Powers of lexsegment ideals with linear resolution
aa r X i v : . [ m a t h . A C ] N ov POWERS OF LEXSEGMENT IDEALS WITH LINEARRESOLUTION
VIVIANA ENE AND ANDA OLTEANU
Abstract.
All powers of lexsegment ideals with linear resolution (equivalently,with linear quotients) have linear quotients with respect to suitable orders of theminimal monomial generators. For a large subclass of the lexsegment ideals thecorresponding Rees algebra has a quadratic Gr¨obner basis, thus it is Koszul. Wealso find other classes of monomial ideals with linear quotients whose powers havelinear quotients too.Keywords: Lexsegment ideals, linear resolution, linear quotients, Rees ring,Koszul algebra, ideals of fiber type, Gr¨obner bases.MSC: Primary 13D02; Secondary 13C15, 13H10, 13P10.
Introduction
Let S = K [ x , . . . , x n ] be the polynomial ring in n variables over a field K. Foran integer d ≥ , we denote by M d the set of all the monomials of S of degree d. A lexsegment ideal of S is a monomial ideal generated by a lexsegment set , that is aset of the form L ( u, v ) = { w ∈ M d : u ≥ lex w ≥ lex v } where u ≥ lex v are two givenmonomials of M d . Lexsegment ideals were introduced in [8]. Their homological properties and in-variants have been studied in several papers. We refer the reader to [1], [2], [4], [5],[6], [9], [10].In [1], lexsegment ideals with linear resolution are characterized in numericalterms on the ends of the generating lexsegment set. In [6] it is shown that, for alexsegment ideal, having a linear resolution is equivalent to having linear quotientswith respect to a suitable order of the elements in the generating lexsegment set.There are known examples [3] which show that, in general, powers of monomialideals with linear quotients may have no longer linear quotients, or even more, theydo not have a linear resolution.In this paper we show that the lexsegment ideals have a nice behavior with respectto taking powers, namely all powers of a lexsegment ideal with linear quotients(equivalently, with linear resolution) have linear quotients too (Theorem 2.11 and
The second author was supported by the CNCSIS-UEFISCSU project PN II-RU PD23/06.08.2010 and by the strategic grant POSDRU/89/1.5/S/58852, Project “Postdoctoral pro-gram for training scientific researchers” co-financed by the European Social Fund within the Sec-torial Operational Program Human Resources Development 2007 - 2013”. orollary 3.9). Therefore, by collecting all the known results, we may now state thefollowing Theorem 1.
Let u = x a · · · x a n n with a > and v = x b · · · x b n n be monomials ofdegree d with u ≥ lex v and let I = ( L ( u, v )) be a lexsegment ideal. Then the followingstatements are equivalent; (1) I has a linear resolution. (2) I has linear quotients. (3) All the powers of I have linear quotients. (4) All the powers of I have a linear resolution. In order to prove (2) ⇒ (3) in the above theorem, we are going to study in the firstplace (Section 2) the completely lexsegment ideals, that is, those whose generatinglexsegment set has the property that its shadows are again lexsegment sets, and,secondly (Section 3), those which are not completely lexsegment ideals. For the firstclass of ideals we need to use and develop some of the techniques introduced in [4].For the second class, we extend some results of [7].It will turn out that the Rees algebras of the lexsegment ideals which are not com-pletely have quadratic Gr¨obner bases, therefore they are Koszul (Corollary 3.11).For showing this property we need to slightly extend the notion of ℓ -exchange prop-erty which was defined in [7] to the notion of σ -exchange property. By exploiting thisextension, we show in the last section that one may find larger classes of monomialideals for which the Gr¨obner basis of the relation ideal of the Rees algebra R ( I )can be determined (Theorem 3.4). Moreover, any monomial ideal I ⊂ S whoseminimal monomial generating set satisfies a σ -exchange property is of fiber type,that is the relations of its Rees algebra R ( I ) consist of the relations of the sym-metric algebra S ( I ) and of the fiber relations (Corollary 3.5). We also show thatthe equigenerated monomial ideals whose minimal monomial generating set satis-fies a σ -exchange property have the nice property that all their powers have linearquotients (Theorem 3.6). 1. Preliminaries
In this section we recall the basic definitions and known results needed for theother sections.Let K be a field and S = K [ x , . . . , x n ] the polynomial ring in n variables over K. For an integer d ≥ , we denote by M d the set of the monomials of degree d in S ordered lexicographically with x > x > · · · > x n . For two monomials u, v ∈ M d such that u ≥ lex v, we denote by L ( u, v ) the lexsegment set bounded by u and v, that is, L ( u, v ) = { w ∈ M d : u ≥ lex w ≥ lex v } . If u = x d , then L ( u, v ) is denoted L i ( v ) and is called the initial lexsegment de-termined by v. Similarly, if v = x dn , then L ( u, v ) is denoted by L f ( u ) and is calledthe final lexsegment determined by u. An ( initial, final ) lexsegment ideal of S is amonomial ideal generated by an (initial, final) lexsegment set. According to [4], wedenote by L u,v the K -subalgebra of S generated by the monomials of L ( u, v ) . In
