Remarks on the Stanley depth of monomial ideals with linear quotients
aa r X i v : . [ m a t h . A C ] F e b Remarks on the Stanley depth of monomial ideals with linearquotients
Mircea Cimpoea¸s
Abstract
We prove that if I is a monomial ideal with linear quotients in a ring of poly-nomials S in n indeterminates and depth( S/I ) = n −
2, then sdepth(
S/I ) = n − S/I ) ≥ depth( S/I ) for a monomial ideal I with linearquotients which satisfies certain technical conditions. Keywords: monomial ideal; Stanley depth; linear quotients.
Introduction
Let K be a field and let S = K [ x , x , . . . , x n ] be the ring of polynomials in n variables.Let M be a Z n -graded S -module. A Stanley decomposition of M is a direct sum D : M = r M i =1 m i K [ Z i ] , as K -vector spaces, where m i ∈ M , Z i ⊂ { x , . . . , x n } such that m i K [ Z i ] is a free K [ Z i ]-module. We define sdepth( D ) = min ri =1 | Z i | andsdepth( M ) = max { sdepth( D ) | D is a Stanley decomposition of M } . The number sdepth( M ) is called the Stanley depth of M . Herzog Vl˘adoiu and Zheng [7]proved that this invariant can be computed in a finite number of steps, when M = I/J ,where J ⊂ I ⊂ S are monomial ideals.We say that the multigraded module M satisfies the Stanley inequality ifsdepth( M ) ≥ depth( M ) . Stanley conjectured in [10] that sdepth( M ) ≥ depth( M ), for any Z n -graded S -module M .In fact, in this form, the conjecture was stated by Apel in [1]. The Stanley conjecture wasdisproved by Duval et. al [4], in the case M = I/J , where (0) = J ⊂ I ⊂ S are monomialideals, but it remains open in the case M = I , a monomial ideal.A monomial ideal I ⊂ S has linear quotients , if there exists u u · · · u m , anordering on the minimal set of generators G ( I ), such that, for any 2 ≤ j ≤ m , the ideal( u , . . . , u j − ) : u j is generated by variables. Given a monomial ideal I ⊂ S , SoleymanJahan [6] noted that I satisfies the Stanley inequality, i.e. sdepth( I ) ≥ depth( I ). However,a similar result for S/I seems more difficult to prove, only some particular cases beingknown. For instance, Seyed Fakhari [5] proved the inequality sdepth(
S/I ) ≥ depth( S/I )for weakly polymatroidal ideals I ⊂ S , which are a class of ideals with linear quotients.1he aim of this paper is to tackle the general problem; however, we are able to obtainonly partial results. In Theorem 1.5, we prove that if I ⊂ S is a monomial ideal with linearquotients with depth( S/I ) = n −
2, then sdepth(
S/I ) = n −
2. In Theorem 1.7, we provethat if I ⊂ S is a monomial ideal with linear quotients which has a Stanley decompositionwhich satisfies certain conditions, then sdepth( S/I ) ≥ depth( S/I ). Also, we conjecturethat for any monomial ideal I ⊂ S with linear quotients, there is a variable x i such thatdepth( S/ ( I, x i )) = depth( S/I ) and in Theorem 1.13 we prove that if this conjecture istrue, then sdepth(
S/I ) ≥ depth( S/I ), for any monomial ideal I ⊂ S with linear quotients. Let I ⊂ S be a monomial ideal and let G ( I ) be the set of minimal monomial generators of I . We recall that I has linear quotients , if there exists a linear order u u · · · u m on G ( I ), such that for every 2 ≤ j ≤ m , the ideal ( u , . . . , u j − ) : u j is generated by a subsetof n j variables.We let I j := ( u , . . . , u j ), for 1 ≤ j ≤ m . Let Z = { x , . . . , x n } and Z j = { x j | x j / ∈ ( I j − : u j ) } for 2 ≤ j ≤ m . Note that, for any 2 ≤ j ≤ m , we have I j /I j − = u j ( S/ ( I j − : u j )) = u j K [ Z j ] , Hence the ideal I has the Stanley decomposition I = u K [ Z ] ⊕ u K [ Z ] ⊕ · · · ⊕ u m K [ Z m ] . (1.1)According to [9, Corollary 2.7], the projective dimension of S/I ispd(
S/I ) = max { n j : 2 ≤ j ≤ m } + 1 . Hence, Ausl¨ander-Buchsbaum formula implies thatdepth(
S/I ) = n − max { n j : 2 ≤ j ≤ m } − . (1.2)Note that, (1.1) and (1.2) implies sdepth I ≥ depth I , a fact which was proved in [6]. Werecall the following results: Proposition 1.1.
Let I ⊂ S be a monomial ideal and u ∈ S a monomial. Then:(1) depth( S/ ( I : u )) ≥ depth( S/I ) . ([8, Corollary 1.3])(2) sdepth( S/ ( I : u )) ≥ sdepth( S/I ) . ([3, Proposition 2.7(2)]) Proposition 1.2.
Let → U → M → N → be a short exact sequence of Z n -graded S -modules. Then:(1) depth( M ) ≥ min { depth( U ) , depth( N ) } . (Depth Lemma)(2) sdepth( M ) ≥ min { sdepth( U ) , sdepth( N ) } . ([8, Lemma 2.2]) roposition 1.3. Let I ⊂ S be a proper monomial ideal with linear quotients with depth( S/I ) = n − s − , where ≤ s ≤ n − . Then there exists a subset τ ⊂ [ n ] with | τ | = s and a monomial u ∈ G ( I ) , such that I + ( x i : i ∈ τ ) = ( u ) + ( x i : i ∈ τ ) .Proof. If I = ( u ) is principal, that is s = 0, then there is nothing to prove. Assume s ≥
1. Since I has linear quotients, we can assume that G ( I ) = { u , . . . , u m } such that(( u , . . . , u j − ) : u j ) is generated by variables, for every 2 ≤ j ≤ m . We consider thedecomposition (1.1), that is I = u K [ Z ] ⊕ u K [ Z ] ⊕ · · · ⊕ u m K [ Z m ] , where Z = { x , . . . , x n } and Z j is generated by the variables which do not belong to(( u , . . . , u j − ) : u j ), for 2 ≤ j ≤ m . From (1.2), it follows that | Z j | ≥ s for 2 ≤ j ≤ m .Note that, for any 2 ≤ j ≤ m , we have( u , . . . , u j − ) ∩ u j K [ Z j ] = { } (1.3)We assume, by contradiction, that for any τ ⊂ [ n ] with | τ | = s , there exists k τ = ℓ τ ∈ [ m ]such that u k τ , u ℓ τ / ∈ ( x i : i ∈ τ ), that is u k τ , u ℓ τ ∈ K [ x i : i / ∈ τ ]. We claim that thereexists i = i ( τ ) ∈ τ such that x i ∈ Z k τ or x i ∈ Z ℓ τ . Indeed, otherwise, since | Z k τ | , | Z k ℓ | ≥ s we would have Z k τ = Z k ℓ = { x i : i / ∈ τ } and hencegcd( u k τ , u ℓ τ ) ∈ u k τ K [ Z k τ ] ∩ u ℓ τ K [ Z ℓ τ ] , a contradiction. Without any loss of generality, we may assume that x i ∈ Z k τ .Let σ = { n − s + 1 , . . . , n } . By reordering the variables, we can assume that k σ ≥ k τ for any τ ⊂ [ n ] with | τ | = s . Note that u k σ = x a · · · x a n − s s . By the above argument, thereexists i ≥ n − s + 1 such that x i ∈ Z k σ . Let A ⊂ [ n ] with | A | = n − s such that i ∈ A and { x ℓ : ℓ ∈ A } ⊂ Z k σ . It follows that τ = [ n ] \ A = σ . On the other hand,gcd( u k τ , u k σ ) ∈ u k τ S ∩ u k σ K [ Z k σ ] , which contradicts (1.3) for j = k σ . Hence, the proof is complete. Corollary 1.4.
Let I ⊂ S be a monomial ideal with linear quotients with depth( S/I ) = n − s − , where ≤ s ≤ n − . Then there exists a subset σ ⊂ [ n ] with | σ | ≤ s + 1 suchthat I ⊂ ( x i : i ∈ σ ) .Proof. If I = ( u ) is principal, that is s = 0, then I ⊂ ( x i ), where x i | u , hence the assertionis true. If sdepth( S/I ) = 0, that is s = n −
1, then the assertion follows from the fact that I ⊂ ( x , . . . , x n ).We assume that 1 ≤ s ≤ n −
2. According to Proposition 1.3, there exists u ∈ G ( I )and τ ⊂ [ n ] with | τ | = n − s such that I + ( x i : i ∈ τ ) = ( u ) + ( x i : i ∈ τ ) . If u ∈ ( x i : i ∈ τ ) then I ⊂ ( x i : i ∈ τ ) and there is nothing to prove. Otherwise, wechoose x ℓ | u with ℓ τ . The assertion holds for σ = τ ∪ { ℓ } .3ote that, a monomial ideal I ⊂ S is principal if and only if depth( S/I ) = n − S/I ) = n − Theorem 1.5.
Let I ⊂ S be a monomial ideal with linear quotients with depth( S/I ) = n − . Then sdepth( S/I ) = n − .Proof. We assume that G ( I ) = { u , . . . , u m } . We use induction on m and d = P mj =1 deg( u i ).If m = 2, then from [2, Proposition 1.6] it follows that sdepth( S/I ) = n −
2. If d = 2, then I is generated by two variables and there is nothing to prove.Assume m > d >
2. Acording to Proposition 1.3, there exists i ∈ [ n ] and u k ∈ G ( I ) such that ( I, x i ) = ( u k , x i ). Since ( I, x i ) = ( u k , x i ), from [2, Proposition 1.2] it followsthat sdepth( S/ ( I, x i )) ≥ n −
2. If ( I : x i ) is principal, then sdepth( S/ ( I : x i )) = n −
1. Notethat, at least one of u j ’s must be disivible with x i , otherwise G ( I, x i ) = { u , . . . , u m , x i } , acontradiction with the fact that ( I, x i ) = ( u k , x i ) and m ≥
2. It follows that d ′ := X u ∈ G ( I : x i ) deg u < d, thus, by induction hypothesis, we have sdepth( S/ ( I : x i )) = n −
2. In both cases, sdepth( S/ ( I : x i )) ≥ n −
2. From Proposition 1 . → S/ ( I : x i ) → S/I → S/ ( I, x i ) → , it follows that sdepth( S/I ) ≥ min { sdepth( S/ ( I : x i )) , sdepth( S/ ( I, x i )) } ≥ n −
2. Since I is not principal, it follows that sdepth( S/I ) = n −
2, as required.
Lemma 1.6.
Let I ⊂ S be a monomial ideal and u ∈ S a monomial with ( I : u ) =( x , . . . , x m ) . Assume that S/I has a Stanley decomposition D : S/I = r M i =1 v i K [ Z i ] , (1.4) such that there exists i with Z i = { x m +1 , . . . , x n } and v i | u . Then: sdepth( S/ ( I, u )) ≥ min { sdepth( D ) , n − m − } . Proof.
If sdepth(
S/I ) = 0 or m = n −
1, then there is nothing to prove. We assume thatsdepth(
S/I ) ≥ m ≤ n −
2. Since S/ ( I : u ) = S/ ( x , . . . , x m ) ∼ = K [ x m +1 , . . . , x n ],from the short exact sequence0 → S/ ( I : u ) · u −→ S/I −→ S/ ( I, u ) −→ , it follows that we have the K -vector spaces isomorphism S/I ∼ = S/ ( I, u ) ⊕ uK [ x m +1 , . . . , x n ] . (1.5)4rom our assumption, uK [ x m +1 , . . . , x n ] = uK [ Z i ] ⊂ v i K [ Z i ]. Hence, from (1.4) and(1.5) it follows that S/ ( I, u ) ∼ = M i = i v i K [ Z i ] ! ⊕ v i K [ Z i ] uK [ Z i ] ∼ = M i = i v i K [ Z i ] ! ⊕ v i K [ x m +1 , . . . , x n ] w K [ x m +1 , . . . , x n ] , (1.6)where w = uv i . On the other hand, sdepth (cid:16) K [ x m +1 ,...,x n ] w K [ x m +1 ,...,x n ] (cid:17) = n − m −
1. Hence (1.6) yieldsthe required conclusion.
Theorem 1.7.
Let I ⊂ S be a monomial ideal with linear quotients, G ( I ) = { u , . . . , u m } , I j = ( u , . . . , u j ) for ≤ j ≤ m , such that ( I j − : u j ) = ( { x , . . . , x n } \ Z j ) , where Z j ⊂ { x , . . . , x n } , for all ≤ j ≤ m .We assume that for any ≤ j ≤ m , there exists a Stanley decomposition D j − of S/I j − such that sdepth( D j − ) ≥ depth( S/I j − ) and there exists a Stanley subspace w j − K [ W j − ] of D j − with w j − | u j and W j − = Z j .Then sdepth( S/I ) ≥ depth( S/I ) .Proof. According to Lemma 1.6, for any 2 ≤ j ≤ m , we havesdepth( S/I j ) ≥ min { depth( S/I j − ) , n − n j − } , for all 2 ≤ j ≤ m, (1.7)where n j = n − | Z j | , 1 ≤ j ≤ m . On the other hand, according to (1.2),depth( S/I ) = m min j =1 { n − n j − } . (1.8)Since sdepth( S/I ) = depth( S/I ) = n −
1, from (1.7) and (1.8) we get the requiredconclusion.
Example 1.8.
Let I = ( x , x x , x x x ) ⊂ S = K [ x , x , x , x ]. Let u = x , u = x x and u = x x x . Since (( u ) : u ) = ( x ) and (( u , u ) : u ) = ( x , x ), it follows that I has linear quotients with repect to the order u u u . Moreover, I = u K [ Z ] ⊕ u K [ Z ] ⊕ u K [ Z ] = x K [ x , x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x x K [ x , x ] . Let I = ( u ) and I = ( u , u ). We consider the Stanley decomposition D : S/I = K [ x , x , x ] ⊕ x K [ x , x , x ] , of S/I with sdepth( D ) = sdepth( S/I ) = 2. Let w = x and W = { x , x , x } . Clearly, W = Z and w | u . As in the proof of Lemma 1.6, we obtain the Stanley decomposition D : S/I = K [ x , x , x ] ⊕ x K [ x , x , x ] x x K [ x , x , x ] = K [ x , x , x ] ⊕ x K [ x , x ] ⊕ x x K [ x , x ] . Let w = x x and W = { x } . Clearly, W = Z and w | u . Hence, according to Theorem1.7, sdepth( S/I ) ≥ depth( S/I ). In fact, we have that sdepth(
S/I ) ≥ depth( S/I ) = 0 and D : S/I = K [ x , x , x ] ⊕ x K [ x , x ] ⊕ x x K [ x ] ⊕ x x x K [ x ] , is a Stanley decomposition of S/I with sdepth( D ) = sdepth( S/I ) = 1.5he following two lemmas are well known. However, for the sake of completness, weprove them:
Lemma 1.9.
Let I ⊂ S be a monomial with linear quotients and u ∈ S a monomial. Then ( I : u ) has linear quotients.Proof. Since ( I : vw ) = (( I : v ) : w ) for any monomials v, w ∈ S , without any loss ofgenerality, we can assume that u is a variable, let’s say u = x n . We consider the order u u · · · u m on G ( I ), such that, for every 2 ≤ j ≤ m , the ideal ( I j − : u j ) isgenerated by a nonempty subset ¯ Z j of variables, where I j = ( u , . . . , u j ) for 1 ≤ j ≤ m .Note that ( I : x n ) is generated by u ′ , . . . , u ′ m , where u ′ j = ( u j /x n , x n | u j u j , x n ∤ u j , 1 ≤ j ≤ m .We let I ′ j := ( I j : x n ) = ( u ′ , . . . , u ′ j ), for 1 ≤ j ≤ m . We fix 2 ≤ j ≤ m . We have two cases:1. x n ∤ u j . Since x n u ′ j = x n u j ∈ I j − , it follows that u ′ j ∈ I ′ j − . Thus I ′ j = I ′ j − .2. x n | u j . If x n ∈ Z j , since x n u j ∈ I j − , there exists some 1 ≤ s ≤ j − x n u j | u s . In particular, x n | u s . Hence u ′ j = u j /x n and u ′ s = u s /x n . It follows that x n u ′ j | u ′ s . Therefore, x n ∈ ( I ′ j − : u ′ j ). If x i ∈ ¯ Z j \ { x n } , then it is obvious that x i u ′ j ∈ I ′ j − . Therefore, ( ¯ Z j ) ⊂ ( I ′ j − : u ′ j ).In order to prove the converse, let w ∈ ( I ′ j − : u ′ j ), that is wu ′ j ∈ I ′ j − . It follows that x n wu ′ j = wu j ∈ I j − , hence w ∈ ( I j − : u j ) = ( ¯ Z j ).In conclusion, I ′ has linear quotients. Lemma 1.10.
Let I ⊂ S be a monomial with linear quotients and x i a variable. Then ( x i , I ) has linear quotients. Moreover, if S ′ = K [ x , . . . , x i − , x i +1 , . . . , x n ] , then ( x i , I ) =( x i , J ) , where J ⊂ S ′ is a monomial ideal with linear quotients.Proof. We consider the order u u · · · u m on G ( I ), such that, for every 2 ≤ j ≤ m ,the ideal ( I j − : u j ) is generated by a nonempty subset ¯ Z j of variables. We assume that u j u j · · · u j p are the minimal monomial generators of I which are not multiple of x i . We have that (( x i ) : u j ) = ( x i ). Also, for 2 ≤ k ≤ p , we claim that(( x i , u j , . . . , u j k − ) : u j k ) = ( x i , ¯ Z j k ) . (1.9)Indeed, since (( u , . . . , u j k − ) : u j k ) = ( ¯ Z j k ) and x i u j k ∈ ( x i , u j , . . . , u j k − ) it follows that( x i , ¯ Z j k ) ⊂ (( x i , u j , . . . , u j k − ) : u j k ). Conversely, assume that v ∈ S is a monomial with vu j k ∈ ( x i , u j , . . . , u j k − ) = ( x i , u , . . . , u j k − ). If x i ∤ v , then vu j k ∈ ( u , . . . , u j k − ), hence v ∈ ( ¯ Z j k ). If x i | v , then v ∈ ( x i , u , . . . , u j k − ). Hence the claim (1.9) is true and therefore( x i , I ) has linear quotients. Now, let J = ( u j , . . . , u j p ). For any k ≥
2, we have that(( u j , . . . , u j k − ) : u j k ) ⊂ (( u , . . . , u j k − ) : u j k ) = ( ¯ Z j k ) . (1.10)From (1.9) and (1.10), one can easily deduce that (( u j , . . . , u j k − ) : u j k ) = ( ¯ Z j k \ { x i } ).Hence, J has linear quotients. 6e propose the following conjecture: Conjecture 1.11. If I ⊂ S is a proper monomial ideal with linear quotients, then thereexists i ∈ [ n ] such that depth( S/ ( I, x i )) ≥ depth( S/I ) . Remark 1.12.
Let I ⊂ S be a monomial ideal with linear quotients, G ( I ) = { u , . . . , u m } , I j = ( u , . . . , u j ) for 1 ≤ j ≤ m , such that ( I j − : u j ) = ( { x , . . . , x n } \ Z j ), where Z j ⊂ { x , . . . , x n } , for all 2 ≤ j ≤ m . I has the Stanley decomposition: I = u K [ Z ] ⊕ u K [ Z ] ⊕ · · · ⊕ u m K [ Z m ] , where Z = { x , . . . , x n } . We have thatdepth( S/I ) = n − s − , where n − s = min {| Z j | | ≤ j ≤ m } . Conjecture 1.11 is equivalent to the fact that there exists i ∈ [ n ] such that there is no1 ≤ j ≤ m with x i ∤ u j , x i ∈ Z j and | Z j | = n − s . Theorem 1.13.
If Conjecture 1.11 is true and I ⊂ S is a monomial ideal with linearquotients, then sdepth( S/I ) ≥ depth( S/I ) .Proof. We use induction on n and d = P v ∈ G ( I ) deg( v ). If n = 1 or d = 1 then there isnothing to prove. Assume n ≥ d ≥
2. Let I ⊂ S be a monomial ideal with linearquotients and let i ∈ [ n ] such that depth( I, x i ) ≥ depth( I ). If x i ∈ I , then, according toLemma 1.10, I = ( x i , J ) where J ⊂ S ′ := K [ x , . . . , x i − , x i +1 , . . . , x n ] is a monomial idealwith linear quotients. From induction hypothesis on n , it follows thatsdepth( S/I ) = sdepth( S ′ /J ) ≥ depth( S ′ /J ) = depth( S/I ) . Assume that x i / ∈ I . Note that there exists at least a minimal monomial generator v of I such that x i | v , otherwise depth( S/ ( I, x i )) = depth( S/I ) −
1, a contradiction. We considerthe short exact sequence 0 → S ( I : x i ) → SI → S ( I, x i ) → . (1.11)According to Lemma 1.9, ( I : x i ) is a monomial ideal with linear quotients. Moreover, d ′ = P v ′ ∈ G (( I : x i )) deg( v ′ ) < d , hence, from induction hypothesis on d , it follows thatsdepth( S/ ( I : x i )) ≥ depth( S/ ( I : x i )).Let S ′ := K [ x , . . . , x i − , x i +1 , . . . , x n ]. According to Lemma 1.10, ( x i , I ) = ( x i , J ) where J ⊂ S ′ is a monomial ideal with linear quotients. Note thatsdepth( S/ ( x i , I )) = sdepth( S/ ( x i , J )) = sdepth( S ′ /J ) and depth( S/ ( I, x i )) = depth( S/J ) . Since P v ∈ G ( J ) deg( v ) < d , by induction hypothesis, we get sdepth( S/J ) ≥ depth( S/J ).Hence sdepth( S/ ( x i , I )) ≥ depth( S/ ( I, x i )) ≥ depth( S/I ). According to Proposition 1.1(2),we have depth( S/ ( I : x i )) ≥ depth( S/I ). From (1.11) and Proposition 1.2(2), it followsthat sdepth(
S/I ) ≥ min { sdepth( S/ ( I : x i )) , sdepth( S/ ( I, x i )) } ≥≥ min { depth( S/ ( I : x i )) , depth( S/ ( I, x i )) } ≥ depth( S/I ) , as required. 7 eferences [1] J. Apel, On a conjecture of R. P. Stanley; Part II - Quotients Modulo Monomial Ideals ,J. of Alg. Comb. (2003), 57–74.[2] M. Cimpoea¸s, Stanley depth of monomial ideals with small number of generators , Cent.Eur. J. Math.
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