Extreme values of the resurgence for homogeneous ideals in polynomial rings
aa r X i v : . [ m a t h . A C ] M a y EXTREME VALUES OF THE RESURGENCE FOR HOMOGENEOUSIDEALS IN POLYNOMIAL RINGS
B. HARBOURNE, J. KETTINGER, AND F. ZIMMITTI
Abstract.
We show that two ostensibly different versions of the asymptotic resurgenceintroduced by Guardo, Harbourne and Van Tuyl in 2013 are the same. We also show thatthe resurgence and asymptotic resurgence attain their maximal values simultaneously, if atall, which we apply to a conjecture of Grifo. For radical ideals of points, we show thatthe resurgence and asymptotic resurgence attain their minimal values simultaneously. Inaddition, we introduce an integral closure version of the resurgence and relate it to the otherversions of the resurgence. In closing we provide various examples and raise some relatedquestions, and we finish with some remarks about computing the resurgence. Introduction
This paper is motivated by wanting to better understand concepts, conjectures and meth-ods introduced in [BH], [GHVT], [G], [DFMS] and [DD] involving various approaches to thecontainment problem. It makes particular use of the groundbreaking methods of [DFMS]and [DD].Let K be a field and let N ≥
1. Then R = K [ P N ] denotes the polynomial ring K [ P N ] = K [ x , . . . , x N ]. Now let (0) = I ( K [ P N ] be a homogeneous ideal; thus I = ⊕ t ≥ I t , where I t is the K vector space span of all homogeneous polynomials of degree t in I . The symbolicpower I ( m ) is defined as I ( m ) = R ∩ ( ∩ P ∈ Ass(
R/I ) I m R P )where the intersections take place in K ( P N ).While I r ⊆ I ( m ) holds if and only if m ≤ r , the containment problem of determining forwhich m and r the containment I ( m ) ⊆ I r holds is much more subtle. If h I is the minimumof N and the bigheight of I (i.e., the maximum of the heights of associated primes of I ), itis known that(1.1) I ( rh I ) ⊆ I r [ELS, HoHu], so given r , the issue is for which m with m < rh I do we have I ( m ) ⊆ I r . The resurgence ρ ( I ), introduced in [BH], gives some notion of how small the ratio m/r can beand still be sure to have I ( m ) ⊆ I r ; specifically, ρ ( I ) = sup n mr : I ( m ) I r o . Date : edited: May 11, 2020; compiled May 12, 2020.2010
Mathematics Subject Classification.
Primary: 14C20, 13B22, 13M10; Secondary: 14N05, 13D40 .
Key words and phrases. resurgence, asymptotic resurgence, symbolic power, integral closure, fat points,polynomial ring, ideals, containment problem.
Acknowledgements : Harbourne was partially supported by Simons Foundation grant
A case of particular interest is that of ideals of fat points . Given distinct points p , . . . , p s ∈ P N and nonnegative integers m i (not all 0), let Z = m p + · · · + m s p s denote the scheme(called a fat point scheme) defined by the ideal I ( Z ) = ∩ si =1 ( I ( p i ) m i ) ⊆ K [ P N ] , where I ( p i ) is the ideal generated by all homogeneous polynomials vanishing at p i . Notethat I ( Z ) is always nontrivial (i.e., not (0) nor (1)). Symbolic powers of I ( Z ) take the form I ( Z ) ( m ) = I ( mZ ) = ∩ si =1 ( I ( p i ) mm i ). We say Z is reduced if m i is either 0 or 1 for each i (i.e.,if I ( Z ) is a radical ideal).Subsequent to [BH], two asymptotic notions of the resurgence were introduced by [GHVT].The first is ρ ′ ( I ) = lim sup t ρ ( I, t ) = lim t →∞ ρ ( I, t ) , where ρ ( I, t ) = sup n mr : I ( m ) I r , m ≥ t, r ≥ t o . The second is b ρ ( I ) = sup n mr : I ( mt ) I rt , t ≫ o . A useful new perspective on b ρ ( I ) is given by [DFMS, Corollary 4.14], which shows that b ρ ( I ) = sup n mr : I ( m ) I r o , where I r is the integral closure of I r (defined below). Since we always have I r ⊆ I r , this newperspective makes it clear that we always have b ρ ( I ) ≤ ρ ( I ), and that we have b ρ ( I ) = ρ ( I ) if I r = I r for all r ≥ b ρ ( I ) = ρ ′ ( I ) (Theorem 2.2), that b ρ ( I ) = h I if and onlyif ρ ( I ) = h I (Theorem 2.3), and that b ρ ( I ( Z )) = 1 if and only if ρ ( I ( Z )) = 1 when Z ⊂ P N is a reduced scheme of points (Theorem 2.4) and for every fat point subscheme Z ⊂ P (Corollary 2.9). We also introduce a new version of the resurgence, ρ int , based on integralclosure, and relate it to the original resurgence. We then discuss the relevance of our resultsto a conjecture of Grifo. Finally we provide some examples and raise some questions, andinclude a discussion of the computability of the resurgence.1.1. Background.
Let I ⊆ K [ P N ] be a nontrivial homogeneous ideal. Given the commentsand definitions above we have (see [GHVT])(1.2) 1 ≤ b ρ ( I ) ≤ ρ ′ ( I ) ≤ ρ ( I ) ≤ h I . By [BH] and [GHVT] we also have(1.3) α ( I ) b α ( I ) ≤ b ρ ( I ) ≤ ρ ′ ( I ) ≤ ρ ( I ) , where α ( I ) is the least degree of a nonzero element of I and b α ( I ) is the Waldschmidt constant ,defined as b α ( I ) = lim m →∞ α ( I ( m ) ) m . For a nontrivial fat point subscheme Z ⊆ P N , by [BH] we have in addition(1.4) ρ ( I ( Z )) ≤ reg( I ( Z )) b α ( I ( Z )) , where reg( I ( Z )) is the Castelnuovo-Mumford regularity of I ( Z ). ALUES OF RESURGENCES 3
A version of resurgence can be defined with integral closure replacing symbolic powers.We pause to briefly discuss the concept of integral closure. Given an ideal I ⊆ R = K [ P N ],we recall (see [SH]) that the integral closure I of I consists of all elements c ∈ R satisfyingfor some n ≥ c n + a c n − + · · · + a n = 0where a j ∈ I j . We say I is integrally closed if I = I . We note that I is monomial (resp.homogeneous) if I is [SH, Proposition 1.4.2] (resp. [SH, Corollary 5.2.3]). If I r is integrallyclosed for all r ≥
1, we say I is normal .For example, the ideal I ( p i ) of a point p i ∈ P N is normal. Likewise, M = ( x , . . . , x N )is normal. This is because M is a monomial prime ideal and I ( p i ) is also, up to choiceof coordinates, but monomial primes are normal. (Apply the usual criterion for integralclosure for monomial ideals, that the integral closure of a monomial ideal I is the monomialideal associated to the convex hull of the Newton polygon of I [SH].) As further examplesof integrally closed ideals, we note I ( Z ) is integrally closed for all Z , as is M t ∩ I ( Z ) forevery t ; this is because intersections of integrally closed ideals are integrally closed. Nowassume α ( I ( Z )) = reg( I ( Z )); then ( I ( Z ) r ) t = ( I ( rZ )) t for t ≥ α ( I ( Z ) r ) (apply [BH, Lemma2.3.3(c)] using the fact that α ( I ( Z ) r ) = rα ( I ( Z ))). Thus we have I ( Z ) r = M rα ( I ) ∩ I ( rZ )and hence I ( Z ) is normal, if α ( I ( Z )) = reg( I ( Z )). Thus, when α ( I ( Z )) = reg( I ( Z )), wehave b ρ ( I ( Z )) = ρ ( I ( Z )) by the normality, but in fact (1.3) and (1.4) give us more, namely α ( I ( Z )) b α ( I ( Z )) = b ρ ( I ( Z )) = ρ ( I ( Z )) . Now we define the integral closure resurgence. Given any nontrivial homogeneous ideal I ⊂ K [ P N ] but replacing symbolic power by integral closure in the definition of resurgencegives us the integral closure resurgence , ρ int ( I ) = sup n mr : I m I r o . If symbolic powers of I are integrally closed (as is the case when I is the ideal of a fat pointsubscheme), we have ρ int ( I ) ≤ ρ ( I ), and if moreover I ( m ) = I m for all m ≥
1, it follows that ρ int ( I ) = ρ ( I ) (see Theorem 2.7). A lower bound such as ρ int ( I ) ≤ ρ ( I ) is of interest since forany c > ρ int ( I ) ≥ c , and if so to compute ρ int ( I )exactly. In a further analogy of ρ int with ρ , by Lemma 2.6 we have 1 ≤ ρ int ( I ) ≤ N , and, if I = I , we have I m ⊆ I r for all m ≥ N r .2.
