Theorem 3.1. With the above notations, then we have the following assertions
(1) If G is the empty set, then a2(R/I) = 0.
(2) If G is a union of disjoint edges, then a2(R/I) = 1.
(3) If G does not contain any triangles and G has at least two distinct edges which have a
common vertex, then a2(R/I) = 2.
(4) If G contains at least a triangle, then a2(R/I) = 3.
Proof. Let a := a2(R/I), we is now choosing α ∈ Zn such that Hm2(R/I)α 6= 0 and |α| = a. We
have four cases as follows:
Case 1: G is empty. By Lemmas 3.1, 3.2, 3.3, a ≤ 0. Put α = (0, 0, ., 0). Since Lemma
2.1, one can see that the facets of ∆α are {{i, j}|1 ≤ i < j ≤ n}. So, ∆α contains at least a
cycle, such as (1, 2, ., n). By Lemma [4, Lemma 1.5], He1(∆α; k) 6= 0. Using Theorem 2.1, then
H2
m(R/I)α 6= 0, whence a ≥ 0. Thus a = 0.
Case 2: G is an union of disjoint edges. By Lemmas 3.1, 3.2, 3.3, a ≤ 1. We assume that
G = {{1, 2}}. Put α = (0, 0, 1, 0, ., 0). It is easy to check that the facets of ∆α are
{12, 13, 23, 3i, jh | 4 ≤ i, j, h ≤ n and {j, h} ∈ G}.
Therefore, ∆α contains the cycle C = (1, 2, 3). Thus, we have He1(∆α; k) 6= 0, whence a ≥ 1.
Hence, a = 1
Case 3: G does not contain any triangles but has at least two edges which have a
common vertex. Without loss of generality, we may assume that 12, 13 ∈ G and 23 ∈/ G. Put
α = (0, 1, 1, 0, ., 0). By Lemma 2.1, ∆α contains the cycle C = (1, 2, 3). So, He1(∆α; k) 6= 0,
whence a ≥ 2. Thus, a = 2.
Case 4: G contains a triangle, assume 12, 13, 23 ∈ G. Put α = (1, 1, 1, 0, ., 0). With a
similar argument as in the Case 3, we get a = 3.

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JOURNAL OF SCIENCE OF HNUE DOI: 10.18173/2354-1059.2017-0026
Mathematical and Physical Sci., 2017, Vol. 62, Iss. 8, pp. 3-8
This paper is available online at
REGULARITY OF IDEALS ASSOCIATED WITH A COMPLETE GRAPH
IN TWO WEIGHTS (2,1)
Phan Thi Thuy
Faculty of Mathematics, Hanoi National University of Education
Abstract. In this paper, we will determine regularity of ideals associated with a complete
graph in two weight. The value of this function depends on the structure of an associated
simplex graph.
Keywords: Regularity, connected graphs, monomial ideals.
1. Introduction
Let R = k[x1, x2, ..., xn] be the polynomial ring over a field k and n ≥ 3. Let Pi,j be the
prime ideal generated by variables {x1, x2, ..., xn}\{xi, xj} for 1 ≤ i < j ≤ n. In this article, we
take ideals in R of the form
I :=
⋂
1≤i<j≤n
P
ωi,j
i,j , where ωi,j ∈ {2, 1}.
Then dim(R/I) = 2. We know that R/I is Cohen-Macaulay by Minh and Nakamura [1]. In this
article, we consider in computing the regularity of ideals I . Firstly, let us recall this notation. Let
m = (x1, x2, ..., xn) be the maximal homogeneous ideal of R. For a proper homogeneous ideal J
of R, let
ai(J) = sup{j | H im(J)j 6= 0},
where H i
m
(J) is the local cohomology module of J with respect m. The Castelnuovo-Mumford
regularity (or regularity) of J , denoted by reg(J), is defined by
reg(J) = max{ai(J) + i | 0 ≤ i ≤ dim(J)}.
Regularity of J is a kind of universal bound for important invariants, such as the maximum degree
of the syzygies and the maximum non-vanishing degree of the local cohomology modules of J .
