex y such that the inner interval [e, q] has r, y as
the antidiagonal corners. Therefore,

S(F,G) = (abs− dyf)r + (−df)(qe− yr)

is reduced to zero.

a

b

c

d

e

f

s qy

r

Figure 20: Case p = c and e <P r.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2635

If r <P e, note that there is no inner 2-minor whose initial monomial divides the ini-
tial monomial of S(F,G), which is abrs. Hence the binomials whose initial monomial
may divide abrs are the elements of S3. From Theorem 1, the only initial monomials
of the element in S3 that may divide abrs are ars, abr, or abs. Let x, y, z be the
vertices such that ars − qxf , abr − dey, abs − dfz be the corresponding binomials,
respectively.

a

x

b

c

d

e

f

s q z

r

y

Figure 21: Case p = c and r <P e.

• The binomial ars− qxf is not in S3 since [a, c] is not an inner interval.

• The binomial abr−dey is contained in S3 if and only if the interval determined
by {q, f} is an inner interval. If the interval is an inner interval then

S(F,G) = (abr − dey)s+ de(ys− fq)

is reduced to zero.

• The binomial abs−dfz is contained in S3 if and only if the interval determined
by {q, e} is an inner interval. If the interval is an inner interval then

S(F,G) = (abs− dfz)r + df(zr − eq)

is reduced to zero.

We summarize our discussion above to the following theorem.

Theorem 2. Let F = a1a3a5 − a2a4a6 be the element in S3 as in Definition 1 and
G = pq − rs be the element in S2 associated to the inner interval [p, q] with lower-right
and upper-left corners r and s, respectively, then S(F,G) is not reduced to zero by S2 ∪S3

if and only if one of the following statement holds:

• p = a3, a4 > s and both intervals determined by {q, a2} and {q, a4} are not inner
intervals

• p = a5, a4 > r and both intervals determined by {q, a4} and {q, a6} are not inner
intervals.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2636

a2

a1

a4a5

a6

p = a3
r

qs

a1

a2 a3

a4

a6

r

p = a5

qs

: every cell is contained in the polyomino

: some cells are not in the polyomino

Figure 22: Binomial in Theorem 2.

3.2.2. The Case F ∈ S3 and G ∈ S3

Let F = A1A2A3 −B1B2B3 and G = a1a2a3 − b1b2b3 with initial monomials A1A2A3 and
a1a2a3, respectively. We assume that A1 <P A2 <P A3, B1 <P B2 <P B3, a1 <P a2 <P
a3, b1 <P b2 <P b3, and |{A1, A2, A3} ∩ {a1, a2, a3}| = 2. We consider every possibility of
{A1, A2, A3} ∩ {a1, a2, a3}

(i) If {A1, A2, A3} ∩ {a1, a2, a3} = {A1, A2}.
This case is only possible if A1 = a1 and A2 ∈ {a2, a3}.

• If A2 = a2 then B2 = b2 and

S(F,G) = (−B2)(a3B1B3 −A3b1b3).

Without loss of generality, we may assume B1 <P b1 (for the possibility B1 = b1,
we have the interval determined by {A3, b3} is an inner interval and hence
S(F,G) = B2B1(A3b3 − a3B3) is reduced to zero).

– If b3 = B3 then
S(F,G) = (−B2B3)(a3B1 −A3b1)

is reduced to zero since A3b3 − a3B3 is an inner 2-minor.

– If B3 < b3 (see Figure 23 in the left side) then a3B1B3 − A3b1b3 ∈ S3

since the interval determined by {b1, B3} and the interval determined by
{A3, a3}, both are not inner intervals. Therefore S(F,G) is reduced to zero.

– IfB3 < b3 (see Figure 23 in the right side), note that the interval determined
by {a3, B3} is an inner interval and is the same with the interval determined
by {b3, y} for some vertex y. Therefore,

S(F,G) = (−B2b3)(B1y −A3b1) + (−B2B1)(a3B3 − b3y)

is reduced to zero.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2637

A1

A2

A3

B1

B2

B3

b1

a3 b3

A1

A2

A3

B1

B2

B3

b1

a3 b3

y y

Figure 23: Case A1 = a1 and A2 = a2.

• If A2 = a3 then
S(F,G) = A3b1b2b3 − a2B1B2B3

has initial monomial a2B1B2B3.

