The behavior of Stanley depth under polarization

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by  Bogdan Ichim

THE BEHAVIOR OF STANLEY DEPTH UNDER POLARIZATION
BOGDANICHIM,LUKASKATTHÄN,ANDJULIOJOSÉMOYANO-FERNÁNDEZ
4
1
0 ABSTRACT. Let K be a field, R=K[X1,...,Xn] be the polynomialring and J (I two
2 monomialidealsinR. Inthispaperweshowthat

p sdepth I/J−depth I/J=sdepth Ip/Jp−depth Ip/Jp,
e
S wheresdepthI/J denotestheStanleydepthandIp denotesthepolarization. Thissolves
aconjecturebyHerzog[Her13]andreducesthefamousStanleyconjecture(formodules
4
2 of the form I/J) to the squarefree case. As a consequence, the Stanley conjecture for
algebrasoftheformR/Iandthewell-knowncombinatorialconjecturethateveryCohen-

] Macaulaysimplicialcomplexispartitionableareequivalent.
C
A
.
h
t 1. INTRODUCTION
a
m
In 1982, R. Stanley conjectured in his celebrated paper [Sta82]an upperbound for the
[
depth of a multigraded module of combinatorial nature, called Stanley depth later on. A

3 proofofthisconjectureturnedouttobeadifficultproblem: itbegansoontobecalledthe
v
Stanley conjecture. Since then, several authors began to study intensively this problem,
9
0 startingwiththereformulationbyApel ofthemostimportantcases oftheconjecture, i.e.
3
the Stanley conjecture for a monomial ideal I and for the factor ring R/I, see [Ape03a,
4
. Conjecture2]and[Ape03b,Conjecture1]. Afterwards,mostoftheresearchconcentrates
1
on the particular case of a module of the form I/J for two monomial ideals J (I in the
0
4 polynomialringR=K[X ,...,X ]oversomefield K; motivatedbyworks ofHerzog and
1 n
1
Popescu [HP06, Pop09], the Stanley conjecture became one important open problem in
:
v algebraand combinatorics.
i
X
AnaturalfirststeptoapproachtheStanleyconjectureistotrytoreduceittosquarefree
r
a monomial ideals. The arguable most straightforward method for this is via polarization.
This is a process which replaces an arbitrary monomial ideal I with a certain squarefree
monomial ideal Ip, such that I can be recovered from Ip by dividing out a regular se-
quence. The behavior of many invariants of I under polarization is well understood. In
particular, as polarization preserves the projective dimension, the change in the depth is
just the change in the number of variables. In view of the Stanley conjecture, one would
2010MathematicsSubjectClassification. Primary:05E40;Secondary:16W50.
Key words and phrases. Monomial ideal; Stanley depth; Stanley decomposition; poset map;
polarization.
ThefirstauthorwaspartiallysupportedbyprojectPN-II-RU-TE-2012-3-0161,grantedbytheRomanian
NationalAuthorityforScientificResearch,CNCS-UEFISCDI.Thesecondauthorwaspartiallysupported
bytheGermanResearchCouncilDFG-GRK1916. ThethirdauthorwaspartiallysupportedbytheSpan-
ish Government Ministerio de Economía y Competitividad (MEC), grant MTM2012–36917–C03–03in
cooperationwith the EuropeanUnionin the frameworkof the founds“FEDER”, as wellas by the DFG-
GRK1916.
1
2 BOGDANICHIM,LUKASKATTHÄN,ANDJULIOJOSÉMOYANO-FERNÁNDEZ
hope for a similarbehavior of the Stanley depth. This was formulated as a conjecture by
Herzog in thesurvey[Her13]as follows:
Conjecture 1.1(Conjecture 62,[Her13]). Let I ⊂R bea monomialideal. Then
sdepthR/I−depthR/I =sdepthRp/Ip−depthRp/Ip
where Rp istheringwhereIp isdefined.