Let
G := {{i, j} | 1 ≤ i < j ≤ n, ωi,j = 2},
Received April 15, 2017. Accepted August 17, 2017.
Contact Phan Thi Thuy, e-mail: thuysp1@gmail.com.
3
Phan Thi Thuy
which forms a simplex graph of the vertex set [n]. We can see that the regularity of ideals I is
computed in the main result of the paper as follows:
Theorem 3.2. For the above notations, the following assertions holds true
(1)If G is the empty set, then reg(I) = 3.
(2) If G is a union of disjoint edges, then reg(I) = 4.
(3) If G does not contain any triangles and G has at least two distinct edges which have a common
vertex, then reg(I) = 5.
(4) If G contains at least a triangle, then reg(I) = 6.
Now let us explain the organization of this paper. In Section [2], we recall Takayama’s
formula to compute local cohomology modules of monomial ideals, and then give description of
associated simplicial complex.
2. Preliminaries
We will use some notations on graphs according to [2] . For the detailed information about
combinatorial and algebraic background, we refer the reader [1].
Let ∆ be a simplicial complex on [n] = {1, . . . , n} that is a collection of subsets of [n]
closed under taking subsets. We put dimF = |F | − 1, where |F | is the cardinality of F , and
dim∆ = max{dimF | F ∈ ∆}, which is called the dimension of ∆. It is clear that ∆ can be
uniquely determinate by the set of its maximal elements under inclusion, are called by facets. The
complex ∆ is said that pure if all its facets have the same cardinality. For a face F ∈ ∆, we define
the link of F in a simplicial complex ∆ as follows:
lk∆ F = {G ∈ ∆ | F ∪G ∈ ∆, F ∩G = ∅}.
Let I be a monomial ideal in R. We denote by G(I) the minimal set of monomial generators
of I . Let ∆ be the simplicial complex corresponding to the radical ideal of I , i.e.,
∆ = {{i1, i2, ..., ik} | xi1 . . . xik /∈
√
I}.
For every α = (α1, α2, ..., αn) ∈ Zn let xα = xα11 . . . xαnn and Gα = {i | αi < 0}. The degree
complex ∆α is the simplicial complex whose faces are sets of form F\Gα, where Gα ⊆ F ⊆ [n],
so that for every minimal generator xβ of I there exists an index i /∈ F with αi < βi. To present
∆α in a more compact way, for a subset F of [n] let RF := R[x
−1
i |i ∈ F ]. Then, by [3, Lemma
1.2] we have
∆α := {F ⊆ [n]\Gα | xα /∈ IRF∪Gα}.
The following theorem takes an important role in our argument.
Theorem 2.1 (Takayama’s formula). For all i ≥ 0, we have
dimkH
i
m(R/I)α =
{
dimkH˜i−|Gα|−1(∆α; k) if Gα ∈ ∆
0 otherwise.
4
Regularity of ideals associated with a complete graph in two weights (2,1)
From the general property of the reduced homology group, we have
(1) H˜i(∅; k) = 0 for all i;
(2) H˜i({∅}; k) = 0 for all i 6= −1;
(3) H˜−1({∅}; k) = k;
(4) H˜0(∆α; k) = 0 if and only if ∆α is a connected.
From now on, we consider ideals of form
I :=
⋂
1≤i<j≤n
P
ωi,j
i,j , where ωi,j ∈ {2, 1}.
For α ∈ Zn, we put |α| =∑ni=1 αi and σi,j = |α| − αi − αj , where 1 ≤ i < j ≤ n.
Lemma 2.1. For α ∈ Nn, then the following assertions hold true
(1) dim∆α ≤ 1.
(2) {i, j} ∈ ∆α if and only if σi,j < ωi,j for any 1 ≤ i < j ≤ n.
(3) If ∆α 6= {∅}, then dim∆α = 1 and ∆α is pure.
Proof. (1) See [3, Lemma 2.2].
(2) See [3, Lemma 2.3].
