aa r X i v : . [ m a t h . A C ] A ug DEPTH AND STANLEY DEPTH OF MULTIGRADED MODULES
ASIA RAUF
Abstract.
We study the behavior of depth and Stanley depth along short exact se-quences of multigraded modules and under reduction modulo an element.
Introduction
Let K be a field and S = K [ x , . . . , x n ] a polynomial ring in n variables over K . Let M be a finitely generated multigraded (i.e. Z n -graded) S -module. Let m ∈ M be ahomogeneous element in M and Z ⊆ { x , . . . , x n } . We denote by mK [ Z ] the K -subspaceof M generated by all elements mv , where v is a monomial in K [ Z ]. The multigraded K -subspace mK [ Z ] ⊂ M is called Stanley space of dimension | Z | , if mK [ Z ] is a free K [ Z ]-module. A Stanley decomposition of M is a presentation of the K -vector space M as afinite direct sum of Stanley spaces D : M = L ri =1 m i K [ Z i ]. Set sdepth D = min {| Z i | : i = 1 , . . . , r } . The numbersdepth( M ) := max { sdepth( D ) : D is a Stanley decomposition of M } is called Stanley depth of M . In 1982, Richard P. Stanley [14, Conjecture 5.1] conjecturedthat sdepth( M ) ≥ depth( M ) for all finitely generated Z n -graded S -modules M . Theconjecture is discussed in some special cases in [7], [2], [9], [13], [1], [12] [4], [5].If M is a finitely generated module over a Noetherian local ring ( R, m ) and x ∈ m thenit is well-known that dim M/xM ≥ dim M −
1. Our Proposition 1.2 and Lemma 1.7, showthat the above inequality is preserved for depth and sdepth when M = S/I and I ⊂ S isa monomial ideal and x = x k for any k ∈ [ n ]. If M is a general multigraded S -module,then we might have depth M/x k M < depth M − M/x k M < sdepth M − x k is regular on M , as shows Example 1.8.As we know depth decreases by one if we reduce modulo a regular element. In [13,Theorem 1.1], it is proved that the corresponding statement holds for the Stanley depthin the case of M = S/I where I ⊂ S is a monomial ideal and f a monomial in S which isregular on M . The next question arises whether this is true for any multigraded module?The answer is no, see Example 1.8. Let F : 0 = M ⊂ M ⊂ . . . ⊂ M r = M be a chain of Z n -graded submodules of M . Then F is called a prime filtration of M if M i /M i − ∼ = ( S/P i )( − a i ) where a i ∈ Z n and P i is a monomial prime ideal for all i . We denote Supp F = { P , . . . , P r } . A finitely generated module M is called almostclean if there exists a prime filtration F of M such that Supp( F ) = Ass( M ). Weshow in Lemma 1.9 that for almost clean module M and x k ∈ S being regular on M ,we have sdepth M/x k M ≥ sdepth M − . However, we show in Proposition 1.10 that
Mathematics Subject Classification.
Primary 13H10, Secondary 13P10, 13C15, 13F20.
Key words and phrases.
Multigraded modules, depth, Stanley decompositions, Stanley depth.The author wants to acknowledge Higher Education Commission Pakistan for partial financial supportduring preparation of this work. depth M/x k M ≤ sdepth M −
1, if x k is regular on M . As an application we get thatStanley’s conjecture holds for M if it holds for the module M/x k M (see Corollary 1.11).Moreover if M has a maximal regular sequence given by monomials then Stanley’s con-jecture holds for M (see Corollary 1.13).Given a short exact sequence of finitely generated multigraded S -modules, then theStanley depth of the middle one is greater than or equal to the minimum of Stanleydepths of the ends (see Lemma 2.2). Several examples show that the ”Depth lemma” ismainly wrong in the frame of sdepth (see Examples 2.5 and 2.6). However, we prove inLemma 2.7 that if I is any monomial complete intersection of S , then sdepth I is greaterthan or equal to sdepth S/I + 1. But in general for any monomial ideal this inequality isstill an open question.In the last section, we prove that if I ⊂ S = K [ x , . . . , x n ], J ⊂ S = K [ y , . . . , y m ] aremonomial ideals and S = K [ x , . . . , x n , y , . . . , y m ], then the Stanley depth of the tensorproduct of S /I and S /J (over K ) is greater than or equal to the sum of sdepth S /I and sdepth S /J . This inequality could be strict as shows Example 3.2.I am grateful to Professor J¨urgen Herzog for useful discussions and comments duringthe preparation of the paper.1. The behavior of depth and sdepth under reduction modulo element
In dimension theory it is well known the following result (see e.g. [3, Proposition A.4],or [6, Corollary 10.9])
Theorem 1.1. If ( R, m ) is a Noetherian local ring and M is finitely generated R-module,then for any x ∈ m we have dim M/xM ≥ dim M − . If we consider reduction by a regular element, then the depth decreases by one. Butwhat happens if we take reduction by a non-regular element?
