On the Stanley depth of the path ideal of a cycle graph

0.1 MB·

by  Mircea Cimpoeas

On the Stanley depth of the path ideal of a cycle graph
Mircea Cimpoea¸s
Abstract
We give tight bounds for the Stanley depth of the quotient ring of the path ideal
of a cycle graph. In particular, we prove that it satisfies the Stanley inequality.
Keywords: Stanley depth, cycle graph, path ideal.
2010 Mathematics Subject Classification: 13C15, 13P10, 13F20.
6
1
0
Introduction
2

n
a Let K be a field and S = K[x ,...,x ] the polynomial ring over K. Let M be a Zn-graded
1 n
3 J S-module. A Stanley decomposition of M is a direct sum D : M = ri=1miK[Zi] as a
Zn-graded K-vector space, where m ∈ M is homogeneous with respect to Zn-grading,
i
L
] Z ⊂ {x ,...,x } such that m K[Z ] = {um : u ∈ K[Z ]} ⊂ M is a free K[Z ]-submodule
C i 1 n i i i i i
of M. We define sdepth(D) = min |Z | and sdepth(M) = max{sdepth(D)| D is a
A i=1,...,r i
Stanley decomposition of M}. The number sdepth(M) is called the Stanley depth of M.
.
h
Herzog, Vladoiu and Zheng show in [10] that sdepth(M) can be computed in a finite
t
a
number of steps if M = I/J, where J ⊂ I ⊂ S are monomial ideals. In [13], Rinaldo give a
m
computer implementation for this algorithm, in the computer algebra system CoCoA [6]. In
[
[2], J. Apel restated a conjecture firstly given by Stanley in [14], namely that sdepth(M) ≥

1
depth(M) for any Zn-graded S-module M. This conjecture proves to be false, in general,
v
1 for M = S/I and M = J/I, where 0 6= I ⊂ J ⊂ S are monomial ideals, see [7]. For a
6
friendly introduction in the thematic of Stanley depth, we refer the reader [11].
2
0 Let ∆ ⊂ 2[n] be a simplicial complex. A face F ∈ ∆ is called a facet, if F is maximal
0
with respect to inclusion. We denote F(∆) the set of facets of ∆. If F ∈ F(∆), we denote
.
1
x = x . Then the facet ideal I(∆) associated to ∆ is the squarefree monomial ideal
0 F j∈F j
6 I = (xF : F ∈ F(∆)) of S. The facet ideal was studied by Faridi [8] from the depth
Q
1 perspective.
:
v Theline graph oflenght n,denotedbyL ,isagraphwiththevertex set V = [n] andthe
n
i
X edge set E = {{1,2},{2,3},...,{n−1,n}}. Let ∆ be the simplicial complex with the
n,m
r set of facets F(∆ ) = {{1,2,...,m},{2,3,...,m+1},...,{n−m+1,n−m+2,...,n}},
a n,m
where 1 ≤ m ≤ n. We denote I = (x x ···x ,x x ···x ,...,x x ···x )
n,m 1 2 m 2 3 m+1 n−m+1 n−m+2 n
, the associated facet ideal. Note that I is the m-path ideal of the graph L , provided
n,m n
with the direction given by 1 < 2 < ... < n, see [9] for further details.
According to [9, Theorem 1.2],
2(n−d), n ≡ d(mod (m+1)) with 0 ≤ d ≤ m−1,
pd(S/I ) = m+1
n,m (2n−m+1, n ≡ m(mod (m+1)).
m+1
1The support from grant ID-PCE-2011-1023 of Romanian Ministry of Education, Research and Inno-
vation is gratefully acknowledged.
1
ByAuslander-Buchsbaum formula(see[15]),itfollowsthatdepth(S/I ) = n−pd(S/I )
n,m n,m
and, by a straightforward computation, we can see depth(S/I ) = n + 1 − n+1 −
n,m m+1
n+1 =: ϕ(n,m). We proved in [5] that sdepth(S/I ) = ϕ(n,m).
m+1 n,m (cid:4) (cid:5)
The cycle graph of lenght n, denoted by C , is a graph with the vertex set V =
n
(cid:6) (cid:7)
[n] and the edge set E = {{1,2},{2,3},...,{n − 1,n},{n,1}}. Let ∆¯ be the sim-
n,m
¯
plicial complex with the set of facets F(∆ ) = {{1,2,...,m},{2,3,...,m + 1},··· ,
n,m
{n − m + 1,n − m + 2,...,n},{n − m + 2,...,n,1},...,{n,1,...,m − 1}}. We denote
J = (x x ···x ,x x ···x ,...,x x ···x ,...,x x ···x ), the associ-
n,m 1 2 m 2 3 m+1 n−m+1 n−m+2 n n 1 m−1
ated facet ideal. Note that J is the m-path ideal of the graph C .
n,m n
Let p = n and d = n−(m+1)p. According to [1, Corollary 5.5],
m+1
(cid:4) (cid:5)
2p+1, d 6= 0,
pd(S/J ) =
n,m
(2p, d = 0.