A1

A2

A3

B1

B2

B3

a2
b2

b3

z

y

b1

Figure 24: Case A1 = a1 and A2 = a3.

Note that there is no inner 2-minor whose initial monomial divides a2B1B2B3.
Moreover, the binomials in S3 whose initial monomial divides a2B1B2B3 are
the binomial with initial monomial a2B1B3. This can only happen when the
interval determined by {B3, b3} is an inner interval and the binomial in S3 that
satisfies the property is a2B1B3 − b1yz where z is a vertex such that [A1, z]
is the inner interval determined by {b2, B1}. Therefore, S(F,G) is reduced to
zero since

S(F,G) = −B2(a2B1B3 − b1yz) + b1(A3b2b3 −B2yz)

and A3b2b3 −B2yz is an element in S3.

(ii) If {A1, A2, A3} ∩ {a1, a2, a3} = {A1, A3}. This case is only possible if A1 = a1 and
A3 ∈ {a2, a3}.
For the case A3 = a2, since S(F,G) = −S(G,F ) then it is similar with the case
A1 = a1 and A2 = a3. We conclude that S(F,G) is reduced to zero if and only if
the interval determined by {B3, b3} is an inner interval.

For the case A3 = a3, we have B1 = b1 and

S(F,G) = B1(A2b2b3 − a2B2B3).



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2638

By similar argument with the case A1 = a1 and A2 = a2, we may assume B3 <P b3
and we have three subcases

• If b2 = B2 then S(F,G) is reduced to zero since A2b3−a2B3 is an inner 2-minor.

• If B2 < b2 then S(F,G) is reduced to zero since a2B2B3 −A2b2b3 ∈ S3.

• If b2 < B2, notice that the interval determined by {B3, a2} is an inner interval
that is the same with [y, b3] for some vertex y. Therefore, S(F,G) is reduced
to zero since we can write

S(F,G) = B1b3(A2b2 −B2y) +B1B2(b3y − a2B3).

A1

A2

A3

B1

B2

B3

A1

A2

A3

B1

B2

B3

b2
a2

b3

y

b3

a2

b2 y

Figure 25: Case A1 = a1 and A3 = a3.

(iii) If {A1, A2, A3} ∩ {a1, a2, a3} = {A2, A3} then A2 = a2 and A3 = a3. Thus, B3 = b3
and

S(F,G) = B3(A1b1b2 − a1B1B2).

A1

A2

A3

B1

B2

B3

A1

A2

A3

B1

B2

B3

b2

a1 b1

b2

a1 b1

y

y

Figure 26: Case A2 = a2 and A3 = a3.

Similarly, we may assume b2 <P B2, and this implies that [a1, B2] is an inner interval
having {b2, y} as the antidiagonal corners for some vertex y. Therefore, S(F,G) is
reduced to zero since

S(F,G) = −B1B3(a1B2 − b2y) +B3b2(A1b1 −B1y).

We summarize this discussion with the following theorem.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2639

Theorem 3. Let F = A1A2A3−B1B2B3 and G = a1a2a3−b1b2b3 with initial monomials
A1A2A3 and a1a2a3, respectively, in S3 with A1 <P A2 <P A3, B1 <P B2 <P B3,
a1 <P a2 <P a3, b1 <P b2 <P b3, and |{A1, A2, A3} ∩ {a1, a2, a3}| = 2. The binomial
S(F,G) is not reduced to zero by S2 ∪ S3 if and only if

• A1 = a1,

• the interval determined by {B3, b3} is not inner interval, and

• A3 = a2 or A2 = a3 holds.

A1 = a1 B1 b1

B2

A2 = a3
b3

A3 B3

a2b2

A1 = a1

B2 A2

B3
A3 = a2

B1

b2

b3a3

b1

: every cell is contained in the polyomino

: some cells are not in the polyomino

Figure 27: Binomials in Theorem 3.

Note that there are no elements in S2∪S3 whose initial monomial divides the binomials
S(F,G) discussed in Theorem 2 and 3. We conclude that all binomials of degree four from
the Buchberger Algorithm is of the form a1a2a3a4 − b1b2b3b4 with a1 <P a2 <P a3 <P a4,
b1 <P b2 <P b3 <P b4, and initial monomial a1a2a3a4, and can be illustrated in the
following figure.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2640

b2

a1

b4a4

b1

a2

b3
a3

b2 a2

b3

b1

a3

a4b4

a1

: every cell is contained in the polyomino

: some cells are not in the polyomino

Figure 28: Binomials in Theorem 2 and Theorem 3.