Main results
We begin with the result that started this paper.
Proposition 2.1.
Let Z be a fat point subscheme of P N with an integer c such that I ( cZ ) t = I ( ctZ ) for all t ≥ . Then b ρ ( I ( Z )) = ρ ′ ( I ( Z )) .Proof. Let b be a rational such that b ρ ( I ( Z )) < c/b . Pick any integer d > db is an integer. Since b ρ ( I ( Z )) < c/b = cd/ ( db ), we have by definition of b ρ ( I ( Z )) that I ( cdtZ ) ⊆ I ( Z ) dbt for t ≫
0, Note that I ( cdZ ) t = I ( cZ ) dt = I ( cdtZ ) for all t ≥
1. Henceby [GHVT, Theorem 1.2(3)] we have ρ ′ ( I ( Z )) ≤ cdt/ ( dbt ) = c/b . Since this holds for all b with c/b > b ρ ( I ( Z )), we have b ρ ( I ( Z )) ≥ ρ ′ ( I ( Z )). But by [GHVT, Theorem 1.2(1)] we have b ρ ( I ( Z )) ≤ ρ ′ ( I ( Z )), hence b ρ ( I ( Z )) = ρ ′ ( I ( Z )). (cid:3) B. HARBOURNE, J. KETTINGER, AND F. ZIMMITTI
A much stronger result can be proved based on an argument in [DD].
Theorem 2.2.
Let I be a nontrivial homogeneous ideal of K [ P N ] . Then b ρ ( I ) = ρ ′ ( I ) .Proof. If b ρ ( I ) = ρ ( I ), then (1.2) gives b ρ ( I ) = ρ ′ ( I ), so assume b ρ ( I ) < ρ ( I ). Thus there is an ǫ > b ρ ( I ) + ǫ < ρ ( I ). As in the proof of [DD, Proposition 2.6] (which in turn is aconsequence of [DFMS, Lemma 4.12]), we have sr + N < b ρ ( I )whenever I ( s ) I r .This means there are only finitely many s and r for which b ρ ( I )+ ǫ ≤ s/r holds but I ( s ) ⊆ I r fails. (This is because b ρ ( I ) + ǫ ≤ s/r and sr + N < b ρ ( I ) implies r ( b ρ ( I ) + ǫ ) ≤ s < b ρ ( I )( r + N )and r < N b ρ ( I ) /ǫ .)Hence for all s and r sufficiently large which have I ( s ) not contained in I r , we will have s/r < b ρ ( I ) + ǫ and hence ρ ′ ( I ) ≤ b ρ ( I ) + ǫ . This is true for every ǫ >
0, and so we get ρ ′ ( I ) ≤ b ρ ( I ). Since we already have ρ ′ ( I ) ≥ b ρ ( I ), we conclude that ρ ′ ( I ) = b ρ ( I ). (cid:3) An alternate statement of the next result is that b ρ ( I ) = h I if and only if ρ ( I ) = h I . Wehave learned that Theorem 2.3 was also obtained independently by DiPasquale and Drabkin,but not included in [DD]. Theorem 2.3.
Let I be a nontrivial homogeneous ideal of K [ P N ] . Then b ρ ( I ) < h I if andonly if ρ ( I ) < h I .Proof. If ρ ( I ) < h I , then we have b ρ ( I ) ≤ ρ ( I ) < h I by (1.2). So assume b ρ ( I ) < h I .If b ρ ( I ) = ρ ( I ), then ρ ( I ) < h I . If b ρ ( I ) < ρ ( I ), then by [DD, Proposition 2.6], ρ ( I ) isthe maximum of finitely many ratios s/r with I ( s ) I r . Thus ρ ( I ) = h I would imply that s/r = h I , which contradicts the result of [ELS, HoHu] that I ( m ) ⊆ I r whenever m ≥ rh I . (cid:3) The other extreme is also of interest. Let Z ⊂ P N be a fat point subscheme. The nextresult, Theorem 2.4, shows that the question of when ρ ( I ( Z )) = 1 is related to two concepts:to analytic spreads (see [SH]) and to symbolic defects (see [GGSV]).We can define the analytic spread ℓ ( I ) of a homogeneous ideal I ⊆ R = K [ P N ] as being theminimum number of elements of I such that, after localizing at the irrelevant ideal, the ideal J they generate has J = I [SH, Corollary 1.2.5, Proposition 8.3.7]. The analytic spread ℓ ( I )of an ideal I ⊆ R is at least the height of I , since J and I have the same minimal primes,and the minimal number of generators of an ideal is at least the height of its minimal primeof minimal height. By [SH, Proposition 5.1.6], it is also at most the dimension of R (whichis N + 1). For I ( Z ) ⊂ K [ P N ] we thus have N ≤ ℓ ( I ( Z )) ≤ N + 1.We say an ideal B ⊆ K [ P N ] is a complete intersection if B is generated by a regularsequence; equivalently, B is a complete intersection if all associated primes of B have height c , where c is the minimal number of generators of B . If for a fat point subscheme Z ⊂ P N its ideal I ( Z ) is a complete intersection (i.e., I ( Z ) has N generators), then ℓ ( I ( Z )) = N ,but ℓ ( I ( Z )) = N can occur even when I ( Z ) is not a complete intersection (for example, ℓ ( I ( mZ )) = N when I ( Z ) is a complete intersection but m > symbolic defect sdefect( I ( Z ) , m ) is the minimum number of generators of the module I ( mZ ) /I ( Z ) m [GGSV]. Thus sdefect( I ( Z ) , m ) = 0 if and only if I ( mZ ) = I ( Z ) m . If I ( Z ) is acomplete intersection, then sdefect( I ( Z ) , m ) = 0 for all m ≥ I ( mZ ) = I ( Z ) m (indeed, if I is generated by a regular sequence, then I r = I ( r ) for all r ≥ ALUES OF RESURGENCES 5
5, Appendix 6]). When N = 2 and Z is reduced (i.e., I ( Z ) is radical), [GGSV, Theorem 2.6]gives a converse: if sdefect( I ( Z ) , m ) = 0 for all m ≥
1, then I ( Z ) is a complete intersection(see also [CFGLMNSSV, Remark 2.5] and [HU, Theorem 2.8]).Our next result gives a number of equivalent conditions for sdefect( I ( Z ) , m ) = 0 for all m ≥ Z ⊂ P N , thereby extending from N = 2 to all N the result that I ( mZ ) = I ( Z ) m for all m ≥ I ( Z ) is a complete intersection.We thank Seceleanu and Huneke for the implications (d) ⇒ (e), and, for Z reduced, (e) ⇒ (a). We do not know if (e) ⇒ (a) holds for nonreduced fat point schemes Z (i.e., when I ( Z )is not radical), but see Corollaries 2.9 and 2.10, and also Examples 4.1 and 4.3 and Question4.4. Theorem 2.4.
