MONOMIAL IDEALS, ALMOST COMPLETE INTERSECTIONS AND THE WEAK LEFSCHETZ PROPERTY

Transcription

1 MONOMIAL IDEALS, ALMOST COMPLETE INTERSECTIONS AND THE WEAK LEFSCHETZ PROPERTY JUAN C MIGLIORE, ROSA M MIRÓ-ROIG, UWE NAGEL + Abstact Many algebas ae expected to have the Weak Lefschetz popety though this is often vey difficult to establish We illustate the subtlety of the poblem by studying monomial and some closely elated ideals Ou esults exemplify the intiguing dependence of the popety on the chaacteistic of the gound field, and on aithmetic popeties of the exponent vectos of the monomials Contents 1 Intoduction 1 2 Tools fo studying the WLP A class of monomial ideals 5 4 Monomial almost complete intesections in any codimension 8 5 An almost monomial almost complete intesection 14 6 Monomial almost complete intesections in thee vaiables 17 7 A poof of half of Conjectue Final Comments 26 Refeences 27 1 Intoduction Let A be a standad gaded Atinian algeba ove the field K Then A is said to have the Weak Lefschetz popety (WLP if thee is a linea fom L (A 1 such that, fo all integes j, the multiplication map L : (A j 1 (A j has maximal ank, ie it is injective o sujective In this case, the linea fom L is called a Lefschetz element of A (We will often abuse notation and say that the coesponding ideal has the WLP The Lefschetz elements of A fom a Zaiski open, possibly empty, subset of (A 1 Pat of the geat inteest in the WLP stems fom the fact that its pesence puts sevee constaints on the possible Hilbet functions (see [6], which can appea in vaious disguises (see, eg, [1] Though many algebas ae expected to have the WLP, establishing this popety is often athe difficult Fo example, it is open whethe evey Pat of the wok fo this pape was done while the fist autho was sponsoed by the National Secuity Agency unde Gant Numbe H Pat of the wok fo this pape was done while the second autho was patially suppoted by MTM Pat of the wok fo this pape was done while the thid autho was sponsoed by the National Secuity Agency unde Gant Numbe H The authos thank Fabizio Zanello fo useful and enjoyable convesations elated to some of this mateial They also thank David Cook II fo useful comments 1

2 2 J MIGLIORE, R MIRÓ-ROIG, U NAGEL complete intesection of height fou ove a field of chaacteistic zeo has the WLP (This is tue if the height is at most by [6] In some sense, this note pesents a case study of the WLP fo monomial ideals and almost complete intesections Ou esults illustate how subtle the WLP is In paticula, we investigate its dependence on the chaacteistic of the gound field K The following example (Example 77 illustates the supising effect that the chaacteistic can have on the WLP Conside the ideal I = (x 10, y 10, z 10, x y z R = K[x, y, z] Ou methods show that R/I fails to have the WLP in chaacteistics 2, and 11, but possesses it in all othe chaacteistics One stating point of this pape has been Example 1 in [4], whee Benne and Kaid show that, ove an algebaically closed field of chaacteistic zeo, any ideal of the fom (x, y, z, f(x, y, z, with deg f =, fails to have the WLP if and only if f (x, y, z, xyz In paticula, the latte ideal is the only such monomial ideal that fails to have the WLP This pape continues the study of this question The example of Benne and Kaid satisfies seveal inteesting popeties In this pape we isolate seveal of these popeties and examine the question of whethe o not the WLP holds fo such algebas, and we see to what extent we can genealize these popeties and still get meaningful esults Some of ou esults hold ove a field of abitay chaacteistic, while othes show diffeent ways in which the chaacteistic plays a cental ole in the WLP question (Almost none ae chaacteistic zeo esults Most of ou esults concen monomial ideals, although in Section 5 and Section 8 we show that even mino deviations fom this popety can have dastic effects on the WLP Most of ou esults deal with almost complete intesections in thee o moe vaiables, but we also study ideals with moe geneatos (genealizing that of Benne and Kaid in a diffeent way Moe specifically, we begin in Section 2 with some simplifying tools fo studying the WLP These ae applied thoughout the pape We also ecall the constuction of basic double linkage In Section we conside the class of monomial ideals in K[x 1,, x ] of the fom (x k 1, x k 2,, x k + (all squaefee monomials of degee d Note that the example of Benne and Kaid is of this fom Ou main esult in this section (Theoem says that when d = 2 we always have the WLP, but if d = and k 2 then we have two cases: if K has chaacteistic 2 then we neve have the WLP, but if the chaacteistic is not 2 then we have the WLP if and only if k is even In Section 4, we conside almost complete intesections of the fom (x 1,, x, x 1 x with (note that the esult of Benne and Kaid dealt with the case = in chaacteistic zeo Ou main esult fo these algebas is that they always fail to have the WLP, egadless of the chaacteistic The poof is supisingly difficult In Section 5 we explicitly illustate the fact that even a minuscule change in the ideal can affect the WLP Specifically, we conside the ideals of the fom (x 1,, x, x 1 x (x 1 + x We show that this has the same Hilbet function as the coesponding ideal in the pevious section, but the WLP behavio is vey diffeent Fo example, the two ideals (x 4 1,, x 4 4, x 1 x 2 x x 4 and (x 4 1,, x 4 4, x 1 x 2 x (x 1 + x 4 have the same Hilbet function, but the fome neve has the WLP while the latte has the WLP if and only if the chaacteistic of K is not two o five

3 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY In Section 6 we tun to monomial almost complete intesections in thee vaiables, genealizing the Benne-Kaid example in a diffeent diection To facilitate this study, we assume that the algeba is also level (as is the case fo Benne and Kaid s example We give a numbe of esults in this section, which depend on the exponent vectos of the monomials We end with a conjectued classification of the level Atinian monomial ideals in thee vaiables that fail to have the WLP (Conjectue 68 The wok in Sections 6 and 7 poves most of this conjectue We end the pape in Section 8 with some suggestive computations and natual questions coming fom ou wok 2 Tools fo studying the WLP In this section we establish vaious geneal esults that help to study the WLP and that ae used thoughout the emainde of this pape Thoughout this pape we set R = K[x 1,, x ], whee K is a field Sometimes we will have specific values of (usually and sometimes we will have futhe estictions on the field K Ou fist esults singles out the cucial maps to be studied if we conside the WLP of a level algeba Recall that an Atinian algeba is called level if its socle is concentated in one degee Poposition 21 Let R/I be an Atinian standad gaded algeba and let L be a geneal linea fom Conside the homomophisms φ d : (R/I d (R/I d+1 defined by multiplication by L, fo d 0 (a If φ d0 is sujective fo some d 0 then φ d is sujective fo all d d 0 (b If R/I is level and φ d0 is injective fo some d 0 0 then φ d is injective fo all d d 0 (c In paticula, if R/I is level and dim(r/i d0 = dim(r/i d0 +1 fo some d 0 then R/I has the WLP if and only if φ d0 is injective (and hence is an isomophism Poof Conside the exact sequence 0 [I : L] I R/I L (R/I(1 (R/(I, L(1 0 whee L in degee d is just φ d This shows that the cokenel of φ d is just (R/(I, L d+1 fo any d If φ d0 is sujective, then (R/(I, L d0 +1 = 0, and the same necessaily holds fo all subsequent twists since R/I is a standad gaded algeba Then (a follows immediately Fo (b, ecall that the K-dual of the finite length module R/I is a shift of the canonical module of R/I, which we will denote simply by M Since R/I is level, M is geneated in the fist degee But now if we conside the gaded homomophism of M to itself induced by multiplication by L, a simila analysis (ecalling that M is geneated in the fist degee gives that once this multiplication is sujective in some degee, it is sujective theeafte The esult on R/I follows by duality Pat (c follows immediately fom (a and (b If the field is infinite and the K-algeba satisfies the WLP fo some linea fom, then it does fo a geneal linea fom Fo monomial ideals thee is no need to conside a geneal linea fom Poposition 22 Let I R be an Atinian monomial ideal and assume that the field K is infinite Then R/I has the WLP if and only if x x is a Lefschetz element fo R/I

