The Least Generating Degree of Symbolic Powers of Fermat-like Ideals of Planes and Lines Arrangements
aa r X i v : . [ m a t h . A C ] F e b THE LEAST GENERATING DEGREE OF SYMBOLIC POWERS OFFERMAT-LIKE IDEALS OF PLANES AND LINES ARRANGEMENTS
TH ´AI TH `ANH NGUY˜ˆEN
Abstract.
We compute explicitly the least degree of generators of almost all symbolicpowers of the defining ideal of Fermat-like configuration of lines in P C and provide efficientbounds for remaining cases. We will also compute explicitly those numbers for ideal deter-mining the singular locus of the arrangement of lines given by the pseudoreflection group A . As direct applications, we verify Chudnovsky’s(-like) Conjecture, Demailly’s(-like) Con-jecture and Harbourne-Huneke Containment problem as well as calculate the Waldschmidtconstant and (asymptotic) resurgence number. Introduction
Let n ≥ I n be the ideal in C [ x, y, z, w ] generated by: g = ( x n − y n )( z n − w n ) xy, g = ( x n − y n )( z n − w n ) zwg = ( x n − z n )( y n − w n ) xz, g = ( x n − z n )( y n − w n ) ywg = ( x n − w n )( y n − z n ) xw, g = ( x n − w n )( y n − z n ) yz which we will refer as Fermat-like ideal . This ideal corresponds to the restricted Fermatarrangement of planes in P , where the correspondent variety is the union of all lines withmultiplicity at least 3, i.e, there are at least 3 planes passing through each line. Fermat-likeideals were first introduced by Malara and Szpond in their work [MS17] in an effort to pro-vide counterexamples in higher dimension to the famous containment I (3) ⊆ I . It can beseen as an analog of Fermat ideals in higher dimension. For more information about
Fermatideals , we refer interested readers to [NS16], [DSTG13], [HS15], [Szp19],[Ngu21].In this paper, we will discover many similarities of these ideals and
Fermat ideals in termsof the least degree of generators of their symbolic powers,
Waldschmidt constants , (asymp-totic) resurgence numbers and especially, the fact that the verification to Harbourne-Hunekecontainment of them can also be decided purely by the knowledge of the aforementionednumerical invariants. This work is a continuation of our paper [Ngu21] where we answerthose questions for Fermat ideals. Our results on Fermat-like ideals are the following:
Theorem 1.1.
Let n ≥ be an integer and Fermat-like ideal I n in C [ x, y, z, w ] as describedabove. Then:(1) For n ≥ , α ( I ( m ) n ) = 2 nm for all m ≥ .(2) α ( I ( m )4 ) = 8 m for all m ≥ and m = 5 . Mathematics Subject Classification.
Key words and phrases.
Fermat Ideals, Fermat Lines Configuration, Resurgence Number, WaldschmidtConstant, Ideals of Points, Ideals of Lines, Symbolic Powers, Containment problem, Stable Harbourne–Huneke Conjecture, Interpolation Problem. α ( I (3 k )3 ) = 18 k for k ≥ , α ( I (3 k +1)3 ) = 18 k + 8 and α ( I (3 k +2)3 ) = 18 k + 16 for k ≥ .(4) n + 2 ≤ α ( I (2) n ) ≤ n + 4 = α ( I n ) .(5) b α ( I n ) = 2 n . This gives the almost complete answer to the question of the least degree of generatorsof almost all symbolic powers of the defining ideal of Fermat-like configuration of lines in P C and bounds only for the case when m = 2. Note that for each m, n the least generatingdegree of symbolic power of the Fermat-like ideal is twice as large as that of Fermat ideal inalmost all cases (and should be in all cases as suggested by Macaulay2 computation [GS]).One important problem that has attracted a lot of attention recently is the containmentproblem , namely, to determine the set of pairs ( m, r ) for which I ( m ) ⊆ I r for a given ideal I .Following the celebrated results in [ELS01] and [HH02] that I ( m ) ⊆ I r whenever m ≥ N r ,the resurgence ρ ( I ) is introduced in [BH10], and the asymptotic resurgence b ρ ( I ) is introducedin [GHVT13] in order to study the pairs ( r, m ) numerically and has turned out to be veryuseful invariants. In term of these invariants, we prove the followings. Theorem 1.2.
For ideal I n where n ≥ :(1) b ρ ( I n ) = n + 1 n for all n ≥ .(2) ρ ( I ) = 32 .(3) For n ≥ , the resurgence number ρ ( I n ) can only be one of the following numbers: , , , or . Another motivation for computing the least degree of generators of their symbolic powersis to provide more evidence for ideals that satisfy
Chudnovsky-like inequality and
Demailly-like inequality as well as
Harbourne-Huneke Containment and stable containment. Thismotivation stems from the previous work by Bisui, Grifo, H`a and the author, see [BGHN20b,Section 3] and containment in [BGHN20a]. Similar to the Fermat ideals, the verificationand failure of the containment for I n can be checked purely by their numerical invariantsincluding the least degree of generators of symbolic powers, the regularity, or the resurgencenumber and the maximal degree of generators. As direct applications, we will give anevidence to a similar question as question 3 . Harbourne-Huneke Containment , stable Harbourne Containment and somestronger containment. Corollary 1.3.
Fermat-like ideals verifies the following containment(1) Harbourne-Huneke containment (see [HH13, Conjecture 2.1] for ideal of points) I (2 r ) n ⊆ m r I rn , ∀ r (2) Harbourne-Huneke stable containment (see [HH13, Conjecture 4.1.5] for ideal ofpoints) I (2 r − n ⊆ m r − I rn , ∀ r ≫ (3) A stronger stable containment I (2 r − n ⊆ m r I rn , ∀ r ≫ s another direct application of these computations, we will give an evidence to question3 . Chudnovsky-like inequality and
Demailly-like inequality . Corollary 1.4.
Fermat-like ideals satisfy Demailly-like inequality (and hence, Chudnovsky-like inequality) b α ( I n ) ≥ α ( I ( m ) n ) + h − m + h − for all m , where h = 2 . Along the way, we give the free resolution, Castelnuovo-Mumford regularity formula for I n and use them to improve the stable version of the above containment to a condition on r for I .In previous work [Ngu21], we investigate the above questions for Fermat ideals. In particu-lar, the Fermat ideal for 7 points can be seen as ideal of the singular locus of the arrangementof lines given by the pseudoreflection group G (2 , ,
3) = D . We continue to study the idealcorresponding to the group G (1 , ,
4) = A in this paper and give some discussion for group G (2 , ,
3) = B . These are 3 groups that have correspondent ideals with small degree of gen-erators, that had to be considered separately in the result involving Harbourne containmentin [DS20, Proposition 6.3]. It turns out interestingly that the ideal corresponding to thegroup A is the same as that of group D in term of least degree of generators of symbolicpowers, Waldschmidt constant, (asymptotic) resurgence numbers; and hence, satisfies Chud-novsky’s Conjecture and Demailly’s Conjecture as well as Harbourne-Huneke Containment,stable Harbourne Containment and some stronger containment. Theorem 1.5.
Let J be the ideal of the singular locus of the arrangement of lines given bythe group G (1 , ,
4) = A . Then(1) α ( J (2 k ) ) = 5 k for all k ≥ and α ( J (2) ) = 6 .(2) α ( J (2 k +1) ) = 5 k + 3 for all k ≥ .(3) b α ( J ) = 52 .(4) ρ ( J ) = b ρ ( J ) = 65 . Similar to the Fermat ideals and Fermat-like ideals, the verification and failure of thecontainment J can be checked purely by their numerical invariants including the least degreeof generators of symbolic powers, the regularity, or the resurgence number and the maximaldegree of generators. Corollary 1.6.
Ideal J verifies the following containment and stable containment(1) Harbourne-Huneke containment (see [HH13, Conjecture 2.1] ) J (2 r ) ⊆ m r J r , ∀ r (2) Harbourne-Huneke containment (see [HH13, Conjecture 4.1.5] ) J (2 r − ⊆ m r − J r , ∀ r ≥ (3) A stronger containment J (2 r − ⊆ m r J r , ∀ r ≥ e work over the field of complex numbers but our results hold over any algebraicallyclosed field of characteristic 0. Acknowledgements.
The author would like to thank his advisor, Dr. T`ai H`a, for intro-ducing him this subject and giving many helpful suggestions and comments. He also thanksDr. Alexandra Seceleanu and Abu Chackalamannil Thomas for comments on an early draftof this manuscript. 2.
Preliminary
Let R = C [ x , . . . , x N ] be the homogeneous coordinate ring of P N , and let m be its maximalhomogeneous ideal. For a homogeneous ideal I ⊆ R , let α ( I ) denote the least degree of ahomogeneous polynomial in I , and let I ( m ) := \ p ∈ Ass(
R/I ) I m R p ∩ R denote its m -th symbolic power .The Waldschmidt constant of I is defined to be the limit and turned out to be the infimum b α ( I ) := lim m →∞ α ( I ( m ) ) m = inf m →∞ α ( I ( m ) ) m In studying ideals defining sets of points in P N , Chudnovsky’s Conjecture gives a lowerbound for α ( I ( m ) ) in term of α ( I ) , N and m as follows. Conjecture 2.1 (Chudnovsky) . Let I be the defining ideal of a set of points in P N C . Then,for all n ≥ α ( I ( n ) ) n ≥ α ( I ) + N − N It is also natural to ask if the
Chudnovsky-like inequality is still true for a homogeneousradical ideal I , for example, if we replace N by bigheight h of I . Many partial results areknown for Chudnovsky’s Conjecture, for example, in [EV83, BH10, HH13, GHM13, Dum15,DTG17, FMX18, BGHN20a]. Recently, the conjecture was proved for a very general set ofpoints in [DTG17, FMX18] and for a general set of sufficiently many points in [BGHN20a].On the other hand, Chudnovsky-like inequality was verified for a set of very general lines in P in [DFSTG19]. The following generalization dues to Demailly [Dem82]. Conjecture 2.2 (Demailly) . Let I be the defining ideal of a set of points in P N C . Let m ∈ N be any fixed integer. Then, for all n ≥ α ( I ( n ) ) n ≥ α ( I ( m ) ) + N − m + N − Demailly’s Conjecture for N = 2 was proved by Esnault and Viehweg [EV83]. In higherdimension, Demailly’s Conjecture holds for a very general set of sufficiently many points (thenumber of points depends on m ) in [MSS18] and for a general set of k N points in [CJ20].In [BGHN20b], the conjecture was proved for a general set of sufficiently many points (alsodepends on m ). We also raise a question [BGHN20b, Question 3.2] for ideal that verifiedDemailly-like inequality and show some examples of such ideals including: defining ideal ofa codimension h star configuration in P N , generic determinantal ideals, determinantal ideals f symmetric matrices and pfaffian ideals of skew symmetric matrices.The resurgence number ρ ( I ) is introduced in [BH10] as ρ ( I ) := sup { mr : I ( m ) I r } and the asymptotic resurgence b ρ ( I ) is introduced in [GHVT13] as b ρ ( I ) := sup { mr : I ( mt ) I rt for t ≫ } in effort to study the containment problem numerically, namely, the pairs ( m, r ) such that I ( m ) ⊆ I r . It is a celebrated result in [ELS01] and [HH02] that I ( hr ) ⊆ I r for codimension h ideal I . In another effort to improve this containment for ideal of points and study Chud-novsky’s Conjecture and Demailly’s Conjecture, Harbourne and Huneke in [HH13] conjec-tured that the defining ideal I for any set of points in P N satisfies some stronger containment,namely, I ( Nm ) ⊆ m m ( N − I m and I ( Nm − N +1) ⊆ m ( m − N − I m , for all m >
1. We also raisea question [BGHN20b, Question 3.1] for other classes of ideals that satisfies some version of
Harbourne-Huneke Containment (replacing N by h ) and showed some examples includingthe ideals mentioned in previous paragraph. Note that suitable version of stable Harbourne-Huneke Containment (or even infinitely many such containment) would imply Chudnovsky’sand Demailly’ Conjecture (or Demailly-like inequality), for example, see [BGHN20b]. We re-fer interested readers to [CHHVT20] for more information about the Waldschmidt constant,resurgence number, containment between symbolic and ordinary powers of ideals.3. Ideal of restricted Fermat configuration of lines in P Recall that Fermat-like ideal I n is the ideal generated by g = ( x n − y n )( z n − w n ) xy, g = ( x n − y n )( z n − w n ) zwg = ( x n − z n )( y n − w n ) xz, g = ( x n − z n )( y n − w n ) ywg = ( x n − w n )( y n − z n ) xw, g = ( x n − w n )( y n − z n ) yz From [MS17], geometrically, I n is the defining ideal of the union of lines with multiplicityat least 3 of the Fermat arrangement of flats (planes) in P , that is defined by the vanishingof the polynomial F n = ( x n − y n )( z n − w n )( x n − z n )( y n − w n )( x n − w n )( y n − z n )There are 4 n + 6 lines in the above restricted Fermat configuration of lines. Notice also thatfrom this description, each of 6 n planes H j passes through exactly 2 n + 1 lines, for example,the plane x = ǫy passes through the line defined by ( x, y ), n lines defined by ( x − ǫy, x − ǫ k z )and n lines defined by ( x − ǫy, y − ǫ k w ) for k = 0 , . . . n − ǫ in an n − root of 1.Algebraically, let f = ( x n − y n )( z n − w n ) , g = ( x n − z n )( y n − w n ) , h = ( x n − w n )( y n − z n )and K n = ( f, g ), we can write I n = K n ∩ ( x, y ) ∩ ( x, z ) ∩ ( x, w ) ∩ ( y, z ) ∩ ( y, w ) ∩ ( z, w )and since f, g form a regular sequence, for any m ≥ I ( m ) n = K mn ∩ ( x, y ) m ∩ ( x, z ) m ∩ ( x, w ) m ∩ ( y, z ) m ∩ ( y, w ) m ∩ ( z, w ) m ecall our main strategy in previous work [Ngu21], we will study a subsequence of α ( I ( m ) n ),which gives us information about b α ( I n ), then use this to calculate other α ( I ( m ) n ). Proposition 3.1.
For n ≥ , we have b α ( I n ) = 2 n .Proof. For every m ≥
1, since I ( m ) n ⊂ K mn , we have α ( I ( m ) n ) ≥ α ( K mn ) = 2 nm . Thus, b α ( I n ) ≥ n . On the other hand, since h = g − f ∈ K n , we have F mn = f m g m h m ∈ K mn . Furthermore,since nm ≥ m , F mn ∈ ∩ ( x, y ) m ∩ ( x, z ) m ∩ ( x, w ) m ∩ ( y, z ) m ∩ ( y, w ) m ∩ ( z, w ) m so F mn ∈ I (3 m ) , hence α ( I (3 m ) n ) ≤ nm for every m ≥
1. Therefore b α ( I n ) ≤ n and it followsthat b α ( I n ) = 2 n . (cid:3) Now following exactly the same argument as those of [Ngu21, Section 3] we have thefollowing two results:
Theorem 3.2.
For n ≥ , we have α ( I ( m ) n ) = 2 nm for all m ≥ .Proof. The proof is identical to that of [Ngu21, Theorem 3.2]. (cid:3)
Theorem 3.3.
For n = 4 , for all m ≥ but m = 5 , α ( I ( m )4 ) = 8 m Proof.
The proof is identical to that of [Ngu21, Theorem 3.3]. For m = 5, we can check bythe same argument that the element yzf g h ∈ I (5)4 . Macaulay2 computation [GS] suggeststhat, in fact, α ( I (5)4 ) = 42. (cid:3) Now we calculate the least degree of generators of symbolic powers for I . Theorem 3.4.
For n = 3 and for m ≥ we have the following(1) α ( I (3 m )3 ) = 18 m for m ≥ ,(2) α ( I (3 m +1)3 ) = 18 m + 8 for m ≥ ,(3) α ( I (3 m +2)3 ) = 18 m + 16 for m ≥ .Proof. (1) Since b α ( I ) = 6, α ( I (3 m )3 ) ≥ m for m ≥
1. As we see earlier, F m ∈ I (3 m )3 for m ≥ α ( I (3 m )3 ) ≤ m and thus α ( I (3 m )3 ) = 18 m for m ≥ m ≥ α ( I (3 m +1)3 ) ≤ m + 7. Then there is adivisor D of degree 18 m + 7 vanishing to order at least 3 m + 1 along every line of42 lines L i in the restricted Fermat configuration. Intersecting D with any of the18 planes H j , since each planes H j contains exactly 7 lines, suppose that D doesn’tcontain H j then by the generalized Bezout Theorem, the intersection of D and H j isof dimension 1 and degree 18 m +7, which is a contradiction since 18 m +7 < m +1).Thus we conclude that each H j is a component of D . Hence, there exists a divisor D ′ = D − P j =1 H j of degree 18( m −
1) + 7 vanishing to order at least 3( m −
1) + 1along every line L i . Repeating this argument m times we get a contradiction with α ( I ) = 8. Thus α ( I (3 m +1)3 ) ≥ m + 8 for all m ≥
1. On the other hand, it easy to ee that, f m g m h m +1 yz ∈ I (3 m +1)3 so α ( I (3 m +1)3 ) ≤ m + 8 for all m ≥
1. Therefore, α ( I (3 m +1)3 ) = 18 m + 8 for all m ≥ α ( I (2)3 ) = 16, we have that α ( I (3 m +2)3 ) ≥ m + 16 for all m ≥
1. Since the element f m +1 g m +1 h m xyzw ∈ I (3 m +2)3 we have that α ( I (3 m +2) ) ≤ m + 16 for all m ≥ (cid:3) Proposition 3.5.
For all n ≥ , n + 2 ≤ α ( I (2) n ) ≤ n + 4 .Proof. We know that α ( I (2) n ) ≤ α ( I n ) = 2(2 n + 2). Now suppose that α ( I (2) n ) ≤ n + 1. Thenthere is a divisor D of degree 4 n + 1 vanishing to order at least 2 along every line L i in therestricted Fermat configuration. Since the intersection of D and any plane H j consists of2 n + 1 lines to order at least 2, by generalized Bezout theorem, each H j is a component of D because deg( D ) . deg( H j ) = 4 n + 1 < n + 1). This is a contradiction since there are 6 n planes and 6 n > n + 1 = deg( D ) when n ≥
3. Therefore, 4 n + 2 ≤ α ( I (2) n ) ≤ n + 4 for all n ≥ (cid:3) Remark 3.6.
Macaulay2 computations [GS] for n small suggest that α ( I (2) ) = 2(2 n + 2)for n ≥ I n also gives an example for Demailly-like inequality. Corollary 3.7.
For n ≥ , the ideal I n satisfies Demailly-like inequality b α ( I n ) ≥ α ( I ( m ) n ) + h − m + h − for all m ≥ where h is the big height of I n .Proof. Direct from the above calculation with notice that h = 2. (cid:3) We can easily compute the asymptotic resurgence number of I n as follows. Corollary 3.8. b ρ ( I n ) = n + 1 n .Proof. It is clear from [GHVT13, Theorem 1.2] that2 n + 22 n = α ( I n ) b α ( I n ) ≤ b ρ ( I n ) ≤ ω ( I n ) b α ( I n ) = 2 n + 22 n . (cid:3) The following containment are also direct consequences of the above calculations.
Corollary 3.9.
For every n ≥ , restricted Fermat configuration ideal verifies Harbourne-Huneke containment I (2 r ) n ⊆ m r I rn , ∀ r roof. Since I n has big height 2, I (2 r ) n ⊆ I rn , ∀ r so the above containment once again comefrom the fact that α ( I (2 r ) n ) ≥ r + ω ( I rn )for all n ≥ r ≥ r = 0 is trivial). Indeed, α ( I (2 r ) n ) ≥ rn > r + r (2 n + 2) = r + ω ( I rn ) for n ≥ (cid:3) Corollary 3.10.
For every n ≥ , restricted Fermat configuration ideal verifies stableHarbourne-Huneke containment I (2 r − n ⊆ m r − I rn , ∀ r ≫ Proof.
Since b ρ ( I n ) = n + 1 n <
2, by [HKZ20, Theorem 2.3] or [DD20, Corollary 4.1] we seethat ρ ( I n ) < I n has expected resurgence. Thus, by [Gri20, Remark 2.7] for all r ≫ I (2 r − n ⊆ I rn , the above containment comes from the fact that α ( I (2 r − n ) ≥ r − ω ( I rn )for all n ≥ r ≥
3. Indeed, we have α ( I (2 r − n ) ≥ (2 r − n and r − ω ( I rn ) = r − r (2 n + 2). Since (2 r − n ≥ r − r (2 n + 2) ⇔ n ( r − ≥ r − n ( r − ≥ r − ≥ r − r ≥
3, we get the desired inequality. (cid:3)
Corollary 3.11.
For every n ≥ , restricted Fermat configuration ideal verifies the stablecontainment I (2 r − n ⊆ m r I rn , ∀ r ≫ Proof.
As before, since I n has expected resurgence, I (2 r − n ⊆ I rn for r ≫
0. On the otherhand, the inequalities α ( I (2 r − n ) ≥ r + ω ( I rn )is true when r ≥ (cid:3) We will compute here the resurgence number for I by using an upper bound for regularityof its powers and use it to improve the above results of containment for I . Theorem 3.12. ρ ( I ) = 32 .Proof. By [MS17], I (3) n I n hence, ρ ( I ) ≥
32 . We will show that for all m, r such that mr >
32 we have I ( m ) ⊆ I r and then it follows that ρ ( I ) ≤
32 . In fact, we will show that forall m, r such that mr >
32 then α ( I ( m )3 ) ≥ reg( I r ).We will apply [Cha07, Theorem 0.6], which states the following:Let I be an homogeneous ideal of a polynomial ring R over a field, generated in degrees atmost d , such that dim( R/I ) = 2. Assume that I p ⊆ R p is a complete intersection for everyprime I ⊆ p such that dim R/p = 2. Thenreg( I ) ≤ max { I ) , reg( I sat ) + 2 d − } nd for r ≥ I r ) ≤ max { I ) + ( r − d, reg( I sat ) + rd − } Clearly, dim(
R/I ) = 2. Since the set of associated primes of I is also the set of itsminimal primes which consists of 42 minimal primes P j where each minimal prime is thedefining ideal of one of a line in the configuration. Since any prime ideal P that contains I must contain one P j , such prime ideal P with dim( R/P ) = 2 has to be P j . Thus, for anyprime I ⊂ P such that dim( R/P ) = 2, ( I ) P is a complete intersection. Since reg( I ) = 11, α ( I ) = 8 and I = I sat , by [Cha07, Theorem 0.6] we have for r ≥ I r ) ≤ max { I ) + 8( r − , reg( I ) + 8 r − } = max {
33 + 8( r − ,
11 + 8 r − } = 8 r + 9and reg( I ) ≤ max { ,
11 + 16 − } = 25. Thus, for every r ≥ I r ) ≤ r + 9Now for mr >
32 , i.e, 2 m ≥ r + 1, we have α ( I ( m )3 ) ≥ m ≥ r + 3. Since9 r + 3 ≥ r + 9 ≥ reg( I r )for all r ≥
6, it suffices now to check that for r ≤ m ≥ r + 1 we still have α ( I ( m )3 ) ≥ reg( I r ). Indeed, • When r = 1, for m ≥
2, we have reg( I ) = 11 <
16 = α ( I (2)3 ) ≤ α ( I ( m )3 ). • When r = 2, for m ≥
4, we have reg( I ) ≤ <
26 = α ( I (4)3 ) ≤ α ( I ( m )3 ). • When r = 3, for m ≥
5, we have reg( I ) ≤ <
34 = α ( I (5)3 ) ≤ α ( I ( m )3 ). • When r = 4, for m ≥
7, we have reg( I ) ≤ <
44 = α ( I (7)3 ) ≤ α ( I ( m )3 ). • When r = 5, for m ≥
8, we have reg( I ) ≤ <
52 = α ( I (8)3 ) ≤ α ( I ( m )3 ). (cid:3) As a consequence, we can verify that I satifies Harbourne-Huneke Containment for r ≥ Corollary 3.13.
Ideal I verifies the following containment(1) I (2 r − ⊆ m r − I r , ∀ r ≥ (2) I (2 r − ⊆ m r I r , ∀ r ≥ Proof.
Since ρ ( I ) = 32 , I (2 r − ⊆ I r if and only if r ≥ I (2 r − ⊆ I r if and only if r ≥ α ( I (2 r − ) ≥ r − ω ( I r ) for all r ≥ α ( I (2 r − ) ≥ r + ω ( I r ) for all r ≥ (cid:3) Remark 3.14.
As shown in the proof, the first containment fails if r ≤ r ≤ I n for all n ≥ I n where n ≥ Proposition 3.15.
For all n ≥ , R/I n is Cohen-Macaulay and its minimal graded freeresolution is → R ( − (2 n + 3)) ⊕ R ( − n ) ϕ → R ( − (2 n + 2)) ϕ → R → R/I n → here ϕ = [ g , g , g , g , g , g ] and ϕ = − y − w x n − z n − x − w − y n − z n − z − w − x n − y n − − y z − x n − w n − x z y n − w n − x − y − z n − w n − The maximal minors of ϕ are multiple of g , . . . , g as in Hilbert-Burch theorem.Proof. Direct calculation give the minors of ϕ deleting row 1 , , , , , g , − g , 2 g , − g , 2 g , − g respectively. Applying [Eis95, Theorem 18.18], we see that I n is the idealgenerated by 5 × × ϕ and since I n has codimension 2 which agreeswith (6 − − R/I n is Cohen-Macaulay. In particular, depth( R/I n ) = 2 andby Auslander–Buchsbaum formula R/I n has projective dimension 2. Thus by Hilbert-Burchtheorem, R/I n has such minimal graded free resolution. (cid:3) The immediate consequence of the above free resolution is the regularity of ideal I n . Corollary 3.16.
For all n ≥ , reg( I n ) = 4 n − . We now deduce the following containment as a step towards calculating the resurgencenumber of I n for n ≥ Corollary 3.17. If n ≥ then for all m, r such that mr > and r ≥ , we have theinequality α ( I ( m ) n ) ≥ reg( I rn ) which implies that I ( m ) n ⊆ I rn .Proof. By the same argument as that of in the proof of theorem 3.12, I n satisfies all conditionsof [Cha07, Theorem 0.6], so we havereg( I rn ) ≤ max { n −
1) + ( r − n + 2) , n − r (2 n + 2) − } = r (2 n + 2) + 6 n − r ≥
3. Since α ( I ( m ) n ) ≥ nm , for m, r such that mr >
32 , i.e 2 m ≥ r + 1 we have α ( I ( m ) n ) ≥ n (3 r + 1). Now the inequality n (3 r + 1) ≥ r (2 n + 2) + 6 n − r ≥ n − n − n − r ≥ n ≥ (cid:3) The resurgence number of I n for n ≥ Corollary 3.18.
For n ≥ , the resurgence number ρ ( I n ) can only be one of the followingnumbers: , , , or . More precisely, among the four containment I (9) n ⊆ I n , I (7) n ⊆ I n , I (5) n ⊆ I n and I (8) n ⊆ I n , the ratio mr of the first containment (in that order) that fails is theexact value for ρ ( I n ) , otherwise, ρ ( I n ) = 32 . roof. We know that I (3) n I n by [MS17], hence, ρ ( I n ) ≤
32 . On the other hand, I ( m ) n ⊆ I rn for all m, r such that mr >
32 and r ≥
6. The containment are still true if mr >
32 for r = 1 , mr >
32 implies that m ≥ r in these cases and I n has bigheight 2. On the other hand,if r = 3 then mr >
32 implies m ≥ I ( m ) n ⊆ I n for m ≥
6. Similarly, if r = 4 then mr >
32 implies m ≥ I ( m ) n ⊆ I n for m ≥
8. Lastly, if r = 5 then mr > m ≥ I ( m ) n ⊆ I n for m ≥ < < < <
95 , to compute the exact value of ρ ( I n ), it suffices toexamine whether the following four containment I (9) n ⊆ I n , I (7) n ⊆ I n , I (5) n ⊆ I n and I (8) n ⊆ I n fails or not in this order. In fact, if those containment all hold, then I ( m ) n ⊆ I rn for all m, r such that mr >
32 and it follows that ρ ( I n ) = 32 . Otherwise, suppose I ( m ) n ⊆ I r n is the firstcontainment that fails among the above four containment, then ρ ( I n ) ≥ m r . For all m, r such that mr > m r >
32 , we know that either I ( m ) n ⊆ I rn if mr is not among 53 , ,
85 or 95 byabove result or mr is among 53 , ,
85 or 95 that is greater than m r , in which I ( m ) n ⊆ I rn holdas well. Either way we have ρ ( I n ) ≤ m r , therefore ρ ( I n ) = m r . (cid:3) Remark 3.19.
Macaulay2 calculations [GS] for n, r small suggests that reg( I rn ) = ( r +1) α ( I n ) − r + 1)(2 n + 2) −
5. If we can prove this then α ( I ( m ) n ) ≥ reg( I rn ) for all mr >
32 .We also know that I (3) n I n by [MS17]. Thus, in this case ρ ( I n ) = 32 for all n ≥ n n ≥ m k k + 1 3 k + 2 2 5 ≥ =5 2 ≥ α ( I ( m ) n ) 16 18 k k +8 18 k +16 20 42 4 m ≥ n + 2, ≤ n + 4 2 nm b α ( I n ) 3 4 n b ρ ( I n ) 43 54 n + 1 nρ ( I n ) 32 among 32 , 53 , 74 , 85 , 95 Table 1.
Least degree of generators and other invariants related to symbolicpowers of of I n . Arrangements given by the group A In this section, we will deal with ideal J = ( yz ( y − z ) , zx ( z − x ) , xy ( x − y )), which is theideal of the singular locus of the arrangement of lines given by the group G (1 , ,
4) = A .We will see that its least degree of generators of symbolic powers, Waldschmidt constant,(asymptotic) resurgence number are the same as those of group D , see [Ngu21, Section 4].Geometrically, J is the defining ideal of the singular locus of the line arrangement in P that consists of 6 lines L j whose equations are x = 0 , y = 0 , z = 0 , x = y, y = z, z = x These 6 lines intersect at 7 points P i which are [1 : 0 : 0] , [0 : 1 : 0] , [0 : 0 : 1] , [1 : 1 : 1] , [0 : 1 : 1] , [1 : 0 : 1] and [1 : 1 : 0] such that the first 4 points lie on 3 lines each and the restlie on 2 lines each; and each line contains exactly 3 points.Algebraically, we can write J = (( y − z )( y + z − x ) , ( x − y )( x + y − z )) ∩ ( x, y ) ∩ ( y, z ) ∩ ( z, x )and J ( m ) = (( y − z )( y + z − x ) , ( x − y )( x + y − z )) m ∩ ( x, y ) m ∩ ( y, z ) m ∩ ( z, x ) m ∀ m Proposition 4.1.
For ideal J we have the following(1) b α ( J ) = 52 (2) α ( J (2 k ) ) = 5 k , for all k ≥ .(3) α ( J (2 k +1) ) = 5 k + 3 , for all k ≥ .(4) α ( J (2) ) = 6 Proof.
By [FGH +
17, Theorem 2.3], we can check that b α ( J ) ≥
52 .In particular, we have that α ( J (2 k ) ) ≥ k and α ( J (2 k +1) ) ≥ k +3, for all k ≥
1. We will showthe reverse by showing there exists some element with the desired degree in the symbolicpowers.Denote K = (( y − z )( y + z − x ) , ( x − y )( x + y − z )) and first notice that since − ( z − x )(( z + x − y ) = ( y − z )( y + z − x ) + ( x − y )( x + y − z ) ∈ K e have that(4.1) 2 x ( x − y )( z − x ) = ( z − x )( x − y )( x + y − z ) + ( x − y )( z − x )( z + x − y ) ∈ K Similarly, y ( y − z )( x − y ) , z ( z − x )( y − z ) ∈ K . It follows that(4.2) ( y − z )( z − x ) xyz = ( y − z )( z − x ) xy [ z − ( x − y ) ] + ( y − z )( z − x ) xy ( x − y ) ∈ K since ( y − z )( z − x ) xy [ z − ( x − y ) ] = xy ( y − z )( y + z − x )( z − x )( z + x − y ) ∈ K and by 4.1 ( y − z )( z − x ) xy ( x − y ) = x ( x − y )( z − x ) y ( y − z )( x − y ) ∈ K By 4.1 we also have that(4.3) yz ( y − z ) = ( y − z ) z ( y + z − x ) − ( y − z ) z ( z − x ) ∈ K We have the following cases(1) Case 1: when m = 4 k . Consider the polynomial F = [( y − z )( z − x ) xyz ] k ( x − y ) k ( x + y − z ) k that has degree 10 k . By 4.3, [( y − z )( z − x ) xyz ] k ∈ K k so F ∈ K k . On the other hand, [( y − z ) yz ] k ∈ ( y, z ) k , [( z − x ) xz ] k ∈ ( z, x ) k and( xy ) k ( x − y ) k ∈ ( x, y ) k . Thus F ∈ K k ∩ ( x, y ) k ∩ ( y, z ) k ∩ ( z, x ) k = J (4 k ) , ∀ k ≥ m = 4 k + 2. We first show that G = x y z ( x − y ) ( y − z ) ( z − x ) ( x + y − z )( y + z − x )( z + x − y ) ∈ J (6) In fact, ( x − y )( y − z )( z − x )( x + y − z )( y + z − x )( z + x − y ) ∈ K and by 4.3 x y z ( x − y ) ( y − z ) ( z − x ) ∈ K . It is also clear that G ∈ ( x, y ) ∩ ( y, z ) ∩ ( z, x ) .The polynomial F = G [( y − z )( z − x ) xyz ] k ( x − y ) k ( x + y − z ) k has degree10( k + 1) + 5 and by case 1 F ∈ J (6) J (4 k ) ⊆ J (4( k +1)+2) , ∀ k ≥ m = 4 k + 1. F = x k y k +1 z k +1 ( y − z ) k +1 ( z − x ) k ( x − y ) k ( x + y − z ) k has degree 10 k + 3. By 4.2, x k y k z k ( y − z ) k ( z − x ) k ∈ K k , by 4.3, yz ( y − z ) ∈ K so F ∈ K k +1 . Similar to case 1, F ∈ ∩ ( x, y ) k +1 ∩ ( y, z ) k +1 ∩ ( z, x ) k +1 and hence, F ∈ J (4 k +1) , ∀ k ≥ m = 4 k + 3. F = [( y − z )( z − x ) xyz ] k +1 ( x − y ) k +1 ( x + y − z ) k +1 hasdegree 10 k + 8 and F ∈ K k +3 ∩ ( x, y ) k +3 ∩ ( y, z ) k +3 ∩ ( z, x ) k +3 = J (4 k +3) , ∀ k ≥ α ( J (2 k ) ) ≤ k , for all k ≥ α ( J (2 k +1) ) ≤ k + 3, for all k ≥
0. It follows thatstatements (2) and (3) are true and by taking limit as k goes to ∞ , (1) follows as well. Part(4) can be checked directly by Macaulay2 or by Bezout theorem argument as follows: Weknow that α ( J (2) ) ≤ α ( J ) = 6. Now suppose that α ( J (2) ) ≤
5. Then there is a divisor D of degree 5 vanishing to order at least 2 at every point P i . Since the intersection of D andany line L j consists of 3 points to order at least 2, we get a contradiction to Bezout theorembecause deg( D ) . deg( L j ) = 5 < . (cid:3) xample 4.2. It is worth to point out that the first immediate application of the above cal-culations is the verification of Chudnovsky’s Conjecture and Demailly’s Conjecture, althoughthe general case is already known from [EV83]. Ideal J verifies:(1) Chudnovsky’s Conjecture b α ( J ) ≥ α ( J ) + 12 .(2) Demailly’s Conjecture b α ( J ) ≥ α ( J ( m ) ) + 1 m + 1 for all m . Theorem 4.3.
The resurgence number and asymptotic resurgence number are: ρ ( J ) = b ρ ( J ) = 65 Proof.
Since J has the same minimal degrees of generators of symbolic powers and same freeresolutions of all its ordinary powers as those of the ideal of configuration of D , the proofis identical to that of ideal of configuration of D [Ngu21, Proposition 4.2]. (cid:3) The following corollaries and remark also follow from the fact that ideal of configuration A and D share the same minimal degrees of generators of all symbolic powers, see [Ngu21,Section 4]. Corollary 4.4.
Ideal J verifies Harbourne-Huneke containment(1) J (2 r ) ⊆ m r J r , ∀ r .(2) J (2 r − ⊆ m r − J r , ∀ r ≥ . Remark 4.5.
The above corollary gives a proof for the case A in [DS20, Proposition 6.3].In [BGHN20a, Example 3.7], we showed the stronger containment (which implies bothHarbourne-Huneke) containment J (2 r − ⊆ m r J r for r = 5 (by Macaulay2) and thus for all r ≫ r ≥
10. Here we see that thecontainment hold for all r ≥ Corollary 4.6.
For every n ≥ , Fermat configuration ideal verifies the following contain-ment J (2 r − ⊆ m r J r , ∀ r ≥ Remark 4.7.
For r ≤
4, the above containment fail with the same reason to that of [Ngu21,Remark 4.6]. m k k + 1 α ( J ( m ) ) 6 5 k k + 3 b α ( J ) 5/2 ρ ( J ) = b ρ ( J ) 6/5 Table 2.
Least degree of generators and other invariants related to symbolicpowers of J he table provide a complete answer to the question in the theory of Hermite interpo-lation, that is to determine the least degree of a homogeneous polynomial that vanishes toorder m at the 7 points of the given configuration in P .We end this section by calculating the invariant β ( J ( m ) ). As introduced first in [HH13,Definition 2.2] for homogeneous ideals and later considered for ideal I of a finite set of (fat)points in [HS15], β ( I ) is set to be the smallest integer t such that I t contains a regularsequence of length two, or equivalently, is the least degree t such that the zero locus of I t is 0-dimensional. It is known that for ideals of fat points, β ( I ) ≤ ω ( I ) [Ngu21, Proposition3.9]. By the same argument to that of [Ngu21, Proposition 4.9] we have the following Proposition 4.8.
For all m ≥ , β ( J ( m ) ) = 3 m and ω ( J ( m ) ) ≥ m .Proof. The proof is identical to that of [Ngu21, Proposition 4.9] since J m is generated indegree 3 m and each line in this configuration also passed through exactly 3 points in theconfiguration. (cid:3) Remark 4.9.
Similar to Fermat ideals, it is suggested by Macaulay2 that ω ( J ( m ) ) = 3 m .5. An Additional Example
Consider ideal J = ( yz ( y − z ) , zx ( z − x ) , xy ( x − y )). Geometrically, J is the definingideal of the singular locus of the line arrangement in P corresponds to the group B , [DS20].This arrangement consists of 9 lines L j whose equations are x = 0 , y = 0 , z = 0 , x = ± y, y = ± z, z = ± x These 9 lines intersect at 13 points P i which are [1 : 0 : 0] , [0 : 1 : 0] , [0 : 0 : 1] , [1 : 1 : 1] , [ − , [1 : − , [1 : 1 : − , [1 : 0 : 1] , [1 : 1 : 0], [0 : 1 : − , [ − − J = K ∩ K ∩ K ∩ K ∩ ( x, y ) ∩ ( y, z ) ∩ ( z, x )where K = ( x, y − z ) , K = ( y, z − x ) , K = ( z, x − y ) and K = ( x − y , y − z ) ,orwe can write J = K ′ ∩ K ′ ∩ K ∩ ( x, y ) ∩ ( y, z ) ∩ ( z, x )where K ′ = (( y − z )( y + z − x ) , ( x − y )( x + y − z )) , K ′ = ( x + y + z, yz ( y + z )), then for all m J ( m )2 = K m ∩ K m ∩ K m ∩ K m ∩ ( x, y ) m ∩ ( y, z ) m ∩ ( z, x ) m = K ′ m ∩ K ′ m ∩ K m ∩ ( x, y ) m ∩ ( y, z ) m ∩ ( z, x ) m Note that K ∩ ( x, y ) ∩ ( y, z ) ∩ ( z, x ) is the Fermat ideal with n = 2, so we immediately havethat b α ( J ) ≥
52 . Moreover, computations with Macaulay2 show that α ( J (6)2 ) = 21, hence, α ( J ) ≤
72 . It seems that b α ( J ) would be 72 but the α ( J ( m )2 ) are tricky to deal with. On theother hand, α ( J (3)2 ) = 12, by [EV83], since J satisfies Demailly’s Conjecture, we have: b α ( J ) ≥ α ( J (3)2 ) + 13 + 1 = 134Since reg( J ) = 6, by [GHVT13, Theorem 1.2] we have:87 ≤ α ( J ) b α ( J ) ≤ b ρ ( J ) ≤ ρ ( J ) ≤ reg( J ) b α ( J ) ≤ J has expected resurgence. Further computations by Macaulay2 show that, J (7)2 J , thus, 76 ≤ ρ ( J ). It is interesting to know if ρ ( J ) = 76 .More general, for n ≥
1, consider the ideal J n = ( yz ( y n − z n ) , zx ( z n − x n ) , xy ( x n − y n ))which will capture J and J = J . Geometrically, J n corresponds to the configuration of3 n +3 lines that consists of all 3 n lines in Fermat configuration and 3 lines x = 0 , y = 0 , z = 0;and n + 3 n + 3 intersection points. In particular, each line passes through n + 2 points.Similar to J , note that J n is a subset of the Fermat ideals for n ≥
2, thus b α ( J n ) ≥ n for n ≥ Remark 5.1.
Only by looking at the above lower bound, we easily see that J n verifiesChudnovsky’s Conjecture b α ( J n ) ≥ α ( J n ) + 12 for all n . In fact, the case n = 1 was verified inprevious sections and n = 2 follows from b α ( J ) ≥
52 . When n ≥ b α ( J n ) ≥ n ≥ n + 2 + 12 .On the other hand, by [NS16, Theorem 2.5], since J n is a strict almost complete intersectionideal with minimal generators of degree n + 2 and its module sygyzies is generated in degree1 and n + 1, the minimal free resolution of J rn is:0 → R ( − ( n + 2)( r + 1))( r ) ψ → R ( − ( n + 2) r − r +12 ) ⊕ R ( − ( n + 2) r − ( n + 1))( r +12 ) ϕ → R ( − ( n + 2) r )( r +22 ) → J rn → r ≥
2, in particular, reg( J rn ) = ( n + 2) r + n for all r ≥
2. It would be interestingto know if we can determine α ( J ( m ) n ) and use them with the knowledge of reg( J rn ) to verifyDemailly’s Conjecture as well as Harbourne-Huneke Containment, stable Harbourne Con-tainment as we did for Fermat ideals and Fermat-like ideals.Back to ideal J , the following proposition (part (2)) gives a proof for the case B in [DS20,Proposition 6.3]. Proposition 5.2.
Ideal J verifies Harbourne-Huneke containment(1) J (2 r )2 ⊆ m r J r , ∀ r .(2) J (2 r − ⊆ m r − J r , ∀ r ≥ .Proof. (1) For all r , J (2 r )2 ⊆ J r , hence, the containment follows since for all r , we have α ( J (2 r )2 ) ≥ r ≥ r + 4 r = r + ω ( J r )
2) First, the containment J (2 r − ⊆ J r for r ≥ α ( J (2 r − ) ≥
134 (2 r − ≥ r + 2 = reg( J r )for all r ≥
3. Thus, J (2 r − ⊆ m r − J r for r ≥ α ( J (2 r − ) ≥
134 (2 r − ≥ r − r = r + ω ( J r )for all r ≥ (cid:3) Remark 5.3.
Macaulay2 shows that for r = 2, α ( J (3)2 ) = 12, hence, α ( J (3)2 ) ≥ . J ). Therefore, J (2 r − ⊆ J r for r ≥ α ( J (2 r − ) ≥ r + ω ( J r ) for all r ≥ J (2 r − ⊆ m r − J r , for r ≥
2. The case r = 1 is obvious. Thus, J (2 r − ⊆ m r − J r , for r ≥ β and ω forFermat ideals in [Ngu21] and J = J in the previous section. Proposition 5.4.
For all n ≥ and m ≥ , β ( J ( m ) n ) = m ( n + 2) and ω ( J ( m ) n ) ≥ m ( n + 2) .Proof. The proof is the same to that of [Ngu21, Proposition 3.10] with notice that J mn isgenerated in degree m ( n + 2) and each line in this configuration passed through exactly n + 2points in the configuration. (cid:3) Remark 5.5.
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Tulane University, Department of Mathematics, 6823 St. Charles Ave., New Orleans,LA 70118, USA and Hue University, College of Education, Vietnam.
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