STANLEY DEPTH AND SYMBOLIC POWERS OF MONOMIAL IDEALS
S. A. SEYED FAKHARI∗
Abstract
The aim of this paper is to study the Stanley depth of symbolic powers of a squarefree monomial ideal. We prove that for every squarefree monomial idealI and every pair of integersk, s ≥ 1, the inequalities sdepth(S/I(ks))≤sdepth(S/I(s))and sdepth(I(ks))≤sdepth(I(s))hold. If moreoverIis unmixed of heightd, then we show that for every integerk≥1, sdepth(I(k+d))≤ sdepth(I(k))and sdepth(S/I(k+d))≤sdepth(S/I(k)). Finally, we consider the limit behavior of the Stanley depth of symbolic powers of a squarefree monomial ideal. We also introduce a method for comparing the Stanley depth of factors of monomial ideals.
1. Introduction
LetKbe a field andS=K[x1, . . . , xn] be the polynomial ring innvariables over the fieldK. LetM be a nonzero finitely generatedZn-gradedS-module.
Letu∈Mbe a homogeneous element andZ⊆ {x1, . . . , xn}. TheK-subspace uK[Z] generated by all elementsuvwithv∈K[Z] is called aStanley space of dimension |Z|, if it is a free K[Z]-module. Here, as usual, |Z| denotes the number of elements of Z. A decomposition D of M as a finite direct sum of Stanley spaces is called aStanley decompositionofM. The minimum dimension of a Stanley space inD is called theStanley depth of D and is denoted by sdepth(D). The quantity
sdepth(M):=max
sdepth(D)|D is a Stanley decomposition ofM is called theStanley depthofM. Stanley [10] conjectured that
depth(M)≤sdepth(M)
for allZn-gradedS-modulesM. As a convention, we set sdepth(M)=0, when Mis the zero module. For a reader friendly introduction to the Stanley depth, we refer to [8] and for a nice survey on this topic we refer to [3].
∗This research was in part supported by a grant from IPM (No. 93130422).
Received 24 February 2014.
DOI: https://doi.org/10.7146/math.scand.a-25501
In this paper, we generalize the technique which was used in [9] to introduce a method for comparing the Stanley depth of factors of monomial ideals (see Theorem 2.1). We show that our method implies the known results regarding the Stanley depth of radical, integral closure and colon of monomial ideals (see Propositions 2.2, 2.3, 2.4 and 2.5).
In Section 3, we apply our method to study the Stanley depth of symbolic powers of squarefree monomial ideals. We show that for every pair of integers k, s≥1 the Stanley depth of thekth symbolic power of a squarefree monomial idealI is an upper bound for the Stanley depth of the(ks)th symbolic power ofI (see Theorem 3.2). If moreoverI is unmixed of heightd, then we show that for every integerk ≥1, the Stanley depth of thekth symbolic power ofIis an upper bound for the Stanley depth of the(k+d)th symbolic power ofI(see Theorem 3.7). Finally, in Theorem 3.10 we show that the limit behavior of the Stanley depth of unmixed squarefree monomial ideals can be very interesting.
Indeed, we show that there exist finite setsL1andL2such that sdepth(S/I(k))∈ L1and sdepth(I(k))∈L2, for everyk0.
2. A comparison tool for the Stanley depth
The following theorem is the main result of this section. Using this result, we deduce some known results regarding the Stanley depth of the radical, the integral closure and the colon of monomial ideals. We should mention that in the following theorem we use Mon(S)to denote the set of all monomials in the polynomial ringS.
Theorem2.1.LetI2 I1andJ2 J1be monomial ideals inS. Assume that there exists a functionφ: Mon(S) → Mon(S), such that the following conditions are satisfied:
(i) for every monomialu∈Mon(S),u∈I1if and only ifφ(u)∈J1; (ii) for every monomialu∈Mon(S),u∈I2if and only ifφ(u)∈J2; (iii) for every Stanley spaceuK[Z] ⊆Sand every monomialv ∈Mon(S),
v∈uK[Z]if and only ifφ(v)∈φ(u)K[Z].
Then
sdepth(I1/I2)≥sdepth(J1/J2).
Proof. Consider a Stanley decomposition D:J1/J2=
m i=1
tiK[Zi]
ofJ1/J2, such that sdepth(D)=sdepth(J1/J2). By our assumptions, for every monomialu∈I1\I2, we have
φ(u)∈J1\J2.