4 4 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Poof Set A = R/I and let L = a 1 x a x be a geneal linea fom in R Thus, we may assume that each coefficient a i is not zeo and, in paticula, a = 1 Let J S := K[x 1,, x ] be the ideal that is geneated by elements that ae obtained fom the minimal geneatos of I afte substituting a 1 x a x fo x Then A/LA = S/J Each minimal geneato of J is of the fom x j 1 1 x j (a 1 x 1 + +a x j Replacing it by (a 1 x 1 j1 (a x j ( a 1 x 1 + a x j does not change the ideal J because a 1 a 0 Using the isomophism K[y 1,, y ] S, y i a i x i, we see that A/LA and A/(x x A have the same Hilbet function Since we can decide whethe L is a Lefschetz element fo A by solely looking at the Hilbet function of A/LA, the claim follows If A is an Atinian K-algeba with the WLP and E is an extension field of K, then also A K E has the WLP Howeve, the convese is not clea We pose this as a poblem Poblem 2 Is it tue that A has the WLP if and only if A K E has the WLP? Poposition 22 shows that the answe is affimative in the case of monomial ideals Coollay 24 Let E be an extension field of the infinite field K If I R is an Atinian monomial ideal, then R/I has the WLP if and only if (R/I K E does The following esult applies if we can hope that the multiplication by a linea fom is sujective Poposition 25 Let I R = K[x 1,, x ], whee K is a field and A = R/I is Atinian Let d be any degee such that h A (d 1 h A (d > 0 Let L be a linea fom, let R = R/(L and let Ī be the image of I in R Denote by Ā the quotient R/Ī Conside the minimal fee R-esolution of Ā: 0 p i=1 R( b i p 1 j=1 R( a j R Ā 0 whee a 1 a p1 and b 1 b p Then the following ae equivalent: (a the multiplication by L fom A d 1 to A d fails to be sujective; (b Ād 0; (c b p d + 1; (d Let G 1,, G be a egula sequence in Ī of degees c 1,, c espectively that extends to a minimal geneating set fo Ī Then thee exists a fom F R of degee c c (d + 1, non-zeo modulo (G 1,, G, such that F Ī (G 1,, G Poof Fom the exact sequence A d 1 L A d (R/(I, L d 0 it follows that the multiplication fails to be sujective if and only if Ād = (R/(I, L d 0 The latte holds if and only if d socle degee of Ā = b p ( 1, fom which the equivalence of (a, (b and (c follows To show the equivalence of (c and (d we invoke liaison theoy Let J = (G 1,, G : Ī

5 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 5 A fee esolution fo J can be obtained fom that of Ī and (G, G by a standad mapping cone agument (see fo instance [9], as follows We have the following commutative diagam (whee the second one is the Koszul esolution fo (G 1,, G : 0 p i=1 R( b i p 1 R( a j=1 j Ī 0 0 R( c 1 c R( c k=1 k (G, G 0 whee the ightmost vetical aow is an inclusion This yields a fee esolution fo J (afte splitting R( c k=1 k and e-numbeing the a j, and setting c := c c : 0 p 1 ( j=1 R(a j c p i=1 R(b i c k=1 R( c k J 0 Clealy b p d + 1 if and only if J has a minimal geneato, F, of degee c (d + 1 The esult then follows fom the definition of J as an ideal quotient We conclude this section by ecalling a concept fom liaison theoy, which we do not state in the geatest geneality Let J I R = K[x 1,, x ] be homogeneous ideals such that codim J = codim I 1 Let l R be a linea fom such that J : l = J Then the ideal I := l I + J is called a basic double link of I The name stems fom the fact that I can be Goenstein linked to I in two steps if I is unmixed and R/J is Cohen-Macaulay and geneically Goenstein ([8], Poposition 510 Howeve, hee we only need the elation among the Hilbet functions Lemma 26 Fo each intege j, dim K (R/I j = dim K (R/I j 1 + dim K (R/J j dim K (R/J j 1 Poof This follows fom the exact sequence (see [8], Lemma 48 0 J( 1 J I( 1 I 0 A class of monomial ideals We now begin ou study of a cetain class of Atinian monomial ideals Let I,k,d be the monomial ideal defined by (1 (x k 1, x k 2,, x k + (all squaefee monomials of degee d Ou fist obsevation follows immediately be detemining the socle of R/I,k,d It shows that we may apply Poposition 21 Poposition 1 The invese system fo I,k,d is geneated by the module geneated by all monomials of the fom x k 1 i 1 x k 1 i d 1 Coollay 2 The algeba R/I,k,d is level of socle degee (k 1(d 1 and socle type ( d 1 Concening the WLP we have: Theoem Conside the ing R/I,k,d (a If d = 2, then it has the WLP (b Let d = and k 2 Then: (i If K has chaacteistic two, then R/I,k,d does not have the WLP

6 6 J MIGLIORE, R MIRÓ-ROIG, U NAGEL (ii If the chaacteistic of K is not two, then R/I,k,d has the WLP if and only if k is even Poof Fo simplicity, wite I = I,k,d and A = R/I,k,d Claim (a follows easily fom the obsevation that A has socle degee k 1 and that up to degee k 1 the ideal I is adical, so multiplication by a geneal linea fom is injective in degee k 1 To show claim (b we fist descibe bases of (A k 1 and (A k, espectively We choose the esidue classes of the elements in the following two sets B k 1 = {x j i xk 1 j m 1 i < m, 1 j k 2} {x k 1 i 1 i } B k = {x j i xk j m 1 i < m, 1 j k 1} Counting we get ( ( (2 h A (k 1 = (k 2 + (k 1, 2 2 whee the inequality follows fom d = Now we assume that k is odd In this case we claim that A does not have the WLP Because of Inequality (2, this follows once we have shown that, fo each linea fom L L R, the multiplication map φ k : (A k 1 (A k is not injective To show the latte assetion we exhibit a non-tivial element in its kenel Wite L = a 1 x a x fo some a 1,, a K We define the polynomial f R as f = ( 1 max{j i,j m} (a i x i j i (a m x m jm j i +j m=k 1 Note that f is not in I We claim that L f is in I Indeed, since all monomials involving thee distinct vaiables ae in I, a typical monomial in L f mod I is of the fom (a i x i j i (a m x m k j i It aises in exactly two ways in Lf, namely as (a i x i (a i x i ji 1 (a m x m k j i and as (a m x m (a i x i j i (a m x m k 1 j i Using that k 1 is even, it is easy to see that these two monomials occu in f with diffeent signs It follows that the above multiplication map is not injective If k is even, but cha K = 2, then the same analysis again shows that φ k is not injective Hence, fo the emainde of the poof we may assume that the chaacteistic of K is not two Assume k is even Then we claim that L = x x is a Lefschetz element To L this end we fist show that the multiplication map φ k : (A k 1 (A k is injective Let f be any element in the vecto space geneated by B k 1 Pick thee of the vaiables x 1,, x and call them x, y, z Below we explicitly list all the tems in f that involve only the vaiables x, y, z: f = a 0 x k 1 + a 1 x k 2 y + + a k 2 xy k 2 + b 1 x k 2 z + + b k 2 xz k 2 + b k 1 z k 1 + c 0 y k 1 + c 1 y k 2 z + + c k 2 yz k 2 + As above, we see that each monomial in L f aises fom exactly two of the monomials in f Hence the condition L f I leads to the following thee systems of equations