By Auslander-Buchsbaum formula, it follows that depth(S/J ) = n − pd(S/J ) =
n,m n,m
n − n − n =: ψ(n,m). Note that ψ(n,m) = ϕ(n − 1,m). Our main result is
m+1 m+1
Theorem 1.4, in which we prove that ϕ(n,m) ≥ sdepth(S/J ) ≥ ψ(n,m). We also prove
n,m
(cid:4) (cid:5) (cid:6) (cid:7)
that, sdepth(J /I ) = depth(J /I ) = ψ(n,m)+m−1, see Proposition 1.6. These
n,m n,m n,m n,m
results generalize [4, Theorem 1.9] and [4, Proposition 1.10].
1 Main results
First, we recall the well known Depth Lemma, see for instance [15, Lemma 1.3.9].
Lemma 1.1. (Depth Lemma) If 0 → U → M → N → 0 is a short exact sequence of
modules over a local ring S, or a Noetherian graded ring with S local, then
0
a) depthM ≥ min{depthN,depthU}.
b) depthU ≥ min{depthM,depthN +1}.
c) depthN ≥ min{depthU −1,depthM}.
In [12], Asia Rauf proved the analog of Lemma 1.1(a) for sdepth:
Lemma 1.2. Let 0 → U → M → N → 0 be a short exact sequence of Zn-graded S-
modules. Then: sdepth(M) ≥ min{sdepth(U),sdepth(N)}.
The following result is well known. However, we present an original proof.
Lemma 1.3. Let I ⊂ S be a nonzero proper monomial ideal. Then, I is principal if and
only if sdepth(S/I) = n−1.
Proof. Assume sdepth(S/I) = n−1 and let S/I = r u K[Z ] be a Stanley decomposi-
i=1 i i
tion with |Z | = n−1 for all i, and u ∈ S monomials. Since 1 ∈/ I, we may assume that
i i
L
u = 1. Let x be the variable which is not in Z . If x ∈ I, since S/(x ) = K[Z ] and
1 j1 1 j1 j1 1
K[Z ] ⊂ S/I, then I = (x ). Otherwise, we may assume that u = x .
1 j1 2 j1
2
Let x be the variable which is not in Z . If x x ∈ I, then, one can easily see that
j2 2 j1 j2
I = (x x ). If x x ∈/ I, then we may assume u = x x and so on. Thus, we have
j1 j2 j1 j2 3 j1 j2
u = x ···x , for all 1 ≤ i ≤ r+1, where x is the variable which is not in Z . Moreover,
i j1 ji−1 ji i
I = (u ), and therefore I is principal.
r+1
In order to prove theother implication, assume that I = (u)andwrite u = r x . We
i=1 ji
letu = i−1 x andZ = {x ,...,x }\{x },forall1 ≤ i ≤ r.Then,S/I = r u K[Z ]
i k=1 jk i 1 n ji Qi=1 i i
is a Stanley decomposition with |Z | = n−1 for all i. Therefore sdepth(S/I) = n−1.
i
Q L
Our main result, is the following Theorem.
Theorem 1.4. ϕ(n,m) ≥ sdepth(S/J ) ≥ depth(S/J ) = ψ(n,m).
n,m n,m
Proof. If n = m, then J = (x ...x ) is a principal ideal, and, according to Lemma 1.3
n,n 1 n
we are done. Also, if m = 1, then J = (x ,...,x ) and so there is nothing to prove, since
n,1 1 n
S/J = K. The case m = 2 follows from [4, Proposition 1.8] and [4, Theorem 1.9].
n,1
Assume n > m ≥ 3. If n = m + 1, then we consider the short exact sequence
0 → S/(J : x ) → S/J → S/(J ,x ) → 0. Note that (J : x ) =
n,n−1 n n,n−1 n,n−1 n n,n−1 n
∼
(x ···x ,x ···x ,x ···x x ,··· ,x x ···x ) = J S. Therefore, by in-
1 n−2 2 n−1 3 n−1 1 n−1 1 n−3 n−1,n−2
duction hypothesis and [10, Lemma 3.6],
sdepth(S/(J : x )) = depth(S/(J : x )) = 1+ψ(n−1,n−2) = n−2.
n,n−1 n n,n−1 n
∼
Also,(J ,x ) = (x ···x ,x )andthusS/(J ,x ) = K[x ,...,x ]/(x ···x ).
n,n−1 n 1 n−1 n n,n−1 n 1 n−1 1 n−1
Therefore, by Lemma 1.3, we have sdepth(S/(J ,x )) = n−2 = depth(S/(J ,x )).
n,n−1 n n,n−1 n
Now, assume n > m + 1 > 3. We consider the ideals L = J , L = (L : x )
0 n,m k+1 k n−k
and U = (L ,x ), for 0 ≤ k ≤ m − 2. Note that L = (J : x ···x ) =
k k n−k m−1 n,m n−m+2 n
(x ,x ···x ,...,x ···x ,x ).
1 2 m+1 n−2m+1 n−m n−m+1
Ifn−2m ≤ 2,thenL = (x ,x )andthussdepth(S/L ) = depth(S/L ) =
m−1 1 n−m+1 m−1 m−1
n − 2 = ϕ(n,m), since n+1 = 1 and n+1 = 2. If n − 2m > 2, then S/L ∼=
m+1 m+1 m−1
K[x ,...,x ,x ,...,x ]/(x ···x ,...,x ···x ) and therefore, by [10,
2 n−m n−m+2 n 2 m+1 n−2m+1 n−m
(cid:4) (cid:5) (cid:6) (cid:7)
Lemma 3.6] and [5, Theorem 1.3], we have sdepth(S/L ) = depth(S/L ) = n −
m−1 m−1
1 − n−m − n−m = ϕ(n,m). On the other hand, for example by [3, Proposition 2.7],
m+1 m+1
sdepth(S/L ) ≥ sdepth(S/J ). Thus, sdepth(S/J ) ≤ ϕ(n,m).
m−1 n,m n,m
(cid:4) (cid:5) (cid:6) (cid:7)
For any 0 < k < m, we have L = (x ···x ,x ···x ,...,x ···x ,
k 1 m−k 2 m+1 n−m−k n−k−1
x ···x ,x ···x x ,...,x x ···x ). Therefore, U = (x ···x ,
n−m+1 n−k n−m+2 n−k 1 n−k 1 m−k−1 k 1 m−k
x ···x ,...,x ···x ,x ), for k ≤ m−2. We consider two cases:
2 m+1 n−m−k n−k−1 n−k
(i) If n−m−k < 2 and 0 ≤ k ≤ m−2, then U = (x ···x ,x ) and therefore
k 1 m−k n−k
sdepth(S/U ) = depth(S/U ) = n−2 = ϕ(n,m), since n+1 = 1 and n+1 = 2.
k k m+1 m+1
(ii) If n − m − k ≥ 2, then, for any 0 ≤ j ≤ k ≤ m − 2, we consider the ideals
(cid:4) (cid:5) (cid:6) (cid:7)
V := (x ···x ,x ···x ,...,x ···x ) in S := K[x ,...,n ]. Note
k,j 1 m−j 2 m+1 n−m−k n−k−1 k 1 n−k−1
∼
that S/U = (S /V )[x ,...,x ] and thus, by [10, Lemma 3.6], depth(S/U ) =
k k k,k n−k+1 n k
depth(S /V )+k and sdepth(S/U ) = sdepth(S /V )+k.
k k,k k k k,k
For any 0 ≤ j < k ≤ m−2, we claim that V /V is isomorphic to
k,j k,j+1
(K[x ,...,x ]/(x ···x ,...,x ···x ))[x ,...,x ].
m−j+2 n−k−1 m−j+2 2m−j+1 n−m−k n−k−1 1 m−j
3
Indeed, if u ∈ V \ V is a monomial, then x ···x |u and x ∤ u. Also,
k,j k,j+1 1 m−j m−j+1
x ···x ∤ u, ..., x ···x ∤ u. Denoting v = u/(x ···x ), we can
m−j+2 2m−j+1 n−m−k n−k−1 1 m−j
writev = v′v′′,withv′ ∈ K[x ,...,x ]\(x ···x ,...,x ···x )
m−j+2 n−k−1 m−j+2 2m−j+1 n−m−k n−k−1
and v′′ ∈ K[x ,...,x ].
