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Open AccessArticle

Cohen Macaulay Bipartite Graphs and Regular Element on the Powers of Bipartite Edge Ideals

1
Ramakrishna Mission Vivekananda Educational and Research Institute, Belur, West Bengal 711202, India
2
Department of Mathematics, University of Virginia, Charlottesville, VA 22902, USA
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(8), 762; https://doi.org/10.3390/math7080762
Received: 8 August 2019 / Revised: 8 August 2019 / Accepted: 13 August 2019 / Published: 20 August 2019
(This article belongs to the Special Issue Current Trends on Monomial and Binomial Ideals)

Abstract

In this article, we discuss new characterizations of Cohen-Macaulay bipartite edge ideals. For arbitrary bipartite edge ideals I ( G ) , we also discuss methods to recognize regular elements on I ( G ) s for all s 1 in terms of the combinatorics of the graph G.
Keywords: Cohen Macaulay; Bipartite graphs; regular elements on powers of bipartite graphs; colon ideals; depth of powers of bipartite graphs; dstab; associated graded rings Cohen Macaulay; Bipartite graphs; regular elements on powers of bipartite graphs; colon ideals; depth of powers of bipartite graphs; dstab; associated graded rings

1. Introduction

The interplay between the combinatorics of finite simple graphs G and the algebra of the underlying edge ideals I ( G ) has been studied by various researches during the last few decades. The algebraic invariants that have been particularly prone to combinatorial interpretation are regularity, projective dimension, depth, and Betti numbers. In this article, we study the depth of powers of edge ideals of bipartite graphs. Combinatorics of bipartite graphs have been particularly ripe with interesting algebraic counterparts in the edge ideals and their powers. Interested readers are referred to [1,2,3], etc. In this paper, we continue the study pursued by the same authors in [3]. We study the closely related topics of combinatorial characterization of regular elements and Cohen-Macaulayness of various powers of bipartite edge ideals.
In section two of this paper, we offer a new characterization of Cohen-Macaulay bipartite edge ideals. We characterize it using colon ideals of the form ( I ( G ) 2 : e ) , where e is an edge/generator of I ( G ) , somehow in the same way as it is done in [3,4], etc., in the study of regularity. An often quoted and important characterization of Cohen-Macaulay bipartite edge ideals is due to Herzog-Hibi in [2]. In this article, we also give a new proof of this characterization ([2]). One important feature of our proof is that it does not use Hall’s marriage theorem or any variant of it as it was done in [2]. Throughout this article, we refer to S as the polynomial ring k [ x 1 , , x n , y 1 , , y n ] . Our main results in this section are as follows:
Theorem 1.
Let G be a bipartite graph with partition V 1 = { x 1 , , x n } and V 2 = { y 1 , , y n } . Then the following are equivalent
1.
S / I ( G ) is Cohen-Macaulay
2.
n = n and there exists a re-ordering of the vertex sets V 1 , V 2 such that
(a) 
x i y i I ( G ) for all i
(b) 
If x i y j I ( G ) then i j .
(c) 
If x i y j , x j y k I ( G ) then x i y k E .
3.
I ( G ) is unmixed and S / I ( G ) is connected in codimension one.
4.
n = n and there exists exactly n edges e 1 , , e n such that ( I ( G ) 2 : e i ) = I ( G ) and for i j , e i and e j are disjoint.
5.
n = n and there exists exactly n edges e 1 , , e n such that ( I ( G ) 2 : e i ) is Cohen-Macaulay and for i j , e i and e j are disjoint.
For arbitrary bipartite edge ideals, it is often hard to compute the depth of powers of its edge ideals I ( G ) s for all s 1 . Even if G is Cohen-Macaulay, it is not so easy to compute the depth S / I ( G ) s for s 2 . It is well known that depth S / I ( G ) s is asymptotically equal to the number of connected components of G ([5]). An important invariant related to the study of depth S / I ( G ) s is the dstab I ( G ) which measures the minimal t for which depth S / I ( G ) t equals the number of connected components of G. To study such invariants the same authors in [3] characterized regular elements on I ( G ) s for any unmixed bipartite graphs G. In the third section of this paper we characterize elements of the form x ν y μ that are regular on the powers I ( G ) s of a bipartite edge ideal G. This is a generalization of the similar result proved in [3]. Our characterization turns out to be the exactly same as the ☆-condition proved there. To signify its usefulness we call it the neighborhood properties (we refer to the definition in Definition 12) Our main result proved here is as follows:
Theorem 2.
Let G be a bipartite graph and suppose that x μ V 1 and y ν V 2 satisfies the neighborhood properties. Then x μ y ν is an regular element on S / I ( G ) s for all s.