7 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 7 Focussing only on the vaiables x, y we get: It follows that a i = ( 1 i a 0 and a 0 + a 1 = 0 a 1 + a 2 = 0 a k + a k 2 = 0 a k 2 + c 0 = 0 ( c 0 = ( 1 k 2 a 0 = a 0 because k is even Consideing the vaiables x, z we obtain: hence a 0 + b 1 = 0 b 1 + b 2 = 0 b k 2 + b k 1 = 0, (4 b i = ( 1 i a 0 Finally, using the vaiables y, z we get: Combining this, it follows that c 0 + c 1 = 0 c k 1 + c k 2 = 0 c k 2 + b k 1 = 0 a 0 = b k 1 = c 0 = a 0 Since we assumed that the chaacteistic of K is not two, we conclude that the thee linea systems above have only the tivial solution Since the vaiables x, y, z wee chosen abitaily, we see that the map φ k is injective, as claimed Accoding to Lemma 21 it emains to show that the multiplication map φ k+1 : (A k L (A k+1 is sujective Note that the esidue classes of the elements of the fom x j i xk+1 j m with 2 j k 1, 1 i < m fom a basis of (A k+1 Setting fo simplicity x := x i, y = x m it is enough to show that, fo each j = 2,, k 1, the esidue class of x j y k+1 j is in the image of φ k+1 We induct on j 2 If j = 2, then we get modulo I that L xy k 1 x 2 y k 1, thus x 2 y k 1 im φ k+1, as claimed Let j k 1, then, modulo I, we get L x j 1 y k j x j y k j + x j 1 y k j+1 Since by induction x j 1 y k j+1 im φ k+1, we also obtain x j y k j im φ k+1 This completes the poof The above esult and ou compute expeiments suggest that the lage d becomes, the ae it is that R/I,k,d has the WLP Based on compute expeiments we expect the following to be tue Conjectue 4 Conside the algeba R/I,k,d Then (a If d = 4, then it has the WLP if and only if k mod 4 is 2 o (b If d = 5, then the WLP fails

8 8 J MIGLIORE, R MIRÓ-ROIG, U NAGEL (c If d = 6, then the WLP fails We summaize ou esults in case k = d = Example 5 Conside the ideal I,, = (x 1, x 2,, x, (all squaefee monomials of degee Then the coesponding invese system is (x 2 1x 2 2, x 2 1x 2,, x 2 x 2 Futhemoe, the Hilbet function of R/I,, is ( ( ( and R/I,, fails to have the WLP because the map fom degee 2 to degee by a geneal linea fom is not injective Remak 6 By tuncating, we get a compessed level algeba with socle degee that fails to have the WLP We expect that thee ae compessed level algebas with lage socle degee that fail to have the WLP Howeve, we do not know such an example 4 Monomial almost complete intesections in any codimension In the pape [10] the fist and second authos asked the following question (Question 42, page 95: Fo any intege n, find the minimum numbe A(n (if it exists such that evey Atinian ideal I K[x 1,, x n ] with numbe of geneatos µ(i A(n has the WLP In Example 710 below, we show that A(n does not exist in positive chaacteistic In any case, in [4] it was shown fo n = and chaacteistic zeo that A( = (also using a esult of [6], as noted in the intoduction A consequence of the main esult of this section, below, is that in any numbe of vaiables and any chaacteistic thee is an almost complete intesection that fails to have the WLP Hence the main open question that emains is whethe, in chaacteistic zeo, all complete intesections have the WLP (as was shown fo n = in [6], ie whethe A(n = n in chaacteistic zeo We begin by consideing ideals of the fom (41 I,k = (x k 1,, x k, x 1 x K[x 1,, x ] Notice that this is a special case of the class of ideals descibed in Section It is not too difficult to detemine the gaded Betti numbes Poposition 41 The minimal fee esolution of I,k has the fom 0 R( + ( 1(1 k ( R( ( 1k ( R( + ( 2(1 k ( 2 R( k ( R( + 2(1 k ( 2 R( 2k ( 2 R( + (1 k R( k R( I,k 0 Poof Since I,k is an almost complete intesection, we can link it using the complete intesection a = (x k 1,, x k to an Atinian Goenstein ideal, J Howeve, since both I,k and a ae monomial, so is J But it was fist shown by Beintema [2] that any monomial Atinian Goenstein ideal is a complete intesection Hence we get by diect computation that (x k 1,, x k : x 1 x 2 x = (x k 1 1,, x k 1 Then use the mapping cone and obseve that thee is no splitting

9 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 9 Befoe we come to the main esult of this section, we pove a peliminay esult about the Hilbet function of complete intesections that will allow us to apply Poposition 25 Lemma 42 Let R = K[x 1,, x s ] with s 2, and let I s = (x s 1,, x s s and J s = (x s+1 1, x s+1 2, x s,, x s s Note that the midpoint of the Hilbet function of R/I s is ( s 2 and that of R/Js is ( s Then h R/Is 2 hr/is 2 1 hr/js hr/js Poof The lemma is tivial to veify when s = 2 o s =, so we assume s 4 fo this poof Obseve that both quantities ae positive, but one involves a diffeence to the left of the midpoint of the Hilbet function, while the othe involves a diffeence to the ight We will use this fomulation, although thee exists othes thanks to the symmety of the Hilbet function of an Atinian complete intesection Ou appoach will be via basic double linkage We will use the fomula in Lemma 26 without comment In fact, J s is obtained fom I s by a sequence of two basic double links: I s x 1 I s + (x s 2,, x s s := G = (x s+1 1, x s 2,, x s s x 2 G + (x s+1 1, x s,, x s s = J s Note that G is a complete intesection of codimension s and that the ideals C 1 := (x s 2,, x s s and C 2 := (x s+1 1, x s,, x s s ae complete intesections of codimension s 1 The midpoints of the h-vectos of R/C 1 and R/C 2 ae (s 12 and (s espectively 2 2 We now compute Hilbet functions (and notice the shift, and that the lines fo R/C 1 and R/C 2 ae the fist diffeence of those Hilbet functions, ie the h-vectos: (( R/I s 1 s h s (( R/Is 2 1 h s (( R/Is 2 R/C 1 1 s 1 h s (( C1 h s 2 C R/G 1 s A B whee (42 and h R/G 2 = A = hr/is hr/c1 2, h R/G = B = hr/is 2 + hr/c , R/G 1 s A B R/C 2 1 s 1 h C hc R/J s 1 s C D whee h R/Js = C = hr/is hr/c1 2 + hr/c , h R/Js = D = hr/is 2 + hr/c hr/c Now obseve that the complete intesection G has odd socle degee s(s 1 + 1; hence A = B Then it follows fom (42 that (4 h R/Is 2 hr/is 2 1 = hr/c1 2 hr/c