4. The Socket Wrench Polyominoes

Consider the following polyomino constructed from 8 unit squares forming a 3 × 3
square without the unit square in the center and continued by adding n unit squares to
the left of the unit square on the leftmost cell on the middle row. We call this polyomino
a socket wrench polyomino, because it looks like the socket wrenches used by mechanics
to tighten or loosen nuts and bolts. We use for this polyomino the same labelling with
reference to Section 3.

1 2 3 4

5 5 + n 6 + n 7 + n 8 + n

9 + n 9 + 2n 10 + 2n 11 + 2n 12 + 2n

13 + 2n 14 + 2n 15 + 2n 16 + 2n

Figure 29: A Socket Wrench Polyomino with n Additional Unit Squares.

Theorem 4. Let P be a socket wrench polyomino then the polyomino ideal IP is a radical.

Proof. We claim that according to this labelling and lexicographic order, the ideal IP
has S2 ∪ S3 as the Gröbner bases. Since the initial monomial of every element in S2 ∪ S3



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2641

is square-free then we have the result.
By Theorem 1, every element of S3 is one of the following forms:

• xix11+2nx14+2n − xi+4+nx15+2nx6+n, 5 ≤ i ≤ 4 + n

• xix12+2nx14+2n − xi+4+nx16+2nx6+n, 5 ≤ i ≤ 4 + n

• xix11+2nx13+2n − xi+4+nx15+2nx5+n, 5 ≤ i ≤ 4 + n

• xix12+2nx13+2n − xi+4+nx16+2nx5+n, 5 ≤ i ≤ 4 + n

To complete our claim, we observe the following.

(i) S(F,G) with F,G ∈ S2 is either reduced to zero or contained in S3.

(ii) Suppose there are F ∈ S3 and G ∈ S2 such that S(F,G) is not reduced to zero. Write
F = a1a3a5 − a2a4a6 as in Definition 1 and G = pq − rs. Here, a1 <P a3 <P a5,
a6 <P a2 <P a4, p <P q, and r <P s. From the structure of elements in S3, we have
a1 ∈ [5, 4 + n], a3 ∈ {11 + 2n, 12 + 2n}, and a5 ∈ {13 + 2n, 14 + 2n}. By Theorem
2, we have p = a3 or p = a5. Suppose p = a5, then [p, q] can not be an inner
interval. Therefore p = a3. Since [p, q] is an inner interval, then p = a3 = 11 + 2n
and q = 16 + 2n. But then s = a4 = 15 + 2n, contradiction with a4 > s.

(iii) Suppose there are F,G ∈ S3 such that S(F,G) is not reduced to zero. By the
definition of S3, since the non-initial monomial of a binomial in S3 is completely
determined by its initial monomial then we can eliminate the cases when the initial
monomials of F,G are relatively prime or equal.

(a) If the greatest common divisor of their initial monomial is a monomial of degree
two, we will have a similar argument as in the previous case but using Theorem
3 that it will come to a contradiction.

(b) If the greatest common divisor of their initial monomial is a monomial of degree
one, by our classifications above, we need to consider several cases of S(F,G).

• S(xix11+2nx14+2n−xi+4+nx15+2nx6+n, xjx12+2nx14+2n−xj+4+nx16+2nx6+n),
5 ≤ i < j ≤ 4 + n.
The above expression is equal to

xix11+2nxj+4+nx16+2nx6+n − xi+4+nx15+2nx6+nxjx12+2n

= x6+nx11+2nx16+2n(xixj+4+n − xi+4+nxj)

+x6+nxi+4+nxj(x11+2nx16+2n − x15+2nx12+2n)

and is reduced to zero.

• S(xix11+2nx14+2n−xi+4+nx15+2nx6+n, xjx11+2nx13+2n−xj+4+nx15+2nx5+n),
5 ≤ i < j ≤ 4 + n.
The above expression is equal to

xix14+2nxj+4+nx15+2nx5+n − xjx13+2nxi+4+nx15+2nx6+n



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2642

= x15+2nx14+2nx5+n(xixj+4+n − xjxi+4+n)

+x15+2nxjxi+4+n(x14+2nx5+n − x6+nx13+2n)

and is reduced to zero.