However, despite the naturality of the question and a considerable effort, it remained
open for quite some time. The main result of our paper (Corollary 4.4) is the proof of
Conjecture 1.1. We show it even more generally for modules of the form I/J for two
monomialidealsJ (I ⊆R.
This has two important consequences. First, it immediately follows that I/J satisfies
theStanleyconjectureifandonlyifitspolarizationIp/Jpdoesso,cf. Corollary4.5. Thus
theStanleyconjectureformodulesoftheformI/J—inparticular[Ape03a,Conjecture2]
and [Ape03b, Conjecture1]—iseffectivelyreduced tothesquarefree case.
For the second consequence, as noted by Stanley himself in [Sta82, p. 191], the Stan-
ley conjecture was formulated such that "the question raised in [Sta79, p. 149, line 6] or
[Gar80, Rmk. 5.2]would followaffirmatively". This questionwas reformulated by Stan-
ley[Sta83,Conjecture2.7],andaskswhethereveryCohen-Macaulaysimplicialcomplex
is partitionable. While it is clear that the Stanley conjecture implies the Garsia-Stanley
conjectureonCohen-Macaulaysimplicialcomplexes,inthispaperweshowthatthecon-
verse is also true, that is, the Garsia-Stanley conjecture on Cohen-Macaulay simplicial
complexes implies what is (arguably) the most important case of the Stanley conjecture.
More precisely, we show that Stanley’s conjecture on Cohen-Macaulay simplicial com-
plexesisequivalentto[Ape03b, Conjecture1](seeCorollary 4.7).
The content of the paper is organized as follows. Section 2 is devoted to explain the
prerequisites and to fix notations. Section 3 introduces a suitable tool for the measure of
theStanleydepthofaquotientI/Jasabove,namelythemapschangingtheStanleydepth.
Helpfulproperties ofthosemapsare recorded inbothLemma3.4and Lemma3.6.
Themostremarkable applicationofSection 3 isthat tothepolarizationofthequotient
I/J; this is exactly the content of Section 4, which also includes the main results of the
paper—thealready mentionedCorollaries 4.4, 4.5 and4.7.
Further applications of the maps changing Stanley depth are described in Section 5,
closingthepaper.
The reader is referred to Bruns and Herzog [BH96], and Miller and Sturmfels [MS05]
forgeneral definitions,notation,and backgroundmaterial.
2. PREREQUISITES
Let K be a field. We considerthe polynomialring R=K[X ,...,X ] overK, endowed
1 n
with the fine Zn-grading (i.e. the Zn-grading with degX =e being the i-th vector of the
i i
canonical basis).
Let M be a finitely generated graded R-module, and m ∈ M homogeneous. Let Z ⊂
{X ,...,X } be a subset of the set of indeterminates of R. The K[Z]-submodule mK[Z]
1 n
THEBEHAVIOROFSTANLEYDEPTHUNDERPOLARIZATION 3
of M is called a Stanley space of M if mK[Z] is free (as K[Z]-submodule). A Stanley
decompositionofM isafinitefamily
D =(K[Zi],mi)i∈W
in whichZ ⊂{X ,...,X } andm K[Z ]isa StanleyspaceofM foreach i∈W with
i 1 n i i
M = m K[Z ]
i i
i∈W
M
as a graded or multigraded K-vector space. This direct sum carries the structure of R-
module and has therefore a well-defined depth. The Stanley depth sdepth M of M is
definedtobethemaximaldepthofaStanleydecompositionofM. TheStanleyconjecture
statesthefollowinginequality:
Conjecture 2.1(Stanley). sdepth M ≥depth M.
Forarecent account onthesubject,thereaderis referred toHerzog’s survey[Her13].
WeendowZn (orNn)withthecomponentwise(partial)order: Givena,b∈Zn, wesay
thata≤bifandonlyifa ≤b fori=1,...,n. NotethatthispartialorderturnsZn intoa
i i
distributivelatticewithmeeta∧bandjoina∨bbeingthecomponentwiseminimumand
maximum,respectively. Fora,b∈Zn theintervalbetween a andb isdefined to be
[a,b]:={c∈Zn | a≤c≤b}.
Remark here thatforn∈N wewillusethenotation[n]:={1,...,n}.
Monotonic poset maps will play a prominent role in Section 3. Let P ⊂Zn, P′ ⊂Zn′
beposets. We calla mapf :P→P′ monotonicifit preservestheorder. Moreover,af is
said to preserve joins resp. meets if it satisfies f (a∨b) =f (a)∨f (b) resp. f (a∧b) =
f (a)∧f (b)forall a,b∈P.
Fora=(a ,...,a )∈Nn wedenoteby Xa themonomialXa1···Xan. LetJ (I ⊂Rbe
1 n 1 n
two monomial ideals. The quotient I/J is a graded or multigraded R-module. Following
Herzog, Vladoiu and Zheng [HVZ09], we fix a vector g∈Nn satisfying a≤g for all Xa
g
inminimalsetsofgeneratorsforI andJ. ThecharacteristicposetP ofI/J withrespect
I/J
to gis defined tobethe(finite)subposet
Pg :={a∈Nn :Xa ∈I\J, a≤g}
I/J
ofZn. A partitionofafiniteposetP isa disjointunion
r
P : P= [ai,bi]
i=1
[
ofintervals. Akeyresultin[HVZ09]describesawaytocomputesdepth I/J fromaStan-
g
leydecompositionofI/JcomingfromapartitionoftheposetP . Moreprecisely,byset-
I/J
tingZ :={X :b =g }foreachb∈Pg ,andthefunctionr =r g :Pg →Z , r (c)=
b j j j I/J I/J ≥0
#(Z ), Theorem2.1 in[HVZ09]says:
c
4 BOGDANICHIM,LUKASKATTHÄN,ANDJULIOJOSÉMOYANO-FERNÁNDEZ
Theorem 2.2(Herzog, Vladoiu,Zheng).
(a)Let P : Pg = r [ai,bi]bea partitionofPg , then
I/J i=1 I/J
S
r
D(P): I/J = XcK[Z ]
bi
i=1 c !
M M
is a Stanleydecomposition of I/J, where the inner direct sum is taken over all c∈[ai,bi]
forwhich c =ai for all j withX ∈Z . Moreover, we have
j j j bi
sdepth D(P)=min{r (bi):i=1,...,r}.
(b) Let D be a Stanley decomposition of I/J. Then there exists a partition P of Pg
I/J
such that sdepth D(P)≥ sdepth D. In particular, sdepth I/J can be computed as the
maximumofthenumberssdepth D(P),whereP runsoverthe(finitelymany)partitions
g
of P .
I/J
The main topic of this paper is the behavior of Stanley depth under polarization. We
recall the definition following Herzog and Hibi [HH11]. Let I ⊂R be a monomial ideal
with generators u ,...,u , where u = (cid:213) n Xaij for i = 1,...,m. For each j let a =
1 m i j=1 j j
max{a :i=1,...,m}. Set a=(a ,...,a ) and choosea g∈Nn such that a≤g. Set Rp
ij 1 n
to bethepolynomialring
Rp :=K[X : 1≤ j≤n,1≤k≤g ].
jk j
Then the polarization of I is the squarefree monomial ideal Ip ⊂ Rp generated by
v ,...,v ,where
1 m
n aij
(cid:213) (cid:213)
v = X for i=1,...,m.
i jk
j=1k=1
3. POSET MAPS
Let I,J be monomial ideals of R such that J ( I. We are interested in measuring the
StanleydepthofthedeformationsofthequotientI/J. Thefollowingkindofmapsreveals
to bea usefultoolforthataim:
Definition3.1. Letℓ∈Zandn,n′∈N. Amonotonicmapf :Nn →Nn′ issaidtochange
theStanleydepthbyℓwithrespecttog∈Nn andg′ ∈Nn′,ifitsatisfiesthefollowingtwo
conditions:
(1) f (g)≤g′
(2) Foreachinterval[a′,b′]⊂[0,g′],the(restricted)preimagef −1([a′,b′])∩[0,g]can
bewrittenas afinitedisjointunion [ai,bi] ofintervals,suchthat
i
#{j∈[n] : bi =g }≥#{jS∈[n′] : b′ =g′}+ℓ foralli.
j j j j
THEBEHAVIOROFSTANLEYDEPTHUNDERPOLARIZATION 5
Notation 3.2. Let R=K[X ,...,X ], R′ =K[X ,...,X ]. A map f :Nn →Nn′ givesrise
1 n 1 n′
to anaturalK-linear mapF :R→R′, which isdefined on monomialsas
F (Xa):=Xf (a).
and extendedtoR linearly. Notethat thisisnot aring homomorphism.
NextpropositionjustifiesthenameforamonotonicmapchangingtheStanleydepthof
thepreviousdefinition:
Proposition 3.3. Let n,n′ ∈N, R=K[X ,...,X ], R′ =K[X ,...,X ] be two polynomial
1 n 1 n′
rings and let J′ (I′ ⊂R′ be monomial ideals. Consider a monotonic map f :Nn →Nn′
andsetI :=F −1(I′),J :=F −1(J′). Chooseg∈Nn andg′ ∈Nn′,suchthateveryminimal
generatorofI andJ dividesXg,andeveryminimalgeneratorofI′ andJ′ dividesXg′. Let
ℓ∈Z andassumethatf changestheStanleydepthbyℓwith respect tog andg′. Then
(i) I and J aremonomialideals,and
(ii) sdepth I/J ≥sdepth I′/J′+ℓ.
Proof. It is clear that I and J are monomial ideals (since monomial ideals correspond
to subsets of Nn that are closed under “going up”, and taking the preimage under a
monotonic map preserves this property). For the second claim, we compute the Stan-
ley depth of I′/J′ via an interval partition of Pg′ (cf. Theorem 2.2). Note that the as-
I′/J′
sumption f (g)≤g′ together with the equalities F −1(I′)=I and F −1(J′)=J imply that
f −1(Pg′ )∩[0,g]=Pg . Hence, taking thepreimages of the intervalsin the partitionof
I′/J′ I/J
Pg′ yieldsan intervalpartitionofPg oftherequired Stanley depth. (cid:3)
I′/J′ I/J
The following results are useful for constructing maps f which satisfy the conditions
ofthepreviousproposition:
Lemma3.4. Letn1,n′1,n2,n′2∈N. Fori=1,2,letf i:Nni →Nn′i bemonotonicmapsthat
change the Stanley depth by ℓi with respect to gi ∈ Nni and g′i ∈ Nn′i. Then the product
map
(f 1,f 2):Nn1+n2 →Nn′1+n′2
changestheStanleydepthbyℓ +ℓ with respectto (g ,g )and (g′,g′).
1 2 1 2 1 2
Proof. Let us denote the product map by f := (f ,f ). It is enough to consider one
1 2
interval[(p1,p2),(q1,q2)]⊂Nn′1+n′2. Byassumption,thepreimagef i−1([pi,qi]) [0,gi]=
T
6 BOGDANICHIM,LUKASKATTHÄN,ANDJULIOJOSÉMOYANO-FERNÁNDEZ
[pj,qj]⊂Nni isa disjointfiniteunionofintervals. Then
j i i
Sf −1([(p ,p ),(q ,q )])∩[0,(g ,g )]=
1 2 1 2 1 2
=f −1(([p ,q ]×Nn2) ∩(Nn1×[p ,q ]))∩[0,(g ,g )]
1 1 2 2 1 2
=(f −1([p ,q ])×Nn2) ∩(Nn1×f −1([p ,q ]))∩([0,g ]×Nn2) ∩(Nn1×[0,g ])
1 1 1 2 2 2 1 2
= (f −1([p ,q ])∩[0,g ])×Nn2 ∩ Nn1×(f −1([p ,q ])∩[0,g ])
1 1 1 1 2 2 2 2
(cid:0) (cid:1) (cid:0) (cid:1)
= ( [pj,qj])×Nn2 ∩ Nn1×( [pj,qj])
1 1 2 2
! !
j j
[ [
= [(pj,pk),(qj,qk)],
1 2 1 2
j,k
[
which impliesthestatement. (cid:3)
Remark 3.5. The most important special case of this lemma is when one of the maps is
theidentity. Inthiscase,iff isamapchangingtheStanleydepthbyℓ,thenwecanpadit
withidentitiestogetanewmapfˆ =(id,...,id,f )thatstillchangestheStanleydepthby
ℓ. In the sequel, fˆ will always denote the padded version of f and Fˆ the padded version
ofF .
Lemma 3.6. Everymonotonicmapf :N→Nn′ changestheStanleydepthby1−n′ with
respect tog∈N andg′ :=f (g).
Proof. Let Q′ = [a′,b′] ⊂ [0,g′] ⊂ Nn′ be an interval and let Q := f −1(Q′)∩[0,g] be
its (restricted) preimage. Let a resp. b ∈ N be the minimal resp. maximal element in
Q. Then Q ⊂ [a,b] and we claim that we have equality. For c ∈ [a,b], it follows from
f (a) ≤ f (c) ≤ f (b) that f (c) ∈ [a′,b′], because f (a),f (b) ∈ [a′,b′]. Thus c ∈ Q. So
the preimage of an interval is again an interval. It remains to verify the condition (2) of
Definition3.1,namely
#{j∈[1] : b =g }≥#{j∈[n′] : b′ =g′}+1−n′ foralli.
j j j j
Ifb′<g′,thentherighthandsideisnonpositive,sotheconditionistriviallysatisfied. On
theotherhand, ifb′ =g′, then b=g and thustheconditionisalsosatisfied. (cid:3)
Weclosethis sectionwith aprecise descriptionofthebehaviorof posetmaps preserv-
ing joins and meets with respect to the property of changing Stanley depth. This result
willbenotneeded in thesequel.
Theorem3.7. Letn,n′∈Nandletf :Nn→Nn′ beamapthatpreservesjoinsandmeets.
Then f changestheStanleydepthbyn−n′ withrespect tog∈Nn and g′ :=f (g).
Proof. Weassumethatf (0)=0,asweotherwisereplacef byf −f (0)withoutchanging
thevalidityofthe statement. First we showthat f is monotonic. Consider a,b∈Nn with
a≤b. Then
f (a)≤f (a)∨f (b)=f (a∨b)=f (b),
hencef (a)≤f (b).
THEBEHAVIOROFSTANLEYDEPTHUNDERPOLARIZATION 7
In view of Lemma 3.4 and Lemma 3.6, it is sufficient to show that f is a product of
monotonicmapsf i :N→Nn′i with1≤i≤nand (cid:229) in′i =n′.
For this, let e denote the i-th unit vector of Nn. Every vector v = (cid:229) l e ∈ Nn can
i i i i
be written as a sum v=l e +(0,v′) with v′ ∈Nn−1. As f is monotonic, the support of
1
f (l e ) can only increase with l . Hence, without loss ofgenerality, we may assumethat
1
themaximalsupportoff (l e ) forl ≫0 isthesetofthefirst n′ coordinates. That is
1 1
f (l e1)∈Nn′1×(0,...,0).
Sincef preserves meets,foralll ∈N and v′ ∈Nn−1 wehave
0=f (0)=f (l e ∧(0,v′))=f (l e )∧f ((0,v′)).
1 1
It follows that for each vector (0,v′) with vanishing first coordinate, the support of its
imageiscontainedin thelastn′−n′ coordinates. That is
1
f ((0,v′))∈(0,...,0)×Nn′−n′1.