2. Structure of Cohen-Macaulay and Unmixed Bipartite Graphs

A characterization theorem for Cohen-Macaulay bipartite graphs was given by Herzog-Hibi in [2].
Theorem 3.
(Herzog-Hibi, [2]) Let G be a bipartite graph with partition V 1 = { x 1 , , x n } and V 2 = { y 1 , , y n } . Then the following are equivalent
1.
S / I ( G ) is Cohen-Macaulay
2.
n = n and there exists a re-ordering of the vertex sets V 1 , V 2 such that
(a) 
x i y i I ( G ) for all i
(b) 
If x i y j I ( G ) then i j .
(c) 
If x i y j , x j y k I ( G ) then x i y k E .
The following theorem is an improvement of the Herzog-Hibi characterization (Theorem 3). We are grateful to Prof. Huneke for the ideas presented in this proof. It is important to notice here that the following theorem does not make use of the Halls marriage theorem which is an important element of any proofs known to us of Theorem 3.
Definition 1.
(Definition, p. 498, [6]) Let I be an ideal in a polynomial ring S such that I = P 1 P k , P i Spec ( S ) , 1 i k . We say that the ring S / I is connected in codimension one if for any two primes Q , Q Min ( S / I ) , there is a sequence of minimal primes Q = Q 1 , , Q r = Q Min ( S / I ) such that for each i = 1 , 2 , , r 1 , ht ( Q i + Q i + 1 ) = 1 in S / I .
Theorem 4.
Let G be a bipartite graph with partition V 1 = { x 1 , , x n } and V 2 = { y 1 , , y n } . Then the following are equivalent
1.
S / I ( G ) is Cohen-Macaulay
2.
n = n and there exists a re-ordering of the vertex sets V 1 , V 2 such that
(a) 
x i y i I ( G ) for all i
(b) 
If x i y j I ( G ) then i j .
(c) 
If x i y j , x j y k I ( G ) then x i y k E .
3.
I ( G ) is unmixed and S / I ( G ) is connected in codimension one.
Proof. 
First we show ( 2 ) ( 1 ) . We prove by induction on n. If n = 1 , then I ( G ) = ( x 1 y 1 ) and hence clearly S / I ( G ) is Cohen-Macaulay. Now assume that the result is true for n 1 and let G be a graph which satisfies the conditions ( a ) ( c ) of ( 2 ) on 2 n vertices (with partition V 1 = { x 1 , , x n } and V 2 = { y 1 , , y n } ). Consider
0 S ( I ( G ) : x 1 ) S I ( G ) S ( ( I ( G ) , x 1 ) 0
Notice that ( I ( G ) , x 1 ) = ( I ( G ) , x 1 ) , where G is the graph obtained by deleting x 1 and y 1 from G. Clearly G satisfies the conditions ( a ) ( c ) of ( 2 ) and hence S / I ( G ) is Cohen-Macaulay (on 2 n 2 vertices) by induction. So S / ( I ( G ) , x 1 ) is Cohen-Macaulay of dimension n. Let { y 1 , y i 1 , . . . . , y i k } ( I ( G ) : x 1 ) for some i 1 , . . . . , i k . Let x i j y l I ( G ) for some 1 j k . As x 1 y i j I ( G ) by the condition ( c ) , x 1 y l I ( G ) and hence l { 1 , i 1 , . . . . , i k } . So ( I ( G ) : x 1 ) = ( I ( G ) , y 1 , . . . . , y i k ) , where G is the graph obtained from G by deleting x 1 , y 1 , x i 2 , y i 2 , , x i k , y i k . But by induction, S / I ( G ) is Cohen-Macaulay of dimension n k . Hence S / ( I ( G ) : x 1 ) is Cohen-Macaulay of dimension n. Now in (1), both S / ( I ( G ) : x 1 ) and S / ( I ( G ) , x 1 ) are Cohen-Macaulay of dimension n, we have S / I ( G ) is also Cohen-Macaulay of dimension n ((Proposition 1.2.9, [7]) and the fact that dimension of S / I ( G ) is the maximum of the dimensions of S / ( I ( G ) : x 1 ) and S / ( I ( G ) , x 1 ) ).
The implication ( 1 ) ( 3 ) is a consequence of (Corollary 2.4, [6]).
We finally show ( 3 ) ( 2 ) . We first observe that n = n as I ( G ) is unmixed and both ( x 1 , . . . , x n ) and ( y 1 , . . . , y n ) are minimal primes. Next, we prove that the existence of conditions ( a ) and ( b ) by induction. Let L { 1 , . . . . , n } and define
y L = i L y i x L = i L x i T L = { j | x j y i I ( G ) for any i L } u L = y L x T L .
Note that u L I ( G ) for any subset S { 1 , , n } . We now consider the ideals ( I ( G ) : u L ) . If L = { 1 , . . . . , n } then ( I ( G ) : u L ) = ( x 1 , . . . . , x n ) which shows that ( x 1 , , x n ) Ass ( I ( G ) ) . Since I ( G ) is unmixed, we have ht I ( G ) = n . Clearly for any L { 1 , . . . . , n } , ( I ( G ) : u L ) = ( x j 1 , . . . , x j t , y l 1 , . . . . , y l t ) where for each 1 i t , x j i y r i I ( G ) for some r i S and for each 1 k t , x w k y l k I ( G ) for some w k T s . Since I ( G ) is unmixed of height n and ( x j 1 , . . . , x j t , y l 1 , . . . . , y l t ) Ass ( I ( G ) ) , we have t + t = n .
Now choose y i with minimum vertex degree. Without loss of generality we may assume i = 1 . Let x 1 , . . . , x t be neighbors of y 1 and L = { 1 } . Then as in the previous paragraph, consider ( I ( G ) : u L ) = ( x 1 , , x t , y l 1 , , y l n t ) . After relabeling, we may assume y 1 , , y t are only connected to x 1 , , x t . Let G be the induced subgraph on x 1 , . . . , x t , y 1 , . . . . , y t . By our choice of y 1 , of minimal vertex degree t, notice that every other vertex y j has to have vertex degree at least t. In other words, since t is minimal, each vertex y i , 1 i t in G has at least t neighbors and hence G is a complete bipartite graph.