Proposition 1.2.
Let S = K [ x , . . . , x n ] be a polynomial ring over a field K , I ⊂ S amonomial ideal and R = S/I . Then depth(
R/x n R ) ≥ depth( R ) − . (1) Proof.
Let ¯ R = R/x n R ≃ S/ ( I, x n ). We denote ¯ S = K [ x , . . . , x n − ] and let x ′ = { x , . . . , x n − } , x = { x , . . . , x n } .Let ϕ be the canonical map from R to ¯ R and α be the composite map¯ S −→ S −→ R = S/I, where the first map is the canonical embedding and the second map is the canonicalsurjection. It is clear that ker( α ) = I ∩ ¯ S. Let α be the composite map¯ S −→ S −→ R = S/I −→ S/ ( I, x n ) . It is clear that α is surjective. We claim that ker( α ) = I ∩ ¯ S . One inclusion is obvious. Toprove other inclusion, we consider a monomial v ∈ ker( α ), that is, v ∈ ( I, x n ). Since v ∈ ¯ S and I is a monomial ideal, it follows that v ∈ I . Let ker( α ) = ¯ I then ¯ S/ ¯ I ≃ S/ ( I, x n ). Itfollows that the composition ¯ R → R → ¯ R of the natural maps is the identity. Therefore,the S -module ¯ R is a direct summand of the S -module R . This implies that the S -module H i ( x ′ ; ¯ R ) is a direct summand of H i ( x ′ ; R ) for all i , where H i ( x ′ ; ¯ R ) and H i ( x ′ ; R ) are thei-th Koszul homology modules of x ′ with respect to ¯ R and R respectively. In particular,if H i ( x ′ ; ¯ R ) = 0, then H i ( x ′ ; R ) = 0. Let k = max { i | H i ( x ′ : ¯ R ) = 0 } . Then depth ¯ R = − − k , by [3, Theorem 1.6.17]. Since H k ( x ′ ; ¯ R ) = 0, it follows that H k ( x ′ ; R ) = 0 whichimplies that H k ( x ; R ) = 0 by [3, Lemma 1.6.18]. Therefore applying again [3, Theorem1.6.17] it follows that depth R ≤ n − k = depth ¯ R + 1. (cid:3) Corollary 1.3.
Let I ⊂ S be a monomial ideal. Then depth S/ ( I : u ) ≥ depth S/I for allmonomials u I .Proof. Since ( I : uv ) = (( I : u ) : v ), where I is a monomial ideal and u and v aremonomials, we may reduce to the case u = x n , and apply recurrence. Then we have theexact sequence 0 −→ S/ ( I : x n ) −→ S/I −→ S/ ( I, x n ) −→ . By Depth Lemma [15, Lemma 1.3.9] and Proposition 1.2, we obtain the required result. (cid:3)
This Corollary does not hold (and so Proposition 1.2) if u is not a monomial, as wehave the following example: Example 1.4.
Let S = K [ x, y, z, t ] and I = ( x, y ) ∩ ( y, z ) ∩ ( z, t ) and u = y + z . Then J := ( I : u ) = ( x, y ) ∩ ( z, t ) and depth S/J = 1 < S/I .The Proposition 1.2 is not true in general. If M is a finitely generated graded R -moduleand x ∈ R then we might havedepth( M/xM ) < depth( M ) − , as shows the following example: Example 1.5.
Let S = K [ x, y, z, t ], M = ( x, y, z ) / ( xt ). We have depth M = 2 and M/xM = ( x, y, z ) / ( x , xy, xz, xt ). Since the maximal ideal is an associated prime ideal of M/xM , we get depth
M/xM = 0. Hence depth(
M/xM ) < depth( M ) − R/x n R ) > depth( R ) − , as shows the following example: Example 1.6.
Let I = ( x , x x , . . . , x x n ) ⊂ S = K [ x , . . . , x n ] be a monomial ideal of S and R = S/I . Then depth( R ) = 0 since the maximal ideal ( x , x , . . . , x n ) ∈ Ass( R ). Since R/x R = S/ ( x ) ≃ K [ x , . . . , x n ], we get depth( R/x R ) = n −
1. Hence depth(
R/x R ) > depth( R ) − S/ ( I, x n ) = sdepth S/I − x n is regular on S/I , we actually showed the following (without any assumption on x n ): Lemma 1.7.
Let S = K [ x , . . . , x n ] be a polynomial ring over the field K . Let I ⊂ S beany monomial ideal. Then sdepth( S/ ( I, x n )) ≥ sdepth( S/I ) − . This lemma can not be extended to general multigraded modules M as shows thefollowing example, where the variable is even regular on M . Example 1.8.
Let M = ( x, y, z ) be an ideal of S = K [ x, y, z ]. Consider a Stanleydecomposition M = zK [ x, z ] ⊕ xK [ x, y ] ⊕ yK [ y, z ] ⊕ xyzK [ x, y, z ]. Since sdepth M ≤ dim S = 3 and M is not a principle ideal, it follows sdepth M = 2. Note that x induces a on-zero element in the socle of M/xM which cannot be contained in any Stanley spaceof dimension greater or equal with one. Hence sdepth
M/xM = 0. Thus sdepth
M/xM < sdepth M − M is almost clean (see [9]), that is there exists aprime filtration F of M such that Supp( F ) = Ass( M ), we have the following: Lemma 1.9.
Let M be a finitely generated multigraded S -module. If M is almost cleanand x k ∈ S is regular on M , then sdepth M/x k M ≥ sdepth M − . Proof.
Suppose that F is given by0 = M ⊂ M ⊂ . . . ⊂ M r = M with M i /M i − ∼ = S/P i ( − a i ) for some a i ∈ N n and some monomial prime ideals P i . SinceAss M = { P , . . . , P r } and x k is regular on M , we get x k P i and so x k is regular on M i /M i − . Set ¯ M i = M i /x k M i . Then ¯ M i ⊂ ¯ M i +1 and { ¯ M i } define a filtration ¯ F of¯ M = ¯ M r with ¯ M i / ¯ M i − ∼ = S/ ( P i , x k )( − a i ). Thus sdepth ¯ M ≥ min i dim S/ ( P i , x k ) =sdepth M − (cid:3) The above Example 1.8 hints that if x is a regular element on M , then sdepth M/xM ≤ sdepth M −
1. This is the subject of our next proposition.
Proposition 1.10.
Let M be finitely generated Z n -graded S -module and let x k be regularon M . If D : M/x k M = L ri =1 ¯ m i K [ Z i ] , is a Stanley decomposition of M/x k M , where m i ∈ M is homogeneous and ¯ m i = m i + x k M . Then M = r M i =1 m i K [ Z i , x k ](2) is a Stanley decomposition of M . In particular sdepth M/x k M ≤ sdepth M − . Proof.
Let N = P ri =1 m i K [ Z i , x k ]. Then N ⊆ M . Since D is a Stanley decompositionof M/x k M it follows that ψ ( N ) = M/x k M where ψ : M → M/x k M is the canonicalepimorphism. This implies that M = x k M + N as Z n -graded K vector spaces. Weshow that M = N . First we observe that M = x dk M + N for all d . This follows byinduction on d , because if we have M = x d − k M + N , then M = x d − k ( x k M + N ) + N = x dk M + x d − k N + N = x dk M + N since x d − k N ⊂ N . This completes the induction. Since M is finitely generated there exists an integer c such that deg x k ( m ) ≥ c for all homogeneouselements m ∈ M . Now let m ∈ M be a homogeneous element with deg x k ( m ) = a and let d > a − c be an integer. Since M = x dk M + N , there exist homogeneous elements v ∈ M and w ∈ N such that m = x dk v + w , where a = deg x k v + d = deg x k w . It follows thatdeg x k v = a − d < c , a contradiction. It implies that v = 0, hence m = w ∈ N .Now we show that