• S(xix11+2nx14+2n−xi+4+nx15+2nx6+n, xix12+2nx13+2n−xi+4+nx16+2nx5+n),
5 ≤ i ≤ 4 + n.
The above expression is equal to

x11+2nx14+2nxi+4+nx16+2nx5+n − x12+2nx13+2nxi+4+nx15+2nx6+n

= xi+4+nx11+2nx16+2n(x5+nx14+2n − x6+nx13+2n)

+xi+4+nx6+nx13+2n(x11+2nx16+2n − x12+2nx15+2n)

and is reduced to zero.

• S(xix12+2nx14+2n−xi+4+nx16+2nx6+n, xix11+2nx13+2n−xi+4+nx15+2nx5+n),
5 ≤ i ≤ 4 + n.
The above expression is equal to

x12+2nx14+2nxi+4+nx15+2nx5+n − xi+4+nx16+2nx6+nx11+2nx13+2n

= xi+4+nx12+2nx15+2n(x5+nx14+2n − x6+nx13+2n)

+xi+4+nx6+nx13+2n(x12+2nx15+2n − x11+2nx16+2n)

and is reduced to zero.

• S(xix12+2nx14+2n−xi+4+nx16+2nx6+n, xjx12+2nx13+2n−xj+4+nx16+2nx5+n),
5 ≤ i < j ≤ 4 + n.
The above expression is equal to

xix14+2nxj+4+nx16+2nx5+n − xi+4+nx16+2nx6+nxjx13+2n

= x16+2nx14+2nx5+n(xixj+4+n − xjxi+4+n)

+x16+2nxjxi+4+n(x14+2nx5+n − x13+2nx6+n)

and is reduced to zero.

• S(xix11+2nx13+2n−xi+4+nx15+2nx5+n, xjx12+2nx13+2n−xj+4+nx16+2nx5+n),
5 ≤ i < j ≤ 4 + n.
The above expression is equal to

xix11+2nxj+4+nx16+2nx5+n − xi+4+nx15+2nx5+nxjx12+2n

= x5+nx11+2nx16+2n(xixj+4+n − xjxi+4+n)

+x5+nxjxi+4+n(x11+2nx16+2n − x12+2nx15+2n)

and is reduced to zero.

And we are done with the proof.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2643

Remark 1. Note that we can rotate the socket wrench polyominoes 90◦, 180◦, and 270◦

and get the same conclusion since we also can rotate the labelling and using the same
monomial order.

We also can prove a stronger result by using the similar argument with [5, Section 4].

Theorem 5. Let P be a socket wrench polyomino then the ideal IP is prime.

Proof. We label P according to Figure 29. Let {Vi}i∈I be the set of maximal vertical
edge intervals of P and {Hj}j∈J be the set of maximal horizontal edge intervals of P,
where I = {1, 2, . . . , n + 4} and J = {1, 2, 3, 4}. Let {vi}i∈I and {hj}j∈J be the set of
variables associated respectively to {Vi}i∈I and {Hj}j∈J , respectively. Let w be another
variable different from vi and hj . Let A = {3, 4, 7 + n, 8 + n}. Define

α : V (P) → K[{vi, hj , w} : i ∈ I, j ∈ J ]

r 7→ vihjw
k

with r ∈ Vi ∩Hj , k = 0 if r /∈ V (A), and k = 1 if r ∈ V (A).
Consider the following surjective ring homomorphism

ϕ : K[xr : r ∈ V (P)] → K[α(v) : v ∈ V (P)]

xr 7→ α(r)

The toric ideal JP is the kernel of ϕ. We will prove that IP = JP .
We start by proving IP ⊆ JP . Let f = xpxq−xrxs be a generator of IP that associated

to the inner interval [p, q]. We may assume that p, r and q, s, respectively, are on the same
maximal vertical edge interval. Then, p, s and q, r, respectively, are on the same maximal
horizontal edge interval. If [p, q] ∩ A = ∅ then f ∈ JP . Consider the case [p, q] ∩ A ̸= ∅.
If [p, q] = A then f ∈ JP . If [p, q] ̸= A, by the construction of P, then either s, q or p, s
must be two vertices of A. In the first case, p, r are not the vertices of A. In the second
case, r, q are not the vertices of A. In both cases, we conclude that f ∈ JP .

Now, it remains to prove that JP ⊆ IP . We will prove this by showing that every
binomial of degree two in JP belongs to IP and every irredundant binomial in JP is of
degree two (or for some cases, it is in IP).