10 10 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Thus we obtain h R/Js ( ( s hr/js ( ( s = hr/is ( ( s 2 1 hr/is ( ( s 2 (44 + h R/C1 ( ( s 2 hr/c1 ( ( s h R/C2 ( ( s hr/c2 ( ( s = h R/C2 ( ( s hr/c2 ( ( s Combining (4 and (44, we see that it emains to show that (( (45 h s (( R/C1 2 s hr/c hr/c2 ( ( s hr/c2 ( ( s By the symmety of the h-vectos of R/C 1 and R/C 2 we see that this is equivalent to showing (46 h R/C1 1 2 hr/c hr/c2 ( ( s 1 2 hr/c2 ( ( s Now, h R/Ci (i = 1, 2 is the Hilbet function of an Atinian monomial complete intesection in R, namely R/C i, whee C i is obtained fom C i by adding the missing vaiable Futhemoe, if we eplace C 2 by D 2 = (x 1, x s+1 2, x s,, x s s, we have that R/C 2 and R/D 2 have the same Hilbet function, and D 2 C 1 But such ideals have the Weak Lefschetz popety ([12], [14] In paticula, if L is a geneal linea fom, then the left-hand side of (46 is the Hilbet function of R/(C 1 +(L in degee ( s 1 2 and the ight-hand side is the Hilbet function of R/(D2 + (L in the same degee Because of the inclusion of the ideals, (46 follows and so the poof is complete We now come to the main esult of this section The case = was poven by Benne and Kaid [4] Note that when 2, all quotients of R have the WLP by a esult of [6] Theoem 4 Let R = K[x 1,, x ], with, and conside I, = (x 1,, x, x 1 x 2 x Then the level Atinian algeba R/I, fails to have the WLP Poof Specifically, we will check that the multiplication on R/I, by a geneal linea fom fails sujectivity fom degee ( ( 2 1 to degee 2, even though the value of the Hilbet function is non-inceasing between these two degees The poof is in two steps Step 1 We fist pove this latte fact, namely that h R/I, (d 1 h R/I, (d fo d = ( 2

11 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 11 To do this, we again use basic double linkage We obseve that J 1 := (x 1, x 2,, x, x 1 x 2 (x 1, x 2,, x, x 1 + (x 1, x,, x = (x 1, x 2, x,, x, x 1 x 2 := J 2 x (x 1, x 2, x,, x, x 1 x 2 + (x 1, x 2, x 4,, x = (x 1, x 2, x, x 4,, x, x 1 x 2 x := J x (x 1,, x, x, x 1 x + (x 1,, x = (x 1,, x, x 1 x := J = I, and we note that the fist ideal, J 1, is just (x 2,, x, x 1 Futhemoe, this ideal is a complete intesection with socle degee ( 1( 2 The midpoint of the Hilbet function is in degee ( 2 Let C 1 = (x 1, x,, x and C 2 = (x 1, x 2, x 4,, x We note that the fist diffeence of the Hilbet function of R/C 1 is symmetic with odd socle degee, so the values in degees ( ( 2 and 2 +1 ae equal The key point, though, is that by applying Lemma 42 with s = 1, we obtain that (47 h R/J1 ( ( 2 hr/j1 ( ( 2 1 hr/c2 ( ( hr/c2 ( ( We ae inteested in the values of the Hilbet function of R/I, in degees ( ( 1 and 2 Since I, is obtained fom J 1 by a sequence of 1 basic double links, and since each one involves a shift by 1 of the Hilbet function, this coesponds (fist to an examination of the Hilbet function of R/J 1 in degees ( ( 2 1 ( 1 = ( 2 1 and 2 The fome is smalle than the latte, but we do not need to know the pecise values Ou obsevation in the paagaph peceding (47 shows that when we add the fist diffeence of the (shifted Hilbet function of R/C 1 to get h R/J2 ( ( 2 and hr/j2 ( ( 2 +1, the diffeence between these two values is the same as the diffeence between the values of the Hilbet function of R/J 1 in degees ( ( 2 1 and 2, with the latte being lage Howeve, the point of (47 is that when we then add the (shifted fist diffeence of the Hilbet function of R/C 2, we ovecome this diffeence and aleady have a Hilbet function with the value in degee ( ( 2 +1 lage than that in degee 2 +2 Since each subsequent Hilbet function has the (shifted value in the smalle of the coesponding degees lage than the value in the second, we finally obtain the same fo the desied Hilbet function, namely that of R/I, This concludes step 1 Step 2 To pove that R/I, fails sujectivity fom degee ( 2 1 to degee ( 2, we will use Poposition 25 (d Note that Ī, = (x 1,, x, (x x, x 1 x (x x We now claim that it is enough to veify that thee is a homogeneous fom F R = K[x 1,, x ] of degee ( 2 such that (48 F x 1 x 2 x (x x (x 1,, x and (49 F (x x (x 1,, x

12 12 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Indeed, clealy x 1,, x is a egula sequence in Ī, that extends to a minimal geneating set fo I, So Poposition 25 (d shows that it is enough to find a fom F of degee ( 1 (( ( = 2, non-zeo modulo (x 1,, x, such that (48 and (49 hold Hence ou claim holds The heat of the poof is to show that the specific polynomial F = ( 1 sgn(i 1,,i x i 1 1 x i i 1 + +i = ( 2 0 i j 2, i j i l satisfies (48 and (49 Note that F is the deteminant of a Vandemonde matix F simply consists of a sum of tems, all with coefficient 1 o 1, obtained as follows Each tem consists of a poduct of diffeent powes of the 1 vaiables (emembe that we ae in the quotient ing Namely, fo each pemutation, σ, of (0, 1, 2,, 2, we look at the tem ( 1 sgn(σ A whee A is the monomial obtained by taking the i-th vaiable to the powe given by the i-th enty in σ Fo example, if = 5 and σ = (2, 0,, 1 then sgn(σ = 1 so we have the tem x 2 1x x 4 We fist check (48 In ode fo the poduct to be contained in (x 1,, x, we need that evey tem in the poduct that does not contain at least one exponent be canceled by anothe tem in the poduct That is, we have F x 1 x 2 x (x 1 ++x (x 1,, x if and only if m=1 i 1 + +i =( 2 0 i j 2, i j i l, im 2 ( 1 sgn(i 1,,i x i 1+1 1, x im+2 m,, i i +1 = 0 Notice that we have uled out i m = 2 since othewise i m + 2 = and that tem is automatically in the desied ideal But then the hypotheses i i = ( 2, 0 i j 2 and i j i l imply that thee exists a unique intege n with 1 n 1 such that i m + 2 = i n + 1 Hence the summand ( 1 sgn(i 1,,m,,n,,i x i 1+1 1, x im+2 m,, x i +1 is cancelled against the summand ( 1 sgn(i 1,,m+1,,n 1,,i x i 1+1 1, x im+2 m,, x i +1 (Fo notational convenience we have assumed m < n but this is not at all impotant We now pove (49 We have j 1 + +j = j i 0 F (x x = F! j 1! j! xj 1 1 x j j 1 + +j = j i 0 i 1 + +i =( 2 0 i j 2, i j i l! j 1! j! xj 1 1 x j = ( 1 sgn(i 1,,i x i 1 1 x i =