1 m−j
By [10, Lemma 3.6] and [5, Theorem 1.3], sdepth(V /V ) = depth(V /V ) =
k,j k,j+1 k,j k,j+1
m − j + ϕ(n − k − m + j − 2,m) = n − k − 1 − n−m−1−k+j − n−m−1−k+j = n −
m+1 m+1
k + 1 − n−k+j − n−k+j ≥ ϕ(n,m) − k. On the other hand, V = I for any
m+1 m+1 (cid:4) (cid:5) k,0(cid:6) n−k−1,m(cid:7)
0 ≤ k ≤ m − 2 and therefore, by [5, Theorem 1.3], sdepth(S/V ) = depth(S/V ) =
k,0 k,0
(cid:4) (cid:5) (cid:6) (cid:7)
ϕ(n − k − 1,m) = n − k − n−k − n−k ≥ ϕ(n,m) − k, for any k ≥ 1. From the
m+1 m+1
short exact sequences 0 → V /V → S/V → S/V → 0, 0 ≤ j < k, Lemma 1.1
k,j k,j+1 k,j+1 k,j
(cid:4) (cid:5) (cid:6) (cid:7)
and Lemma 1.2, it follows that sdepth(S/V ) ≥ depth(S/V ) = ϕ(n,m)−k, for all
k,j+1 k,j+1
0 ≤ j < k ≤ m−2. Thus sdepth(S/U ) ≥ depth(S/U ) ≥ ϕ(n,m),forall0 < k ≤ m−2.On
k k
the other hand, sdepth(S/V ) = depth(S/V ) = ϕ(n−1,m) = ψ(n,m) −m, and thus
0,0 0,0
sdepth(S/U ) = depth(S/U ) = ψ(n,m).
0 0
Now, we consider short exact sequences
0 → S/L → S/L → S/U → 0. for 0 ≤ k < m.
k+1 k k
By Lemma 1.1 and Lemma 1.2 we get sdepth(S/L ) ≥ depth(S/L ) = ϕ(n,m), for any
k k
0 < k ≤ m−2, and sdepth(S/L ) ≥ depth(S/L ) = ψ(n,m).
0 0
Corollary 1.5. If n+1 = n and n+1 = n , then
m+1 m+1 m+1 m+1
(cid:4) sd(cid:5)epth(cid:4)(S/J(cid:5)n,m) =(cid:6)dept(cid:7)h(S/(cid:6)Jn,m(cid:7)) = ϕ(n,m).
Proposition 1.6. sdepth(J /I ) ≥ depth(J /I ) = ψ(n,m)+m−1.
n,m n,m n,m n,m
Proof. We claim that J /I is isomorphic to
n,m n,m
xn−m+2···xnx1(K[x2,...,xn−m]/(x2···xm,x3···xm+2...,xn−2m+1···xn−m))[xn−m+2,...,xn,x1]⊕
⊕xn−m+3···xnx1x2(K[x3,...,xn−m+1]/(x3···xm,x4···xm+3,...,xn−2m+2···xn−m+1))[xn−m+3,...,xn,x1,x2]⊕
···⊕xnx1···xm−1(K[xm,...,xn−2]/(xm,xm+1···x2m,...,xn−m−1···xn−2))[xn,x1...,xm−1].
Indeed, let u ∈ J \ I be a monomial. If x ···x x |u, then x ∤ u and
n,m n,m n−m+2 n 1 n−m+1
x ···x ∤ u. It follows that:
2 m
u∈xn−m+2···xnx1(K[x2,...,xn−m]/(x2···xm,x3···xm+2...,xn−2m+1···xn−m))[xn−m+2,...,xn,x1].
If x ···x x ∤ u and x ···x x x |u then x ∤ u and x ···x ∤ u. Thus:
n−m+2 n 1 n−m+3 n 1 2 n−m+2 3 m
u∈xn−m+3···xnx1x2(K[x3,...,xn−m+1]/(x3···xm,x4···xm+3,...,xn−2m+2···xn−m+1))[xn−m+3,...,xn,x1,x2].
Finally,ifx ···x x ∤ u,..., x x x ···x ∤ uandx x ···x |u,thenitfollows
n−m+2 n 1 n−1 n 1 m−2 n 1 m−1
that x ∤ u and x ∤ u. Therefore:
n−1 m
u∈/ xnx1···xm−1(K[xm,...,xn−2]/(xm,xm+1···x2m,...,xn−m−1···xn−2))[xn,x1...,xm−1].
AsintheproofofTheorem3.1(seethecomputationsforV ’s),byapplyingLemma1.1and
k,j
Lemma 1.2, it follows that sdepth(J /I ) ≥ depth(J /I ) = ϕ(n−m−2,m)+m =
n,m n,m n,m n,m
ψ(n,m)+m−1, as required.
Inspiredby[4,Conjecture1.12]andcomputerexperiments [6],weproposethefollowing:
Conjecture 1.7. For any n ≥ 3(m+1)+1, we have sdepth(S/J ) = ϕ(n,m).
n,m
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Mircea Cimpoea¸s, Simion Stoilow Institute of Mathematics, Research unit 5, P.O.Box 1-764,
Bucharest 014700,Romania, E-mail: [email protected]
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