Then for each monomialu∈I1\I2, we defineZu :=Zi andtu :=ti, where i∈ {1, . . . , m}is the uniquely determined index, such thatφ(u)∈tiK[Zi]. It is clear that
I1\I2⊆
uK[Zu],
where the sum asK-vector space is taken over all monomialsu∈I1\I2. For the converse inclusion note that for every u ∈ I1\I2 and every monomial h∈K[Zu], we clearly haveuh∈I1. By the choice oftuandZu, we conclude thatφ(u)∈tuK[Zu] and therefore, by (iii),
φ(uh)∈φ(u)K[Zu]⊆tuK[Zu].
This implies thatφ(uh) /∈J2and it follows from (ii) thatuh /∈I2. Thus I1/I2=
uK[Zu], where the sum is taken over all monomialsu∈I1\I2.
Now for every 1≤i≤m, let
Ui = {u∈I1\I2:uis a monomial,Zu=Zi andtu =ti}.
Without loss of generality we may assume thatUi = ∅for every 1 ≤i ≤ andUi = ∅for every+1≤i≤m. Note that
I1/I2=
i=1
uK[Zi],
where the second sum is taken over all monomialsu∈Ui. For every 1≤i≤, letuibe the greatest common divisor of elements ofUi. We claim that for every 1≤i≤, we haveui ∈Ui.
Proof of claim. It is enough to show thatφ(ui)∈tiK[Zi]. This, together with (i) and (ii) implies thatui ∈I1\I2,Zui =Zi,tui =ti and henceui ∈Ui. So assume thattidoes not divideφ(ui). Then there exists 1≤j ≤n, such that degxj(φ(ui)) <degxj(ti), where for every monomialv∈S, degxj(v)denotes the degree ofvwith respect to the variablexj. Also by the choice ofui, there exists a monomialu∈Ui, such that degx
j(u)=degx
j(ui). We conclude that u∈uiK[x1, . . . , xj−1, xj+1, . . . , xn],
and hence by (iii) that
φ(u)∈φ(ui)K[x1, . . . , xj−1, xj+1, . . . , xn]. This shows that
degxj(φ(u))=degxj(φ(ui)) <degxj(ti).
It follows thatti does not divideφ(u), which is a contradiction, sinceφ(u)∈ tiK[Zi]. Henceti dividesφ(ui). On the other hand, since ui divides every monomialu ∈ Ui, (iii) implies that for every monomial u ∈ Ui, φ(ui)di- videsφ(u). Note that by the definition ofUi, for every for every monomial u∈Ui,φ(u)∈tiK[Zi]. It follows that
φ(ui)∈tiK[Zi] and this completes the proof of our claim.
Our claim implies that for every 1≤i≤, we have uiK[Zi]⊆
u∈Ui
uK[Zi].
On the other hand (iii) implies that, for every monomialu∈Ui,φ(ui)divides φ(u). Since
φ(ui)∈tiK[Zi] and φ(u)∈tiK[Zi], we conclude that
φ(u)∈φ(ui)K[Zi] and it follows from (iii) that
u∈uiK[Zi] and thus
uiK[Zi]=
u∈Ui
uK[Zi]. Therefore
I1/I2=
i=1
uiK[Zi].
Next we prove that for every 1 ≤ i, j ≤ with i = j, the summands uiK[Zi] andujK[Zj] intersect trivially. For a contradiction, letvbe a monomial inuiK[Zi]∩ujK[Zj]. Then there existhi ∈K[Zi] andhj ∈K[Zj] such that
uihi = v = ujhj. Thereforeφ(uihi) = φ(v) = φ(ujhj). Butui ∈ Ui and henceφ(ui)∈tiK[Zi], which by (iii) implies that
φ(uihi)∈φ(ui)K[Zi]⊆tiK[Zi]. Similarlyφ(ujhj)∈tjK[Zj]. Thus
φ(v)∈tiK[Zi]∩tjK[Zj], which is a contradiction, becausem
i=1tiK[Zi] is a Stanley decomposition of J1/J2. Therefore
I1/I2=
i=1
uiK[Zi]
is a Stanley decomposition ofI1/I2which proves that sdepth(I1/I2)≥min
i=1 |Zi| ≥sdepth(J1/J2).
Using Theorem 2.1, we are able to deduce many known results regarding the Stanley depth of factors of monomial ideals. For example, it is known that the Stanley depth of the radical of a monomial idealI is an upper bound for the Stanley depth ofI. In the following proposition we show that this result follows from Theorem 2.1.