For the first part, let f = xpxq − xrxs be a binomial in JP . If p, q are in horizontal or
vertical position, since ϕ(f) = 0 then we can easily argue that {p, q} = {r, s} and f = 0 ∈
IP . We consider the case p, q are the diagonal corners of an interval (the case p, q are the
antidiagonal corners can be done similarly). Let vp and hp be the variables associated to
the maximal vertical and horizontal edge intervals that contain p, respectively. We define
vq, vr, vs, hq, hr, hs similarly. We will prove that r, s are the antidiagonal corners of [p, q]
and argue that f ∈ IP . We divide into three cases:

• If p, q ∈ A. Since ϕ(xpxq) = vpvqhphqw
2 then w2 divides ϕ(xrxs). Thus, r, s ∈ A. If

r = p or r = q then {r, s} = {p, q} and f = 0 ∈ IP . Therefore r is an antidiagonal
corner of [p, q]. If ϕ(xr) = vphqw then ϕ(xs) = vqhpw and s is also an antidiagoal
corner of [p, q]. The same conclusion for ϕ(xr) = vqhpw. Clearly, [p, q] = [3, n+8] is
an inner interval and thus f ∈ IP .



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2644

• If exactly one of p, q belongs to A. We consider the case p ∈ A. By the construction
of P, we have p ∈ {3, 7 + n} and q ∈ {12 + 2n, 16 + 2n}. Since w divides ϕ(xp) then
r ∈ A or s ∈ A. We may assume that r ∈ A. If r = p then s = q and f = 0 ∈ IP .
If p, r are not in horizontal position then Hr contains an edge of A but Hp ̸= Hr

and Hq does not contain any edge of A. Therefore hr divides ϕ(xrxs) but does
not divide ϕ(xpxq), a contradiction. Now, p, r are in horizontal position. By the
construction of P, we conclude that r, q are in vertical position. Thus, ϕ(xs) = vphq
and therefore r, s are the antidiagonal corners of [p, q]. Since p ∈ {3, 7 + n} and
q ∈ {12 + 2n, 16 + 2n} then [p, q] is an inner interval and thus f ∈ IP . The case
q ∈ A is also true by symmetry.

• If both p, q do not belong to A. Similarly, we have that f = 0 ∈ IP or r, s are the
antidiagonal corners of [p, q]. Let P ′ be a polyomino obtained by removing the cells
that has common vertices with A. Note that P ′ is a simple polyomino. Let ϕ′ be the
restriction of ϕ onK[xa : a ∈ V (P)\A] and JP ′ be the kernel of ϕ′. Note that f ∈ JP ′

By [36, Theorem 2.2], we have that IP ′ = JP ′ . Therefore f ∈ JP ′ = IP ′ ⊂ IP .

For the second part, let f be an irredundant binomial in JP . Clearly, f has degree at
least two. Suppose that f has degree at least three and choose f with the least degree.
Suppose that every variable of f is in K[xa : a ∈ V (P)\A]. Define P ′ as the previous
case then f is a binomial in JP ′ and f is irredundant in JP ′ . Since IP ′ = JP ′ then f is
an irredundant binomial in IP ′ which means that f must be a binomial of degree two, a
contradiction. Now, suppose that xv1 is a variable of f with v1 ∈ A. Write f = f+ − f−.
We may assume that xv1 divides f+. If xv1 divides f− then f = xv1(g

+ − g−). Since
JP is prime then g = g+ − g− ∈ JP . If the degree of g is at least three then g must be
irredundant. But, this contradict the choice of f . If the degree of g is two then by the
previous part, we conclude that g ∈ IP and f = xv1g ∈ IP . Now, suppose that xv1 does
not divide f−. We may assume that no xv divides both f+ and f− for v ∈ A. Since
w divides ϕ(f+) and ϕ(f+) = ϕ(f−) then there exists v′1 ∈ A such that xv′1 divides f−.
Let Vv1 and Hv1 be the maximal vertical and horizontal edge intervals, respectively, that
contain v1. Since vv1 divides ϕ(f+) and ϕ(f+) = ϕ(f−) then there exists v′2 ∈ Vv1 such
that xv′2 divides f−. Similarly, there exists v′3 ∈ Hv1 such that xv′3 divides f−. Define Vv′1
and Hv′1

similarly. We also get that there exists v2 ∈ Vv′1
and v3 ∈ Hv′1

such that both xv2
and xv3 divide f+. Consider the following cases:

• If v1 and v′1 are on the same horizontal edge interval of P. By the construction of P
then the interval determined by v1, v2 is an inner interval. By [5, Lemma 2.2] with
three vertices v1, v2 ∈ V +

f dan v′1 ∈ V −
f , we get a contradiction.