the sum P ri =1 m i K [ Z i , x k ] is direct, that is m i K [ Z i , x k ] ∩ r X j =1 j = i m j K [ Z j , x k ] = (0) . Let u = m i q i = P r j =1 j = i m j q j ∈ M be homogeneous for some q j monomials in K [ Z j , x k ]such that deg( u ) = deg( m j q j ) for all j . Let p be the biggest power of x k dividing q i . If = 0, then we have ¯ u = ¯ m i q i = 0 in M/x k M since ¯ m i K [ Z i ] is a Stanley space. It followsthat ¯ u ∈ ¯ m i K [ Z i ] ∩ P r j =1 j = i ¯ m j K [ Z j ], a contradiction. In the case of p >
0, then in
M/x k M we get ¯ u = 0 = P r j =1 j = i ¯ m j ¯ q j . It follows that ¯ q j = 0, since D is a Stanley decomposition of M/x k M . Thus q j = x k q ′ j for some q ′ j ∈ K [ Z j , x k ] and we get x k ( m i q ′ i − P r j =1 j = i m j q ′ j ) = 0,which implies m i q ′ i − P r j =1 j = i m j q ′ j = 0 since x k is regular on M . Applying the same argumentby recurrence we get q j = x pk s j for some s j ∈ K [ Z j , x k ], and m i s i = P r j =1 j = i m j s j . We set v = m i s i . Since ¯ s i = 0, we get ¯ v = 0 because ¯ m i K [ Z i ] is a Stanley space. On the otherhand ¯ v ∈ ¯ m i K [ Z i ] ∩ P r j =1 j = i ¯ m j K [ Z j ]. It implies that ¯ v = 0, a contradiction.Finally we show that each m i K [ Z i , x k ] is a Stanley space. Indeed, suppose that m i f = 0for some f ∈ K [ Z i , x k ] where f = P aj =0 f j x jk such that x k does not divide f j for all j then P aj =0 m i f j x jk = 0 implies that ¯ m i f = 0 in M/x k M . We get f = 0 since ¯ m i K [ Z i ] is aStanley space. It follows that f = x k g where g = P aj =1 f j x j − k and from x k m i g = m i f = 0we get m i g = 0, x k being regular on M . Then induction on the degree of f concludes theproof since deg x k g < deg x k f . (cid:3) Corollary 1.11.
If Stanley’s conjecture holds for the module
M/x i M , where x i ∈ S isregular on M , then it also holds for M . Corollary 1.12.
The equality holds in Lemma 1.7, if x n is regular on S/I . The proof follows from Lemma 1.7 and Proposition 1.10 for M = S/I . Corollary 1.13.
Let depth M = t . If there exists u = u , . . . , u t ∈ M on ( S ) such that u is regular sequence on M then Stanley’s conjecture holds for M .Proof. For any regular sequence u = u , . . . , u t ∈ M on ( S ) we may choose u such that u i = x i j for all 1 ≤ i ≤ t , where x i j ∈ supp( u i ), since for any monomial u i ∈ S beingregular on M implies that each x i j ∈ supp( u i ) is regular on M , because if x i j belong tothe set of zero divisors of M then x i j ∈ P for some P ∈ Ass( M ), so u i ∈ P , which isnot true as u i is regular on M . Since u is a maximal regular sequence on M , we havedepth M/ ( u , . . . , u t ) M = 0. Applying Proposition 1.10 by recurrence we get sdepth M ≥ sdepth M/ ( u , . . . , u t ) M + t ≥ t = depth M . Hence Stanley’s conjecture holds for M . (cid:3) Example 1.14.
Let S = K [ x, y, z, t ] and M = ( x, y, z ) / ( xy ). Since depth M = 2 and { z, t } is a M -regular sequence, we may apply Corollary 1.13 to see that Stanley’s conjectureholds for M . Theorem 1.15.
Let M be a finitely generated multigraded S -module. If M is almost cleanand x k ∈ S is regular on M , then sdepth M/x k M = sdepth M − . The proof follows from Lemma 1.9 and Proposition 1.10.
Theorem 1.16.
Let M be a finitely generated multigraded S -module. If M is almost cleanand u ∈ S is a monomial, which is regular on M , then sdepth M/uM ≥ sdepth M − . Proof.