(3) See [3, Lemma 2.4].
By the above lemma, ∆a can be considered as a graph for α ∈ Nn.
3. Proof of the main result
Since dim(R/I) = 2 and R/I is Cohen-Macaulay by [4, Theorem 3.1], it follows that
H i
m
(R/I) = 0 with 0 ≤ i ≤ 1. Hence
reg(R/I) = a2(R/I) + 2.
Next, we are computing a2(R/I). Let
G := {{i, j} | 1 ≤ i < j ≤ n, ωi,j = 2},
which forms a simplex graph of the vertex set [n], and it provides an important role. Regularity of
the ideals I depends on terms of G.
Let α ∈ Zn such that H2
m
(R/I)α 6= 0. By Lemma 2.1,
dimkH
2
m
(R/I)α = dimk H˜1−|Gα|(∆α; k) 6= 0,
and Gα ∈ ∆. Hence, we must have |Gα| ≤ 2 and distinguish some cases.
Lemma 3.1. Assume that |Gα| = 2. Then |α| ≤ −1.
5
Phan Thi Thuy
Proof. We may assume Gα = {1, 2}. Since H˜−1(∆α; k) 6= 0, we have ∆α = {∅}. Define β ∈ Nn
as follows:
βi =
{
αi if αi ≥ 0
0 else.
Then, by definition of the degree complex we conclude that ∆α = lk∆β{1, 2}. So lk∆β{1, 2} =
{∅}. Hence, by Lemma 2.1, ∆β = {12}. This statement implies that σ1,2 ≤ 1. Therefore, |α| ≤
1 + α1 + α2 ≤ −1.
Lemma 3.2. Assume that |Gα| = 2. Then we have
(1) If G is the empty set, then |α| ≤ −1.
(2) If G has an element, then |α| ≤ 0.
(3) If G contains at least two elements, then |α| ≤ 1.
Proof. We may assume that Ga = {1}. Because H˜0(∆α; k) 6= 0, it follows that ∆α is a connected
and is not the void complex {}.
Put β ∈ Nn such that βi = αi if i 6= 1 and β1 = 0. One can see that ∆α = lk∆β{1}. So,
by Lemma 2.1, dim∆α = 0 and ∆α must contain at least two points. We can assume 2, 3 ∈ ∆α.
This statement implies that 12, 13 belong to different components of ∆β . Hence, by Lemma 2.1,
we have a system of inequalities {
σ1,2 < ω1,2
σ1,3 < ω1,3.
Therefore, |α| ≤ 2|α| − α2 − α3 − α1 ≤ ω1,2 + ω1,3 + α1 − 2 ≤ ω1,2 + ω1,3 − 3. From this, we
obtain as required.
Lemma 3.3. Assume that |Gα| = 0. Then we have
(1) If G is the empty set, then |α| = 0.
(2) If G is a union of disjoint edges, then |α| ≤ 1.
(3) If G does not contain any triangles but G has at least two distinct edges which have a
common vertex, then |α| ≤ 2.
Proof. From [4, Lemma 1.5], one can see that H˜1(∆α; k) 6= 0 if and only if ∆α must contain
a cycle, say C = (1, 2, ..., t), where t ≥ 3. Therefore, we have inequalities σi,i+1 < ωi,i+1 and
σ1,t < ω1,t for all i < t. Hence
(t− 2)|α| ≤
t−1∑
i=1
ωi,i+1 + ω1,t − t.
Thus, |α| ≤
[
t
t− 2
]
≤ 3. We distinguish some cases.
6
Regularity of ideals associated with a complete graph in two weights (2,1)
Case 1: If G is the empty set, then |α| = 0 is obvious.
Case 2: If |G| = 1, then |α| ≤
[
1
t− 2
]
≤ 1.
Case 3: If |G| = 2, then |α| ≤
[
2
t− 2
]
≤ 2. Assume |α| = 2. Then it follows that t = 3
and ω1,2 + ω1,3 + ω2,3 = 4. It mean that G contains two distinct edges which have a common
vertex.
Case 4: If |G| ≥ 3 and G is a union of disjoint edges, then |α| ≤
[
t
2t− 4
]
≤ 1. Assume
|α| = 3. By |α| ≤
[
t
t− 2
]
, we must have t = 3. Hence, ω1,2 = ω1,3 = ω2,3 = 2. This implies
that G must contain a triangle.
Since the above arguments, the proof is complete.
Finally, we can state and prove the main results of this paper.
Theorem 3.1. With the above notations, then we have the following assertions
(1) If G is the empty set, then a2(R/I) = 0.
(2) If G is a union of disjoint edges, then a2(R/I) = 1.
(3) If G does not contain any triangles and G has at least two distinct edges which have a
common vertex, then a2(R/I) = 2.
(4) If G contains at least a triangle, then a2(R/I) = 3.
Proof. Let a := a2(R/I), we is now choosing α ∈ Zn such that H2m(R/I)α 6= 0 and |α| = a. We
have four cases as follows:
Case 1: G is empty. By Lemmas 3.1, 3.2, 3.3, a ≤ 0. Put α = (0, 0, ..., 0). Since Lemma
2.1, one can see that the facets of ∆α are {{i, j}|1 ≤ i < j ≤ n}. So, ∆α contains at least a
cycle, such as (1, 2, ..., n). By Lemma [4, Lemma 1.5], H˜1(∆α; k) 6= 0. Using Theorem 2.1, then
H2
m
(R/I)α 6= 0, whence a ≥ 0. Thus a = 0.
Case 2: G is an union of disjoint edges. By Lemmas 3.1, 3.2, 3.3, a ≤ 1. We assume that
G = {{1, 2}}. Put α = (0, 0, 1, 0, ..., 0). It is easy to check that the facets of ∆α are
{12, 13, 23, 3i, jh | 4 ≤ i, j, h ≤ n and {j, h} ∈ G}.
Therefore, ∆α contains the cycle C = (1, 2, 3). Thus, we have H˜1(∆α; k) 6= 0, whence a ≥ 1.
Hence, a = 1
Case 3: G does not contain any triangles but has at least two edges which have a
common vertex. Without loss of generality, we may assume that 12, 13 ∈ G and 23 /∈ G. Put
α = (0, 1, 1, 0, ..., 0). By Lemma 2.1, ∆α contains the cycle C = (1, 2, 3). So, H˜1(∆α; k) 6= 0,
whence a ≥ 2. Thus, a = 2.
Case 4: G contains a triangle, assume 12, 13, 23 ∈ G. Put α = (1, 1, 1, 0, ..., 0). With a
similar argument as in the Case 3, we get a = 3.
7
Phan Thi Thuy
We have the result on Castelnuovo-Mumford regularity as follows:
Theorem 3.2. For the above notations, the following assertions hold true
(1) If G is the empty set, then reg(I) = 3.
(2) If G is a union of disjoint edges, then reg(I) = 4.
(3) If G does not contain any triangles and G has at least two distinct edges which have a
common vertex, then reg(I) = 5.
(4) If G contains at least a triangle, then reg(I) = 6.
Proof. By the definition of reg(I), we have
reg(I) = reg(R/I) + 1.
Therefore, using Theorem 3.1, this theorem is obvious.
REFERENCES
[1] R. Stanley, 1996. Combinatorics and Commutative Algebra, 2. Edition, Birkha¨user.
[2] R. Diestel, 2010. Graph theory, 2nd. edition, Springer: Berlin/Heidelberg/New York/Tokyo,
2000.
[3] N. C. Minh and Y. Nakamura, 2012. Unmixed monomial ideal of dimension two and their
Cohen-Macaulay properties, Communications in Algebra, 38, pp.1699-1714.
[4] L.T.Hoa and T. N. Trung, 2016. Catelnuovo-Mumford regularity of symbolic power of
two-dimensional square-free monomial ideal, J. Commut. Algebra 8, No. 1, pp. 77- 88.
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