• If v1 and v′1 are on the same vertical edge interval of P. Similarly we get a contra-
diction by [5, Lemma 2.2] and three vertices v1, v3 ∈ V +

f dan v′1 ∈ V −
f .

• If v1 and v′1 are the diagonal corners of [3, 8 + n]. We may assume that v1 = 3
and v′1 = 8 + n. Consider v′3. If v′3 = 4 then v2 ∈ {12 + 2n, 16 + 2n} and we get
a contradiction by [5, Lemma 2.2] and three vertices v1, v2 ∈ V +

f dan v′3 ∈ V −
f .



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2645

Therefore v′3 ̸= 4. In particular, v′3 is not the antidiagonal corner of [3, 8 + n]. With
the similar arguments, we conclude that v′2 is not the antidiagonal corner of [3, 8+n].
Looking at the construction of P, we see that the vertices v1, v

′
1, v

′
2, v

′
3 lie on P as

the following figure:

v1 4

v′1

v′2

v′3

h

Figure 30: Illustration for v1, v
′
1, v

′
2, v

′
3 on P.

Note that [v′3, v
′
1] is an inner interval with 4 as one of the antidiagonal. Let h be the

other antidiagonal. Notice that

f =

(
f+ − f−

xv′1xv′3
xhx4

)
− f−

xv′1xv′3
(xv′1xv′3 − xhx4).

Since xv′1xv′3 − xhx4 ∈ IP ⊆ JP then f+ − f−

xv′1
xv′3

xhx4 ∈ JP . But, both x4 and xv′2

divide f−

xv′1
xv′3

xhx4 and xv1 divide f+. By [5, Lemma 2.2] and three vertices 4, v′2, v1

we get f+ − f−

xv′1
xv′3

xhx4 is redundant and f is also redundant, a contradiction.

• We argue similarly for the case v1 and v′1 are the antidiagonal corners of [3, 8 + n].

Corollary 1. Let P be a socket wrench polyomino then K[P] is a normal Cohen-Macaulay
domain.

Proof. By the previous theorem, we have IP is a toric ideal and has square-free
quadratic Gröbner bases for the suitable monomial order. By a theorem of Sturmfels [23,
Corollary 4.26] we conclude thatK[P] is normal and by a theorem of Hochster [4, Theorem
6.3.5] we have that K[P] is Cohen-Macaulay. Therefore K[P] is a normal Cohen-Macaulay
domain.

Next we compute the h-polynomial of socket wrench polyominoes and prove that K[P]
is Gorenstein if and only if there is no unit square that we add in the definition of the
socket wrench polyominoes. We refer the definition of (L, C)-polyomino in [8]. The socket
wrench polyominoes are (L, C)-polyominoes by the following figure

Here, we take symmetry to the definition of (L, C)-polyomino so it is suitable to the
socket wrench. The results in [8] do not change. We also can rotate the socket wrench
polyominoes by 180◦ to see that the socket wrench polyominoes are (L, C)-polyominoes.

We recall some terminologies from [8] and [37].

(i) For a polyimino P, the rook number r(P) is the maximum number of non-attacking
rooks that can be placed in P.



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2646

a2 d2

a1
d1

b2 b1 b
d

c2 c1 c a

C

L

Figure 31: Socket wrench polyominoes are (L, C)-polyominoes.

(ii) For a polyomino P, denote by rk the number of ways to placed k rook in P in
non-attacking position, conventionally r0 = 1.

(iii) The polyomino P is thin if it does not contain square tetromino.

(iv) Let P be a simple thin polyomino. A cell C of P is single if there exists a unique
maximal inner interval of P containing C. If any maximal inner interval of P has
exactly one single cell, we say that P has the S-property.

We also use some terminologies from [5] and [31].