13 Theefoe if and only if G := MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 1 j 1 + +j = j i 0 j 1 + +j = j i 0 i 1 + +i =( 2 0 i j 2, i j i l ( 1 sgn(i 1,,i! x i 1+j 1 j 1! j! F (x x (x 1,, x i 1 + +i = ( 2 0 i l min( 2, j l, i j i l ( 1 sgn(i 1,,i! x i 1+j 1 j 1! j! 1 x i +j 1 x i +j = 0 Given an ( 1-uple of non-negative integes j := (j 1,, j such that j j =, we set! C j := j 1! j! Notice that two ( 1-uples of non-negative integes (j 1,, j and (j 1,, j with j j = = j j veify (410! j 1! j! =! j 1! j! {j 1,, j } = {j 1,, j } Given an ( 1-uple of non-negative integes j := (j 1,, j such that j j = and an ( 1-uple α := (α 1,, α, we define (fom now on #(B means the cadinality of the set B: N α,j := #(A(α j whee A(α j is the set of monomials ±C j x α 1 1 x α in G of multidegee α and coefficient ±C j To pove (49, it is enough to see that N α,j is even and half of the elements of A(α j have coefficient +C j and the othe half have coefficient C j Let us pove it Without loss of geneality we can assume that α 1 α 2 α (we e-ode the vaiables, if necessay We will see that fo any monomial in A(α j thee is a well detemined way to associate anothe monomial in A(α j with the opposite sign Indeed, the monomials in A(α j have degee ( ( 2 + = and, moeove, 0 αl 1 fo all 1 l 1 Theefoe, thee exist integes 1 p < q 1 such that α p = α q We define p 0 := min{p α p = α p+1 } Now, we take an abitay monomial ±C j x α 1 1 x αp 0 p 0 x α p 0 +1 p 0 +1 x α A(α j whee α 1 = j 1 + i 1,, α p0 = j p0 + i p0, α p0 +1 = j p i p0 +1,, α = j + i It will be cancelled with C j x α 1 1 x αp 0 p 0 x α p 0 +1 p 0 +1 x α A(α j whee α 1 = j 1 + i 1,,α p0 = j p i p0 +1, α p0 +1 = j p0 + i p0,, α = j + i and we ae done

14 14 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Example 44 We illustate the constuction in Step 1 of the poof of Theoem 4 fo the case = 5 In the following table of Hilbet functions and h-vectos, we have J 1 = (x 1, x 4 2, x 4, x 4 4, x 4 5 J 2 = x 2 J 1 + (x 5 1, x 4, x 4 4, x 4 5 = (x 5 1, x 5 2, x 4, x 4 4, x 4 5, x 1 x 2 J = x J 2 + (x 5 1, x 5 2, x 4 4, x 4 5 = (x 5 1, x 5 2, x 5, x 4 4, x 4 5, x 1 x 2 x J 4 = x 4 J + (x 5 1, x 5 2, x 5, x 4 5 = (x 5 1, x 5 2, x 5, x 5 4, x 4 5, x 1 x 2 x x 4 J 5 = x 5 J 4 + (x 5 1, x 5 2, x 5, x 5 4 = (x 5 1, x 5 2, x 5, x 5 4, x 5 5, x 1 x 2 x x 4 x 5 In the following calculation, we have put in boldface the citical ange of degees Ideal Hilbet function/h-vecto J (5,4,4, J (5,5,4, J (5,5,5, J (5,5,5, J It is inteesting to note that expeimentally we have veified that R/J 1 and R/J 2 have the WLP, while R/J, R/J 4 and R/J 5 do not The algebas that fail to have the WLP all fail sujectivity in the ange indicated in boldface Only R/J 5 fails to have the WLP in any othe degee, namely it fails injectivity in the peceding degee As mentioned above, we now have a patial answe to Question 42 of [10] Recall that A(n is defined to be the minimum numbe (if it exists such that evey Atinian ideal I K[x 1,, x n ] with numbe of geneatos µ(i A(n has the WLP Coollay 45 If A(n exists then it equals n 5 An almost monomial almost complete intesection In ode to illustate the subtlety of the Weak Lefschetz Popety, we now descibe a class of ideals that is vey simila to the class of ideals discussed in Section 4 That is, we conside, fo each codimension, the ideal J = (x 1,, x, x 1 x (x 1 + x We will compae the popeties of this ideal with those of the ideal I, = (x 1,, x, x 1 x Included in this subtlety is the fact that the WLP behavio changes with the chaacteistic Notice that ou esults in positive chaacteistic do not depend on whethe the field is finite o not Ou fist esult shows that we cannot distinguish the two ideals by solely looking at thei Hilbet functions Lemma 51 R/J and R/I, have the same Hilbet function Poof We will show that J aises via a sequence of basic double links which ae numeically equivalent to the one that poduced I, in the fist pat of Theoem 4 Notice fist that the ideals (x 1, x 2,, x, x 1 = (x 2,, x, x 1 and (x 1,, x, x 1 + x

15 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 15 have the same Hilbet function Notice also that this latte ideal is equal to (x 1,, x, x, x 1 + x In Step 1 of Theoem 4 we saw a sequence of basic double links stating with the ideal (x 1, x 2,, x, x 1 and ending with (x 1,, x, x 1,, x We will now poduce a paallel sequence of basic double links stating with (x 1,, x, x, x 1 + x and ending with (x 1,, x, x 1 x (x 1 + x, such that at each step the two sequences ae numeically the same, and hence the esulting ideals at each step have the same Hilbet function (x 1,, x, x, x 1 + x x (x 1,, x, x, x 1 + x + (x 1,, x 2, x = (x 1, x 2,, x 2, x, x, (x 1 + x x x 2 (x 1, x 2,, x 2, x, x, (x 1 + x x + (x 1,, x, x, x = (x 1, x 2,, x, x 2, x, x, (x 1 + x x 2 x := J x 1 (x 1, x 2,, x, x, (x 1 + x x 2 x + (x 2,, x = (x 1,, x, x 1 x (x 1 + x = J This completes the poof We will now show that the two algebas behave diffeently with espect to the WLP Recall that R/I, does not have the WLP if Studying R/J when = is not too difficult: Poposition 52 Fo evey field K, the algeba R/J = K[x, y, z]/(x, y, z, xy(x + z has the WLP if and only if the chaacteistic of K is not thee Poof If the chaacteistic of K is thee then fo evey linea fom l R, l is in (x, y, z Thus the esidue class of l 2 is in the kenel of the multiplication map l : (R/J 2 (R/J This shows that R/J does not have the WLP if cha K = Now assume that cha K Conside the linea fom L = x + y + z Then one checks that (J, L/(L = ((x, y, L/(L, which implies that the multiplication map L : (R/J 2 (R/J is sujective Hence R/J has the WLP in this case The case when = 4 is consideably moe complicated