Proposition2.2 (see [1], [6]).LetJ I be monomial ideals inS such that√
I =√ J. Then
sdepth(I /J )≤sdepth√ I /√
J . Proof. LetG(√
I ) = {u1, . . . , us}be the minimal set of monomial gen- erators of√
I. For every 1 ≤ i ≤ s, there exists an integerki ≥1 such that ukii ∈I. LetkI =lcm(k1, . . . , ks)be the least common multiple ofk1, . . . , ks. Now for every 1≤ i ≤s, we haveukiI ∈I and this implies thatukI ∈I, for every monomialu∈√
I. It follows that for every monomialu∈S, we have u∈√
Iif and only ifukI ∈I. Similarly there exists an integerkJ, such that for every monomialu∈ S,u∈√
J if and only ifukJ ∈J. Letk = lcm(kI, kJ) be the least common multiple ofkI andkJ. For every monomialu∈ S, we defineφ(u) = uk. It is clear that φ satisfies the hypothesis of Theorem 2.1.
Hence it follows from Theorem 2.1 that sdepth(I /J )≤sdepth√
I /√ J
.
LetI ⊂Sbe an arbitrary ideal. An elementf ∈SisintegraloverIif there exists an equation
fk+c1fk−1+ · · · +ck−1f +ck =0 with ci ∈Ii.
The set of elementsIinSwhich are integral overIis theintegral closureofI. It is known that the integral closure of a monomial idealI ⊂Sis a monomial ideal generated by all monomialsu ∈ S for which there exists an integerk such thatuk ∈Ik(see [4, Theorem 1.4.2]).
Let I be a monomial ideal inS and let k ≥ 1 be a fixed integer. Then for every monomialu ∈ S, we have u∈ I if and only ifus ∈ Is, for some s ≥1, if and only ifuks ∈Iks, for somes ≥1, if and only ifuk ∈Ik. This shows that by settingφ(u)=ukin Theorem 2.1 we obtain the following result from [9]. We should mention that the method used in the proof of Theorem 2.1 is essentially a generalization of the one used in [9].
Proposition2.3 ([9, Theorem 2.1]).LetJ Ibe two monomial ideals in Ssuch thatI =J. Then for every integerk ≥1,
sdepth(Ik/Jk)≤sdepth(I /J ).
Similarly, using Theorem 2.1 we can deduce the following result from [9].
Proposition2.4 ([9, Theorem 2.8]).LetI2 I1be two monomial ideals inSsuch thatI1 =I2. Then there exists an integerk ≥1, such that for every s≥1,
sdepth(I1sk/I2sk)≤sdepth(I1/I2).
Proof. Note that by [9, Remark 1.1], there exist integersk1, k2 ≥1, such that for every monomialu∈S, we haveuk1 ∈I1k1(resp.uk2 ∈I2k2) if and only ifu∈I1(resp.u∈I2). Letk= lcm(k1, k2)be the least common multiple of k1andk2. Then for every monomialu∈S, we haveuk ∈I1k (resp.uk ∈ I2k) if and only ifu ∈ I1 (resp.u ∈ I2). Hence for every monomial u ∈ S and everys ≥1, we haveusk ∈I1sk (resp.usk ∈ I2sk) if and only ifu∈ I1(resp.
u∈I2). Setφ(u)=usk, for every monomialu∈Sand everys ≥1. Now the assertion follows from Theorem 2.1.
LetIbe a monomial ideal inSandv ∈Sbe a monomial. It can be easily seen that(I :v)is a monomial ideal. Popescu [7] proves that sdepth(I :v)≥ sdepth(I ). On the other hand, Cimpoea¸s [2] proves that sdepth(S/(I :v))≥ sdepth(S/I ). Using Theorem 2.1, we prove a generalization of these results.
Proposition2.5.LetJ I be monomial ideals inS and letv ∈ Sbe a monomial such that(I :v) =(J :v). Then
sdepth(I /J )≤sdepth((I :v)/(J :v)).
Proof. It is enough to use Theorem 2.1 setting φ(u) = vu, for every monomialu∈S.
3. Stanley depth of symbolic powers
LetIbe a squarefree monomial ideal inSand suppose thatIhas the irredundant primary decomposition
I =ᒍ1∩. . .∩ᒍr,
where everyᒍi is an ideal ofS generated by a subset of the variables ofS. Letk be a positive integer. Thekthsymbolic power ofI, denoted byI(k), is defined to be
I(k)=ᒍk1∩. . .∩ᒍkr.
As a convention, we define thekth symbolic power ofSto be equal toS, for everyk≥1.
We now use Theorem 2.1 to compare the Stanley depth of symbolic powers of squarefree monomial ideals.
Theorem 3.1.Let J ⊆ I be squarefree monomial ideals inS. Then for every pair of integersk, s≥1
sdepth(I(ks)/J(ks))≤sdepth(I(s)/J(s)).
Proof. Suppose thatI = ri=1ᒍi is the irredundant primary decomposi- tion ofI and letu∈Sbe a monomial. Thenu∈I(s) if and only if for every
1≤i≤r
xj∈Pi
degxju≥s
if and only if
xj∈Pi
degxj uk ≥sk
if and only if uk ∈ I(sk). By a similar argument, u ∈ J(s) if and only if uk ∈ J(sk). Thus for proving our assertion, it is enough to use Theorem 2.1, settingφ(u)=uk, for every monomialu∈S.
The following corollary is an immediate consequence of Theorem 3.1.
Corollary3.2.LetIbe a squarefree monomial ideal inS. Then for every pair of integersk, s ≥1, the inequalities
sdepth(S/I(ks))≤sdepth(S/I(s)) and
sdepth(I(ks))≤sdepth(I(s)) hold.
Remark 3.3. Let t ≥ 1 be a fixed integer. Also let I be a squarefree monomial ideal inSand suppose thatI = ri=1ᒍi is the irredundant primary decomposition ofI. Assume thatA⊆ {x1, . . . , xn}is a subset of variables of S, such that
|ᒍi ∩A| =t,
for every 1 ≤ i ≤ r. We setv = xi∈Axi. It is clear that for every integer k ≥1 and every integer 1 ≤ i ≤ r, a monomialu∈ Mon(S)belongs toᒍki if and only ifuvbelongs toᒍk+ti . This implies that for every integerk ≥ 1, a monomialu∈Mon(S)belongs toI(k)if and only ifuvbelongs toI(k+t). This shows
(I(k+t) :v)=I(k) and thus Proposition 2.5 implies that
sdepth(I(k+t))≤sdepth(I(k)) and sdepth(S/I(k+t))≤sdepth(S/I(k)).
In particular, we conclude the following result.
Proposition3.4.LetI be a squarefree monomial ideal inSand suppose there exists a subsetA⊆ {x1, . . . , xn}of variables ofS, such that for every prime idealᒍ∈Ass(S/I ),
|ᒍ∩A| =1. Then for every integerk ≥1, the inequalities
sdepth(I(k+1))≤sdepth(I(k)) and
sdepth(S/I(k+1))≤sdepth(S/I(k)) hold.
As an example of ideals which satisfy the assumptions of Proposition 3.4, we consider the cover ideal of bipartite graphs. LetGbe a graph with vertex
setV (G)= {v1, . . . , vn}and edge setE(G). A subsetC⊆V (G)is aminimal vertex coverofGif, first, every edge ofGis incident with a vertex inCand, second, there is no proper subset ofC with the first property. For a graphG thecover idealofGis defined by
JG=
{vi,vj}∈E(G)
xi, xj.
For instance, unmixed squarefree monomial ideals of height two are just cover ideals of graphs. The name cover ideal comes from the fact thatJGis generated by squarefree monomialsxi1. . . xir with{vi1, . . . , vir}a minimal vertex cover ofG. A graphGisbipartiteif there exists a partitionV (G) = U ∪W with U∩W =∅such that each edge ofGis of the form{vi, vj}withvi ∈U and vj ∈W.
Corollary3.5.LetGbe a bipartite graph andJGbe the cover ideal ofG. Then for every integerk ≥1, the inequalities
sdepth(JG(k+1))≤sdepth(JG(k)) and sdepth(S/JG(k+1))≤sdepth(S/JG(k)) hold.
Proof. LetV (G)=U∪Wbe the partition for the vertex set of the bipartite graphG. Note that
Ass(S/JG)=
xi, xj:{vi, vj} ∈E(G) . Thus for everyᒍ∈Ass(S/JG), we have|ᒍ∩A| =1, where
A= {xi :vi ∈U}.
Now Proposition 3.4 completes the proof of the assertion.
It is known [5, Theorem 5.1] that for a bipartite graphGwith cover idealJG, we have JG(k) = JGk, for every integer k ≥ 1. Therefore we conclude the following result from Corollary 3.5.
Corollary3.6.LetGbe a bipartite graph andJGbe the cover ideal ofG. Then for every integerk ≥1, the inequalities
sdepth(JGk+1)≤sdepth(JGk) and sdepth(S/JGk+1)≤sdepth(S/JGk) hold.
LetGbe a non-bipartite graph and letJGbe its cover ideal. We do not know whether the inequalities
sdepth(JG(k+1))≤sdepth(JG(k)) and sdepth(S/JG(k+1))≤sdepth(S/JG(k))
hold for every integerk ≥ 1. However, we will see in Corollary 3.8 that we always have the following inequalities:
sdepth(JG(k+2))≤sdepth(JG(k)) and sdepth(S/JG(k+2))≤sdepth(S/JG(k)).
In fact, we can prove something stronger as follows.
Theorem3.7.LetIbe an unmixed squarefree monomial ideal and assume thatht(I )=d. Then for every integerk≥1the inequalities
sdepth(I(k+d))≤sdepth(I(k)) and
sdepth(S/I(k+d))≤sdepth(S/I(k)) hold.
Proof. LetA= {x1, . . . , xn}be the whole set of variables. Then for every prime idealᒍ∈Ass(S/I ), we have|ᒍ∩A| =d. Hence the assertion follows from Remark 3.3.
Since the cover ideal of every graphGis unmixed of height two, we con- clude the following result.
Corollary3.8. LetGbe an arbitrary graph andJG be the cover ideal ofG. Then for every integerk ≥1, the inequalities
sdepth(JG(k+2))≤sdepth(JG(k)) and
sdepth(S/JG(k+2))≤sdepth(S/JG(k)) hold.
Corollary3.9.Let I be an unmixed squarefree monomial ideal and as- sume thatht(I )=d. Then for every integer1≤≤dthe sequences
sdepth(S/I(kd+))
k∈Z≥0 and
sdepth(I(kd+))
k∈Z≥0
converge.
Proof. Note that by Theorem 3.7, the sequences sdepth(S/I(kd+))
k∈Z≥0 and
sdepth(I(kd+))
k∈Z≥0
are both nonincreasing and so stabilize.
We do not know whether the Stanley depth of symbolic powers of a square- free monomial ideal stabilizes. However, Corollary 3.9 shows that one can expect a nice limit behavior for the Stanley depth of symbolic powers of square- free monomial ideals. Indeed it shows that for unmixed squarefree monomial ideals of heightd, there exist two setsL1,L2of cardinalityd, such that
sdepth(S/I(k))∈L1 and sdepth(I(k))∈L2,
for everyk0. The following theorem shows that the situation is even better.
Theorem3.10.LetIbe an unmixed squarefree monomial ideal and assume thatht(I )=d. Suppose thatt is the number of positive divisors ofd. Then
(i) There exists a setL1of cardinalityt, such thatsdepth(S/I(k))∈L1, for everyk0.
(ii) There exists a setL2 of cardinalityt, such thatsdepth(I(k)) ∈ L2, for everyk0.
Proof. (i) Based on Corollary 3.9, it is enough to prove that for every couple of integers 1≤1, 2≤d, with gcd(d, 1)=2, we have
k→∞lim sdepth(S/I(kd+1))= lim
k→∞sdepth(S/I(kd+2)).
Setm=1/2. Then by Corollary 3.2,
k→∞lim sdepth(S/I(kd+2))≥ lim
k→∞sdepth(S/I(mkd+m2))
= lim
k→∞sdepth(S/I(mkd+1))
= lim
k→∞sdepth(S/I(kd+1)), where the last equality holds, because the sequence
sdepth(S/I(mkd+1))
k∈Z≥0
is a subsequence of the convergent sequence sdepth(S/I(kd+1))
k∈Z≥0.
On the other hand, since gcd(d, 1)= 2, there exists an integerm ≥ 1, such thatm1is congruent to 2 modulod. Now by a similar argument as above, we have
k→∞lim sdepth(S/I(kd+1))≥ lim
k→∞sdepth(S/I(mkd+m1))
= lim
k→∞sdepth(S/I(kd+2)), and hence
k→∞lim sdepth(S/I(kd+1))= lim
k→∞sdepth(S/I(kd+2)).
(ii) The proof is similar to the proof of (i).
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S. A. SEYED FAKHARI
SCHOOL OF MATHEMATICS, STATISTICS AND COMPUTER SCIENCE COLLEGE OF SCIENCE
UNIVERSITY OF TEHRAN TEHRAN
IRAN
E-mail:aminfakhari@ut.ac.ir