(i) Let P be a polyomino. A sequence of distinct inner interval W : I1, . . . , Iℓ of P such
that vi, zi are diagonal (resp. antidiagonal) corners and ui, vi+1 are antidiagonal
(resp. diagonal) corners of Ii, for i = 1, . . . , ℓ, is a zig-zag walk of P, if

(a) I1 ∩ Iℓ = {v1 = vℓ+1} and Ii ∩ Ii+1 = {vi+1} for i = 1, 2, . . . , ℓ− 1;

(b) vi and vi+1 are on the same edge interval of P, for i = 1, . . . , ℓ.

(c) for any i, j ∈ {1, . . . , ℓ}, with i ̸= j, does not exist an inner interval J of P such
that zi, zj ∈ J .

(ii) A polyomino is called closed path if P is a sequence of cells A1, A2, . . . , An, An+1,
n > 5 such that

(a) A1 = An+1;

(b) Ai ∩Ai+1 is a common edge, for i = 1, 2, . . . , n;

(c) Ai ̸= Aj for all i ̸= j and i, j ∈ {1, 2, . . . , n};
(d) For all i ∈ {1, 2, . . . , n} and for all j /∈ {i− 2, i− 1, i, i+ 1, i+ 2} then V (Ai) ∩

V (Aj) = ∅, where A−1 = An−1, A0 = An, An+1 = A1, An+2 = A2.

In [31, Corollary 3.6], the authors prove that if there exists a zig-zag walk in P then
the ideal IP is not prime. For socket wrench polyominoes, since IP is prime then P has
no zig-zag walk. Now, we are ready to prove the next theorem.

Theorem 6. Let P be a socket wrench polyomino with n additional unit squares. Then:

(i) the h-polynomial of K[P] is

hK[P](t) = 1 + (n+ 8)t+ (7n+ 16)t2 + (11n+ 8)t3 + (3n+ 1)t4;



Y. Y. Hamonangan, I. Muchtadi-Alamsyah / Eur. J. Pure Appl. Math, 17 (4) (2024), 2621-2650 2647

(ii) reg(K[P]) = 4;

(iii) K[P] is Gorenstein if and only if n = 0.

Proof. Since P is a (L, C)-polyomino and C is a simple and thin polyomino then by [8,
Theorem 5.2], we obtain that

hK[P](t) =

r(P)∑
k=0

rkt
k

and reg(K[P]) = r(P). Note that r(P) = 4 since the first, the second and the third row
can not contain more than one rook, two rooks, one rook, respectively, and we can place
four rooks like illustrated in the figure below

R R

R

R
Figure 32: r(P) = 4.

We also can easily get r1 = n+8, r2 = 7n+16, r3 = 11n+8, and r4 = 3n+1 by some
counting arguments. Then

hK[P](t) = 1 + (n+ 8)t+ (7n+ 16)t2 + (11n+ 8)t3 + (3n+ 1)t4

and reg(K[P]) = 4, (i) and (ii) are proven.
For (iii), by [39, Theorem 4.2] since 1 ̸= 3n+1 for n > 0 then K[P] is not Gorenstein.

If n = 0, then P is a closed path having no zig-zag walks. We notice that the single cells
of P are the four cells in the middle of the first row, the third row, the first column, and
the third column. We also notice that the maximal intervals of P are the four intervals
containing three cells in the first row, the third row, the first column, and the third column.
Each of them only has one single cell, and thus P has the S-property. By [8, theorem 5.7],
we conclude that K[P] is Gorenstein.

5. Conclusion

In this paper, we classify some few-degree binomials that arise from the Buchberger
Algorithm on polyomino ideal. Based on the labelling and the monomial order that were
explained at the beginning of the third section, we obtain that the Buchberger Algorithm
produces binomials of degree three (Theorem 1) and binomials of degree four (Theorem
2 and 3). We also give a class of polyominoes (the socket wrench polyominoes) that has
Gröbner bases of degree at most three with respect to the previous labelling and monomial
order, and hence the polyomino ideal is radical. We also study some properties of the



REFERENCES 2648

polyomino ideal IP of the socket wrench polyominoes. The ideal IP is prime (Theorem
4). The quotient ring KP is a normal Cohen-Macaulay domain (Corollary 1). The h-
polynomial, regularity, and Goreinsteness are given in Theorem 6. The problem about
Gröbner bases and radicality of polyomino ideal are still open.

Acknowledgements

This research is funded by PMDSU Program 2015. We also thank to Ayesha Qureshi
for suggesting the problem, for the guidance and for the useful advice.

The authors also thank the referees of European Journal of Pure and Applied Mathe-
matics, for their careful reading and helpful suggestions.

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