Let Z be a nontrivial fat point subscheme of P N . Then each of the followingcriteria implies the next.(a) I ( Z ) m = I ( mZ ) (i.e., sdefect( I ( Z ) , m ) = 0 ) for all m ≥ .(b) ρ ( I ( Z )) = 1 .(c) b ρ ( I ( Z )) = 1 .(d) I ( Z ) m = I ( mZ ) for all m ≥ .(e) The analytic spread of I ( Z ) is N .Moreover, if Z is reduced or N = 1 , then (e) implies (a). In fact, if Z is reduced or N = 1 ,then each of the conditions (a)-(e) is equivalent to I ( Z ) being a complete intersection.Proof. That (a) implies (b) is clear (since I ( Z ) m ⊆ I ( Z ) r if and only if m ≥ r ), and (b)implies (c) since 1 ≤ b ρ ( I ( Z )) ≤ ρ ( I ( Z )). Next (c) is equivalent to (d) by [DFMS, Corollary4.16]. Now we show (d) implies (e). By [SH, Proposition 5.4.7], we have that (d) implies ℓ ( I ( Z )) = N + 1, hence ℓ ( I ( Z )) = N .If N = 1, then I ( Z ) is principal, hence (a) always holds and I ( Z ) is a complete intersection.Finally assume that Z is reduced. Thus the primary components of I ( Z ) are ideals ofpoints of multiplicity 1, hence are complete intersections (i.e., I ( Z ) is locally a completeintersection). By [CN], ℓ ( I ( Z )) = N implies that the localization I ( Z ) M of I ( Z ) at theirrelevant ideal M ⊂ R = K [ P N ] is a complete intersection, and hence ( I ( Z ) m ) M = ( I ( Z ) M ) m is saturated for all m ≥
1, so I ( Z ) m itself is saturated so (a) holds. Moreover, since (e) impliesthat I ( Z ) M is a complete intersection when Z is reduced, the number of generators of I ( Z ) M is N = dim I ( Z ) M /I ( Z ) M M M = dim I ( Z ) /I ( Z ) M , hence I ( Z ) also has N generators, so isitself is a complete intersection and hence (a) holds. (cid:3) We can now give a characterization of those Z for which I ( mZ ) = I ( Z ) m for all m ≥ I ( Z ) is not alwaysnormal (see Example 4.9), we do not know any examples with b ρ ( I ( Z )) = 1 where I ( Z ) isnot normal. Corollary 2.5.
Let Z ⊂ P N be a fat point subscheme. Then I ( mZ ) = I ( Z ) m for all m ≥ if and only if b ρ ( I ( Z )) = 1 and I ( Z ) is normal.Proof. Having I ( mZ ) = I ( Z ) m for all m ≥ b ρ ( I ( Z )) = 1 by Theorem 2.4, and itimplies I ( Z ) is normal since I ( Z ) m ⊆ I ( Z ) m ⊆ I ( mZ ). Conversely, b ρ ( I ( Z )) = 1 implies I ( mZ ) = I ( Z ) m for all m ≥ I ( Z ) m = I ( Z ) m for all m ≥
1, hence I ( mZ ) = I ( Z ) m holds for all m ≥ (cid:3) B. HARBOURNE, J. KETTINGER, AND F. ZIMMITTI
We now consider ρ int ( I ). Note for a nontrivial homogeneous ideal I ⊂ K [ P N ] that an idealindependent version of (1.1) can be stated as I ( Nr ) ⊆ I r , and the corresponding bounds (1.2)on ρ ( I ) are 1 ≤ ρ ( I ) ≤ N . In analogy with this, we have the following lemma. (See Example4.5 for an example showing that the assumption I = I is needed both for the second part ofLemma 2.6 and for Theorem 2.7(b).) Lemma 2.6.
Let I ⊂ K [ P N ] be a nontrivial homogeneous ideal. Then ≤ ρ int ( I ) ≤ N .Moreover, if N > and t > , or if I = I and t ≥ , then I Nt ⊆ I t .Proof. Since I t ⊆ I t but I t I t +1 , we see that I t I t +1 , so 1 ≤ ρ int ( I ).The Brian¸con-Skoda Theorem [SH] asserts I t + N ⊆ I t for each t ≥
1. For t, N ≥ tN ≥ t + N , hence I Nt ⊆ I t + N and thus I Nt ⊆ I t + N ⊆ I t . This also tells us that if I m I t , we must have either m < N t (and so m/t < N ) or t or N must be equal to 1. ByBrian¸con-Skoda, I m I t implies m < t + N , so if t = 1, then m/t = m ≤ N , while if N = 1,then we have m < t + 1, so m/t ≤ t/t = 1 = N . Thus in all cases we have m/t ≤ N , hence ρ int ( I ) ≤ N .We already saw that I Nt ⊆ I t if N, t >
1. Assume
N > t = 1. Then I N ⊆ I so if I = I we have I N ⊆ I . Finally assume N = 1 and I ⊂ K [ x , x ] is a nontrivial homogeneousideal. Thus I = ( G , . . . , G s ) for some nonzero homogeneous generators G i . Let F be thegreatest common divisor of the G i , and for each i let H i F = G i . Then I = ( F ) Q where Q = ( H , . . . , H s ). If deg H i = 0 for some i , we have Q = (1) and so I = ( F ). If deg H i > i , then Q is primary for ( x , x ).If I = ( F ), then I is normal, so we have I t = I t for all t ≥
1. So say I = ( F ) Q where Q is primary for ( x , x ), hence I t = ( F t ) Q t for each t ≥ I t = I t if and only if Q t = Q t . (Here’s why. Assume I t = I t . By [SH, Remark1.3.2(2)], we have I t : J ⊆ I t : J for any ideal J . Take J = ( F r ). Then Q t ⊆ Q t = I t : J ⊆ I t : J = I t : ( F r ) = Q t . Alternatively, say x ∈ Q t . Then x n + a x n − + · · · + a n = 0 for some n and some a i ∈ Q ti . Multiplying by F tn gives ( F t x ) n + F t a ( F t x ) n − + · · · + ( F t ) n a n = 0so F t x ∈ ( F t ) Q t = I t = I t = ( F t ) Q t , so x ∈ Q t . Now assume Q t = Q t . Say x ∈ I t . Then x n + F t a x n − + · · · + ( F t ) n a n = 0 for some n with a i ∈ Q ti . Say F t = AP m where P is anirreducible factor of F and P does not divide A . Then P divides x . Let yP = x . Dividingout gives y n + AP m − a y n − + · · · + A n P n ( m − a n = 0. We can keep dividing out until wehave z n + Aa z n − + · · · + A n a n = 0, where zP m = x and z ∈ ( A ) Q t . Continuing in thisway, dividing out irreducible factors of F t , we eventually see that we get an element w with wF t = x , where w ∈ Q t = Q t , hence x ∈ ( F t ) Q t = I t .)Now we claim that Q is normal if and only if Q = Q . Given this, if I = I , then Q = Q ,hence Q t = Q t so I t = I t (i.e., I is normal), which is what we needed to show. But notethat Q normal implies by definition that Q = Q . Conversely, assume Q = Q . It suffices toshow Q t = Q t for each t . By [AM, Proposition 4.8] and [SH, Proposition 1.1.4], it is enoughto check this after localizing at ( x , x ). But by results of Zariski, Q = Q implies Q t = Q t for each t in the local case [SH, Theorem 14.4.4]. (cid:3) Theorem 2.7.
Let I ⊂ K [ P N ] be a nontrivial homogeneous ideal, N ≥ .(a) We have ≤ ρ int ( I ) = max n(cid:8) mr : I m I r (cid:9) ∪ { } o . (b) If I = I and N > , then ρ int ( I ) < N . ALUES OF RESURGENCES 7 (c) If I ( m ) = I ( m ) for all m ≥ (as for example is the case for I = I ( Z ) for a fat pointsubscheme Z ⊂ P N ), then ρ int ( I ) ≤ ρ ( I ) .(d) If I ( m ) = I m for all m ≥ , then ρ int ( I ) = ρ ( I ) .Proof. (a) By Lemma 2.6 we have 1 ≤ ρ int ( I ). In order for ρ int ( I ) > r , m ) with both m /r > I m I r (and hence m < r + N by Brian¸con-Skoda). Set c = m /r . If ρ int ( I ) > c , then as before we have ( r, m ) withboth m/r > c and I m I r (and hence m < r + N ). But there are only finitely manypairs ( r, m ) with cr < m and m < r + N (in particular, we have r < N/ ( c −
1) and cr < m < r + N ). Thus either ρ int ( I ) = 1 or ρ int ( I ) = max { mr : I m I r } , hence ρ int ( I ) = max n(cid:8) mr : I m I r (cid:9) ∪ { } o . (b) Now assume I = I and N >
1. By Lemma 2.6 we have I Nr ⊆ I r for r ≥
1, so I m I r implies m/r < N , hence ρ int ( I ), being either 1 or a maximum of values m/r less than N , isless than N .(c) Here we assume I ( m ) = I ( m ) for all m ≥
1. Then since I m ⊆ I ( m ) we have I m ⊆ I ( m ) ,so I m I r implies I ( m ) I r , and hence ρ int ( I ) ≤ ρ ( I ).(d) Assuming I m = I ( m ) for all m ≥
1, we have ρ int ( I ) = sup n mr : I m I r o = sup n mr : I ( m ) I r o = ρ ( I ) . (cid:3) We do not know any examples with ρ int ( I ( Z )) >
1. For Z ⊂ P , there are none, by thenext result, which is an immediate consequence of [AH, Theorem 3.3]. We thank Huneke foralerting us to this result. Corollary 2.8.
Let Z ⊂ P N be a nontrivial fat point subscheme and let I = I ( Z ) . Then wehave the following:(a) I N + m − ⊆ I m for m ≥ ;(b) ρ int ( I ( Z )) ≤ N/ for N ≥ ; and(c) ρ int ( I ( Z )) = 1 for N = 1 , .Proof. (a) Let M = ( x , . . . , x N ) ⊂ K [ P N ]. Using I as a reduction for I and ℓ for theanalytic spread of I , [AH, Theorem 3.3] states (after localizing at the irrelevant ideal M )that I ℓ + m ⊆ I ℓ − N + m +1 for m ≥
0. But N ≤ ℓ ≤ N + 1, so for ℓ = N we have I N + m ⊆ I m +1 ,while for ℓ = N + 1 we have I N + m +1 ⊆ I m +2 , both for m ≥
0. Either way, we have I N + m − ⊆ I m for m ≥
1, and hence I N + m − ⊆ I m holds without localizing, since all of theideals are homogeneous.(b) Assume N ≥
2. For r = 1 we have I m I r for all m ≥
1, so consider r >
1. Thenwe have I m ⊆ I r for all m ≥ N + r −
1, so the fractions m/r for which we have I m I r are contained in the set { m/r : 1 ≤ m ≤ N + r − , r ≥ } . The supremum occurs for m = N + r − r = 2, hence the supremum is N/
2, so ρ int ( I ( Z )) ≤ N/ N = 1 we have ρ int ( I ( Z )) = 1 from Lemma 2.6, and when N = 2 we have ρ int ( I ( Z )) = 1 from (b). (cid:3) If ρ int ( I ( Z )) is always 1, then the next result would imply that b ρ ( I ( Z )) = 1 if and only if ρ ( I ( Z )) = 1. In particular, it shows that b ρ ( I ( Z )) = 1 if and only if ρ ( I ( Z )) = 1 for everyfat point subscheme Z ⊂ P . B. HARBOURNE, J. KETTINGER, AND F. ZIMMITTI
Corollary 2.9.
Let Z ⊂ P N be a nontrivial fat point subscheme. If b ρ ( I ( Z )) = 1 , then ρ int ( I ( Z )) = ρ ( I ( Z )) , hence b ρ ( I ( Z )) = 1 if and only if ρ ( I ( Z )) = 1 when N = 2 .Proof. By Theorem 2.4, b ρ ( I ( Z )) = 1 implies I ( mZ ) = I ( Z ) m , so ρ int ( I ( Z )) = ρ ( I ( Z )). Since ρ int ( I ( Z )) = 1 when N = 2 by Corollary 2.8, and since ρ ( I ( Z )) = 1 implies b ρ ( I ( Z )) = 1, wehave b ρ ( I ( Z )) = 1 if and only if ρ ( I ( Z )) = 1 when N = 2. (cid:3) We now recover a version of [DFMS, Corollary 4.17].
Corollary 2.10.
Let Z ⊂ P N be a nontrivial fat point subscheme. Then ρ ( I ( Z )) = 1 if andonly if b ρ ( I ( Z )) = ρ int ( I ( Z )) = 1 .Proof. The proof is immediate from1 ≤ ρ int ( I ( Z )) ≤ ρ ( I ( Z )) , ≤ b ρ ( I ( Z )) ≤ ρ ( I ( Z ))and Corollary 2.9. (cid:3) Remark 2.11.
For a nontrivial homogeneous ideal I ⊂ K [ P N ], it is also of interest to define ρ int ( I ) = sup n m + 1 r : I m I r o . This is exactly what [DFMS] denotes as K ( I ). Clearly we have ρ int ( I ) ≤ ρ int ( I ), and byapplying Theorem 2.7 we see we have equality if and only if ρ int ( I ) = 1, in which case I isnormal. By [DFMS, Proposition 4.19] we have ρ ( I ) ≤ b ρ ( I ) ρ int ( I ). Thus we have ρ int ( I ( Z )) ≤ ρ ( I ( Z )) ≤ b ρ ( I ( Z )) ρ int ( I ( Z ))for every fat point subscheme Z ⊂ P N .3. Grifo’s Conjecture
We now discuss Grifo’s containment conjecture [G, Conjecture 2.1]. In our context it saysthe following (we note it is true and easy to prove for N = 1). Conjecture 3.1.
Let I ⊆ K [ P N ] be a radical homogeneous ideal. Then I ( h I r − h I +1) ⊆ I r forall r ≫ . Remark 2.7 of [G] shows the conjecture holds for I if ρ ( I ) < h I (whether I is radical ornot), or if ρ ′ ( I ) < h I , and raises the question of whether it holds when b ρ ( I ) < h I . Thiswas answered affirmatively by [GHM, Proposition 2.11] with a direct proof (we thank Grifoand Huneke for bringing this result to our attention). Our results also answer this questionaffirmatively, in two ways. By Theorem 2.2, b ρ ( I ) < h I implies ρ ( I ) ′ < h I , hence Conjecture3.1 holds for I by the results of [G]. And by Theorem 2.3, b ρ ( I ) < h I implies ρ ( I ) < h I , soagain Conjecture 3.1 holds for I by the results of [G]. Remark 3.2.
When I = I ( Z ) for a fat point scheme Z ⊂ P N we have h I = N . Noexamples of a fat point scheme Z ⊂ P N (radical or not) are known for which it is not truethat I ( Z ) ( Nr − N +1) ⊆ I ( Z ) r for all r ≫
0. By Remark 2.7 of [G], one approach to provingthat I ( Z ) ( Nr − N +1) ⊆ I ( Z ) r for all r ≫ Z is to show ρ ( I ( Z )) < N whenever N >
1. This raises the question of: for which Z is it known that ρ ( I ( Z )) < N ?Let Z = m p + · · · + m s p s . If gcd( m , . . . , m s ) >
1, then ρ ( I ( Z )) < N by [TX, Proposition2.1(2)]. Thus it is the cases with gcd( m , . . . , m s ) = 1 that remain of interest. ALUES OF RESURGENCES 9
Another approach is to apply our Theorem 2.3: ρ ( I ( Z )) < N holds if b ρ ( I ( Z )) < N .Aiming to show b ρ ( I ( Z )) < N has the advantage that the results of [DFMS, DD] suggestthat b ρ ( I ( Z )) is more accessible computationally than is ρ ( I ( Z )). Another advantage is thatin most cases where b ρ ( I ( Z )) is known we have b ρ ( I ( Z )) = α ( I ( Z )) b α ( I ( Z )) (but see Example 4.10) andin all known cases we have b α ( I ( Z )) ≥ α ( I ( mZ ))+ N − m + N − for all m ≥
1. Assuming both we have b ρ ( I ( Z )) = α ( I ( Z )) b α ( I ( Z )) ≤ α ( I ( Z )) α ( I ( Z ))+ N − N = N α ( I ( Z )) α ( I ( Z )) + N − < N and thus we would have ρ ( I ( Z )) < N .The previous paragraph merits further discussion. First we recall [HaHu, Conjecture 2.1],which if true would refine the containment I ( rN Z ) ⊆ I ( Z ) r of [ELS, HoHu]: Conjecture 3.3.
Let Z ⊂ P N be a fat point scheme. Let M = ( x , . . . , x N ) . Then I ( rN Z ) ⊆ M r ( N − I ( Z ) r holds for all r > . A further refinement of I ( rN Z ) ⊆ I ( Z ) r is that I ( r ( m + N − Z ) ⊆ I ( mZ ) r [ELS, HoHu].This suggests a refinement of Conjecture 3.3 (cf. [HaHu, Question 4.2.3]), namely(3.1) I ( r ( m + N − Z ) ⊆ M r ( N − I ( mZ ) r . If (3.1) were true, then the following conjecture would also be true:
Conjecture 3.4.
Let Z ⊂ P N be a fat point scheme. Then b α ( I ( Z )) ≥ α ( I ( mZ )) + N − m + N − for all m ≥ . The proof that Conjecture 3.3 implies Conjecture 3.4 when m = 1 is given in [HaHu]. Thesame argument shows that (3.1) implies Conjecture 3.4.The first version of Conjecture 3.4 was posed by Chudnovsky [Ch] over the complex num-bers for the case that m = 1 and Z is reduced. He also sketched a proof of his conjecture for N = 2 which works over any algebraically closed field (see [HaHu] for a proof). Conjecture3.4 assuming Z reduced was posed by Demailly [Dem].The best result currently known is by Esnault and Viehweg [EV]. It is over the complexnumbers and says that if Z is reduced with N >
1, then b α ( I ( Z )) ≥ α ( I ( mZ )) + 1 m + N − m ≥
1. Thus for reduced Z over the complex numbers, taking m = 1, we have α ( I ( Z )) b α ( I ( Z )) ≤ α ( I ( Z )) α ( I ( Z ))+1 N = N α ( I ( Z )) α ( I ( Z )) + 1 < N. Hence ρ ( I ( Z )) < N holds over the complex numbers whenever Z is reduced and b ρ ( I ( Z )) = α ( I ( Z )) b α ( I ( Z )) . Examples and questions
For this section assume Z is a fat point subscheme of P N (so I ( Z ) is a nontrivial ideal of K [ P N ]). Example 4.1. If I ( Z ) is a complete intersection or a power thereof, then I ( mZ ) = I ( Z ) m for all m ≥
1, hence ρ ( I ( Z )) = 1 by Theorem 2.4. Again by Theorem 2.4, if Z is reduced,then ρ ( I ( Z )) = 1 if and only if I ( Z ) is a complete intersection. However, when Z is notreduced, ρ ( I ( Z )) = 1 does not imply I ( Z ) is a complete intersection, or even a power of acomplete intersection ideal (homogeneous or not), as is shown by Example 4.14 and Remark4.15. (Additionally, let p , p , p ∈ P N be noncollinear points and let Z = m p + m p + m p with 1 ≤ m ≤ m ≤ m . If either m + m ≤ m or if m + m + m is even, then by[BZ, Theorem 2] we have I ( mZ ) = I ( Z ) m for all m ≥ ρ ( I ( Z )) = 1. See [BH2,Example 5.1] for additional examples in P of Z for which all powers of I ( Z ) are symbolic.By Remark 4.16, in none of these cases is there a homogeneous complete intersection ideal J such that I ( Z ) = J r for some r ≥
1. See [HaHu, Proposition 3.5] for a criterion for Z ⊂ P such that I ( mZ ) = I ( Z ) m for all m ≥
1; this gives further examples for which I ( Z ) is not apower of a homogeneous complete intersection.)It is an interesting problem to clarify which fat point subschemes Z ⊂ P N have I ( mZ ) = I ( Z ) m for all m ≥
1. We do not know any Z for which the analytic spread ℓ ( I ( Z )) = N butfor which I ( mZ ) = I ( Z ) m fails for some m ≥
1. Nor do we know any Z for which either ρ ( I ( Z )) = 1 or b ρ ( I ( Z )) = 1 but for which I ( mZ ) = I ( Z ) m fails for some m ≥
1. Thus wehave the following question.
Question 4.2.
Is it true that all powers of I ( Z ) are symbolic (i.e., I ( mZ ) = I ( Z ) m for all m ≥ ) if the analytic spread of I ( Z ) is N , or if ρ ( I ( Z )) = 1 or b ρ ( I ( Z )) = 1 ? Example 4.3.
It is worth noting here that there is a monomial ideal I with ρ ( I ) = b ρ ( I ) = 1where I m ( I ( m ) for every m > I given in [DD, Remark 3.5] is a radicalmonomial ideal and powers of monomial primes are primary, we have I ( m ) = ∩ P ∈ Ass(I) P m .But monomial primes are normal, so I ( m ) is integrally closed. Thus I m ⊆ I ( m ) , and since b ρ ( I ) = 1 we have by [DFMS, Corollary 4.16] that I ( m ) ⊆ I m , so I ( m ) = I m for all m ≥ I m is integrally closed only for m = 1, but nonetheless ρ int ( I ) = ρ ( I ) = 1.Answering Question 4.2 is closely related to whether having ρ ( I ( Z )) = 1 or b ρ ( I ( Z )) = 1gives a complete solution to the containment problem for I ( Z ). If ρ ( I ( Z )) = 1 or b ρ ( I ( Z )) =1, then by Theorem 2.4 when I ( Z ) is radical, we do have a complete solution to the contain-ment problem for I ( Z ): I ( mZ ) I ( Z ) r for m < r and otherwise we have I ( mZ ) ⊆ I ( Z ) r .When I ( Z ) is not radical, we do not know if having either ρ ( I ( Z )) = 1 or b ρ ( I ( Z )) = 1 solvesthe containment problem for I ( Z ). For example, having ρ ( I ( Z )) = 1 means b ρ ( I ( Z )) = 1and it means I ( mZ ) I ( Z ) r for m < r and I ( mZ ) ⊆ I ( Z ) r for m > r , but we do not knowfor which m ≥ I ( mZ ) ⊆ I ( Z ) m .We also do not know any examples with b ρ ( I ( Z )) = 1 but ρ ( I ( Z )) >
1. This raises thefollowing question.
Question 4.4.
Does b ρ ( I ( Z )) = 1 always imply ρ ( I ( Z )) = 1 ? ALUES OF RESURGENCES 11
Example 4.5.
The assumption I = I is needed in both Lemma 2.6 and Theorem 2.7(b).For example, take t = 1 and any N ≥
1. Let I = ( x N +10 , x N +11 , . . . , x N +1 N ) ⊂ K [ P N ]. Then x N · · · x NN ∈ I N = ( x , . . . , x N ) N ( N +1) but x N · · · x NN I , so I tN I t and ρ int ( I ) = N .Although I ( Z ) is not always normal (see Example 4.9) and hence I ( Z ) m = I ( Z ) m canfail, we do not have an example with I ( Z ) m I ( Z ) r when m > r , so we do not know of any Z for which ρ int ( I ( Z )) = 1. If ρ int ( I ( Z )) = 1 were always true, then by Corollary 2.10 wewould have that ρ ( I ( Z )) = 1 if and only if b ρ ( I ( Z )) = 1, thus answering Question 4.4. Thisraises the following question. Question 4.6.
Is it ever true that ρ int ( I ( Z )) > ? The next example shows that ρ int ( I ( Z )) = 1 does not force ρ ( I ( Z )) = 1 or b ρ ( I ( Z )) = 1,even if Z is reduced. In contrast, we know that (b) (and hence (a)) of Theorem 2.4 implies ρ int ( I ( Z )) = 1, but we do not know if any of the other criteria of Theorem 2.4 imply ρ int ( I ( Z )) = 1, unless Z is reduced or N = 2. Example 4.7.
Examples occur with ρ int ( I ( Z )) = 1 but with b ρ ( I ( Z )) >
1. Let Z ⊂ P N be a star configuration , meaning we have s > N general hyperplanes, and Z is the reduced schemeconsisting of (cid:0) sN (cid:1) points, where each point is the intersection of N of the s hyperplanes;see [HaHu, Definition 3.8]. Then the ideal I ( Z ) has α ( I ( Z )) = reg( I ( Z )) = s + N − b α ( I ( Z )) = s/N [BH, Lemma 2.4.1]. As noted in § α ( I ( Z )) =reg( I ( Z )) implies that I ( Z ) is normal and hence ρ int ( I ( Z )) = 1, but α ( I ( Z )) = reg( I ( Z ))also implies that α ( I ( Z )) b α ( I ( Z )) = b ρ ( I ( Z )) = ρ ( I ( Z )), and in the case of a star configuration wehave α ( I ( Z )) b α ( I ( Z )) = N ( s − N + 1) /s > N >
1. However we do not know of any Z with b ρ ( I ( Z )) < ρ int ( I ( Z )). Question 4.8.
Is it ever true that ρ int ( I ( Z )) > b ρ ( I ( Z )) ? Do any of (c), (d) or (e) ofTheorem 2.4 imply ρ int ( I ( Z )) = 1 , when Z is not reduced and N > ? Example 4.9.
When b ρ ( I ( Z )) >
1, it is known that b ρ ( I ( Z )) < ρ ( I ( Z )) can occur. Forexample, let I = ( x ( y n − z n ) , y ( z n − x n ) , z ( x n − y n )) ⊂ C [ x, y, z ] = C [ P ], so I = I ( Z ) where Z is a certain set of n + 3 points. Then by [DHNSST, Theorem 2.1], we have b ρ ( I ( Z )) = α ( I ( Z )) b α ( I ( Z )) = ( n + 1) /n < / ρ ( I ( Z )) for n ≥
3. Moreover, since b ρ ( I ( Z )) < ρ ( I ( Z )), it followsthat I ( Z ) cannot be normal. Example 4.10.
Moreover, examples of Z occur with α ( I ( Z )) b α ( I ( Z )) < b ρ ( I ( Z )). The results of[DFMS] suggest that this should occur, but up to now no explicit examples have beengiven. For one such explicit example (indeed, the first we are aware of), let Z consist of8 points in the plane, where 3 of the points (say p , p , p ) are general (we may as wellassume they are the coordinate vertices) and the other 5 (say p , . . . , p ) are on a generalline L (defined by a linear form F ) and are general on that line. Then one can showthat α ( I ( Z )) = 3, b α ( I ( Z )) = 5 / I (25 sZ ) ( I ( Z ) s +1 for all s ≥
1, and hencethat α ( I ( Z )) / b α ( I ( Z )) = 6 / < / ≤ b ρ ( I ( Z )). (We now sketch the justification ofthese claims. The key is that ( I (25 sZ )) s vanishes on L with order 15 s , but ( I ( Z ) s +1 ) s vanishes on L with order 15 s + 3, hence ( I (25 sZ )) s ( I ( Z ) s +1 ) s and so I (25 sZ ) I ( Z ) s +1 . It is easy to check that α ( I ( Z )) = 3. Using Bezout’s Theorem, one can showthat ( I (2 tZ )) t = ( xyzF ) t K ⊂ K [ P ] = K [ x, y, z ] and hence (since dim(( I ( mZ )) i ) > I ( mZ )) j ) > j > i ) that α ( I (2 tZ )) = 5 t . (In more detail, note that ( xyzF ) t ∈ ( I (2 tZ )) t . If ( I (2 tZ )) t contained anything more than scalar multiples of( xyzF ) t , then ( I (8 tZ )) t would contain more than scalar multiples of ( xyzF ) t , so it’senough to consider ( I (8 tZ )) t . But by Bezout, we have( I (8 tZ )) t = F t (( I (8 t ( p + p + p ) + 3 t ( p + · · · + p ))) t )= ( xyz ) t F t (( I (3 t (2( p + p + p ) + ( p + · · · + p )))) t )= ( xyz ) t F t +1 (( I ((3 t (2( p + p + p )) + (3 t − p + · · · + p )))) t − )= ( xyz ) t +1 F t +1 (( I ((3 t − p + p + p ) + ( p + · · · + p )))) t − )= · · · = ( xyz ) t F t K . )It follows that b α ( I ( Z )) = 5 /
2. We also have by Bezout that ( I (25 sZ )) s = F s ( I (5 sZ ′ )) s ,where Z ′ is the fat point subscheme obtained by taking the three coordinate vertices withmultiplicity 5 and the five points on L with multiplicity 2. Moreover, one can check that( I ( Z ′ )) has greatest common divisor 1 (i.e., it defines a linear system which is fixed com-ponent free); to see this, write elements of ( I ( Z ′ )) in two different ways using conics andlines, where the two ways have no factors in common. Thus ( I (5 sZ ′ )) s also defines a linearsystem that is fixed component free, so ( I (25 sZ )) s = F s ( I (5 sZ ′ )) s has 15 sL as thedivisorial part of its base locus. But ( I ( Z )) t has L in its base locus for t = 3 , t ≥
5. Now ( I ( Z )) s +1 is spanned by products of a homogeneouselements of I ( Z ) of degree 5 or more and b elements of degree 4 or less with a + b = 19 s + 1.Such an element vanishes on L to order at least b . To minimize b we want a as large aspossible, and hence we want to take all a elements to have degree 5 and as many of the b elements as possible to have degree 3, and thus we have the inequality 5 a + 3 b ≤ s . Solving5 a + 3 b ≤ s given a + b = 19 s + 1 gives 2 a ≤ s − a as large as possiblewe have a = 4 s −
2, so b = 15 s + 3. I.e., every element of (( I ( Z )) s +1 ) s vanishes on L toorder at least 15 s + 3.) We do not know if b ρ ( I ( Z )) = 25 /
19, nor do we know the value of ρ ( I ( Z )). Question 4.11.
For the example Z in the previous paragraph, what are the exact values of b ρ ( I ( Z )) , ρ ( I ( Z )) and ρ int ( I ( Z )) ? As Remark 3.2 explains, if the next question has a negative answer, then Conjecture 3.1holds for I = I ( Z ) whenever b ρ ( I ( Z )) = α ( I ( Z )) / b α ( I ( Z )). Question 4.12.
Is it ever false that b α ( I ( Z )) ≥ α ( I ( Z )) + 1 N ?By [DFMS], b ρ ( I ( Z )) is equal to the maximum of v ( I ( Z )) / b v ( I ( Z )) for valuations v sup-ported on I ( Z ), and this maximum always occurs for what is known as a Rees valuation.Thus we have v ( I ( Z )) b v ( I ( Z )) ≤ b ρ ( I ( Z )) ≤ N, hence v ( I ( Z )) N ≤ b v ( I ( Z )) . This raises the question of whether Chudnovsky-like bounds occur for valuations:
ALUES OF RESURGENCES 13
Question 4.13.
Is it always true for
N > that v ( I ( Z )) + 1 N ≤ b v ( I ( Z ))?If the answer is affirmative, then for some v we would have b ρ ( I ( Z )) = v ( I ( Z )) b v ( I ( Z )) ≤ v ( I ( Z )) v ( I ( Z ))+1 N < N which would confirm Grifo’s Conjecture for I ( Z ). Example 4.14.
Consider Z = p + · · · + p N + 2 p N +1 ⊂ P N where the points p i are thecoordinate vertices. Here we show that I ( Z ) m = I ( mZ ) for all m ≥
1. For N = 1, I ( Z ) m = I ( mZ ) holds for every Z (since fat point ideals are principal), so assume N >
1. (Wenote that the case N = 2 is covered by the results of [BZ].) Since I ( Z ) is a monomial ideal, wemerely have to check for each monomial f = x a · · · x a N N ∈ I ( mZ ) that f ∈ I ( Z ) m ; i.e., that f can is divisible by a product of m monomials, each in I ( Z ). But if we choose coordinates x i such that x j = 0 at p j for each j , then f ∈ I ( mZ ) is equivalent to a + · · · + a N ≥ m and a + a + · · · + a N − a i ≥ m for each 0 < i ≤ N . Without loss of generality we mayassume a ≤ a ≤ · · · ≤ a N , so the inequalities a + a + · · · + a N − a i ≥ m reduce to thesingle inequality a + a + · · · + a N − ≥ m . Let us set b = a + · · · + a N − . Then f ∈ I ( mZ )if and only if a + b ≥ m and b + a N ≥ m , so we want to show a + b ≥ m and b + a N ≥ m implies f ∈ I ( Z ) m .First suppose b ≥ m . We have two cases: b + a N = 2 m and b + a N > m . First assume b + a N = 2 m , so 0 < b/ ( N − ≤ a N − ≤ a N ≤ b . Let e = ( e , . . . , e N ) = ( a , · · · , a N ) so e is the exponent vector of g = x e · · · x e N N = x a · · · x a N N . Let b = b = e + · · · + e ,N − .Note that g ∈ I ( mZ ) and g divides f , so it’s enough to show g ∈ I ( Z ) m .Starting with e and b , we will recursively define a sequence e i of exponent vectors e i =( e i , . . . , e iN ) and nonnegative integers b i for 0 ≤ i ≤ ω with b i > ≤ i < ω and b ω = 0,satisfying the following conditions:(i) the entries of e i are nonnegative and nondecreasing;(ii) b i + e i,N is even, where b i = e i + · · · + e i,N − ;(iii) b i ≥ e iN .For e we have b = b ≥ m >
0. Moreover, conditions (i)-(iii) hold for e : (i) holds byassumption; (ii) holds since b + e ,N = b + a N = 2 m by assumption; (iii) holds since wenoted above that a N ≤ b , but e N = a N and b = b .Given that conditions (i)-(iii) hold for some e i with b i >
0, we now define e i +1 =( e i +1 , , . . . , e i +1 ,N ) with b i > b i +1 . In brief, j i and k i are chosen to be as large as possi-ble such that j i < k i and so that the entries of e i +1 = ( e i +1 , , . . . , e i +1 ,N ) are nondecreasing,where e i +1 ,l = e il for all l except that e i +1 ,j i = e ij i − e i +1 ,k i = e ik i −
1. More precisely, j i is the least index t such that e it = e i,N − . Then k i = N if e iN > e i,N − and k i = j i + 1 if e iN = e i,N − . Since b i +1 is either b i − b i − b i > b i +1 .It is easy to check that the construction ensures that the entries of e i +1 are nonnegative andnondecreasing, so (i) holds for e i +1 . Since b i + e i,N is even and b i +1 + e i +1 ,N = b i + e i,N − e i +1 ,N = e iN , then k i < N so 0 < e i,N − = e i,N − = e i,N , andwe cannot have e i,N − = 1 with e i,N − = 0 since then the sum of the entries of e i wouldbe odd. Thus either e i,N − > e i,N − >
0; either way b i ≥ e i,N − = 2 + e iN , so b i +1 = b i − ≥ e iN = e i +1 ,N . If e i +1 ,N = e iN −
1, then b i +1 = b i −
1, so b i ≥ e iN implies b i +1 ≥ e i +1 ,N (and hence b i +1 ≥ g i be the monomial whose exponent vector is e i , so g i = x e i · · · x e iN N . We have g ω = 1,since 0 = b ω ≥ e ωN , so e ω = (0 , . . . , i < ω , then g i = g i +1 x j i x k i . Thus g =( x j x k ) · · · ( x j ω − x k ω − ), but each factor x j i x k i is in I ( Z ), and since deg g = b + e N = 2 m ,we see that there are m factors, so g ∈ I ( Z ) m , as we wanted to show.We now consider case 2, b + a N > m (still under the assumption that b ≥ m ). Startingwith ( a , . . . , a N − , a N ), replace a N by the smallest integer a satisfying a ≥ a N − and b + a ≥ m . We then have that g = x a · · · x a N − N − x aN divides f and is still in I ( mZ ). Thus it is enoughto show that g ∈ I ( Z ) m . If b + a = 2 m , we are done by case 1, so assume b + a > m .Then by construction we have a = a N − . Since b ≥ m , we must have a ≤ m . It is now nothard to find new exponents 0 ≤ a ′ ≤ a ′ ≤ · · · ≤ a ′ N − such that a ′ i ≤ a i and b ′ + a = 2 m ,where b ′ = a ′ + · · · + a ′ N − . Let g = x a ′ · · · x a ′ N − N − x aN ; then g divides f and by case 1 we have g ∈ I ( Z ) m .We are left with considering the case that b < m . We have a + b ≥ m and b + a N ≥ m . Clearly we can reduce a and a N so that a + b = m and b + a N = 2 m (since theassociated monomial g divides f and still is in I ( mZ )). So we may assume a + b = m and b + a N = 2 m , in which case f = ( x a · · · x a N − N − x bN ) x a x a N − bN = ( x a · · · x a N − N − x bN ) x m − b x m − b ) N =( x a · · · x a N − N − x bN )( x x N ) m − b . But x X N ∈ I ( Z ) and x a · · · x a N − N − x bN is a product of b factorsof the form x i x N for 1 ≤ i ≤ N −
1, and each of these is in I ( Z ). Thus f is a product of m = b + ( m − b ) elements of I ( Z ), hence f ∈ I ( Z ) m . Remark 4.15.
Again consider Z = p + · · · + p N + 2 p N +1 ⊂ P N where the points p i are thecoordinate vertices and N >
1. Then there is no N -generated ideal J , homogeneous or not,such that I ( Z ) = J s for some s ≥
1. Suppose there were; say J = ( F , . . . , F N ). The ideal I ( Z ) defines the N + 1 coordinate lines in affine N + 1 space, all taken with multiplicity 1except one taken with multiplicity 2. Since each F i vanishes on each coordinate line, none ofthe F i can have any terms which are a power of a single variable (i.e., every term involves aproduct of two or more variables). Therefore, none of the F i have terms of degree less than2. Since ( F , . . . , F N ) s = J s = I ( Z ) and since I ( Z ) has elements with terms of degree 2, wesee that s = 1 (otherwise the least degree of a term of an element of J s would be at least2 s ≥ s = 1 implies that J = I ( Z ) is homogeneous, and an easy argument shows thatthe least number of homogeneous generators of a homogeneous ideal is the least number ofgenerators possible, which in this case is (cid:0) N (cid:1) + N > N (since, for a particular ordering ofthe points p i we have I ( Z ) = ( x i x j : i > , j >
0) + ( x x i : i > Remark 4.16.
Let Z = m p + · · · + m r p r ⊂ P N with N > m i > i and K [ P N ] = K [ x , . . . , x N ]. Here we show that there is an ideal J generated by N forms suchthat I ( Z ) = J m for some m ≥ m = · · · = m r = m where I ( p + · · · + p r )is generated by N forms. The reverse implication is known [ZS, Lemma 5, Appendix 6],so assume I ( Z ) = J m for some m ≥ F i such that J = ( F , . . . , F N ). Choosecoordinates x , . . . , x N such that none of the points lies on x = 0 and p = (1 , , . . . , U be the affine neighborhood defined by x = 0, and let f i ( x , . . . , x N ) = F i (1 , x , . . . , x N )be the polynomial obtained by setting x to 1 in F i . Then on U , Z is defined by the ideal I ′ ( Z ) obtained from I ( Z ) by setting x = 1 in each element of I ( Z ), so I ′ ( Z ) = ( J ′ ) m where J ′ = ( f , . . . , f N ). Localizing at P = I ( p ) = ( x , . . . , x N ) gives ( P ) m P = ( I ′ ( Z )) P = ( J ′ ) mP ALUES OF RESURGENCES 15 and modding out by P m +11 gives graded isomorphisms P m P m +11 ∼ = ( P ) m P ( P ) m +1 P = ( J ′ ) mP ( P ) m +1 P ∼ = ( J ′ ) m + P m +11 ( P ) m +1 . Thus each f i can have no terms of degree less than d = m /m and there must be terms ofdegree exactly d , so d is an integer, hence m = dm and m ≤ m . But the vector spacedimension of P m P m is (cid:0) m + N − N − (cid:1) , and the vector space dimension of ( J ′ ) m + P m ( P ) m is at most (cid:0) m + N − N − (cid:1) , so we have (cid:0) m + N − N − (cid:1) ≤ (cid:0) m + N − N − (cid:1) , which implies m ≤ m , hence m = m and d = 1.With a change of coordinates, the same argument works for each point p i , so m i = m for all i , and, at each point p i , the linear terms of the f j must be linearly independent (otherwisewe would have dim K ( J ′ ) m + P m ( P ) m < (cid:0) m + N − N − (cid:1) ). Thus J ′ = I ( p + · · · + p r ) on U , hence J = I ( p + · · · + p r ) ⊆ K [ P N ]. Question 4.17.
In Remark 4.16, do we need to assume a priori that J is homogeneous? Computational estimates of resurgences
It might be possible to address some of the foregoing questions computationally. Moregenerally, it is of interest to consider to what extent quantities like resurgences can becomputed.5.1.
Denkert’s thesis.
In an unpublished part of her thesis [Den], Denkert gives an algo-rithm for computing ρ ( I ( Z )) arbitrarily accurately when the symbolic Rees algebra R s ( I ( Z )) = ⊕ t I ( tZ ) x t ⊆ K [ P N ][ x ]is Noetherian (equivalently, when for some a ≥
1, all powers of I ( aZ ) are symbolic [HHT,Theorem 2.1]).Let I = I ( Z ) ⊆ K [ P N ] be nontrivial and assume I ( at ) = ( I ( a ) ) t for all t >
0. For each s ≥
1, let b s be the largest b such that I ( asN ) ⊆ I b and let ǫ >
0. Since b s ≥ as , we have asNb s − asNb s +1 = asNb s ( b s +1) ≤ Nas +1 . So, by picking s ≫ ǫ small, we can make Nas +1 + ǫ arbitrarily small. Denkert’s algorithm either computes ρ ( I ) exactly or gives an estimatewhich is accurate to less than Nas +1 + ǫ .Assume we have picked s and ǫ . Let A = asN and let B = b s . Thus we have I ( At ) = ( I ( A ) ) t ⊆ ( I B ) t = I Bt for all t ≥
1, and for m ≥ At and r ≤ Bt we have I ( m ) ⊆ I ( At ) ⊆ I Bt ⊆ I r . If r ≥ B ⌈ ABǫ ⌉ and mr ≥ AB + ǫ , we now show for the least t such that r < Bt that m ≥ At ,and hence I ( m ) ⊆ I r . Indeed, we have B ( t − ≤ r < Bt , so t − ⌊ rB ⌋ ≥ ⌈ ABǫ ⌉ . Thus( t − Bǫ ≥ A , which is equivalent to ( AB + ǫ )( t − B ≥ At , hence we have m ≥ r ( AB + ǫ ) ≥ B ( t − AB + ǫ ) ≥ At .In summary, we have I ( m ) ⊆ I r for all r ≥ B ⌈ ABǫ ⌉ and mr ≥ AB + ǫ , and we have I ( A ) I B +1 .Thus AB +1 ≤ ρ ( I ) and either ρ ( I ) < AB + ǫ or ρ ( I ) = m/r for some r < B ⌈ ABǫ ⌉ with N > m/r ≥ AB + ǫ (and there are only finitely many such r and m ). Moreover, since we have I ( m ) ⊆ I r for all but finitely many m and r with m/r ≥ AB + ǫ ,we have b ρ ( I ) ≤ AB + ǫ for each ǫ >
0, and hence we have b ρ ( I ) ≤ AB . This also follows byTheorem 1.2(3) of [GHVT].5.2. DiPasquale-Drabkin method.
An alternate approach is based on [DD]. Again as-sume the symbolic Rees algebra of a homogeneous ideal I ⊂ K [ P N ] is finitely generated. Then[DD] shows that we can compute b ρ ( I ) exactly (assuming we know the Rees valuations). Let ǫ >
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Department of Mathematics, University of Nebraska, Lincoln, NE 68588-0130 USA
E-mail address : [email protected] Department of Mathematics, University of Nebraska, Lincoln, NE 68588-0130 USA
E-mail address : [email protected] Department of Mathematics, University of Nebraska, Lincoln, NE 68588-0130 USA
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