Let u = x a i . . . x a t i t . Since u is regular on M , it follows that each x i k ∈ supp( u ) isregular on M , where we denote by supp( u ) the set of all variables x j such that x j divides he monomial u . We consider an ascending chain of submodules of M between uM and M where two successive members of the chain are of the form x b i · · · x b k i k · · · x b t i t M ⊂ x b i · · · x b k − i k · · · x b t i t M, and where b i ≤ a i for all i = 1 , . . . , t .We obtain x b i · · · x b k − i k · · · x b t i t M/x b i · · · x b k i k · · · x b t i t M ≃ M/x i k M, since each x i k ∈ supp( u ) is regular on M . Therefore Lemma 1.9 and Corollary 2.3 implythat sdepth( M/uM ) ≥ sdepth( M/x i k M ) = sdepth M − . (cid:3) The behavior of sdepth on short exact sequence of multigraded modules
The following ”Depth Lemma” is well-known.
Lemma 2.1. [15, Lemma 1.3.9] If → U → M → N → is a short exact sequence of modules over a local ring R , then (a) If depth M < depth N , then depth U = depth M . (b) If depth M > depth N , then depth U = depth N + 1 . We will show that most of the statements of the ”Depth Lemma” are wrong if we replacedepth by sdepth. We first observe
Lemma 2.2.
Let → U f −→ M g −→ N → be an exact sequence of finitely generated Z n -graded S -modules. Then sdepth M ≥ min { sdepth U, sdepth N } Proof.
Let D : U = L ri =1 u i K [ Z i ] be a Stanley decomposition of U with sdepth( D ) =sdepth U and let D ′ : N = L sj =1 n j K [ Z ′ j ] be a Stanley decomposition of N with sdepth( D ′ )= sdepth N . Since f is injective map, we may suppose that f is an inclusion. Let n ′ j ∈ M be a Z n homogeneous element such that g ( n ′ j ) = n j . Clearly, M = P ri =1 u i K [ Z i ] + P sj =1 n ′ j K [ Z ′ j ] . We prove that the sum P ri =1 u i K [ Z i ] + P sj =1 n ′ j K [ Z ′ j ] is direct. Set V = P j n ′ j K [ Z ′ j ]. Since the exact sequence splits as linear spaces we see that U ∩ V = { } .Clearly D is already a Stanley decomposition of U and remains to show only that if y ∈ n ′ j K [ Z ′ j ] ∩ P s k =1 k = j n ′ k K [ Z ′ k ] then y = 0. As g ( y ) ∈ n j K [ Z ′ j ] ∩ P s k =1 k = j n k K [ Z ′ k ] = { } , wesee that y ∈ U , that is y ∈ U ∩ V = { } . (cid:3) Corollary 2.3.
Let (0) = M ⊂ M ⊂ . . . ⊂ M r − ⊂ M r = M be an ascending chain of Z n -graded submodules of M . Then sdepth M ≥ min { sdepth M i /M i − : i ∈ { , . . . , r }} (3) for all i ∈ [ r ] . roof. We consider the exact sequence of Z n -graded submodules of M such that0 → M i − → M i → M i /M i − → . By Lemma 2.2, we get sdepth M i ≥ min { sdepth M i − , sdepth M i /M i − } . We apply induc-tion to prove the inequality (3). For i = 1 this holds clearly. We suppose (3) is true for i = t then we have sdepth M t ≥ min { sdepth M i /M i − : i ∈ { , . . . , t }} . Let i = t + 1 then we have sdepth M t +1 ≥ min { sdepth M t , sdepth M t +1 /M t } , which isenough. (cid:3) The analogue of Lemma 2.1( a ) only holds under an additional assumption. Corollary 2.4.
In the hypothesis of Lemma 2.2 suppose that sdepth
M < sdepth N . Then sdepth M ≥ sdepth U .Proof. If sdepth
M < sdepth U , we get sdepth M < min { sdepth U, sdepth N } contradict-ing Lemma 2.2. (cid:3) The analogue of 2.1( b ) is wrong. Example 2.5.
Let S = K [ x, y, z ], M = ( x, y, z ). In the exact sequence 0 → M → S → K →
0, we have sdepth S = 3 > sdepth K = 0 but sdepth M = 2 = sdepth K + 1.Note that the case treated in Proposition 1.10, that is the short exact sequence 0 → M x k −→ M → M/x k M → , and Lemma 2.7 apparently hints that some analogue of ( b )from ”Depth Lemma” in the frame of sdepth might be true. Unfortunately, this is not thecase as shows the following: Example 2.6.
We have a resolution 0 → Ω m → S → m →
0, where S = K [ x, y, z ] and m = ( x, y, z ). Then Ω m is not free because otherwise proj dim S m should be 1, which isnot true. If sdepth Ω m = 3 then follows Ω m free by the elementary Lemma 2.9. Thussdepth Ω m ≤ m .However it remains still the problem in general that if for an exact sequence 0 → U → M → N →
0, sdepth
M > sdepth N implies sdepth U ≥ sdepth N + 1. In general this isfalse (see Example 2.6) but we prove this result in a special case. Lemma 2.7. If I ⊂ S = K [ x , . . . , x n ] is a monomial complete intersection, then sdepth I ≥ sdepth S/I + 1 . Proof.
Let { v , . . . , v m } be the regular sequence of monomials generating I . Since sdepth S/I = n − m , by applying [13, Theorem 1.1] recursively, and sdepth I ≥ n − m + 1, by [7], or[9, Proposition 3.4], it follows the desired result. (cid:3) In general for any monomial ideal the inequality in above lemma is still an open question.This inequality motivates that sdepth I ≥ sdepth J/I + 1 for any two monomial ideals I ⊂ J ⊂ S . But this inequality does not hold as shows the following example: Example 2.8.
Let S = K [ x, y ] , I = ( xy, y ) , J = I + ( x ) . Then we have sdepth J/I =1 = sdepth I = sdepth J . Lemma 2.9. If M is multigraded S -module, S = K [ x , . . . , x n ] with sdepth M = n then M is free.Proof. If sdepth M = n , then we have a Stanley decomposition of the form M = ⊕ i u i S and u i S are free S -modules. The direct sum is of linear spaces but it turns out to be offree S -modules. (cid:3) . The behavior of sdepth on algebra tensor product
Theorem 3.1.
Let I ⊂ S = K [ x , . . . , x n ] , J ⊂ S = K [ y , . . . , y m ] be monomial idealsand S = K [ x , . . . , x n , y , . . . , y m ] . Then sdepth S /I + sdepth S /J ≤ sdepth S/ ( IS, J S ) . Proof.
Let D : S /I = r M i =1 u i K [ Z i ]be a Stanley decomposition of S /I such that sdepth D = sdepth S /I and D : S /J = s M j =1 v j K [ W j ]be a Stanley decomposition of S /J such that sdepth D = sdepth S /J . Then we have S/IS = S [ y , . . . , y m ] /IS = ( S /I )[ y , . . . , y m ]= r M i =1 u i K [ Z i ][ y , . . . , y m ]= r M i =1 u i K [ Z i , y , . . . , y m ]and S/J S = S [ x , . . . , x n ] /J S = ( S /J )[ x , . . . , x n ]= s M j =1 v j K [ W j ][ x , . . . , x n ]= s M j =1 v j K [ W j , x , . . . , x n ] . We claim that S/ ( IS, J S ) = M i,j u i v j K [ Z i , W j ] . Let w ∈ ( IS, J S ) c = S/ ( IS, J S ) be a monomial; that is, w ∈ S and w ( IS, J S ). Wehave w IS and w J S . It follows that w ∈ ( IS ) c and w ∈ ( J S ) c . Hence there exist i and j such that w ∈ u i K [ Z i , y , . . . , y m ] and w ∈ v j K [ W j , x , . . . , x n ]. So we have w ∈ u i K [ Z i , y , . . . , y m ] ∩ v j K [ W j , x , . . . , x n ] and u i K [ Z i , y , . . . , y m ] ∩ v j K [ W j , x , . . . , x n ] = u i v j K [ Z i , W j ], since u i ∈ S and v j ∈ S .In order to prove the opposite inclusion, consider a monomial v ∈ u i v j K [ Z i , W j ]. Then v ∈ u i K [ Z i , y , . . . , y n ] ⊂ ( IS ) c and similarly v ∈ ( J S ) c . Thus v ∈ ( IS, J S ) c . So S/ ( IS, J S ) = P i,j u i v j K [ Z i , W j ].Now we prove that this sum is direct. Let i , i ∈ [ r ] and j , j ∈ [ s ] be such that( i , j ) = ( i , j ), let us say i = i . Then u i v j K [ Z i , W j ] ∩ u i v j K [ Z i , W j ] ⊂ u i K [ Z i , y , . . . , y m ] ∩ u i K [ Z i , y , . . . , y m ] = { } , which shows our claim. It followsthat sdepth S /I + sdepth S /J ≤ sdepth S/ ( IS, J S ) . (cid:3) he following example shows that the inequality in the above theorem can be strict. Example 3.2.
Let S = K [ x , x , x , x , x , x , x , x ] be a polynomial ring over the field K . Let I = ( x x , x x , x x , x x ) ⊂ S = K [ x , x , x , x ] be the ideal of the polynomialring S and J = ( x x , x x , x x , x x ) ⊂ S = K [ x , x , x , x ] be the ideal of thepolynomial ring S . Consider the ideal ( IS, J S ) ⊂ S , then a Stanley decomposition D of S/ ( IS, J S ) is D : S/ ( IS, J S ) = K [ x , x , x ] ⊕ x K [ x , x , x ] ⊕ x K [ x , x , x ] ⊕ x K [ x , x , x ] ⊕ x K [ x , x , x ] ⊕ x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x K [ x , x , x ] ⊕ x x x K [ x , x , x , x ] ⊕ x x x K [ x , x , x , x ] ⊕ x x x K [ x , x , x , x ] ⊕ x x x x K [ x , x , x , x ] , hence sdepth D = 3. Note that sdepth S /I = 1 with a Stanleydecomposition S /I = K [ x , x ] ⊕ x K [ x ] ⊕ x K [ x , x ]. We observe that sdepth S /I can not be greater than one. Similarly we have sdepth S /J = 1. Hence we obtain thatsdepth S /I + sdepth S /J < sdepth S/ ( IS, J S ).The following corollary is a particular case of [13, Theorem 1.1].
Corollary 3.3.
Let I ⊂ S be a monomial ideal and u ∈ S is a monomial, which is regularon S/I . Then sdepth S/ ( I, u ) ≥ sdepth S/I − .Proof. Renumbering x i ∈ supp( u j ) for all u j ∈ G ( I ), we may suppose that I is generatedby a monomial ideal J ⊂ S = K [ x , . . . , x r ] and u ∈ S = K [ x r +1 , . . . , x n ] for some1 < r < n . Then sdepth S/ ( I, u ) ≥ sdepth S /J + sdepth S / ( u ), by Theorem 3.1. Sincesdepth S / ( u ) = n − r − S/I = sdepth S /J + n − r , by [13, Lemma 1.2], itfollows that sdepth S/ ( I, u ) ≥ sdepth S/I − (cid:3) In the analogue of Theorem 3.1 for depth we have equality, that isdepth S/ ( IS, J S ) = depth S /I + depth S /J (see [15, Theorem 2.2.21]). We note that ifStanley’s conjecture hold for the modules S /I and S /J it holds also for S/ ( IS, J S ). References [1] I. Anwar, Janet’s algorithm, Bull. Math. Soc. Sc. Math. Roumanie (99), (2008), 11–19.[2] I. Anwar, D. Popescu, Stanley conjecture in small embedding dimension, J. Alg., (2007), 1027–1031.[3] W. Bruns, J. Herzog, Cohen Macaulay Rings , Revised Edition, Cambridge: Cambridge UniversityPress, 1996.[4] M. Cimpoeas, Stanley depth of complete intersection monomial ideals, Bull. Math. Soc. Sc. Math.Roumanie, (99)(2008), 205–211.[5] M. Cimpoeas, Stanley depth of monomial ideals in three variables. arXiv:math. AC/0807.2166,preprint (2008).[6] D. Eisenbud, Commutative algebra; with a view towards algebraic geometry , Graduate Text in Maths.,New York: Springer-Verlang, 1995.[7] J. Herzog, A. S. Jahan, S. Yassemi, Stanley decompositions and partitionable simplicial complexes.J. Alg. Comb., (1), (2007), 113–125.[8] J. Herzog, D. Popescu, Finite filtrations of modules and shellable multicomplexes, Manuscripta Math., (2006), 385–410.[9] J. Herzog, M. Vladoiu, X. Zheng, How to compute the Stanley depth of a monomial ideal, J. Alg., inpress.[10] A. S. Jahan, Prime filtrations of monomial ideals and polarizations, J. Alg., (2), (2007), 1011–1032.[11] S. Nasir, Stanley decompositions and localization, Bull. Math. Soc. Sc. Math. Roumanie, (99),(2008), 151–158.[12] D. Popescu, Stanley depth of multigraded modules, arXiv:math. AC/0801.2632, preprint (2008).
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Asia Rauf, Abdus Salam School of Mathematical Sciences, GC University, Lahore
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