16 16 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Poposition 5 Fo evey field K, the algeba R/J 4 = K[w, x, y, z]/(w 4, x 4, y 4, z 4, wxy(w + z has the WLP if and only if the chaacteistic of K is not two o five Poof The Hilbet function of R/J 4 is 1, 4, 10, 20, 0, 6, 4, Hence, by Poposition 21, R/J 4 has the WLP if and only if, fo a geneal fom L, the multiplication maps L : (R/J 4 4 (R/J 4 5 and L : (R/J 4 5 (R/J 4 6 ae injective and sujective, espectively We fist show that the latte map is sujective if L := 2w + x + y + z, povided the chaacteistic of K is neithe 2 no 5 Notice that this map is sujective if and only if (R/(J 4, L] 6 = 0 Since R/(J 4, L = K[w, x, y]/(w 4, x 4, y 4, (2w + x + y 4, wxy(w + x + y this is equivalent to the fact that dim K ((w 4, x 4, y 4, (2w + x + y 4, wxy(w + x + y 6 = 28 To compute the dimension of ((w 4, x 4, y 4, (2w + x + y 4, wxy(w + x + y 6, we conside the coefficients of the 28 degee 6 monomials in K[w, x, y] occuing in each of the 0 polynomials fq, whee f is one of the foms w 4, x 4, y 4, (2w + x + y 4, wxy(w + x + y and q is one of the quadics w 2, wx, x 2, wy, xy, y 2 Compute these coefficients assuming, tempoaily, that cha K = 0, and ecod them in a 0 28 matix M whose enties ae integes Using CoCoA we veified that the geatest common diviso of all the maximal minos of M is 20 = This shows that the matix M has ank 28 if and only if cha K 2, 5 We now discuss the map L : (R/J 4 4 (R/J 4 5, whee L is a geneal linea fom This map is injective if and only if dim K (R/(J 4, L 5 = 6, which is equivalent to dim K ((J 4, L/(L 5 = 15 Assume fist that the field K is infinite Then an agument simila to the one in the poof of Poposition 22 shows we may assume that whee t K Then L := tw + x + y z, R/(J 4, L = K[w, x, y]/(w 4, x 4, y 4, (tw + x + y 4, wxy((t + 1 w + x + y To compute the dimension of ((w 4, x 4, y 4, (tw + x + y 4, wxy((t + 1 w + x + y 5, we conside the coefficients of the 21 degee 5 monomials in K[w, x, y] occuing in each of the 15 polynomials fl, whee f is one of the foms w 4, x 4, y 4, (tw+x+y 4, wxy((t+1 w+x+y and l is one of the vaiables w, x, y Compute these coefficients assuming, tempoaily, that cha K = 0, and ecod them in a matix N whose enties ae polynomials in Z[t] A CoCoA computation povides that all maximal minos of N ae divisible by 10 and that one of the minos is 80t 4 (t It follows that the ank of N is 15 if and only if cha K 2, 5 Hence we have shown that ove an infinite field, fo a geneal linea fom L, the map L : (R/J 4 4 (R/J 4 5 is injective if and only if the chaacteistic of K is neithe 2 no 5 This also implies that R/J 4 does not have the WLP if K is a finite field of chaacteistic 2 o 5 Futhemoe, evey field whose chaacteistic is not 2 o 5 contains an element t such that 80t 4 (t is not zeo Hence the above aguments show that in this case thee is a linea fom L such that L : (R/J 4 4 (R/J 4 5 is injective Combining this with the fist pat of the poof, ou assetion follows

17 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 17 Remak 54 (i Fo R/J 4, Poposition 5 povides an affimative answe to Poblem 2 (ii We expect that in chaacteistic zeo, fo each intege 2, the algeba R/J has the WLP and that L = 2x 1 + x x x is a Lefschetz element 6 Monomial almost complete intesections in thee vaiables Now we conside ideals of the fom I = (x a, y b, z c, x α y β z γ in R = K[x, y, z], whee 0 α < a, 0 β < b and 0 γ < c This class of ideals was fist consideed in [], Coollay 7 Poposition 61 If I is as above and is not a complete intesection then (i The invese system fo I is given by (x a 1 y b 1 z γ 1, x a 1 y β 1 z c 1, x α 1 y b 1 z c 1, whee we make the convention that if a tem has an exponent of 1 (eg if γ = 0, that tem is emoved (ii In paticula, if α, β, γ > 0 then R/I has Cohen-Macaulay type Othewise it has Cohen-Macaulay type 2 (iii If α, β, γ > 0, the socle degees of R/I ae b + c + α, a + c + β, a + b + γ In paticula, R/I is level if and only if a α = b β = c γ (iv Suppose that one of α, β, γ = 0; without loss of geneality say γ = 0 Then the coesponding socle degee in (iii, namely a + b + γ, does not occu Now R/I is level if and only if a α = b β, and c is abitay (v Let J = I : x α y β z γ be the ideal esidual to I in the complete intesection (x a, y b, z c Then J = (x a α, y b β, z c γ (vi A fee esolution of R/I is (61 0 R( α b c R( β a c R( γ a b R( α β c R( α γ b R( β γ a R( b c R( a c R( a b R( α β γ R( a R( b R( c This is minimal if and only if α, β, γ ae all positive R R/I 0 Poof Pat (i follows by inspection Then (ii, (iii and (iv follow immediately fom (i As befoe, (v is a simple computation of the colon ideal, based on the fact [2] that J is a complete intesection, so it only emains to check the degees Having (v, it is a staightfowad computation using the mapping cone to obtain (vi Theoem 62 Assume that K = K is an algebaically closed field of chaacteistic zeo Fo I = (x a, y b, z c, x α y β z γ, if the WLP fails then a + b + c + α + β + γ 0 (mod Poof Let E be the syzygy bundle of I and let L = P 1 be a geneal line By [4] Theoem, if the WLP fails then E is semistable Futhemoe, the splitting type of E nom must be (1, 0, 1 (apply [4], Theoem 22 and the Gauet-Mülich theoem Hence the twists of E L ae thee consecutive integes Since the estiction of E to L is the (fee syzygy