Since S / I ( G ) is connected in codimension one and ( x 1 , , x n ) , ( y 1 , , y n ) Ass ( I ( G ) ) , there exists a sequence of minimal primes ( x 1 , , x n ) = P 1 , , P r = ( y 1 , , y n ) such that ht ( P i + P i + 1 ) = 1 in S / I ( G ) . If any minimal prime P l of I ( G ) does not contain some x i , 1 i t then it has to contain every y j , 1 j t (as G ,as defined in the previous paragraph, is a complete bipartite graph). Let 1 l r such that for all 1 i l , P i contains all of x 1 , , x t (alternatively, P i ’s do not contain any of y 1 , , y t ). Now P l + 1 does not contain at least one of x 1 , , x t , hence it has to contain all y 1 , , y t . So ht ( P l + P l + 1 ) t in S / I ( G ) . Thus t = 1 and hence y 1 is only connected to x 1 .
Now consider ( I ( G ) , x 1 ) . Since I ( G ) is an intersection of minimal primes, ( I ( G ) , x 1 ) is an intersection of minimal primes of I ( G ) containing x 1 . Thus any minimal prime of ( I ( G ) , x 1 ) is a minimal prime of I ( G ) , and so ( I ( G ) , x 1 ) is unmixed. We now show that ( I ( G ) , x 1 ) is connected at codimension one. Any minimal prime of I ( G ) has to contain either x 1 or y 1 (as it is minimal it cannot contain both as y 1 is only connected to x 1 ). Let P , P Min ( I ( G ) , x 1 ) . As P , P Min I ( G ) , there exists a sequence of minimal primes P = P 1 , , P r = P such that ht P i + P i + 1 = 1 . For any 1 i r ,
P i = P i if x 1 P i ( P i \ { y 1 } ) { x 1 } if x 1 P i .
The sequence P = P 1 , , P r = P defined as before has the property that ht P i + P i + 1 = 1 and hence ( I ( G ) , x 1 ) is connected in codimension one. Now notice that ( I ( G ) , x 1 ) = ( I ( G ) , x 1 ) where G is the graph obtained from G by deleting x 1 . By induction hypothesis, S / I ( G ) is Cohen-Macaulay. So there exists an ordering { x 2 , . . . . , x n } and { y 2 , . . . . , y n } satisfying ( a ) ( b ) of ( 2 ) . As y 1 is only connected to x 1 , G also satisfies ( a ) ( b ) of ( 2 ) .
To prove that condition ( c ) holds, take x i y j and x j y k in E ( G ) such that i , j , k are distinct. Assume that x i y k is not an edge. Then there is a minimal prime P that does not contain either x i or y k as the ideal generated by all x-variables except x i and all y-variables except y k is a prime ideal that contains I ( G ) and does not contain x i or y k . Now because I ( G ) is unmixed, height of this prime has to be n. Since x i and y k are not in P, we get that y j and x j are both in P. As P contains at least one of x m or y m for all m, one observes that height of P is strictly bigger than n, which is a contradiction. □
The following remark is extremely crucial for our work.
Remark 1.
If G is a bipartite graph and a b is an edge then from (Theorem 6.7, [4]) we get ( I ( G ) 2 : a b ) ) = I ( G ) + ( u v | u N ( a ) , v N ( b ) ) .
Theorem 5.
Let G be a bipartite graph with partition V 1 = { x 1 , , x n } and V 2 = { y 1 , , y n } . Then the following are equivalent
1.
S / I ( G ) is Cohen-Macaulay
2.
n = n and there exists exactly n edges e 1 , , e n such that ( I ( G ) 2 : e i ) = I ( G ) and for i j , e i and e j are disjoint.
3.
n = n and there exists exactly n edges e 1 , , e n such that S / ( I ( G ) 2 : e i ) is Cohen-Macaulay and for i j , e i and e j are disjoint.
Proof. 
First, we show ( 1 ) ( 2 ) . If S / I ( G ) is Cohen-Macaulay, we have ordering x 1 , . . . , x n and y 1 , . . . . , y n of the vertices of G which satisfies the conditions of Theorem 4. Condition ( c ) implies for all i, I ( G ) 2 : x i y i = I ( G ) and conditions ( a ) and ( b ) implies for i j ( I ( G ) 2 : x i y j ) I ( G ) .
Now suppose there exist, after possible reordering, e 1 = x 1 y 1 , . . . . , e n = x n y n which satisfied the conditions of ( 2 ) . First, we show that if G i is the induced subgraph obtained by deleting x i and y i then the edge ideal J i related to G i satisfies the condition with e 1 , , e i 1 , e i + 1 , , e n . Without loss of generality, we prove this for G 1 . Clearly ( J 1 2 : e i ) = J 1 for 2 i n . Suppose there exists an edge x i y j , i j such that ( J 1 2 : x i y j ) = J 1 . Without loss of generality we may assume i = 2 , j = 3 . As ( I ( G ) 2 : x 2 y 3 ) I ( G ) and x 1 y 1 is an edge we can conclude that there exists a minimal generator of ( I ( G ) 2 : x 2 y 3 ) which is an edge that is either of the form x 1 y l or x m y 1 (Theorem 6.7, [4]). Again without loss of generality we may assume it is of the form x 1 y l as the proof for the other follows simply by interchanging roles of x and y. So x 1 y 3 and x 2 y l are edges in G (Theorem 6.7, [4]). As ( J 1 2 : x 2 y 3 ) = J 1 we conclude x 3 y 2 is an edge in G. As ( I ( G ) 2 : x 3 y 3 ) = I ( G ) we observe that x 1 y 2 has to be an edge in G. So l 2 , 3 . Without loss of generality we may assume l = 4 . Now ( I ( G ) 2 : x 2 y 2 ) = I ( G ) so x 3 y 4 has to be an edge in G. Again ( I ( G ) 2 : x 3 y 3 ) = I ( G ) hence x 1 y 4 is an edge in G contradicting the assumption. So we may assume for all i the edge ideal I ( G i ) of the graph G i obtained by deleting x i and y i satisfies the conditions in ( 2 ) .