Then, itholdsthat
f (v)=f (l e +(0,v′))=f (l e ∨(0,v′))=f (l e )∨f ((0,v′))=f (l e )+f ((0,v′)),
1 1 1 1
becausethesumequalsthejoininthiscase. Setf 1:N→Nn′1,f 1(l )=p n′f (l e1)where
1
p n′ is the projection on the first n′1 coordinates. Set f n−1 : Nn−1 → Nn′−n′1, f n−1(v′) =
1
p f ((0,v′)) where p is the projection on the last n′−n′ coordinates. Then f
n′−n′ n′−n′ 1
1 1
splitsintoadirect product
f =(f 1,f n−1):N×Nn−1 →Nn′1×Nn′−n′1.
Sincef isarestrictionoff , itfollowsthat f ismonotonic.
1 1
Iteratingthisconstructionyieldsthedesired decompositionoff . (cid:3)
4. APPLICATION TO POLARIZATIONS
Let J ( I be two monomial ideals in R. Then we can choose e ∈ Zn bigger than or
equal to (in the sense of the order ≤) the join of all generators of I and J, and define the
polarizationIp/Jp ofI/J according toSection 2.
Polarizationcanalsobedonestepbystep. Letu ,...,u bethesetofminimalgenera-
1 m
tors of I. Following[Her13] we define the 1-step polarizationof I (with respect to X) to
i
betheidealI1 ⊂R[Y] generated byv ,...,v , where
1 m
Y u ifX2|u
v := Xi j i j
j
u otherwise.
( j
If the indeterminate of the polynomial ring with respect which we apply partial polar-
izationis notrelevant,wewillomitit.
Proposition4.1. Let J (I ⊂Rbemonomialideals. Then
(1) depth I1/J1 =depth I/J+1;
R[Y] R
(2) H (t)= 1 H (t).
I1/J1 1−t I/J
8 BOGDANICHIM,LUKASKATTHÄN,ANDJULIOJOSÉMOYANO-FERNÁNDEZ
Here H (t)denotestheHilbert series with respect to theZ-grading. This proposition
I/J
is easy and partially known (cf. [Her13, p. 47]). We present a complete proof for the
reader’s convenience.
Proof. For both claims it suffices to show that X −Y is a regular element. By contradic-
n
tion,assumingX −Y isazero-divisoronI1/J1,thenthereexistsanidealp∈Ass(I1/J1)
n
withX −Y ∈p. SincebothJ1 andparemonomialideals,thereexistsamonomialh∈/ J1
n
such that p = Ann h. Then h(X −Y) ∈ J1, and again hY ∈ J1, hX ∈ J1, since J1 is
R′ n n
monomial. Let f , f generators of J1 such that f |hX and f |hY. IfY|f then X |f and
1 2 1 n 2 2 n 2
so X |h. ThereforeX2|f , thenY|f . ThisimpliesthatY dividesh,acontradiction. (cid:3)
n n 1 1
One difficulty in proving our main result is that the ideals I1 and J1 do not arise as
preimagesofamap,buttheyarerather(generatedby)theimageofamap. Thefollowing
Lemmais helpfulforconstructingStanley decompositionsinthissetting.
Lemma4.2. Letn,n′∈N,R=K[X ,...,X ],R′=K[X ,...,X ]betwopolynomialrings
1 n 1 n′
and let J (I ⊂R be monomialideals. Let f :Nn →Nn′ be a map and let J′ (I′ ⊂R′ be
theidealsgeneratedbyF (I)resp. F (J). LetW beafinitesetandletI/J= i∈W XaiK[Zi]
beaStanleydecompositionofI/J. LetZ′,i∈W beacollectionofsubsetsof{X ,...,X }.
i L 1 n′
Assumethatf isinjective, monotonic,andpreserves joins. Assumemoreover, that
F (Xai)K[Z′]∩F (R)=F (XaiK[Z]) (⋆)
i i
foreach i∈W .
SetV =(cid:229) i∈W F (Xai)K[Zi′] asa gradedvector space. Then
V = F (Xai)K[Z′]
i
i∈W
M
isa direct sumand itholdsthatV ⊂I′/J′.
Proof. First, we show that the sum is direct. On the contrary, suppose that there are in-
dicesi6= j suchthatamonomialappears inbothparts. Thismonomialisthenacommon
multiple of F (Xai) and F (Xaj). Then also the least common multiple of F (Xai) and
F (Xaj)appearsinbothparts. AsF preservestheleastcommonmultipleoftwomonomi-
als, it followsthat the least common multipleof F (Xai) and F (Xaj) is F (Xai∨aj) (where
Xai∨aj is the least common multiple of Xai and Xaj). But now, the condition (⋆) and the
fact that f isinjectiveimplythat Xai∨aj ∈XaiK[Z ]∩XajK[Z ],acontradiction.
i j
Next, we show thatV ⊂I′/J′. Every Xai is contained in I, henceV is contained in I′.
It remains to show that no monomial in F (Xai)K[Z′] is contained in J′. Assume on the
i
contrarythatsuchamonomialexists. ThenitisacommonmultipleofF (Xai)andF (Xa)
for a minimal generator of Xa of J. Again, it follows that the least common multiple of
F (Xai) and F (Xa) is contained in F (Xai)K[Z′]∩J′∩F (R). But then the least common
i
multipleofXai and Xa is containedinXaiK[Z]∩J,acontradiction. (cid:3)
i
Theorem 4.3. Let J ( I ⊂ R be two monomial ideals, and let J1 ( I1 ⊂ R[Y] be their
1-step polarizations. Then
sdepth I/J =sdepth I1/J1−1.
THEBEHAVIOROFSTANLEYDEPTHUNDERPOLARIZATION 9
Proof. Withoutlossofgenerality,weassumethatX2 dividesoneofthegeneratorsofI or
n
J and weapplypolarizationwithrespect toX . Considerthemap f :N→N2 defined by
n
(i−1,1) ifi≥2
f (i):=
(i,0) ifi=0,1.
(
FIGURE 1. Theimageoff fromtheproofofTheorem4.3.
Itiseasytoseethatf isinjective,monotonicandpreservesjoins. Further,observethat
f (a)≤f (b)ifandonlyifa≤bfora,b∈N. Moreover,allthesepropertiescarry overto
fˆ :Nn →Nn+1 and Fˆ :R→R[Y] (seealso Remark 3.5).
By definition, it holds that I1 is generated by Fˆ(I), hence Fˆ−1(I1)⊇I. We claim that
Fˆ−1(I1)=I. To see this, consider a monomial Xb ∈Fˆ−1(I1). Then Fˆ(Xb) is a multiple
ofageneratorF (Xa)ofI1,whereXa∈I. Bytheobservationabove,Xb isthenamultiple
ofXa andthuscontained inI. Analogouslyit holdsthat Fˆ−1(J1)=J.
Letg=(g ,...,g )bethejoinoftheexponentsofthemonomialgeneratorsofI and J
1 n
(weassumeg ≥2). Theng′=(g ,...,g ,g −1,1)=fˆ(g)isthejoinoftheexponents
n 1 n−1 n
of the monomial generators of I1 and J1. It follows from Lemma 3.6 that fˆ changes
the Stanley depth by −1 with respect to g and (g −1,1). Hence by Lemma 3.4 and
n n
Proposition3.3 itholdsthatsdepth I/J ≥sdepth I1/J1−1.
Now we turn to the second inequality. Let us consider a Stanley decomposition I/J =
XaiK[Z]ofI/J. Remark thatFˆ(X Xa)/Fˆ(Xa)∈{X ,Y}foreverymonomialXa∈R,
i i n n
which allowsus todefine
L
Z ∪{Y} if X ∈Z
i n i
Zi′ := Z ∪ {X ,Y}\{Fˆ(XnXai)} otherwise.
( i n Fˆ(Xai)
Weclaimthat (cid:16) (cid:17)
V := Fˆ(Xai)K[Z′] (⋆⋆)
i
i
M
isa StanleydecompositionofI1/J1. First, weshowthat
Fˆ(Xai)K[Z′]∩Fˆ(R)=Fˆ(XaiK[Z])
i i
foreach i. Theinclusion“⊇” isclear.
Fortheotherinclusion,letM′ :=Fˆ(Xai)N′ beamonomialintheleft-hand sideand let
M :=Fˆ−1(M′)∈Rbeitspreimage. NotethatFˆ(Xai)|M′ impliesthatXai |M (itfollows
from the observationabove that f (a)≤f (b) if and only if a≤b for a,b∈N), hence we
10 BOGDANICHIM,LUKASKATTHÄN,ANDJULIOJOSÉMOYANO-FERNÁNDEZ
may define N :=M/Xai. It suffices to show that N ∈K[Z ], because then M′ =Fˆ(XaiN)
i
iscontained intheright-handside.
By definition, the map Fˆ may change only the exponents of X and Y. Therefore the
n
equality Fˆ(XaiN) = Fˆ(Xai)N′ implies that N′ and N may differ only by multiplication
and/or divisionby (powers of) X andY. Hence every other variable appearing in N also
n
appearsinN′∈K[Z′],soitiscontainedinZ′ and(thus)inZ. Furthermore,N ∈Rimplies
i i i
thatY ∤N. So weonlyneed toprovethefollowing: If X |N, thenX ∈Z.
n n i
So assume that X | N. Then X Xai | NXai and thus Fˆ(X Xai) | Fˆ(NXai) = M′. This
n n n
furtherimpliesthat
Fˆ(X Xai) M′
n =N′.
Fˆ(Xai) Fˆ(Xai)
As N′ ∈K[Z′],itnowfollowsfromthedefinitionofZ′ thatX ∈Z .
i i n i
ItiseasytoseethatFˆ satisfiestheassumptionsofLemma4.2,soweconcludethatthe
sum in (⋆⋆)is direct and thatV ⊂I1/J1. It remains to show thatV =I1/J1. For this, we
computethegraded Hilbertseries ofV:
(cid:229) it|ai|(1−t)n+1−|Zi′|
H (t)=
V (1−t)n+1
(cid:229) t|ai|(1−t)n−|Zi|
i
=
(1−t)n+1
1
= H (t)=H (t).
1−t I/J I1/J1
Here, |a| denotesthesumofthecomponentsofa . Inthefirst andthirdequalityweused
i i
thedecompositionsofV resp. I/J givenabove. Forthelastequalityweused Proposition
4.1. As wehavealready shownthatV ⊂I1/J1, theclaim follows.
Weconcludethatthesumin(⋆⋆)isaStanleydecompositionofI1/J1. As|Z′|=|Z|+1
i i
foreach i,itfollowsthatsdepth I1/J1 ≥sdepth I/J+1. (cid:3)
Iteration ofTheorem 4.3and Proposition4.1has oneimmediateconsequence:
Corollary 4.4. Let J (I ⊂R be monomial ideals, and let Ip,Jp ⊂Rp be their polariza-
tions. Then
sdepth I/J−depth I/J =sdepth Ip/Jp−depth Ip/Jp.
In particular,[Her13, Conjecture62]istrue.
NotethattheprecedingcorollaryeffectivelyreducesStanley’sconjecturetothesquare-
free case:
Corollary 4.5. Let J (I ⊂R be monomial ideals, and let Ip,Jp ⊂Rp be their polariza-
tions. Then I/J satisfiestheStanleyConjectureifand onlyifIp/Jp satisfiesittoo.
Remark 4.6. (1) We would like to point out that the “only if”-part of Corollary 4.5
already appeared in [Ahm11, Theorem 3.5] in the quite particular case J = (0)
and R/I Cohen-Macaulay. We obtain in contrast a full reduction of the Stanley’s
conjecturetothesquarefree case.