18 18 J MIGLIORE, R MIRÓ-ROIG, U NAGEL module coesponding to the estiction of the geneatos of I to P 1, we see that the sum of the geneatos must be divisible by Coollay 6 Assume that K = K is an algebaically closed field of chaacteistic zeo If R/I is level and the WLP fails then a + b + c 0 (mod and α + β + γ 0 (mod Poof Since R/I is level, by Poposition 61 we can wite a = α + t, b = β + t, c = γ + t fo some t 1 By Theoem 62, we have 2(α + β + γ + t 0 (mod It follows that α + β + γ 0 (mod, so again by Theoem 62 we also get a + b + c 0 (mod Remak 64 The poof of Theoem 62 applies not only to monomial ideals Indeed, fo any almost complete intesection in R = K[x, y, z], if the WLP fails then 4 i=1 d i 0 (mod, whee d 1, d 2, d, d 4 ae the degees of the minimal geneatos A vey inteesting class of ideals is the following, ecalling the notation intoduced in (1 Coollay 65 The algeba A = R/I,k, has the following popeties (a the socle degee is e = 2k 2 (b the peak of the Hilbet function occus in degees k 1 and k, and has value k (c The coesponding invese system is (x k 1 y k 1, x k 1 z k 1, y k 1 z k 1 (d Assuming cha K 2, the WLP fails if and only if k is odd Note that in this case e 0 (mod 4 Poof Pats (a, (b and (c ae immediate fom Poposition 61 Pat (d is a special case of Theoem Fo the emainde of this section we focus on the WLP If the non-pue monomial involves only two of the vaiables, then the ideal always has the WLP Lemma 66 Adopt the above notation If α = 0 and K has chaacteistic zeo, then R/I has the WLP Poof The assumption α = 0 povides that R/I is isomophic to B C, whee B = K[y, z]/(y b, y β z γ, z c and C = K[x]/(x a By Poposition 44 in [6], B and C have the Stong Lefschetz Popety (SLP see Example 81, hence A has the WLP by Poposition 22 of [7] If the non-pue monomial involves all thee vaiables then, due to Theoem 61(iii, the ideal is level if and only if I is of the fom (62 I α,β,γ,t = (x α+t, y β+t, z γ+t, x α y β z γ, whee t 1 and, without loss of geneality, 1 α β γ Next, we analyze when the syzygy bundle of I is semistable (The elevance of semistability to the WLP was intoduced in [6] and genealized in [4] Lemma 67 Assume that K is algebaically closed of chaacteistic zeo Then the syzygy bundle of I is semistable if and only if γ 2(α + β and 1 (α + β + γ t Poof Benne ([], Coollay 7 shows that in geneal fo an ideal I = (x a, y b, z c, x α y β z γ, the syzygy bundle is semistable if and only if (i max{a, b, c, α + β + γ} a + b + c + α + β + γ, and

19 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 19 (ii min{α + β + c, α + b + γ, a + β + γ, a + b, a + c, b + a} a + b + c + α + β + γ Applying this to ou ideal I condition (i eads as Hence, it is equivalent to max{γ + t, α + β + γ} 2(α + β + γ + t (6 γ 2(α + β and (64 Condition (ii eads in ou case as which is equivalent to 1 (α + β + γ t min{α + β + γ + t, α + β + t} t + 2 (α + β + γ, t 2γ α β Using Inequality (6 one checks that the last condition is implied by Inequality (64 This completes the agument Suppose we ae given 1 α β and want to choose γ, t such that the syzygy bundle of I is semistable Then thee is only a finite numbe of choices fo γ since we must have β γ 2(α + β, wheeas we have infinitely many choices fo t as the only condition is t 1 (α + β + γ If the syzygy bundle of I is not semistable, then R/I must have the WLP ([4], Theoem Combining this with Coollay 6 and Lemma 67, ou compute expeiments suggest the following chaacteization of the pesence of the WLP in chaacteistic zeo Conjectue 68 Let I R = K[x, y, z] be a level Atinian almost complete intesection, ie, I is of the fom (x α+t, y β+t, z γ+t, x α y β z γ, whee t > 0 and, without loss of geneality, 0 α β γ algebaically closed field of chaacteistic zeo Then: Assume that K is an (a R/I has the WLP if any of the following conditions is satisfied: (i α = 0, (ii α + β + γ is not divisible by, (iii γ > 2(α + β, (iv t < 1 (α + β + γ (b Assume that 1 α β γ 2(α + β, α + β + γ 0 (mod, and t 1(α + β + γ Then R/I fails to have the WLP if and only if t is even and eithe of the following two conditions is satisfied: (i α is even, α = β and γ α (mod 6; (ii α is odd and α = β and γ α 0 (mod 6, o β = γ and γ α 0 (mod Futhemoe, in all of the above cases, the Hilbet function has twin peaks

20 20 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Note that pat (a is tue by Lemma 66, Coollay 6, and Lemma 67 Pat (b will be discussed in the following section, including the fact that thee ae exactly two known counteexamples, which wee pointed out to us by David Cook II afte eading the fist vesion of this pape 7 A poof of half of Conjectue 68 We ae going to establish sufficiency of the numeical conditions given in Conjectue 68(b fo failue of the WLP Fist we establish the claim about the twin peaks of the Hilbet function Lemma 71 Conside the ideal I α,β,γ,t = (x α+t, y β+t, z γ+t, x α y β z γ which (by Poposition 61 defines a level algeba Assume that t max { } 2γ α β, α+β+γ and that 0 < α β γ Assume futhemoe that α + β + γ 0 (mod Then the values of the Hilbet function of R/I α,β,γ,t in degees 2(α+β+γ + t 2 and 2(α+β+γ + t 1 ae the same Poof By Poposition 61, we know the minimal fee esolution of R/I α,β,γ,t, which we can use to compute the Hilbet function in any degee We fist claim that in the specified degees, this computation has no contibution fom the last and the penultimate fee modules in the esolution To do this, it is enough to check that the degee 2(α+β+γ +t 1 component of any summand in the penultimate fee module is zeo The fist thee summands coespond to the obsevation that α β γ 1 < 0 Since α β γ and a = α + t, b = β + t, and c = γ + t, we have only to check that 2(α + β + γ + t 1 α t β t < 0 This is equivalent to the inequality on t in the hypotheses Now athe than explicitly computing the Hilbet functions in the two degees, it is enough to expess them as linea combinations of binomial coefficients and show that the diffeence is zeo, using the fomula ( ( p 2 p 1 2 = p 1 This is a outine computation Theoem 72 Conside the level algeba R/I α,β,γ,t = R/(x α+t, y β+t, z γ+t, x α y β z γ We make the following assumptions: 0 < α β γ 2(α + β; t α + β + γ ; α + β + γ 0 (mod Then thee is a squae matix, M, with intege enties, having the following popeties ( (a M is a t + α + β 2γ ( t + α + β 2γ matix (b If det M 0 (mod p, whee p is the chaacteistic of K, then R/I α,β,γ,t fails to have the WLP This includes the possibility that det M = 0 as an intege (c If det M 0 (mod p then R/I α,β,γ,t satisfies the WLP Poof We note fist that the second bullet in the hypotheses implies (using the fist bullet that the following inequality also holds: t > 2γ α β We will use this fact without comment in this poof