4] it is proved that L u,v is a Koszul algebra. More precisely, it is shown that thepresentation ideal of L u,v has a Gr¨obner basis of quadratic binomials. We brieflyrecall the basic tools used in [4] in proving this result, since they will be also usefulin the next section.Let V n,d be the Veronese subring of S, that is, V n,d = K [ M d ] . Let w be a monomialin M d . One can write w = x a = x a · · · x a d , where 1 ≤ a ≤ · · · ≤ a d ≤ n. Considerthe set of variables T = { T a : a = ( a , . . . , a d ) ∈ N d , ≤ a ≤ · · · ≤ a d ≤ n } , and let ϕ : K [ T ] → V n,d be the K -algebra homomorphism defined by ϕ ( T a ) = x a = x a · · · x a d . Then V n,d ∼ = K [ T ] / ker ϕ and P = ker ϕ is called the toric or the presentation ideal of V n,d . If a = ( a , . . . , a d ) and b = ( b , . . . , b d ) are vectors with 1 ≤ a ≤ · · · ≤ a d ≤ n and 1 ≤ b ≤ · · · ≤ b d ≤ n, we say that a > b if x a > lex x b , that is, if thereexists s ≥ a i = b i for i ≤ s − a s < b s . In this way, one getsa total order on the variables of T by setting T a > T b if a > b . Let > lex be thelexicographic order on K [ T ] induced by this order of the variables of T . Namely, wehave T a (1) · · · T a ( N ) > lex T b (1) · · · T b ( N ) if there exists 1 ≤ t ≤ N such that T a ( i ) = T b ( i ) for i ≤ t − T a ( t ) > T b ( t ) . A tableau is an N × d -matrix A = [ a (1) , . . . , a ( N )] with entries in { , . . . , n } , withthe property that in every row a ( i ) = ( a i , . . . , a id ) we have a i ≤ · · · ≤ a id and therow vectors are in decreasing lexicographic order, that is a (1) > a (2) > · · · > a ( N )or, equivalently, T a (1) > T a (2) > · · · > T a ( N ) . The support of A is the collectionsupp( A ) of the integers which appear in the tableau with their occurrences. It isclear that one may associate to each tableau A its corresponding monomial T A := T a (1) · · · T a ( N ) in K [ T ]. A tableau A = [ a (1) , . . . , a ( N )] is standard if, for everytableau B = [ b (1) , . . . , b ( N )] of same support, B = A , one has T A = T a (1) · · · T a ( N ) < lex T b (1) · · · T b ( N ) = T B . As follows from [4, Proposition 2.10], this is equivalent to saying that for any 1 ≤ i < j ≤ N, the quadratic monomial T a ( i ) T a ( j ) is standard. In [4, Lemma 2.9] itis shown that a quadratic monomial T a T b it is standard if and only if a = b orthere exists 1 ≤ i ≤ d such that a = b , . . . , a i − = b i − , a i < b i , and, if i < d, then b i +1 ≤ · · · ≤ b d ≤ a i +1 ≤ · · · ≤ a d . If A is a standard tableau, then themonomial T A = T a (1) · · · T a ( N ) is called standard . Given a set A of N d indices inthe set { , . . . , n } , then there exists a unique standard tableau A of size N × d withsupp( A ) = A .We recall the recursive procedure given in [4] to construct a standard tableau A with a given support A = { b , . . . , b Nd } where 1 ≤ b ≤ · · · ≤ b Nd ≤ n . Namely, if A = [ a (1) , . . . , a ( N )], where a ( i ) = ( a i , . . . , a id ) for 1 ≤ i ≤ N , then we proceed asfollows. We put b , . . . , b N on the first column of A , that is, a = b , a = b , . . . , a N = b N . ow we consider the decomposition of ( b , . . . , b N ) in blocks of equal integers andfill in each sub-tableau determined by each block from the bottom to the top in aninductive way. We illustrate this procedure by a concrete example.Let N = 5, d = 3, n = 8 and A = { , , , , , , , , , , , , , , } . We indicate each step of the standard tableau of support A .11233 −→ −→
112 5 63 4 43 4 5 −→ Proposition 1.1. [4, Proposition 2.11]
The set G = { T q T r − T a T b : T a T b is a standardmonomial and supp[ a , b ] = supp[ q , r ] } is a Gr¨obner basis of the presentation idealof V n,d with respect to < lex . Moreover, in [4, Lemma 2.12], it was proved that if [ a , b ] is a standard tableauand [ q , r ] is a non-standard tableau such that supp[ a , b ] = supp[ q , r ], then q > a ≥ b > r . Consequently, if T a T b is a standard monomial and T q T r is such thatsupp[ a , b ] = supp[ q , r ], then x a , x b ∈ L ( u, v ) if x q and x r belong to L ( u, v ). There-fore, the set G ′ = { T q T r − T a T b : T a T b is a standard monomial, supp[ a , b ] = supp[ q , r ] , and x q , x r ∈ L ( u, v ) } . is a Gr¨obner basis of the presentation ideal J L u,v of the toric ring L u,v .2. Powers of completely lexsegment ideals with linear resolution
In order to study the powers of the completely lexsegment ideals with linearquotients, we need to prove some preparatory results.
Definition 2.1.
Let w , . . . , w N be monomials in M d , N ≥
2. We call the product w · · · w N standard if T w · · · T w N is a standard monomial, that is, the correspondingtableau is standard. Definition 2.2. If w , . . . , w N are monomials in M d , and w · · · w N = w ′ · · · w ′ N ,where w ′ , . . . , w ′ n ∈ M d and w ′ · · · w ′ N is a standard product, we call w ′ · · · w ′ N the standard representation of w · · · w N . Remark 2.3.
Let u , . . . , u N ∈ L ( u, v ), where L ( u, v ) ⊂ M d is a lexsegment set. If w · · · w N is a standard representation of u · · · u N , then w , . . . , w N ∈ L ( u, v ). In-deed, let us assume that T u · · · T u N is not a standard monomial, that is T u · · · T u N ∈ in < ( P ), where P ⊂ K [ T ] is the presentation ideal of V n,d . Then there exists 1 ≤ i
Let w · · · w N be a standard monomial and let x d w · · · w N = w ′ · · · w ′ N +1 be the standard representation of x d w · · · w N . Then w ′ ≥ lex w .Proof. We make induction on the number of variables. The case n = 2 is straightfor-ward. Let n >
2. One may assume, by induction on the degree d of the monomials,that ν ( w N ) = 0. If ν ( w ) = 0, then x | w ′ , hence w ′ > lex w . Let ν ( w ) = 1.Therefore, there exists 0 < s < N such that x | w s and x ∤ w s +1 . If s + d ≥ N + 2,then we finished, since x | w ′ by the construction of the standard monomials, and ν ( w ) = 1. Now, let us consider s + d ≤ N + 1. Let q = min (cid:18) w x · · · w s x (cid:19) and q ′ = min (cid:18) w ′ x · · · w ′ s + d x (cid:19) . Then, since w · · · w N and w ′ · · · w ′ N +1 are standard products, we have q = min( w /x ) ,q ′ = min( w ′ /x ), max( w j ) ≤ q ≤ min( w i /x ) for all 1 ≤ i ≤ s < j ≤ N , andmax( w ′ j ) ≤ q ′ ≤ min( w ′ i /x ) for all 1 ≤ i ≤ s + d < j ≤ N + 1 . If q < q ′ , then we get ν
1. Therefore,we must have q ≥ q ′ . If q > q ′ , then we finished since w ′ /x > lex w /x , whence w ′ > lex w . What is left to consider is the case q = q ′ . In this case we have ν 1) + ν q ( w ) + · · · + ν q ( w s ) . This implies that w ′ x · · · w ′ s + d x = x d ( d − q (cid:18) w x · · · w s x (cid:19) . Note that w x · · · w s x is a standard product in the variables x q , . . . , x n . Applyinginduction on the number n of variables, we have, after d steps, that x d ( d − q (cid:18) w x · · · w s x (cid:19) = ¯ w · · · ¯ w s + d , here ¯ w · · · ¯ w s + d , is a standard product and ¯ w ≥ lex w /x . But w ′ x · · · w ′ s + d x is astandard product as well, hence we have ¯ w = w ′ /x ≥ lex w /x , whence w ′ ≥ lex w . (cid:3) Lemma 2.5. Let u · · · u N and w · · · w N be standard products and u · · · u N x n = x w · · · w N . Then we have u ≥ lex w . Proof. We use induction on N . If N = 1, the inequality u ≥ lex w is obvious. Nowwe assume N > u · · · u N = x b · · · x b Nd , where 1 = b ≤ · · · ≤ b Nd ≤ n and min( u j ) = b j for all 1 ≤ j ≤ N . We first notice that we may assume withoutloss of generality that ν ( u i ) ≤ ≤ i ≤ N. If b > b , we obviously have w ≤ lex u since min( w ) = b . Therefore, we may assume b = b = 1. If b < b N ,let k ≤ N be the largest integer such that b k − < b k = · · · = b N . We have k ≥ u · · · u N is a standard product, we get u · · · u k − = x b · · · x b k − x b N +( d − N − k +1)+1 · · · x Nd . Similarly, since w · · · w N is a standard product, we get w · · · w k − = x b · · · x b k − x b N +( d − N − k +2)+2 · · · x Nd x n . Therefore, there exists a monomial w ∈ M d , namely w = x b N +( d − N − k +1)+1 · · · x b N +( d − N − k +2)+1 such that x w · · · w k − w = u · · · u k − x n . One observes that w · · · w k − w and u · · · u k − are standard products. Then, byinduction on N , it follows that w ≤ lex u .It remains to consider b = · · · = b N = 1 < b N +1 ≤ · · · ≤ b Nd , since, by ourassumption on u , . . . , u N , we cannot have b N +1 = 1. If b N +1 < b N + d +1 , then, bythe construction of standard products, we get w < lex u . Let b N +1 = b N +2 = · · · = b N + d +1 . Then we obtain x n · u x · · · u N x = w x · · · w N − x (cid:0) x b N +1 · · · x b N + d (cid:1) , whence x n (cid:18) u x · · · u N x (cid:19) = x b N +1 (cid:18) x d − b N +1 w x · · · w N − x (cid:19) . Let w ′ · · · w ′ N be the standard representation of x d − b N +1 w x · · · w N − x . By Lemma 2.4,we have w ′ ≥ lex w /x . On the other hand, we have x n (cid:18) u x · · · u N x (cid:19) = x b N +1 ( w ′ · · · w ′ N ) , with u x · · · u N x and w ′ · · · w ′ N standard monomials in a number of variables smallerthan n . By induction on n we get u /x ≥ lex w ′ whence u /x ≥ lex w /x , whichyields u ≥ lex w . (cid:3) emma 2.6. Let u ≥ lex · · · ≥ lex u N ≥ lex u N +1 be monomials of degree d with ν ( u i ) ≤ for all ≤ i ≤ N, such that u · · · u N is a standard product and max(supp( u · · · u N )) ≤ min(supp( u N +1 )) . Let v · · · v N +1 be the standard repre-sentation of u · · · u N u N +1 . Then v N +1 ≤ lex u N .Proof. We use induction on N . For N = 1, since v v = u u and v v is a standardproduct, then we have u > lex v ≥ lex v > lex u .Let N > u · · · u N = x b · · · x b Nd and u N +1 = x b Nd +1 · · · x b ( N +1) d with b ≤ · · · ≤ b Nd ≤ b Nd +1 ≤ · · · ≤ b ( N +1) d . Since u · · · u N is a standard product, we have min( u j ) = b j for all 1 ≤ j ≤ N . Since v · · · v N v N +1 is standard, we have min( v j ) = b j for all 1 ≤ j ≤ N + 1. If b N +1 > b N ,we obviously have v N +1 ≤ lex u N . Therefore, it remains to consider that b N = b N +1 .Let 1 ≤ k ≤ N be the largest integer such that b k − < b k = · · · = b N . We have k > ν ( u ) ≥ 2. Since u · · · u N is standard, we get that u k · · · u N = x b k · · · x b N x b N +1 · · · x b N +( d − N − k +1) . Similarly, since v · · · v N +1 is standard, we get v k · · · v N +1 = x b k · · · x b N x b N +1 · · · x b N +( d − N − k +2)+1 . Therefore, there exists a monomial w ∈ M d , namely w = x b N +( d − N − k +1)+1 · · · x b N +( d − N − k +2)+1 , such that v k · · · v N +1 = u k · · · u N w and max(supp( u k · · · u N )) ≤ min(supp( w )). Onemay note that u k · · · u N and v k · · · v N +1 are standard products as well. By theinduction hypothesis, we get v N +1 ≤ lex u N . (cid:3) Lemma 2.7. Let u , . . . , u N , w , . . . , w N be monomials of degree d in S such that x n u · · · u N = x w · · · w N , where u · · · u N , w · · · w N are standard products. Then u N ≥ lex w N .Proof. We may assume that ν ( w N ) = 0 which implies that ν ( u i ) ≤ ≤ i ≤ N. Let u · · · u N = x b · · · x b Nd with 1 = b ≤ · · · ≤ b Nd and min( u j ) = b j for all 1 ≤ j ≤ N . We have min( w j ) = b j +1 for all 1 ≤ j ≤ N . If b N +1 > b N ,then w N ≤ lex u N . Let b N +1 = b N and 1 ≤ k ≤ N be the largest integer such that b k − < b k = · · · = b N . If k = 1 , then b = · · · = b N = b N +1 . Since w · · · w N isa standard product, we get ν ( w N ) > , which is impossible by our assumption.Therefore, it follows that k > . Since u · · · u N is a standard product, we have u k · · · u N = x b k · · · x b N x b N +1 · · · x b N +( d − N − k +1) . Similarly, since w · · · w N is a standard product, we get w k − · · · w N = x b k · · · x b N x b N +1 · · · x b N +( d − N − k +2)+1 . Therefore, if w = x b N +( d − N − k +1)+1 · · · x b N +( d − N − k +2)+1 , we have w k − · · · w N = u k · · · u N w. nd max(supp( u k · · · u N )) ≤ min( w ). Since u k · · · u N and w k − · · · w N are also stan-dard products, by using the previous lemma, we get w N ≤ lex u N . (cid:3) In order to state the main theorem of this section we need to recall the following Theorem 2.8 ([6],[2]) . Let u = x a · · · x a n n , with a > , and v = x b · · · x b n n bemonomials of degree d with u ≥ lex v and let I = ( L ( u, v )) be a completely lexsegmentideal. The following statements are equivalent: (1) u and v satisfy one of the following conditions: (i) u = x a x d − a , v = x a x d − an for some a with < a ≤ d ; (ii) b < a − ; (iii) b = a − and, for the largest monomial w of degree d with w < lex v ,one has x w/x max( w ) ≤ lex u . (2) I has linear quotients. (3) I has a linear resolution. Remark 2.9. It is obviously that, if a completely lexsegment ideal is determined by u and v satisfying condition (i) in the above theorem, then all its powers have linearquotients. Therefore, we only need to study the powers of completely lexsegmentideals which are determined by monomials u and v satisfying condition (ii) or (iii)in Theorem 2.8. Theorem 2.10. Let u = x a · · · x a n n with a > and v = x b · · · x b n n be monomialsof degree d with u ≥ lex v and let I = ( L ( u, v )) be a completely lexsegment ideal withlinear quotients. Then all the powers of I have linear quotients.Proof. By using Remark 2.9, we have to consider only the cases when u and v satisfyone of the following conditions:(a) b < a − b = a − w of degree d with w < lex v , onehas x w/x max( w ) ≤ lex u .We recall (see [6, Theorem 1.2]) that in these cases, I has linear quotients withrespect to the following order on M d . For w, w ′ ∈ M d we set w ≻ w ′ if ν ( w ) <ν ( w ′ ) or ν ( w ) = ν ( w ′ ) and w > lex w ′ .Let N > 1. We show that I N has linear quotients with respect to the order ≻ on the set M Nd . Let u · · · u N , v · · · v N ∈ I N be two standard products suchthat v · · · v N ≻ u · · · u N . We have to show that there exists a monomial w ∈ I N such that w ≻ u · · · u N , w/ gcd( w, u · · · u N ) = x i and x i divides the monomial v · · · v N / gcd( v · · · v N , u · · · u N ). We have to analyze two cases.Case I: ν ( v · · · v N ) = ν ( u · · · u N ). By the definition of the order ≻ , we musthave v · · · v N > lex u · · · u N . Let i ≥ ν i ( v · · · v N ) >ν i ( u · · · u N ). We claim that there exists 1 ≤ q ≤ N such that i < max( u q ). Indeed,otherwise we have i ≥ max( u · · · u N ) and obtain N d = deg( u · · · u N ) = i X k =1 ν k ( u · · · u N ) < i X k =1 ν k ( v · · · v N ) ≤ N d, a contradiction. et, therefore, 1 ≤ q ≤ N be such that i < max( u q ). Then we get x i u q x ∈ L ( u, v ) or x i u q x max( u q ) ∈ L ( u, v )(see also the proof of [6, Theorem 1.2]). We recall the argument which was used in [6,Theorem 1.2] and will be also used in this proof several times. We have x i u q /x < lex u q ≤ lex u and x i u q /x max( u q ) > lex u q ≥ lex v . If we assume that x i u q /x < lex v and x u q /x max( u q ) > lex u , we get b = a − x i u q /x ≤ lex w , where w is the largestmonomial of degree d such that w < lex v . We get x i u q x x max( x i u q /x ) ≤ lex wx max( w ) , which, by using condition (b), leads to x i u q x max( u q ) ≤ lex x wx max( w ) ≤ lex u, a contradiction. Therefore, one of the monomials u ′ q = x i u q /x or u ′′ q = x i u q /x max( u q ) belongs to L ( u, v ). Note that u ′ q ≻ u q and u ′′ q ≻ u q . Then we may take w = u · · · u q − u ′ q u q +1 · · · u N or w = u · · · u q − u ′′ q u q +1 · · · u N . In each case it follows that w ≻ u · · · u N , w/ gcd( w, u · · · u N ) = x i and x i | v · · · v N / gcd( v · · · v N , u · · · u N ).Case II: ν ( u · · · u N ) > ν ( v · · · v N ). Then there exist two monomials m, m ′ ∈ S of same degree, let us say p , such that gcd( m, m ′ ) = 1 and mu · · · u N = m ′ v · · · v N . (2.1)Since ν ( u · · · u N ) > ν ( v · · · v N ), we get x | m ′ and x ∤ m. Let i = min(supp( m )) . If there exists 1 ≤ q ≤ N such that i < max( u q ) , then, as in the proof of Case(I), we may take w = u · · · u ′ q · · · u N where u ′ q = x i u q /x or u ′ q = x i u q /x max( u q ) . Then the following conditions hold: w ≻ u · · · u N , w/ gcd( w, u · · · u N ) = x i and x i divides the monomial v · · · v N / (gcd( v · · · v N , u · · · u N )).Now let max( u q ) ≤ i for all 1 ≤ q ≤ N, that is, supp( u · · · u N ) ⊂ { , . . . , i } . Weshow by induction on p = deg( m ) that there exists j > x j | m and x j u · · · u N = x w · · · w N , (2.2)where w , . . . , w N ∈ L ( u, v ) and w · · · w N is a standard product. If p = 1, there isnothing to prove. Let p > < j < i such that x j | m ′ . There exists 1 ≤ q ≤ N such that j < i ≤ max( v q ) since x i | v · · · v N . As j < max( v q ) , it follows that one of the monomials x j v q /x ∈ L ( u, v ) or x j v q /x max( v q ) ∈ L ( u, v ).Let us consider that v ′ q = x j v q /x ∈ L ( u, v ) . By using (2.1), we get the relation mu · · · u N = ( x m ′ /x j )( v · · · v ′ q · · · v N ) . If v ′′ q = x j v q /x max( v q ) ∈ L ( u, v ) , then, by using again (2.1), we get the relation mu · · · u N = ( x max( v q ) m ′ /x j )( v · · · v ′′ q · · · v N ) . These last two relations show that either there exists a relation of the form mu · · · u N = x p w · · · w N where w · · · w N is a standard product of monomials of L ( u, v ), withdeg( m ) = p and x ∤ m, or we may apply induction on p and reach the desired onclusion. In the first case, let m = x i x i · · · x i p , with i = i ≤ i ≤ · · · ≤ i p ≤ n .For j = 1 , p , let w j · · · w jN be the standard product such that x i u · · · u N = x w w · · · w N ,x i w w · · · w N = x w w · · · w N ,x i w w · · · w N = x w w · · · w N , ... x i p w p − , w p − , · · · w p − ,N = x w p w p · · · w pN . Multiplying these equalities, we get mu · · · u N = x p w p w p · · · w pN , hence w pi = v i , for 1 ≤ i ≤ N , since w p w p · · · w pN and v · · · v N are standardproducts.It is easily seen that supp( w j · · · w jN ) ⊂ { , . . . , i j } for all 1 ≤ j ≤ p . Therefore,we may apply Lemma 2.5 and Lemma 2.7 and get u ≥ lex u ≥ lex w ≥ lex w ≥ lex · · · ≥ lex w p = v ≥ lex v and u ≥ lex u N ≥ lex w N ≥ lex w N ≥ lex · · · ≥ lex w pN = v N ≥ lex v. In particular, we have u ≥ lex w ≥ lex · · · ≥ lex w N ≥ lex v, whence x i u · · · u N = x w · · · w N , and w , . . . , w N ∈ L ( u, v ). Therefore, we have an equality of the form x j u · · · u N = x w · · · w N , where w · · · w N ∈ I N is a standard product and j ≥ . Let w =( x j u · · · u N ) /x . Then w ≻ u · · · u N , w/ gcd( w, u · · · u N ) = x j and x j divides themonomial v · · · v N / gcd( v · · · v N , u · · · u N ) , which ends our proof. (cid:3) Combining the above theorem with [1, Theorem 1.3] and [6, Theorem 1.2], we getthe following equivalent statements. Theorem 2.11. Let u = x a · · · x a n n with a > and v = x b · · · x b n n be monomialsof degree d with u ≥ lex v and let I = ( L ( u, v )) be a completely lexsegment ideal withlinear quotients. The the following statements are equivalent; (1) u and v satisfy one of the following conditions: (i) u = x a x d − a , v = x a x d − an for some a with < a ≤ d ; (ii) b < a − ; (iii) b = a − and, for the largest monomial of degree d with w < lex v , onehas x w/x max( w ) ≤ lex u . (2) I has a linear resolution. (3) I has linear quotients. (4) All the powers of I have linear quotients. (5) All the powers of I have a linear resolution. . Exchange properties and applications We first fix some notations. As in the previous section, let S = K [ x , . . . , x n ] bethe ring of polynomials in n variables over a field K and M d the set of all monomialsof degree d in S . If B ⊂ M d is a nonempty set, we denote by K [ B ] the K -subalgebraof S generated by the monomials of B .Let R = K [ { T u } u ∈ B ] be the polynomial ring in a set of variables indexed over B and π : R → K [ B ] the surjective K -algebra homomorphism defined by π ( T u ) = u ,for all u ∈ B . J K [ B ] := ker π is called the toric ideal of K [ B ].Let < be a monomial order on R and in < ( J K [ B ] ) the initial ideal of J K [ B ] withrespect to < . A monomial T u · · · T u N ∈ R is a standard monomial of J K [ B ] withrespect to < if T u · · · T u N / ∈ in < ( J K [ B ] ). We recall the following definition which wasgiven in [7]. Definition 3.1. [7, Definition 4.1] We say that a nonempty set B ⊂ M d satisfiesthe ℓ -exchange property with respect to a monomial order < on R if B posseses thefollowing property: if T u · · · T u N and T v · · · T v N are standard monomials of J K [ B ] with respect to < such that(a) ν i ( u · · · u N ) = ν i ( v · · · v N ) for 1 ≤ i ≤ q − q ≤ n − ν q ( u · · · u N ) < ν q ( v · · · v N ),then there exist 1 ≤ δ ≤ N , and q < j ≤ n with j ∈ supp( u δ ) and x q u δ /x j ∈ B .Inspired by this definition we consider the following slight generalization. Let < σ be a monomial order on S . Definition 3.2. We say that B satisfies the σ -exchange property with respect to < if B has the following property: if T u · · · T u N and T v · · · T v N are standard monomials of J K [ B ] with respect to < such that u · · · u N < σ v · · · v N , then there exist 1 ≤ δ ≤ N , q ∈ supp( v · · · v N ), and j ∈ supp( u δ ) such that(i) ν q ( u · · · u N ) < ν q ( v · · · v N ),(ii) x j < σ x q ,(iii) x q u δ /x j ∈ B .It is straightforward to show that if B satisfies the ℓ -exchange property withrespect to a monomial order < on R , then B satisfies the σ -exchange property withrespect to < for < σ = < lex on S with x > lex · · · > lex x n . Example 3.3. Let < σ be a monomial order on S defined as follows. For m, m ′ monomials in S , we set m < σ m ′ if deg( m ) < deg( m ′ ) or deg( m ) = deg( m ′ ) and m > revlex m ′ , that is, if m = x a · · · x a n n , m ′ = x b · · · x b n n , then there exists some1 ≤ s ≤ n such that a n = b n , a n − = b n − , . . . , a s +1 = b s +1 , and a s < b s . Inparticular, we have x n > σ x n − > σ · · · > σ x . We call this monomial order thedecreasing revlexicographical order on S. Any final lexsegment set L f ( v ), v ∈ M d , satisfies the σ -exchange property for < σ as above, with respect to any monomial order < on R = K [ { T w : w ∈ L f ( v ) } ]. Inorder to prove this claim, let T u · · · T u N and T v · · · T v N be two standard monomialsof J K [ B ] with respect to < such that u · · · u N < σ v · · · v N , that is u · · · u N > revlex v · · · v N . hen there exists 1 ≤ q ≤ n such that ν i ( u · · · u N ) = ν i ( v · · · v N ) for all i ≥ q + 1and ν q ( u · · · u N ) < ν q ( v · · · v N ). Since deg( u · · · u N ) = deg( v · · · v N ), we musthave at least an index j < q such that ν j ( u · · · u N ) > ν j ( v · · · v N ). Let 1 ≤ δ ≤ N be such that j ∈ supp( u δ ). Then the following conditions hold: x j > revlex x q , thatis x j < σ x q and x q u δ /x j < lex u δ , whence x q u δ /x j ∈ L f ( v ).We also notice that, if we choose < on R to be the monomial order given in the pre-vious section, that is the lexicographical order on the monomials { T w : w ∈ L f ( v ) } induced by T w > T w if w > lex w , then L f ( v ) does not satisfy the ℓ -exchangeproperty with respect to < . For example, let v = x x x ∈ K [ x , x , x , x ]. Let u = x and v = x x x , u , v ∈ L f ( v ) . Then ( T u ) and ( T v ) are standardmonomials with respect to < on R = K [ { T w : w ∈ L f ( v ) } ] and u < lex v . In the ℓ -exchange property, we have to take q = 1. Since supp( u ) = { } , we should have x u /x = x x ∈ L f ( v ), which is not possible.Following closely the ideas from the last section in [7], we may prove a slightgeneralization of [7, Theorem 5.1].Let I ⊂ S be a monomial ideal generated in degree d and let B = G ( I ) its minimalmonomial generating set. Let T = S [ { T u } u ∈ B ] = K [ x , . . . , x n , T u : u ∈ B ] be thepolynomial ring over K . T is bigraded by deg( x i ) = (1 , 0) for all 1 ≤ i ≤ n anddeg( T u ) = (0 , 1) for all u ∈ B .Let R ( I ) = L j ≥ I j t j = S [ { ut } u ∈ B ] ⊂ S [ t ] be the Rees ring of I . R ( I ) is alsonaturally bigraded by deg( x i ) = (1 , 0) for 1 ≤ i ≤ n and deg( ut ) = (0 , 1) forall u ∈ B . There exists a canonical bigraded surjective K -algebra homomorphism ϕ : T → R ( I ) defined b ϕ ( x i ) = x i for 1 ≤ i ≤ n and ϕ ( T u ) = ut for all u ∈ B .Let P R ( I ) := ker ϕ be the toric ideal of R ( I ). P R ( I ) is bihomogeneous and generatedby irreducible bihomogeneous binomials of T . Let < be an arbitrary monomialorder on R and < σ be an arbitrary monomial order on S . By < σ we will denote theproduct of these two orders which is a monomial order on T . More precisely, for mT u · · · T u N , m ′ T v · · · T v N , monomials in T , with m, m ′ monomials in S , we have mT u · · · T u N < σ m ′ T v · · · T v N if m < σ m ′ or m = m ′ and T u · · · T u N < T v · · · T v N .The following theorem generalizes [7, Theorem 5.1]. Theorem 3.4. Let I ⊂ S be a monomial ideal generated in degree d , B = G ( I ) , < a monomial order on R and < σ a monomial order on S . Let G < ( J K [ B ] ) bethe reduced Gr¨obner basis of the toric ideal J K [ B ] with respect to < . Suppose that B satisfies the σ -exchange property with respect to < . Then the reduced Gr¨obnerbasis of the toric ideal P R ( I ) with respect to < σ consists of all binomials belongingto G < ( J K [ B ] ) together with the binomials of the form x i T u − x j T v ∈ P R ( I ) where x j is the smallest variable with respect to < σ such that x i > σ x j and x i u/x j ∈ B .Proof. We closely follow the ideas from the proof of [7, Theorem 5.1].We first show that the set G = G < ( J K [ B ] ) ∪ { x i T u − x j T v ∈ P R ( I ) : x i > σ x j } s a Gr¨obner basis of P R ( I ) with respect to < σ .Let f ∈ P R ( I ) ⊂ T be an irreducible binomial. If in < σ ( f ) ∈ R , then f ∈ P R ( I ) ∩ R = J K [ B ] , hence there is a binomial belonging to G < ( J K [ B ] ) which divides in < σ ( f ).Let in < σ ( f ) / ∈ R , that is, we may write f = x i · · · x i t T u · · · T u N − x j · · · x j t T v · · · T v N with { i , . . . , i t } ∩ { j , . . . , j t } = ∅ and where we assume that x i ≥ σ · · · ≥ σ x i t and x j ≥ σ . . . ≥ σ x j t . By successively reductions modulo the binomials from G < ( J K [ B ] )we may assume that T u · · · T u N and T v · · · T v N are standard monomials with respectto < . Let in < σ ( f ) = x i · · · x i t T u · · · T u N . Then x i · · · x i t > σ x j · · · x j t . By usingthe equality x i · · · x i t u · · · u N = x j · · · x j t v · · · v N , we obtain u · · · u N < σ v · · · v N , ν i s ( u · · · u N ) < ν i s ( v · · · v N ) for 1 ≤ s ≤ t, and ν k ( u · · · u N ) ≥ ν k ( v · · · v N ) for all k / ∈ { i , . . . , i t } . Since B satisfies the σ -exchangeproperty, we have that there exist 1 ≤ δ ≤ N , j ∈ supp( u δ ) and q ∈ supp( v · · · v N )such that ν q ( u · · · u N ) < ν q ( v · · · v N ), x j < σ x q , and x q u δ /x j ∈ B .The first above condition on q shows that q = i s , for some 1 ≤ s ≤ t . Thereforewe have x i s u δ = x j v for some v ∈ B and the proof of our claim is finished.To end the proof, let us take some binomial x i T u − x j T v , where u, v ∈ B , x i u = x j v and x j < σ x i is the smallest variable with respect to < σ such that x i u/x j ∈ B .Assume that x j T v is not reduced, hence there exists some binomial x j T v − x l T w with x l < σ x j , which belongs to P R ( I ) . Then x i T u − x l T w ∈ P R ( I ) and x l < σ x j < σ x i , acontradiction. (cid:3) Corollary 3.5. Let I ⊂ S be a monomial ideal generated in degree d and B = G ( I ) .Let < be a monomial order on R and < σ a monomial order on S . If B satisfiesthe σ -exchange property with respect to < , then I is of fiber type. We recall (see [7]) that an ideal I ⊂ S is called of fiber type if the fiber relationstogether with the relations of the symmetric algebra of I generate all the relationsof the Rees algebra of I. The above corollary may be used to find equigenerated monomial ideals of fibertype. Let < σ be an arbitrary graded monomial order on S , u ∈ M d and I =( L i< σ ( u )), where L i< σ ( u ) = { w ∈ M d : w > σ u } . Then it is easily seen that ( L i< σ ( u ))satisfies the σ -exchange property for any monomial order on R = K [ { T w : w ∈ L i< σ ( u ) } ], hence I is of fiber type.We prove now a significant property of the monomial ideals whose minimal mono-mial generating system satisfies a σ -exchange property. Theorem 3.6. Let I ⊂ S be a monomial ideal generated in degree d and B = G ( I ) .Let < be a monomial order on R = K [ { T u : u ∈ B } ] and < σ a monomial orderon S . If B satisfies the σ -exchange property with respect to < , then I N has linearquotients with respect to > σ for N ≥ .Proof. Let G (cid:0) I N (cid:1) = { w , · · · , w r } , where w > σ · · · > σ w r and let T w , . . . , T w r bestandard monomials of J K [ B ] with respect to < . Let 1 ≤ j < i ≤ r be two integersand assume that w j = v · · · v N and w i = u · · · u N for u , . . . , u N , v , . . . , v N ∈ G ( I ), ≥ σ · · · ≥ σ u N , v ≥ σ · · · ≥ σ v N . We have to prove that there exist 1 ≤ k < i and1 ≤ q ≤ n such that w k gcd( w k , w i ) = x q and x q | w j gcd( w j , w i ) . Since w j > σ w i , by using the σ -exchange property of B , there exist 1 ≤ δ ≤ N , l ∈ supp( u δ ), and q ∈ supp( v · · · v N ) such that ν q ( u · · · u N ) < ν q ( v · · · v N ), x l < σ x q ,and x q u δ /x l ∈ B . Let w k = u · · · u δ − x q u δ x l u δ +1 · · · u N = x q w i x l . Then w k satisfies the required conditions. (cid:3) In the sequel we show that the lexsegment ideals with a linear resolution whichare not completely satisfies an exchange property.We first recall the following Theorem 3.7 ([1]) . Let I = ( L ( u, v )) be a lexsegment ideal with x | u and x ∤ v which is not a completely lexsegment ideal. Then I has a linear resolution if andonly if u and v have the following form: u = x x a l +1 l +1 · · · x a n n and v = x l x d − n for some l , ≤ l ≤ n − . Theorem 3.8. Let < σ be the decreasing revlexicographical order on S and I =( L ( u, v )) a lexsegment ideal with linear resolution which is not a completely lexseg-ment ideal. Then L ( u, v ) satisfies the σ -exchange property with respect to any mono-mial order on R = K [ { T w : w ∈ L ( u, v ) } ] .Proof. Let u = x x a l +1 l +1 · · · x a n n , v = x l x d − n for some 2 ≤ l ≤ n − 1. Let us as-sume that there exists a monomial order < on R such that L ( u, v ) does not satisfythe σ -exchange property with respect to < . Then there exist two standard mono-mials T u · · · T u N and T v · · · T v N such that u · · · u N > revlex v · · · v N and with theproperty that for all 1 ≤ δ ≤ N , j ∈ supp( u δ ) and q ∈ supp( v · · · v N ) such that ν q ( u · · · u N ) < ν q ( v · · · v N ) and x j > revlex x q , we have x q u δ /x j / ∈ L ( u, v ). Since u · · · u N > revlex v · · · v N there exists some q , 1 ≤ q ≤ n , such that ν i ( u · · · u N ) = ν i ( v · · · v N ) for all i ≥ q + 1and ν q ( u · · · u N ) < ν q ( v · · · v N ). Since deg( u · · · u N ) = deg( v · · · v N ) there existssome s < q such that ν s ( u · · · u N ) > ν s ( v · · · v N ). Let u δ be such that s ∈ supp( u δ ).By our assumption, we must have x q u δ /x s < lex v , that is x q u δ /x s ≤ lex x dl +1 . Thisimplies, in particular, that q ≥ l + 1, and that for all δ , 1 ≤ δ ≤ N , there exists aunique j δ ≤ l such that u δ = x j δ w δ where min( w δ ) ≥ l + 1.Therefore we have u · · · u N = x j · · · x j N x a t t · · · x a n n , where j , · · · , j N < q and t ≥ q . We have a t + · · · + a n = deg( x a t t · · · x a n n ) = N d − N = N ( d − . et v · · · v N = x b · · · x b n n . By hypothesis, we have a q < b q and a i = b i for all i ≥ q + 1. Since each monomial v γ ∈ L ( u, v ) it is divisible by some variable x i with i ≤ l < q , we have b + · · · + b q − ≥ N . Then we have N d = b + · · · + b q − + b q + · · · + b n > b + · · · + b q − + a q + · · · + a t + · · · + a n ≥≥ N + N ( d − 1) = N d, a contradiction. (cid:3) Corollary 3.9. All powers of a lexsegment ideal with a linear resolution which isnot a completely lexsegment ideal have linear quotients with respect to the increasingrevlexicographic order. Corollary 3.10. Any lexsegment ideal with a linear resolution which is not a com-pletely lexsegment ideal is of fiber type. Corollary 3.11. Let I = ( L ( u, v )) be a lexsegment ideal with a linear resolutionwhich is not a completely lexsegment ideal. Then the Rees algebra R ( I ) is Koszul.Proof. Let < be the lexicographical monomial order on R = K [ { T w : w ∈ L ( u, v ) } ]induced by T w > T w if w > lex w and < σ be the decreasing revlexicographicorder on S . By Theorem 3.4, the reduced Gr¨obner basis of P R ( I ) with respect tothe product order < σ on T is formed by the binomials from G < (cid:0) J K [ L ( u,v )] (cid:1) , thereduced Gr¨obner basis of J K [ L ( u,v )] , and by the binomials of the form x i T u ′ − x j T v ′ , where x i > σ x j , x i u ′ = x j v ′ and j is the smallest integer with x i u ′ /x j ∈ L ( u, v ).Since G < (cid:0) J K [ L ( u,v )] (cid:1) is quadratic ([4, Proposition 2.13]), the statement follows. (cid:3) References [1] A. Aramova, E. De Negri, J. Herzog, Lexsegment ideals with linear resolutions, Illinois J.Math., (3), 1998, 509–523.[2] V. Bonanzinga, V. Ene, A. Olteanu, L. Sorrenti, An overview on the minimal free resolu-tions of lexsegment ideals , Combinatorial Aspects of Commutative Algebra, ContemporaryMathematics, AMS, (V. Ene, E. Miller, Eds), , 2009, 5–24.[3] A. Conca, Regularity jumps for powers of ideals , Commutative Algebra, Lect. Notes PureAppl. Math., , 2006, Chapman & Hall/CRC, Boca Raton, FL, 21–32.[4] E. De Negri, Toric rings generated by special stable sets of monomials , MathematischeNachrichten, , 1999, 31–45.[5] E. De Negri, J. Herzog, Completely lexsegment ideals, Proc. Amer. Math. Soc., (12), 1998,3467–3473.[6] V. Ene, A. Olteanu, L. Sorrenti, Properties of lexsegment ideals, arXiv:0802.1279, to appearin Osaka J. Math., (1), 2010.[7] J. Herzog, T. Hibi, M. Vl˘adoiu, Ideals of fiber type and polymatroids, Osaka J. Math., ,2005, 807–829.[8] H. Hulett, H.M. Martin, Betti numbers of lex-segment ideals, J. Algebra, , 2004, 629–638.[9] M. Ishaq, Lexsegment ideals are sequentially Cohen-Macaulay , arXiv:1010.5615v2.[10] A. Olteanu, Normally torsion-free lexsegment ideals , arXiv:1010.1473v1. aculty of Mathematics and Computer Science, Ovidius University, Bd. Mamaia124, 900527 Constanta, Romania, E-mail address : [email protected] Faculty of Mathematics and Computer Science, Ovidius University, Bd. Mamaia124, 900527 Constanta, Romania, E-mail address : [email protected]@gmail.com