Now by induction we may assume the result holds for n 1 . Pick e i = x i y i such that y i has minimum degree. Let G be the induced subgraph on vertices other than x i , y i with edge ideal I ( G ) . As I ( G ) satisfies the condition it is Cohen-Macaulay by induction. Without loss of generality we may assume i = 1 and ordering that gives ordering of previous theorem for I ( G ) is x 2 , . . . , x n , y 2 , . . . , y n . As y 2 has degree one in G it can have at most degree 2 in G. If x 1 y 2 is not an edge, due to minimality degree of y 1 is at most 1. If x 1 y 2 is an edge in G and x i y 1 is an edge in G for i > 2 , as ( I ( G ) 2 : x 1 y 1 ) = I ( G ) , we have x i y 2 is an edge in G and hence in G contradicting the assumption. Now if x 1 y 2 and x 2 y 1 both are edges in G. Notice that x 2 y 1 also satisfies the hypothesis ( I ( G ) 2 : x 2 y 1 = I ( G ) ). For, x 1 has to be connected to any neighbor of x 2 as x 1 y 2 is an edge and x 2 y 2 satisfies the hypothesis ( I ( G ) 2 : x 2 y 2 = I ( G ) ). This leads to a contradiction and hence no x i for i > 1 is connected to y 1 . This guarantees that conditions ( a ) and ( b ) of Theorem 4 ( 2 ) is satisfied. The condition ( c ) is satisfied as for all i, ( I ( G ) 2 : x i y i ) = I ( G ) .
Next we show ( 1 ) ( 3 ) . To prove the if part, we pick, without loss of generality, y 1 with minimum degree and the corresponding edge e 1 = x 1 y 1 . If degree of y 1 more than one then degree of any other vertex is more than one; as ( I ( G ) 2 : e 1 ) is Cohen-Macaulay this will be a contradiction to the fact that any Cohen-Macaulay bipartite graph should have a y-vertex of degree 1 (Theorem 3). So y 1 has degree one. Hence ( I ( G ) 2 : e 1 ) = I ( G ) and I ( G ) is Cohen-Macaulay.
For the only if part let e 1 = ( x 1 y 1 ) , . . . , e n = ( x n y n ) be as the ordering prescribed by the Herzog-Hibi (Theorem 3) characterization. All we need to show is that J = ( I ( G ) 2 : x i y j ) is not Cohen-Macaulay for i > j . This follows as ( J 2 : e ) = J for e = x j y i (which is a minimal monomial generator of J) as well as for e 1 , . . . , e n . To see this first we show that ( J 2 : e k ) = J for all k. Here at every step we use the description of colon ideal provided by (Theorem 6.7, [4]). If x l y m is a minimal monomial generator of ( J 2 : e k ) which is not in J then x l y k and x k y m are in J. Both of them cannot belong to I ( G ) as from ( I ( G ) 2 : e k ) = I ( G ) that will imply x l y m belongs to I ( G ) and as a result will belong to J, contradicting the assumption. Without loss of generality assume x k y m does not belong to I ( G ) . Then x k y j and x i y m is in I ( G ) . If x l y k does not belong to I ( G ) then x l y j and x i y k belong to I ( G ) . If x l y k is in I ( G ) as x k y j is in I ( G ) and ( I ( G ) 2 : e k ) = I ( G ) we have x l y j is in I ( G ) . In either case we have x l y j and x i y m belong to I ( G ) . Hence x l y m belongs to J contradicting our assumption.
Next we show that ( J 2 : x j y i ) = J . If x l y k is a minimal monomial generator of ( J 2 : x j y i ) which is not in J then x j y k and x l y i is in J. As x j y k is in J it is either in I ( G ) or y k is a neighbor of x i in G. If x j y k is in I ( G ) as ( I ( G ) 2 : x j y j ) = I ( G ) we have x i y k is in I ( G ) . By symmetry x l y j is in I ( G ) . Hence x l y k is in J contrary to the assumption. Hence J is not Cohen-Macaulay. □
The next theorem gives insight into the associated graded ring of a Cohen-Macaulay bipartite edge ideal. The proof of this theorem uses the description of the colon of the n th power of an edge ideal with n 1 edges introduced in [4].
Theorem 6.
Let I ( G ) be Cohen-Macaulay bipartite edge ideal with an ordering of vertices satisfying Theorem 3(2) and e i = x i y i for 1 i n . Then for all i and for all k, ( I ( G ) k : e i ) = I ( G ) k 1 . Hence e i s are non zero divisors in the associated graded ring of I ( G ) .
Proof. 
Let f ( I ( G ) k : e i ) ( I ( G ) k 1 : e i ) be a minimal monomial generator of ( I ( G ) k : e i ) . By induction ( I ( G ) k 1 : e i ) = I ( G ) k 2 . So f = g h 1 . . . . h k 2 where h j s are minimal monomial generators of I ( G ) and g any monomial. So e i h 1 . . . . h k 2 g I ( G ) k . As f is a minimal monomial generator, without loss of generality we may assume g is of degree 2 and e i h 1 . . h k 2 g is a minimal monomial generator of I ( G ) k . Let g = x k y l , k l . If g is an edge we are done. Otherwise by ([4], Theorem 6.7), x k and y l are even connected with respect to e i h 1 . . . h k 2 . If x i y l is an edge and for some j , m , p , h j = x m y p and x m y i is an edge. Then by Theorem 4(2(c)) x m y l is an edge and hence proceeding inductively we show g is an edge and the result follows. □
We illustrate this theorem for k = 3 , 4 .
Example 1.
Let S = k [ x 1 , x 2 , x 3 , y 1 , y 2 , y 3 ] and I = ( x 1 y 1 , x 2 y 2 , x 3 y 3 , x 1 y 2 , x 1 y 3 , x 2 y 3 ) . One can check using Macaulay 2, that ( I 3 : x 1 y 1 ) = ( I 3 : x 2 y 2 ) = ( I 3 : x 3 y 3 ) = I 2 and ( I 4 : x 1 y 1 ) = ( I 4 : x 2 y 2 ) = ( I 4 : x 3 y 3 ) = I 3 .
In a private communication, Prof. Villarreal mentioned that results similar to Theorems 5 and 6 can be found in [8,9].

3. Regular Elements in Powers of Bipartite Edge Ideals

This section presents methods to recognize regular elements on the power of bipartite edge ideals based on the combinatorics of the graph. We first present some examples to motivate the definition and the results.
Example 2.
Consider the ring S = k [ x 1 , x 2 , x 3 , y 1 , y 2 , y 3 ] and the bipartite edge ideal I ( G ) = ( x 1 y 1 , x 2 y 2 , x 3 y 3 , x 1 y 2 , x 1 y 3 , x 2 y 3 ) corresponding to
Mathematics 07 00762 i001
Macaulay2 computations show that x 3 y 1 is a regular element on I ( G ) s for 1 s 10 . Notice that I ( G ) is Cohen-Macaulay. This can also be recovered from (Theorem 3.8, [3]).
One would be tempted to generalize that x n y 1 is always a regular element for bipartite graphs. But it is not always the case as it is shown in this example.
Example 3.
Consider the ring S = k [ x 1 , x 2 , x 3 , y 1 , y 2 , y 3 ] and the bipartite edge ideal I ( G ) = ( x 1 y 1 , x 2 y 2 , x 3 y 3 , x 1 y 2 , x 2 y 1 , x 2 y 3 ) corresponding to
Mathematics 07 00762 i002
Macaulay2 computations show that x 3 y 1 is not a regular element S / I ( G ) or S / I ( G ) 2 .
Studying more such examples, we came up with the following definition involving the combinatorial nature of the graphs.
Definition 2.
Let G be a bipartite graph. Then x μ V 1 , y ν V 2 satisfies the neighborhood condition if
N ( x μ ) N ( x a i ) for all i , 1 i p where N ( y ν ) = { x a 1 , , x a p } .
Remark 2.
Condition (2) of Definition 2 is equivalent to the following condition
N ( y ν ) N ( y b j )   for   all   i , 1 j q   where   N ( x μ ) = { y b 1 , , y b q } .
Suppose (2) of Definition 2 is true. Then { y b 1 , , y b q } = N ( x μ ) N ( x a i ) , where N ( y ν ) = { x a 1 , , x a p } . This means x a i N ( y b j ) where 1 i p , 1 j q . In other words, N ( y ν ) N ( y b j ) , where 1 j q . The other direction is analogous.
We show in [3] that x n y 1 is a regular element on S / I ( G ) s for all s 1 when G is an unmixed bipartite graph. Of course, when G is unmixed bipartite, x n and y 1 satisfies the neighborhood conditions. In this section, we show that the difference of vertices which satisfies the neighborhood condition are the right candidates for being a regular element on S / I ( G ) s for any bipartite graph G.
Theorem 8 is the main theorem we study in this section. We break up the proof of this theorem into three main parts, where Theorem 7, Lemma 1 provide all the tools required to prove Theorem 8.
Theorem 7.
Let G be a bipartite graph and suppose that x μ V 1 and y ν V 2 satisfies the neighborhood properties. If m is a monomial such that m x μ k , m y ν k I ( G ) s , then m I ( G ) s for s , k 1 .
Proof. 
We prove by induction on k. Suppose k = 1 . Then m x μ , m y ν I ( G ) s . As m x μ I ( G ) s , then either m I ( G ) s or m = m y t for some y t N G ( x μ ) and m I ( G ) s 1 . If m I ( G ) s , then the claim is obviously true.
Suppose m = m y t with m I ( G ) s 1 \ I ( G ) s . Let m = a e 1 e s 1 for e 1 , , e k I ( G ) , a S . We assume e i = ( x u i y v i ) , 1 i s 1 . Since m y ν I ( G ) s , we have m y t y ν I ( G ) s . Thus
m y t y ν = a e 1 e s 1 y t y ν I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
Suppose a neighbor of y t divides a, then clearly m = m y t I ( G ) s . Now suppose a neighbor of y ν divides a. Since x μ and y ν satisfies the neighborhood properties, any neighbor of y ν is also a neighbor of y t and hence m = m y t I ( G ) s .
Suppose that no neighbor of y t or y ν divide a. Now in the decomposition in (3), if y ν does not divide f 1 f s , then f 1 f s divides m y t = m and hence m I ( G ) s . Now if y t does not divide f 1 f s , then f 1 f s divides m y ν . Thus m y ν = b 1 f 1 f s . If y ν divides b 1 , then f 1 f s divides m and hence m I ( G ) s . Now suppose y ν divides, say f 1 = ( x δ y ν ) . Again, since x μ and y ν satisfy the neighborhood properties, any neighbor of y ν is a neighbor of y t and hence m = m y t = b 1 ( x δ y t ) f 2 f s I ( G ) s .
Now suppose that y t y ν divides f 1 f s . Since y t y ν divides f 1 f s , we assume, without loss of generality, f 1 = x u 1 y t and f 2 = ( x u 2 y ν ) . Thus we have
m y t y ν = a f 1 f 2 e 3 e s 1 y v 1 y v 2 I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
Now a neighbor of y v 1 , say x 0 , divides a, then
m = m y t = a ( x 0 y v 1 ) ( x u 1 y t ) e 2 e s 1 I ( G ) s where a = a x 0
Similarly if a neighbor of y v 2 , say x 0 , divides a, then
m = m y t = a ( x 0 y v 2 ) e 1 ( x u 2 y t ) e 2 e s 1 I ( G ) s where a = a x 0
Now suppose no neighbor of y v 1 or y v 2 divides a. Consider (4). If y v 1 does not divide f 3 f s , then
m y t y ν = a f 1 f 2 e 3 e s 1 y v 1 y v 2 = b y v 1 f 1 f s = b y v 1 ( x u 1 y t ) ( x u 2 y ν ) f 3 f s = b ( x u 1 y v 1 ) ( x u 2 y t ) f 3 f s y ν = b e 1 ( x u 2 y t ) f 3 f s y ν
Deleting y ν on both sides, we get m = m y t = b ( x u 1 y v 1 ) ( x u 2 y t ) f 3 f s I ( G ) s . Thus we assume y v 1 divides f 3 f s and hence assume, without loss of generality f 3 = ( x u 3 y v 1 ) . Now we have
m y t y ν = a f 1 f 2 f 3 e 4 e s 1 y v 2 y v 3 I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
Now if y v 2 does not divides f 4 f s , then
m y t y ν = a f 1 f 2 f 3 e 4 e s 1 y v 2 y v 3 = b y v 2 f 1 f s = b y v 2 ( x u 1 y t ) ( x u 2 y ν ) f 3 f s = b ( x u 1 y t ) ( x u 2 y v 2 ) f 3 f s y ν = b f 1 e 2 f 3 f s y ν
Deleting y ν on both sides we get m = m y t = b f 1 e 2 f 3 f s I ( G ) s .
Thus we assume y v 2 divide f 4 f s and hence assume, without loss of generality, f 4 = ( x u 4 y v 2 ) . We now have
m y t y ν = a f 1 f 2 f 3 f 4 e 5 e s 1 y v 3 y v 4 I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
We continue in the same fashion and arrive at the j-th decomposition
m y t y ν = a f 1 f 2 j 1 f 2 j e 2 j + 1 e s 1 y v 2 j 1 y v 2 j I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
Also f 2 r 1 = ( x u 2 r 1 y v 2 r 3 ) and f 2 r = ( x u 2 r y v 2 r 2 ) for 2 r j . Now if a neighbor of y v 2 j 1 , say x 0 , divides a, then
m = m y t = a ( x 0 y v 2 j 1 ) f 1 e 2 f 3 e 4 e 2 j 2 f 2 j 1 e 2 j e 2 j + 1 e s 1 I ( G ) s where a = a x 0
If a neighbor of y v 2 j , say ( x 0 ) , divides a, then
m = m y t = a ( x 0 y v 2 j ) e 1 ( x u 2 y t ) e 3 f 4 e 5 f 6 e 2 j 1 f 2 j e 2 j + 1 e 2 j + 2 e s 1 I ( G ) s where a = a x 0
Now suppose no neighbor of y v 2 j 1 or y v 2 j divides a. Now consider (7). If y v 2 j 1 does not divide f 2 j + 1 f s , then
m y t y ν = a f 1 f 2 j 1 f 2 j e 2 j + 1 e s 1 y v 2 j 1 y v 2 j = b y v 2 j 1 f 1 f s = b e 1 ( x u 2 y t ) e 3 f 4 e 5 f 6 f 2 j 2 e 2 j 1 f 2 j f 2 j + 2 f s y ν
Deleting y ν on both sides we have m = m y t = b e 1 ( x u 2 y t ) e 3 f 4 e 5 f 6 f 2 j 2 e 2 j 1 f 2 j f 2 j + 2 f s I ( G ) s . Thus we assume y v 2 j 1 divides f 2 j + 1 f s and hence assume, without loss of generality, f 2 j + 1 = ( x u 2 j + 1 y v 2 j 1 ) . We now have
m y t y ν = a f 1 f 2 j 1 f 2 j f 2 j + 1 e 2 j + 2 e s 1 y v 2 j 1 y v 2 j I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
Again, if y v 2 j does not divide f 2 j + 2 f s , then
m y t y ν = a f 1 f 2 j 1 f 2 j f 2 j + 1 e 2 j + 2 e s 1 y v 2 j 1 y v 2 j = b y v 2 j f 1 f s = b f 1 e 2 f 3 e 4 f 5 f 2 j 1 e 2 j f 2 j + 2 f 2 j + 2 f s y ν
Deleting y ν on both sides, we get m = m y t = b f 1 e 2 f 3 e 4 f 5 f 2 j 1 e 2 j f 2 j + 2 f 2 j + 2 f s I ( G ) s . Thus we assume y v 2 j divides f 2 j + 2 f s and hence assume, without loss of generality, f 2 j + 2 = ( x u 2 j + 2 y v 2 j ) .
Continuing in the same fashion we may reach the final decomposition
m y t y ν = a f 1 f s 1 y v s 2 y v s 1 I ( G ) s = b f 1 f s for f 1 , , f s I ( G ) , b S
Recall that every stage we make sure that none of the neighbors of the y’s appearing in f 1 , , f s 1 divide a. Thus a neighbor of y v s 2 or y v s 1 divides a. Now we can use the decomposition in (8) and (9) to show that m I ( G ) s depending on whether s 2 or s 1 is odd or even. This concludes the proof of claim of this theorem in k = 1 case.
Now assume by induction, that if m x μ l , m y ν l I ( G ) s for 1 l k 1 , then m I ( G ) s . Suppose m x μ k , m y ν k I ( G ) s . We also assume that k s . For, if k > s , then m x μ s , m y ν s I ( G ) s and hence by induction hypothesis, we have m I ( G ) s .
We claim that it is enough to show that m x μ k 1 I ( G ) s or m y ν k 1 I ( G ) s . Suppose we show that m x μ k 1 I ( G ) s . We now have m x μ k 1 , m y ν k I ( G ) s . Thus m x μ k 1 y ν , m y ν k 1 y ν I ( G ) s and hence ( m y ν ) x μ k 1 , ( m y ν ) y ν k 1 I ( G ) s . Since m y ν is a monomial, we use induction hypothesis to conclude that m y ν I ( G ) s . Thus we now have m x μ k 1 , m y ν I ( G ) s . As before, we have ( m x μ k 2 ) x μ , ( m x μ k 2 ) y ν I ( G ) s . Again, since m x μ k 2 is a monomial, we use induction hypothesis to conclude that m x μ k 2 I ( G ) s . We now have m x μ k 2 , m y ν I ( G ) s . We continue the process to get m x μ 2 , m y ν I ( G ) s . We still have ( m x μ ) x μ , ( m x μ ) y ν I ( G ) s . Since m x μ is a monomial, by induction hypothesis, we get m x μ I ( G ) s . We now have m x μ , m y ν I ( G ) s . This is the k = 1 case. We now use the induction hypothesis to get m I ( G ) s . On the other hand, if we show that m y ν k 1 I ( G ) s , then we can analogously show that m I ( G ) s .
Now we go to the induction step. We have m x μ k , m y ν k I ( G ) s . Since m x μ k I ( G ) s and m x μ l I ( G ) s for any l < k , we have m = m y t 1 y t k where m I ( G ) s k , y t 1 , , y t k N G ( x μ ) and not all y t 1 , , y t k may be distinct. Suppose a neighbor of y t 1 , , y t k divides a, then m x μ k 1 I ( G ) s .
Now suppose no neighbor of y t 1 , , y t k divide a. Since m y ν k I ( G ) s we have
m y ν k = m y t 1 y t k y ν k I ( G ) s = b f 1 f s where f 1 , , f s I ( G ) , b S
We observe that m may be written divisible by many minimal monomial generators of I ( G ) s k . We can take m = a e 1 . . . e s k such that m e 1 . . . e s k has smallest number of x variables in common with f 1 . . . . f s .
It is clear that y ν k must divide f 1 . . . . f s , otherwise m y ν l I ( G ) s for some l < k and hence m y ν k 1 I ( G ) s . Recall the no neighbor of y t 1 , , y t k divides a. Thus we can assume that no neighbor of y ν , divides a as that will make m x μ k 1 I ( G ) s . So without loss of generality we may assume for 1 i k , f i = x u i y ν where for every j, e j = x u j y v j .
Now we observe that if any neighbor of y v i for 1 i k divide a then, clearly, m x μ k 1 I ( G ) s . For, without loss of generality, say x 0 y v 1 is an edge where x 0 divides a. As x u 1 y ν is an edge, so is x u 1 y t 1 (by neighborhood properties). Thus we have m = ( a x 0 ) ( x 0 y v 1 ) e 2 . . . . e s k ( x u 1 y t 1 ) . . . . y t k I ( G ) s k + 1 . Hence this will force m x μ k 1 I ( G ) s . So we assume no neighbor of y v i for 1 i k divide a.
As there are s many x variables in f 1 f s and k < s , some of the x variables of f 1 f s divides a. We also have that no neighbor of any y t i divides a and y ν k divides f 1 f s . Let f k + 1 = x 0 y v k + 1 where x 0 divides a and e k + 1 = x u k + 1 y v k + 1 . We may write m = a e 1 . . . e k f k + 1 e k + 2 . . . . e s k where a = ( a x 0 x u k + 1 ) . But this is an expression of m with a having less number of x variables in common with f 1 . . . f s than a which is a contradiction. Thus, one of the neighbors of y v i for some 1 i k divides a and hence m I ( G ) s . □
Lemma 1.
Let G be a bipartite graph and suppose that x μ V 1 and y ν V 2 satisfies the neighborhood properties. Now assume m 1 , , m k S are monomials of the same degree such that ( m 1 + + m k ) ( x μ y ν ) I ( G ) s . Further suppose,
m 1 x μ = m 2 y ν
m i x μ = m i + 1 y ν for 2 i k 1
m 1 y ν , m k x μ I ( G ) s
Then m j I ( G ) s for 1 j k .
Proof. 
First, assume that N G ( y ν ) = { x ν 1 , , x ν p } . We prove by induction on k. If k = 1 , then clearly the claim is true by Theorem 7. By induction, assume the claim is true for ( m 1 + + m l ) ( x μ y ν ) I ( G ) s satisfying (15)–(17) and l k 1 . Now suppose we have
( m 1 + + m k ) ( x μ y ν ) I ( G ) s
satisfying (15)–(17). We show that m 1 I ( G ) s . This will show that ( m 2 + + m k ) ( x μ y ν ) I ( G ) s satisfying (15)–(17). Thus by induction hypothesis we have m j I ( G ) s for 2 j k proving the claim.
From (15), we have m 1 = m y ν and m 2 = m x ν where m S , a monomial. From (16), we have m 3 = m 2 x μ y ν = m x μ 2 y ν . Subsequently, we show that
m i = m x μ i 1 y ν i 2 for 2 i k
Since m 1 y ν I ( G ) s , we have m 1 I ( G ) s or m 1 = a e 1 e s 1 x ν t for some t { 1 , , p } where N G ( y ν ) = { x ν 1 , , x ν p } .
Suppose m 1 = a e 1 e s 1 x ν 1 . Since m 1 = m y ν , y ν divides a or one of the e i ’s. If y ν divides a, then m 1 I ( G ) s .
Now suppose y ν divides, say e 1 = x ν b y ν for some b { 1 , , p } . Since m 1 = m y ν , we have m = a e 2 e s 1 x ν 1 x ν b . Using this equality in (18), we have
m k = m x μ k 1 y ν k 2 = a e 2 e s 1 x ν 1 x ν b x μ k 1 y ν k 2
Since y ν does not divide a, then y ν divides some of the e 1 , , e s k and hence we have k 2 s 2 or k s . Without loss of generality, assume y ν divides e 2 , , e k 1 . Thus
m k = a e k e s 1 x ν 1 l 1 x ν p l p x μ k 1 where j = 1 p l j = k
Let u = a e k e s 1 x ν 1 l 1 x ν p l p . Now as m k x μ I ( G ) s , we have u x μ k I ( G ) s . Also, notice that
u y ν k = m k x μ k 1 y ν k = m y ν k 2 y ν k = m y ν 2 = m 1 y ν I ( G ) s
Since u is a monomial, we have u I ( G ) s , by Theorem 7. Now m 1 = u y ν k 1 I ( G ) s and hence we are done. □
We now prove one of the main results of this section. In this theorem, we attempt to rearrange the sum m 1 + + m k into a configuration shown in the previous lemma.
Theorem 8.
Let G be a bipartite graph and suppose that x μ V 1 and y ν V 2 satisfies the neighborhood properties. Then x μ y ν is an regular element on S / I ( G ) s for all s.
Proof. 
Consider ( m 1 + + m k ) ( x μ y ν ) I ( G ) s where m i ’s are monomials of the same degree. We prove m 1 , , m k I ( G ) s by induction on k.
Suppose k = 1 and m 1 ( x μ y ν ) = m 1 x μ m 1 y ν I ( G ) s . Thus m 1 x μ , m 1 y ν I ( G ) s . Now we use Theorem 7, to show that m 1 I ( G ) s proving the base case of induction.
Suppose ( m 1 + + m l ) ( x μ y ν ) I ( G ) s for l k 1 implies m 1 , , m l I ( G ) s . Now consider
( m 1 + + m k ) ( x μ y ν ) = m 1 x μ m 1 y ν + m 2 x μ m 2 y ν + + m k x μ m k y ν I ( G ) s .
where all m i ’s are distinct. We show m i I ( G ) s for 1 i k .
Observe that if m 1 x μ , m 1 y ν I ( G ) s , then we have m 1 ( x μ y ν ) I ( G ) s and ( m 2 + + m k ) ( x μ y ν ) I ( G ) s . Now we use induction hypothesis to show that m i I ( G ) s for 1 i k .
Now we first consider the following configuration, i.e., after possible re-ordering of m i ’s we have
m 1 x μ = m 2 y ν
m i x μ = m i + 1 y ν , for 2 i k 1
m k x μ = m 1 y ν
We refer to this case as the k-cancellation case. Using (20), we get m 1 = m y ν and m 2 = m x μ . Using this and (20), we get
m i = m x μ i 1 y ν i 2 for 3 i k
Thus m k = m x μ k 1 y ν k 2 . Using this description in (22) we get x μ k = y ν k , a contradiction.
Now consider (19). Without loss of generality, after possible reordering, assume that m 1 x μ = m 2 y ν . If m 1 y ν = m 2 x μ , then we get ( m 1 + m 2 ) ( x μ y ν ) I ( G ) s and ( m 3 + + m k ) ( x μ y ν ) I ( G ) s . Now using induction hypothesis, we get m i I ( G ) s .
Suppose, if m 1 y ν = m 3 x μ we introduce the re-ordering
m 1 ( 1 ) = m 3 , m 2 ( 1 ) = m 1 , m 3 ( 1 ) = m 2 m i ( 1 ) = m i for 4 i k 1
Notice that ( m 1 + + m k ) ( x μ y ν ) = m 1 ( 1 ) + + m k ( 1 ) ( x μ y ν ) . Thus it is enough to show that m i ( 1 ) I ( G ) s . Under this re-ordering m 1 ( 1 ) x μ = m 2 ( 1 ) y ν and m 2 ( 1 ) x μ = m 3 ( 1 ) y ν . If m 1 ( 1 ) y ν = m 3 ( 1 ) x μ , then we get ( m 1 ( 1 ) + m 2 ( 1 ) + m 3 ( 1 ) ) ( x μ y ν ) I ( G ) s and ( m 4 ( 1 ) + + m k ( 1 ) ) ( x μ y ν ) I ( G ) s . Now using induction hypothesis, we get m i ( 1 ) I ( G ) s and hence m i I ( G ) s .
Now if m 1 ( 1 ) y ν = m 4 ( 1 ) x μ , we introduce a new ordering
m 1 ( 2 ) = m 4 ( 1 ) m l ( 2 ) = m l 1 ( 1 ) for 2 l 4 m q ( 2 ) = m q ( 1 ) for 5 q k
As before we consider if m 1 ( 2 ) y ν = m 4 ( 2 ) x μ or m 1 ( 2 ) y ν = m 5 ( 2 ) x μ and introduce new ordering, if necessary.
We now continue this process and arrive at the j-th re-ordering defined as follows
m 1 ( j ) = m j + 2 ( j 1 ) m l ( j ) = m l 1 ( j 1 ) for 2 l j + 2 m q ( j ) = m q ( j 1 ) for j + 3 q k
with the following configuration
m i ( j ) x μ = m i + 1 ( j ) y ν for 1 i j + 1
First, suppose j = k 2 . As before, we consider two cases m 1 ( j ) y ν = m j + 2 ( j ) x μ or m 1 ( j ) y ν m j + 2 ( j ) x μ . If m 1 ( j ) y ν = m j + 2 ( j ) x μ , then we arrive at the k-cancellation case discussed above, which leads to a contradiction. So we have m 1 ( j ) y ν m j + 2 ( j ) x μ which is discussed separately in Lemma 1, showing that m i I ( G ) s .
Now we assume j < k 2 and m 1 ( j ) y ν m t ( j ) x μ for 2 t k . If m j + 2 ( j ) x μ m t ( j ) y ν for j + 3 t k , then we have ( m 1 ( j ) + + m j + 2 ( j ) ) ( x μ y ν ) I ( G ) s and ( m j + 2 ( j ) + + m k ( j ) ) ( x μ y ν ) I ( G ) s and we use induction hypothesis to conclude that m i ( j ) I ( G ) s and hence m i I ( G ) s for 1 i k .
Thus assume m j + 2 ( j ) x μ = m t ( j ) y ν for some j + 3 t k . Now we use the ordering
m j + 3 ( j , 1 ) = m t ( j ) , m t ( j , 1 ) = m j + 3 ( j ) m i ( j , 1 ) = m i ( j ) for i j + 3 , t
with the configuration m i ( j , 1 ) x μ = m i + 1 ( j , 1 ) y ν for 1 i j + 2 .
Now if m j + 3 ( j , 1 ) x μ m a ( j , 1 ) y ν for j + 4 a k , then ( m 1 ( j , 1 ) + + m j + 3 ( j , 1 ) ) ( x μ y ν ) I ( G ) s and ( m j + 4 ( j , 1 ) + + m k ( j , 1 ) ) ( x μ y ν ) I ( G ) s and we use induction hypothesis to conclude that m i ( j , 1 ) I ( G ) s and hence m i I ( G ) s for 1 i k .
Now if m j + 3 ( j , 1 ) x μ = m a ( j , 1 ) y ν for some j + 4 a k , then we use the ordering as before
m j + 4 ( j , 2 ) = m a ( j , 1 ) , m a ( j , 2 ) = m j + 4 ( j , 1 ) m i ( j , 2 ) = m i ( j , 1 ) for i j + 4 , a
with the configuration m i ( j , 2 ) x μ = m i + 1 ( j , 2 ) y ν for 1 i j + 3 .
We continue in the same fashion to reach ( j , l ) -th re-ordering to get
( m 1 ( j , l ) + m k ( j , l ) ) ( x μ y ν ) I ( G ) s
with the following configuration
m i ( j , l ) x μ = m i + 1 ( j , l ) y ν for 1 i j + l + 1
Suppose j + l = k 2 , then m 1 ( j , l ) y ν , m k ( j , l ) x μ I ( G ) s . Now using Lemma 1 we have m i ( j , l ) I ( G ) s , 1 i k and hence m i I ( G ) s for 1 i k .
If j + l < k 2 , then there exists a term m b ( j , l ) such that m b ( j , l ) ( x μ y ν ) I ( G ) s and t b m t ( j , l ) ( x μ y ν ) I ( G ) s and hence we are done by induction. □
Corollary 1.
Let G be a bipartite graph. Suppose x μ V 1 , y ν V 2 . Then x μ and y ν satisfies the neighborhood properties, if and only if x μ y ν is regular on S / I ( G ) s for all s.
Proof. 
Suppose x μ and y ν satisfies the neighborhood properties, then x μ y ν is regular on S / I ( G ) s for all s by Theorem 8.
Now if x μ and y ν does not satisfy the neighborhood properties, then there exists y p such that x μ y p E ( G ) and x ν 1 y p E ( G ) where x ν 1 N ( y ν ) . Thus for all s and e = x ν 1 y ν 1 I ( G ) ,
e s 1 ( x ν 1 y p ) ( x μ y ν ) = e s 1 ( ( x ν 1 y p ) x μ ( x ν 1 y p ) y ν ) = e s 1 ( x ν 1 ( y p x μ ) ( x ν 1 y ν ) y p )
Since y p x μ , x ν 1 y ν I ( G ) , we get e s 1 ( x ν 1 y p ) ( x μ y ν ) I ( G ) s . Thus x μ y ν is not a regular element on I ( G ) s . □

Author Contributions

Both the authors contributed equally to the creation of this article.

Funding

The first author was partially supported by DST INSPIRE (India) research grant (DST/INSPIRE/04/2017/000752) and he would like to acknowledge that.

Acknowledgments

The first author is very thankful to Professor C. Huneke for his valuable suggestions especially regarding the proof of Theorem 5.

Conflicts of Interest

The authors declare no conflict of interest.

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