21 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 21 Thanks to Lemma 71, the values of the Hilbet function of R/I α,β,γ,t in degees 2(α+β+γ + t 2 and 2(α+β+γ + t 1 ae the same Hence thanks to Poposition 21, checking whethe o not the WLP holds is equivalent to checking whethe multiplication by a geneal linea fom between these degees is an isomophism o not We will use Poposition 25 Let L be a geneal linea fom, let R = R/(L = K[x, y] and let Ī be the image of I α,β,γ,t in R Note that Ī = (x α+t, y β+t, l γ+t, x α y β l γ, whee l is the estiction to R of z, and thanks to Lemma 22 we will take l = x + y Of couse x α+t, y β+t is a egula sequence Hence it suffices to check whethe o not thee is an element F R of degee [ ] 2(α + β + γ f := α+t+β+t + t = α + β 2γ +t 1 = α + β + γ γ+t 1, non-zeo modulo (x α+t, y β+t, such that F Ī (xα+t, y β+t equivalent to (71 F (x + y γ+t (x α+t, y β+t and F x α y β (x + y γ (x α+t, y β+t The latte condition is Claim: (1 γ 2(α + β if and only if F (x + y γ+t is automatically in (x α+t, y β+t (2 t α + β + γ if and only if F x α y β (x + y γ is automatically in (x α+t, y β+t Indeed, the fist inequality is equivalent to deg(f (x + y γ+t (α + t + (β + t 1, so evey tem of F (x + y γ+t is divisible by eithe x α+t o y β+t The second inequality is equivalent to deg(f x α y β (x + y γ (α + t + (β + t 1 This establishes the claim Thanks to ou hypotheses, then, the conditions in (71 add constaints on the possibilities fo F We want to count these constaints Let F = λ 0 x f + λ 1 x f 1 y + λ 2 x f 2 y λ f 2 x 2 y f 2 + λ f 1 xy f 1 + λ f y f We now conside how many conditions (71 imposes on the λ i Conside the fist poduct, which has degee α+β+γ +2t 1 A typical tem in F (x+y γ+t is some scala times x i y j, whee i + j = α+β+γ + 2t 1 The set of all pais (i, j fo which x i y j is not in (x α+t, y β+t is {( (72 {(i, j} = α + t 1, β + γ 2α ( } α + γ 2β + t,, + t, β + t 1 Since each such tem has to vanish, this imposes a total of 2α+2β γ conditions on the λ i Similaly, conside the second poduct, which has degee 4α+4β+γ + t 1 A typical tem in F x α y β (x + y γ is some scala times x i y j, whee i + j = 4α+4β+γ + t 1 The set of all pais (i, j fo which x i y j is not in the ideal (x α+t, y β+t is {( (7 {(i, j} = α + t 1, α + 4β + γ ( } 4α + β + γ,,, β + t 1 This imposes a total of t α+β+γ conditions, since we need all of these tems to vanish Combining, we have a total of t+ α+β 2γ = f +1 conditions Since thee ae f +1 vaiables λ i, the coefficient matix is the desied squae matix Now it is clea that det M = 0 (egadless of the chaacteistic if and only of the coesponding homogeneous system has a non-tivial solution, ie thee is a polynomial F as desied, if and only if R/I α,β,γ,t fails to have the WLP We can specifically give the matix descibed in the last esult

22 22 J MIGLIORE, R MIRÓ-ROIG, U NAGEL Coollay 7 The matix in Theoem 72 has the fom M = ( γ α+β+γ ( γ α+β+γ 1 ( γ ( γ t 1 t 2 ( γ+t ( γ+t t+β 1 t+β 2 ( γ+t ( γ+t t+β 2 t+β ( γ+t t+ β+γ 2α ( γ ( γ γ t+2 γ t+1 ( γ 2γ α β +1 ( γ+t 2(β+γ α +1 ( γ 2γ α β ( γ+t 2(β+γ α ( γ+t ( γ+t 2(β+γ α 2(β+γ α 1 ( γ+t ( t 1+ γ+t ( γ+t β+γ 2α γ α+2 γ α+1 Poof This is a tedious computation, but is based entiely on the poof of Theoem 72 The top half of the matix coesponds to the second poduct in (71, and the bottom half of the matix coesponds to the fist poduct Each ow in the top half coesponds to one odeed pai in (7, and each ow in the bottom half coesponds to one odeed pai in (72 The following coollay establishes the sufficiency of the numeical conditions given in Conjectue 68 Coollay 74 Let K be an abitay field and R = K[x, y, z] Conside the ideal I α,β,γ,t = (x α+t, y β+t, z γ+t, x α y β z γ, whee 1 α β γ Assume that one of the following thee cases holds: (1 (α, β, γ, t = (α, α, α + λ, t with α even, λ odd, t α + λ even and 1 λ α; (2 (α, β, γ, t = (α, α, α + 6µ, t with α odd, t α + 2µ even, and 0 µ α 1; o 2 ( (α, β, γ, t = (α, α + ρ, α + ρ, t with α odd, t α + 2ρ even, and ρ 0 Then R/I α,β,γ,t fails to have the WLP Poof Possibly afte an extension of the base field, we may assume that K is an infinite field One can veify quickly (using the constaints on the invaiants given in the theoem that the hypotheses of Theoem 72 hold hee in all thee cases (the paity is impotant in some instances Hence it is only a matte of identifying M, via Coollay 7, and checking that in all the cases mentioned, det M = 0 We fist conside Case (1

23 MONOMIAL IDEALS AND THE WEAK LEFSCHETZ PROPERTY 2 By applying Coollay 7, we obtain the (t 2λ (t 2λ matix M below, coesponding to t 2λ homogeneous equations in t 2λ unknowns M = ( α+λ ( α+λ α+λ α+λ 1 ( α+λ ( α+λ α+λ+1 ( α+λ t 1 ( α+t+λ t+λ ( α+t+λ t+λ+1 ( α+t+λ α+t 1 α+λ ( α+λ t 2 ( α+t+λ λ 1 ( α+t+λ t+λ ( α+t+λ α+t 2 ( α+λ α t+λ+2 ( α+λ α t+λ+1 ( α+λ ( α+λ α t+λ+ α t+λ+2 ( α+λ 2λ+1 ( α+t+λ λ+2 ( α+t+λ λ+ ( α+t+λ α+2λ+1 ( α+λ 2λ ( α+t+λ λ+1 ( α+t+λ λ+2 ( α+t+λ α+2λ This system has a non-tivial solution (giving the existence of the desied fom F if and only if M has deteminant zeo We will show that unde ou assumptions, this deteminant is indeed zeo Obseve that if M is flipped about the cental vetical axis, and then the top potion and the bottom ae (sepaately flipped about thei espective cental hoizontal axes, then we estoe the matix M Since t 2λ is even, the fist step can be accomplished with t 2λ intechanges 2 of columns The top potion contains t λ α ows and the bottom potion contains α λ ows Both ae odd, so the second step can be done with t λ 1 α intechanges and 2 the last one with α λ 1 intechanges All togethe we have t 2λ 1 intechanges of 2 ows/columns, which is an odd numbe Theefoe det M = det M, and so det M = 0 Case (2 is simila and is left to the eade Case (, howeve, is somewhat diffeent Now we will estict to K[y, z] athe than K[x, y] We obtain Ī = ((y + z α+t, y α+ρ+t, z α+ρ+t, (y + z α y α+ρ z α+ρ and we have to check whethe thee is a fom F K[y, z] of degee t + 2ρ 1 (obtained afte a shot calculation such that F (y + z α+t (y α+ρ+t, z α+ρ+t and F (y + z α y α+ρ z α+ρ (y α+ρ+t, z α+ρ+t The calculations again follow the ideas of Theoem 72, and we obtain the following (t + 2ρ (t + 2ρ matix of integes: