The Alexander-Hirschowitz theorem and related problems
aa r X i v : . [ m a t h . A C ] J a n THE ALEXANDER–HIRSCHOWITZ THEOREM AND RELATED PROBLEMS
HUY T `AI H `A AND PAOLO MANTEROA
BSTRACT . We present a proof of the celebrated result due to Alexander and Hirschowitz which determineswhen a general set of double points in P n has the expected Hilbert function. Our intended audience are Com-mutative Algebraists who may be new to interpolation problems. In particular, the main aim of our presentationis to provide a self-contained proof containing all details (including some we could not find in the literature).Also, considering our intended audience, we have added (a) short appendices to make this survey more accessi-ble and (b) a few open problems related to the Alexander–Hirschowitz theorem and the interpolation problems. Dedicated toDavid Eisenbud, on the occasion ofhis 75th birthday.C
ONTENTS
1. Introduction: the Alexander–Hirschowitz Theorem 12. The general case ( d ≥ and n ≥ ) 43. The exceptional cases 164. The case of P ( n = 2 ) 195. The case of cubics ( d = 3 ) 216. Open problems 24A. Appendix: Secant varieties and the Waring problem 28B. Appendix: Symbolic powers 31C. Appendix: Hilbert function 32D. Appendix: Semi-continuity of the Hilbert function and reduction to special configurations 35E. Appendix: Hilbert schemes of points and curvilinear subschemes 38References 391. I NTRODUCTION : THE A LEXANDER –H IRSCHOWITZ T HEOREM
The polynomial interpolation problem originates from the simple fact that a polynomial in one variableover C is completely determined by its zeros. In fact, given r ≤ d distinct points x , . . . , x r on the affineline A C and positive integers m , . . . , m r such that m + . . . + m r = d + 1 , a polynomial f ( x ) = a + a x + · · · + a d x d of degree d is uniquely determined by the following ( d + 1) vanishing conditions onits derivatives, namely f ( j ) ( x i ) = 0 for all i = 1 , . . . , r and j = 0 , . . . , m i − . Equivalently, the matrixarising from these vanishing conditions, which determines the parameters a , . . . , a d , has maximal rank . Anatural question, that has been studied for a long time, is: what happens in higher dimension, meaning forpolynomials in several variables? The problem is much more difficult for several variables, even when the multiplicities m , . . . , m r areall equal and the ambient space is a projective space over the complex numbers. The aim of this paper is to xplore a fundamental result due to Alexander and Hirschowitz, obtained in a series of papers [38, 1, 2, 3, 4](and simplification to its proof given by Chandler in [13, 14]), which shows that, if m = · · · = m r = 2 and the points are chosen to be general points in a projective space, then the same phenomenon happens forhomogeneous polynomials in several variables, except for a few identified exceptional cases.Before proceeding, it may be useful to clarify our intention with this survey. There are already in theliterature a few surveys discussing part of this topic. For instance, the surveys by C. Ciliberto [16] and J.Harris [36] introduce (Hermite) interpolation problems and related results, and include brief discussions ofsome of the geometric ideas behind the Alexander–Hirschowitz theorem. A survey by R. A. Lorentz [39]discusses these topics from a more Numerical Analysis perspective. A survey by M. C. Brambilla and G.Ottaviani [8] discusses the history and presents many details of the core arguments needed in the proof ofthe Alexander–Hirschowitz theorem.These existing surveys use advanced tools from Algebraic Geometry, and are written in languages thatmay be more familiar to an algebraic geometer (cf. [8]) or an analyst (cf. [39]). Some of the stated factsfrom these surveys may not appear so obvious for a young reader who is not specifically well trained inalgebraic geometry; for instance, the use of curvilinear subschemes and the semicontinuity of Hilbert func-tion. Furthermore, the recent large body of work on symbolic powers of ideals in commutative algebra hasdrawn our attention and convinced us that it may be a good time to reintroduce the Alexander–Hirschowitztheorem to commutative algebraists.For these reasons, and partly due to a personal interest, our survey is intended for an audience consistingof young commutative algebraists. We aim to present a detailed and self-contained proof of the Alexander-Hirschowitz theorem and, particularly, to provide all details that may not be easy to see for commutativealgebraists who are new to this research area. We will follow an approach similar to the one of [8]. However,our style of presentation reflects our choices in using algebraic notions and techniques. At the same time,we still identify and appreciate the fundamental geometric ideas at the core of the proof.We should mention that, to the best of our knowledge, there is no survey or paper with a completelyself-contained proof of the Alexander–Hirschowitz theorem. While [8] does include many details of thecore argument, its emphasis is geared towards techniques that historically have been used to approach theInterpolation Problem, and the tight connections of this problem with secant varieties. We have also discov-ered in the literature a few computational inaccuracies and incorrect statements; while they are minor, yeta rechecking was required. Additionally, we shall include all necessary tools in a few appendices; there westate basic results on symbolic powers of ideals, secant varieties, Hilbert functions, generic points, curvilin-ear schemes and the semi-continuity of Hilbert function.The proof we present in this survey incorporates all up-to-date simplifications of the arguments in theoriginal proof of the Alexander–Hirschowitz theorem, including, for instance, the work done by K. Chandler[13, 14], and Brambilla and Ottaviani [8] (regarding the case of cubics).We shall now give a number of important notations and terminology needed to state the Alexander–Hirschowitz theorem. Fix a positive integer n and let R = C [ x , . . . , x n ] = C [ P n ] be the homogeneouscoordinate ring of P n = P n C . For a zero-dimensional subscheme X ⊆ P n , let I X ⊆ R denote its definingideal. It is a basic fact the Hilbert function H R/I X of R/I X is bounded above by its multiplicity e ( R/I X ) and the Hilbert function of R (see Corollary C.5). Particularly, H R/I X ( d ) ≤ min n e ( R/I X ) , (cid:0) n + dd (cid:1)o for all d ∈ N . We say that a zero-dimensional subscheme X in P n has maximal Hilbert function in degree d , orsimply is AH n ( d ) , if H R/I X ( d ) = min (cid:26) e ( R/I X ) , (cid:18) n + dd (cid:19)(cid:27) . This is equivalent to what is often referred to as imposing independent conditions on degree d hypersurfacesin P n . This latter name however could be slightly misleading because the given property is equivalent to the inear system of equations associated to the points having maximal rank; it is not equivalent to the strongerproperty that these equations are linearly independent. Thus, we choose to use the notation AH n ( d ) , whichhas essentially been used already in [13, 8].Another basic fact about Hilbert function of zero-dimensional subschemes in P n , see Propositions C.3and C.4, is that H R/I X ( d ) = e ( R/I X ) for all d ≫ . Thus, we say that X is multiplicity d -independent if H R/I X ( d ) = e ( R/I X ) . In the known literature, this property is commonly referred to as being simply d -independent . We add theword “multiplicity” to the terminology to emphasize the fact that the Hilbert function of R/I X at degree d equals its multiplicity, in this case, and to avoid the potential confusion between the similar-soundingproperties of imposing independent conditions in degree d and being d -independent .Let Y = { P , . . . , P r } be a set of distinct points in P n and suppose that the defining ideal of P i is p i ⊆ R for all i = 1 , . . . , r . Then, the defining ideal of Y is I Y = p ∩ · · · ∩ p r . A celebrated theoremof Zariski and Nagata (Theorem B.5) implies that the symbolic power I (2) Y = p ∩ · · · ∩ p r consists ofhomogeneous polynomials in R that vanish at the points in Y of order at least 2. Let X be the zero-dimensional subscheme in P n defined by I (2) Y . We call X the set of r double points supported on Y , andwrite X = 2 Y = { P , . . . , P r } for simplicity of notation. A double point P is called general if itssupport P is (see Definition D.2 for the precise definition of general set of points). Particularly, X is calleda general set of r double points if Y is a general set of r points.We are ready to state the main theorem surveyed in this paper. Theorem 1.1 (Alexander-Hirschowitz) . Let n, d be positive integers. Let X be a general set of r doublepoints in P n C . Then, X is AH n ( d ) with the following exceptions:(1) d = 2 and ≤ r ≤ n ;(2) d = 3 , n = 4 and r = 7 ; and(3) d = 4 , ≤ n ≤ and r = (cid:0) n +22 (cid:1) − . Our proof of Theorem 1.1 follows an outline similar to the one of [8]. Theorem 1.1 is proved by double-induction, on n and d . For sporadic small values of n and d the inductive hypotheses are not satisfied,mostly because of the exceptional cases. Some of these sporadic cases are indeed the exceptions appearingin the statement, but the other ones are not and they are checked to be AH n ( d ) on an ad hoc basis. Ingeneral, for the inductive step, two fundamental ingredients of the proof are the so-called m´ethode d’Horacediff´erentielle and the use of 0-dimensional schemes of prescribed length and with support on a set of points.Their refined and delicate use is at the core of the simplifications of the original proof.We now outline the structure of this survey. In section 2, we present the proof of Theorem 1.1 for d ≥ and n ≥ , when induction works. This is the most technical section of the paper. We will summarize themain ideas behind the core inductive argument before giving the details of this inductive step in Theorem2.8. Theorem 2.8 is then employed to prove Theorem 1.1 as well as other results in the survey. In section3, we discussthe exceptional cases, leaving out some details of when n = 2 and when d = 3 until laterin Sections 4 and 5. In Section 4, we give the proof of Theorem 1.1 when n = 2 , i.e., for points on theprojective plane. We have chosen to write a proof which employs Theorem 2.8 to provide the reader withanother illustration of the use of this core inductive argument. In Section 5, we conclude the proof of theAlexander–Hirschowitz theorem by examining the case when d = 3 , i.e., for cubics. The paper continueswith a list of open problems and questions in Section 6.As mentioned, we end the paper with a number of short appendices to complement the previous sec-tions. In Appendix A, we briefly illustrate the connection between the (homogeneous, Hermite) doubleinterpolation problem and computing the dimension of certain secant varieties and determining the Waring ank of forms. In Appendix B, we recall the definition of symbolic powers and the statement of a funda-mental theorem of Zariski and Nagata drawing the connection between symbolic powers of ideals of pointsand the interpolation problems. Since this paper is largely about the Hilbert function of zero-dimensionalsubschemes in P n , we have included an appendix about Hilbert functions and, especially, the lower semi-continuity property of Hilbert function; see Appendices C and D. The proof of Theorem 1.1 uses a number ofknown facts about Hilbert schemes of points and curvilinear subschemes , which may not be obvious for analgebraist (they were not obvious for us), so we include an appendix about Hilbert schemes and curvilinearsubschemes; see Appendix E.Finally, for sake of clarity, we have chosen to work over C , however, a large number of results wouldstill be valid over any perfect field (and using divided powers rather than the usual derivatives, in case thecharacteristic of the field is positive). Acknowledgements.
The authors would like to thank Irena Peeva for the invitation to write a paper for thisvolume. The first author is partially supported by Louisiana Board of Regents (grant
HE GENERAL CASE ( d ≥ AND n ≥ )In this section, we discuss the core inductive argument for the proof of Theorem 1.1. It is known, seeProposition C.4, that if X is a set of r double points in P n then e ( R/I X ) = r ( n + 1) . Thus, a set X of r double points in P n is AH n ( d ) if and only if H R/I X ( d ) min (cid:26)(cid:18) n + dn (cid:19) , r ( n + 1) (cid:27) . The following observations allow us to specialize , i.e., to deduce the statement of Theorem 1.1 by con-structing a specific set X of r double points in P n which is AH n ( d ) , and to consider at most two values of r . Remark 2.1.
Fix n, d ∈ Z + and let R = C [ x , . . . , x n ] . • (Corollary D.4) If there exists one collection of r double points in P n that is AH n ( d ) , then anygeneral set of r double points in P n is AH n ( d ) . • (Corollary D.5) To prove that any set of r general double points in P n is AH n ( d ) , it suffices to verifythe statement for the following two values of r : (cid:22) n + 1 (cid:18) n + dd (cid:19)(cid:23) ≤ r ≤ (cid:24) n + 1 (cid:18) n + dd (cid:19)(cid:25) . A key ingredient for the inductive argument of Theorem 1.1 is the so–called Castelnuovo’s Inequalitywhich we now recall.
Lemma 2.2 (Castelnuovo’s Inequality) . Let R be a polynomial ring. Let I be a homogeneous ideal and let ℓ be a linear form in R . Set e I = I : ℓ , R = R/ ( ℓ ) , and I = IR . Then, H R/I ( d ) ≥ H R/ e I ( d −
1) + H R/ ( I ) sat ( d ) . (2.1) Additionally, equality holds for every d if and only if I is saturated in R .Proof. From the standard exact sequence −→ R/I : ℓ ( − · ℓ −→ R/I −→ R/ ( I, ℓ ) −→ , and the fact that I ⊆ ( I ) sat , one obtains H R/I ( d ) = H R/ e I ( d −
1) + H R/I ( d ) ≥ H R/ e I ( d −
1) + H R/ ( I ) sat ( d ) . t is also clear that equality holds for every d if and only if H R/I ( d ) = H R/ ( I ) sat ( d ) for all d , which is thecase if and only if I = ( I ) sat . (cid:3) A intuitive natural approach to Theorem 1.1 is to apply Casteluovo’s Inequality to obtain a proof byinduction on n ≥ . Indeed, Terracini already employed this method to study the case of n = 3 by partlyreducing to the case n = 2 . We shall capture the modern version of Terracini’s argument. Theorem 2.3 (Terracini’s Inductive Argument) . Fix integers r ≥ q ≥ and d ∈ Z + satisfying either r ( n + 1) − (cid:18) d + n − n (cid:19) ≤ qn ≤ (cid:18) d + n − n − (cid:19) or (cid:18) d + n − n − (cid:19) ≤ qn ≤ r ( n + 1) − (cid:18) n − dn (cid:19) . Let L be a hyperplane in P n . If(1) a set of q general double points in L ≃ P n − is AH n − ( d ) , and(2) the union of a set of r − q general double points in P n and a set of q general simple points in L isAH n ( d − ,then a set of r general double points in P n is AH n ( d ) .Proof. Without loss of generality, we may assume that x n = 0 is the equation of L . Let R = C [ x , . . . , x n ] ,and let R = C [ x , . . . , x n − ] ≃ R/ ( x n ) . Let Y be a set of q general simple points in L ≃ P n − , withdefining ideal I Y ⊆ R . If we consider Y as a set of points in P n , then its defining ideal is I Y := ( I Y , x n ) R .Let Y be a set of r − q general simple points in P n − L with defining ideal I Y ⊆ R . Let I = I (2) Y ∩ I (2) Y bethe defining ideal of Y = 2 Y ∪ Y , and I = IR . By Remark 2.1 and Corollary C.5, it suffices to showthat H R/I ≥ min n(cid:0) n + dn (cid:1) , r ( n + 1) o .Since Y ⊆ L and none of the points in Y lies on L , then e I := I : x n = ( I (2) Y : x n ) ∩ ( I (2) Y : x n ) = I Y ∩ I (2) Y , which is the defining ideal of the union of a set of q general simple points in H and r − q general doublepoints in P n . Next we show that ( I ) sat = I Y (2) . First, observe that ht( I ) = dim( R ) − , so ( I ) sat isthe intersection of the minimal components of I . These minimal components are the images in R of theminimal components of ( I, x n ) . Now, the primes containing ( I, x n ) = ( I (2) Y ∩ I (2) Y , x n ) are precisely theprimes containing x n and either I Y or I Y . Since for any p ∈ Min( I Y ) we have x n / ∈ p , then ( p , x n ) =( x , . . . , x n ) , thus it is not a minimal prime of ( I, x n ) (which hasn height n ). On the other hand, forany p ∈ Min( I Y ) we have ht( p ) = n and x n ∈ p , so p ∈ Min(
I, x n ) . It follows that the minimalprimes p of ( I, x n ) are precisely the minimal primes of I Y and when we localize at any of them we have ( I, x n ) p = ( I (2) Y , x n ) p . It follows that ( I, x n ) sat = ( I (2) Y , x n ) sat and by taking images in R we find that ( I ) sat = I Y (2) .Now, by assumptions (1) and (2), we have H R/ e I ( d −
1) = min (cid:26)(cid:18) n + d − n (cid:19) , q + ( n + 1)( r − q ) (cid:27) and H R/I sat ( d ) = min (cid:26)(cid:18) ( n −
1) + dn − (cid:19) , qn (cid:27) . These inequalities together with Lemma 2.2 yield H R/I ( d ) ≥ H R/ e I ( d −
1) + H R/I ( d )= min n(cid:0) n + d − n (cid:1) , q + ( n + 1)( r − q ) o + min n(cid:0) n − dn − (cid:1) , qn o . ow, if r ( n + 1) − (cid:0) d + n − n (cid:1) ≤ qn ≤ (cid:0) d + n − n − (cid:1) , then min n(cid:0) n + dn (cid:1) , r ( n + 1) o = r ( n + 1) , and min (cid:26)(cid:18) n + d − n (cid:19) , q + ( n + 1)( r − q ) (cid:27) +min (cid:26)(cid:18) n − dn − (cid:19) , qn (cid:27) = q +( n +1)( r − q )+ qn = r ( n +1) . Thus, H R/I ( d ) ≥ r ( n + 1) = min (cid:26)(cid:18) n + dn (cid:19) , r ( n + 1) (cid:27) . Similarly, if (cid:0) d + n − n − (cid:1) ≤ qn ≤ r ( n + 1) − (cid:0) n − dn (cid:1) holds, then H R/I ( d ) ≥ min n(cid:0) n + d − n (cid:1) , q + ( n + 1)( r − q ) o + min n(cid:0) n − dn − (cid:1) , qn o = (cid:0) n + d − n (cid:1) + (cid:0) n − dn − (cid:1) = (cid:0) n + dn (cid:1) = min n(cid:0) n + dn (cid:1) , r ( n + 1) o . This concludes the proof (cid:3)
Assumption (1) in Theorem 2.3 is usually provided by the inductive hypothesis. Assumption (2) is abit more delicate, because we have a mix of double points and simple points, however Proposition C.13provides the tool to handle this situation. What prevents one from using Theorem 2.3 to prove Theorem1.1 is the fact that there may not be an integer q satisfying both of the numerical assumptions of Theorem2.3. For instance, to prove the case where n = 3 and d = 6 , by Remark 2.1, we need to prove that a setof r = 21 general double points satisfies AH (6) . To apply Theorem 2.3, we need to find q ∈ Z with − ≤ q ≤ , i.e. q = 28 / . Then, Theorem 2.3 is not applicable. There are in fact infinitely manychoices of n and d for which we run into the same problem, i.e. when we cannot apply Theorem 2.3 directly.The m´ethode diff´erentielle of [2] is designed to overcome this difficulty. For a subscheme X ⊆ P n and ahyperplane L defined by a linear form ℓ , we use e X to denote the residue of X with respect to L ; that is, thesubscheme of P n defined by the ideal I X : ℓ . The underlying ideas of the m´ethode diff´erentielle are:Step 1. Fix a hyperplane L ≃ P n − in P n . For a suitable choice of q and ǫ , choose a general collection of r − q − ǫ double points not in L , a general collection of q double points in L , and a generalcollection of ǫ double points in L .Step 2. By induction on the dimension, the sets ∪ | L and Ψ ∪ ∪ | L have maximal Hilbert functionin degree ( d − in L ≃ P n − . One shows that to prove the theorem it suffices to prove that ismultiplicity [ I ∪ ] d -independent (see Definition 2.5 below).Step 3. Instead of proving this latter statement, for t = ( t , . . . , t ǫ ) ∈ K ǫ we take a flat family of generalpoints Γ t lying on a family of hyperplanes { L t , . . . , L t ǫ } having Γ as a limit when t −→ , andthe problem reduces to show that t is multiplicity [ I ∪ ] d -independent for some t .Step 4. To establish this latter fact, the existence of t in Step 3, we argue by contradiction and a deforma-tion argument reduces the problem to understanding the Hilbert function of schemes of the form ∪ ∪ Θ t , for a suitable curvilinear subscheme Θ t supported on Γ t and contained in t (seeAppendix E for basic facts about curvilinear schemes). Since Γ t is a family of curvilinear schemes,the family has a limit which can be used in the process. Finally, arguments employing the semicon-tinuity of the Hilbert function, the Castelnuovo inequality (2.1) and the material developed in Step2 allows us to conclude.The deformation argument in Step 4 of the m´ethode diff´erentiell e is possible by the use of curvilinearsubschemes and, particularly, Lemma 2.6, which we shall now introduce. efinition 2.4. Let V be a C -vector space of homogeneous polynomials of the same degree in R = C [ x , . . . , x n ] and let I ⊆ R be a homogeneous ideal. Let I ∩ V denote the C -vector space of forms(necessarily of degree d ) belonging to both I and V . Definition 2.5.
Let X ⊆ P n be a zero-dimensional subscheme and let V a C -vector space of homogeneouspolynomials of the same degree in R .(1) The Hilbert function of X (or I X ) with respect to V is defined to be h P n ( X, V ) = dim C V − dim C ( I X ∩ V ) . (It is easy to see that h P n ( X, V ) ≤ min { e ( R/I X ) , dim C V } . )(2) We say that X (or I X ) is multiplicity V -independent if h P n ( X, V ) = e ( R/I X ) . Next, we prove that to check whether a scheme X contained in Y (where Y is a set of finitely manysimple points) is V -independent it suffices to consider curvilinear subschemes of X . This reduction and thefact that curvilinear schemes form a dense open subset of the Hilbert scheme (see Proposition E.7) play animportant role in the proof of Theorem 2.8. Lemma 2.6 (Curvilinear Lemma) . Let X ⊆ P n be a zero-dimensional scheme contained in a finite unionof double points and let V be a C -vector space of homogeneous polynomials of degree d in R . Then X ismultiplicity V -independent if and only if every curvilinear subscheme of X is multiplicity V -independent.Proof. Suppose that X is multiplicity V -independent. This implies that dim C V − dim C ( I X ∩ V ) = e ( R/I X ) ≥ h R/I X ( d ) = dim C R d − dim C ( I X ∩ R d ) . Particularly, V contains all homogeneous polynomials of degree d that are not in I X . As a consequence, forany subscheme Y ⊆ X , we have dim C ( I Y ∩ V ) − dim C ( I X ∩ V ) = e ( R/I X ) − e ( R/I Y ) . Thus, h P n ( Y, V ) = e ( R/I Y ) , and Y is multiplicity V -independent.Suppose now that every curvilinear subscheme of X is multiplicity V -independent. We shall use induc-tion on the number r of points in the support of X and e ( R/I X ) to show that h P n ( X, V ) = e ( R/I X ) .C ASE X is supported at a single point P ∈ P n . If e ( R/I X ) = 1 then X = { P } and the statementis trivial. If e ( R/I X ) = 2 then, locally at P , X ∼ = Spec( T ) where T is a local C -algebra of vector spacedimension 2 over C . This implies that the maximal ideal m of T is of vector space dimension 1 over C and m = 0 . It follows that T ∼ = C [ t ] / ( t ) . As a consequence (see Lemma E.2), X is a curvilinear scheme.Therefore, X is multiplicity V -independent by the hypotheses.Assume that e ( R/I X ) > . Let Y ⊆ X be any subscheme with e ( R/I Y ) = e ( R/I X ) − . Clearly, h P n ( Y, V ) ≤ h P n ( X, V ) . Observe that any curvilinear subscheme of X restricts to a curvilinear subschemeof Y . Thus, by the induction hypothesis, we conclude that Y is multiplicity V -independent. That is, h P n ( Y, V ) = e ( R/I Y ) = e ( R/I X ) − . Particularly, this implies that h P n ( X, V ) ≤ e ( R/I X ) = h P n ( Y, V ) + 1 . Thus, to show that X is multi-plicity V -independent it suffices to construct a subscheme Y of X such that e ( R/I Y ) = e ( R/I X ) − and h P n ( X, V ) = h P n ( Y, V ) + 1 (equivalently, h P n ( X, V ) > h P n ( Y, V ) ).To this end, let ζ ⊆ X be a subscheme of multiplicity 2. As shown above ζ is a curvilinear subscheme of X . Thus, ζ is multiplicity V -independent. That is, h P n ( ζ, V ) = 2 . On the other hand, it is easy to see that h P n ( P, V ) ≤ h R/I P ( d ) = 1 . Therefore, there exists a homogeneous polynomial f in V that vanishes at P but not on ζ . Set Z = V ( f ) be the zero locus of f , and define Y = X ∩ Z . Since X is contained in P , by mposing the condition that f = 0 on Y , we have e ( R/I Y = e ( R/I X ) − . Furthermore, f vanishes on Y but not on X , and so h P n ( X, V ) > h P n ( Y, V ) . The assertion follows in this case.C ASE X is supported at r points P , . . . , P r for r ≥ . By induction on r , we may assume that thestatement is true for schemes supported at r − points.Let I := I X = q ∩ . . . ∩ q r be an irredundant primary decomposition of I = I X and let p i = √ q i for every i . Let q := q r , let Q be its associated scheme and let Z ⊆ X be the scheme defined by I Z := q ∩ . . . ∩ q r − .By assumption, every curvilinear scheme contained in X is V -independent, and then so is every curvilinearscheme contained in Z . Since | Ass(
R/I Z ) | = s − , by inductive hypothesis we have h P n ( Z, V ) = e ( R/I Z ) . Claim 1. X is V -independent if one proves that q is V ∩ [ I Z ] d -independent.Indeed, to prove that X is V -independent we compute h P n ( X, V ) = dim C V − H I Z ∩ q ∩ V ( d )= [dim C V − H I Z ∩ V ( d )] + [ H I Z ∩ V ( d ) − H I Z ∩ q ∩ V ( d )]= h P n ( Z, V ) + [ H I Z ∩ V ( d ) − H I Z ∩ q ∩ V ( d )]= e ( R/I Z ) + [ H I Z ∩ V ( d ) − H I Z ∩ q ∩ V ( d )] . If q is V ∩ [ I Z ] d -independent, then H I Z ∩ V ( d ) − H I Z ∩ q ∩ V ( d ) = e ( R/ q ) , thus h P n ( X, V ) = e ( R/I Z ) + e ( R/ q ) = e ( R/I ) . Claim 2.
It suffices to prove that Z ∪ Q ′ is V -independent for any curvilinear scheme Q ′ ⊆ Q .We prove Claim 2. Let q ′ ⊇ q be the defining ideal of Q ′ ⊆ Q . By Claim 1 it suffices to prove that Q is V ∩ [ I Z ] d -independent. By the base case of induction, it suffices to prove that Q ′ is V ∩ [ I Z ] d -independentfor any curvilinear scheme Q ′ ⊆ Q . We compute h P n ( Q ′ , V ∩ [ I Z ] d ) : h P n ( Q ′ , V ∩ [ I Z ] d ) = [dim C V − H I Z ∩ q ′ ∩ V ( d )] − [dim C V − H I Z ∩ V ( d )]= h P n ( Z ∪ Q ′ , V ) − h P n ( Z, V )= h P n ( Z ∪ Q ′ , V ) − e ( R/I Z ) . so if Z ∪ Q ′ is V -independent, then h P n ( Z ∪ Q ′ , V ) = e ( R/I Z ∩ q ′ ) and thus by the above h P n ( Q ′ , V ∩ [ I Z ] d ) = e ( R/I Z ∩ q ′ ) − e ( R/I Z ) = e ( R/ q ′ ) , proving that Q ′ is V ∩ [ I Z ] d -independent. This proves Claim 2. Claim 3.
It suffices to show that for any curvilinear scheme Z ′ ⊆ Z one has Z ′ is V ∩ [ q ′ ] d -independent.Recall that by Claim 2 it suffices to prove Z ∪ Q ′ is V -independent. We observe that h P n ( Z ∪ Q ′ , X ) = dim C V − H I Z ∩ q ′ ∩ V ( d )= [dim C V − H q ′ ∩ V ( d )] + [ H q ′ ∩ V ( d ) − H I Z ∩ q ′ ∩ V ( d )]= h P n ( Q ′ , V ) + h P n ( Z, V ∩ [ Q ′ ] d ) . Since Q ′ ⊇ Q ⊇ X is curvilinear, then by assumption Q ′ is V -independent, so h P n ( Q ′ , V ) = e ( R/ q ′ ) .So it suffices to prove h P n ( Z, V ∩ [ Q ′ ] d ) = e ( R/I Z ) , because then by the above h P n ( Z ∪ Q ′ , X ) = e ( R/ q ′ ) + e ( R/I Z ) = e ( R/I Z ∩ q ′ ) , proving that Z ∪ Q ′ is V -independent.Since | Ass(
R/I Z ) | < s , by inductive hypothesis, to prove that Z is V ∩ [ Q ′ ] d -independent it suffices toprove that Z ′ is V ∩ [ Q ′ ] d -independent for any curvilinear subscheme Z ′ ⊇ Z . This proves Claim 3. e conclude the proof by showing that Z ′ is V ∩ [ Q ′ ] d -independent. First we compute h P n ( Z, V ∩ [ Q ′ ] d ) : h P n ( Z, V ∩ [ Q ′ ] d ) = H q ′ ∩ V ( d ) − H I Z ′ ∩ q ′ ∩ V ( d )= [dim C V − H I Z ′ ∩ q ′ ∩ V ( d )] − [dim C V − H q ′ ∩ V ( d )]= h P n ( Z ′ ∪ Q ′ , V ) − h P n ( Q ′ , V ) . Since Q ′ ⊆ Q ⊆ X is curvilinear, then by assumption h P n ( Q ′ , V ) = e ( R/ q ′ ) . Since Q ′ and Z ′ arecurvilinear and have disjoint support (because Ass( R/ q ′ ) = { p r } and Ass(
R/I Z ′ ) ⊆ Ass(
R/I Z ) = { p , . . . , p r − } ), then Z ′ ∪ Q ′ is locally curvilinear at point of the support, thus it is curvilinear. Since Z ′ ∪ Q ′ ⊆ Z ∩ Q = X , then by assumption h P n ( Z ′ ∪ Q ′ , V ) = e ( R/I Z ′ ∩ q ′ ) . Therefore h P n ( Z ′ , V ∩ [ Q ′ ] d ) = e ( R/I Z ′ ∩ q ′ ) − e ( R/ q ′ ) = e ( R/I Z ′ ) , which concludes the proof. (cid:3) Recall that, by Remark 2.1, to prove Theorem 1.1 for values of n, d not in the list of exceptional cases, itsuffices prove that a general set of r double points has AH n ( d ) for j n + 1 (cid:18) n + dn (cid:19)k ≤ r ≤ l n + 1 (cid:18) n + dn (cid:19)m . Set q and ǫ to be the quotient and remainder of the division of r ( n + 1) − (cid:0) n + d − n (cid:1) by n . For the ease ofreferences, we shall provide the values of q and ǫ for a few special choices of n and d . We start with thecase where d = 4 . n value of r ∆ := r ( n + 1) − (cid:0) n + d − n (cid:1) value of q value of ǫ value of r − q − ǫn = 2 r = 5 ∆ = 5 q = 2 ǫ = 1 r − q − ǫ = 2 n = 3 r = 8 ∆ = 12 q = 4 ǫ = 0 r − q − ǫ = 4 r = 9 ∆ = 16 q = 5 ǫ = 1 r − q − ǫ = 3 n = 4 r = 14 ∆ = 35 q = 8 ǫ = 3 r − q − ǫ = 3 n = 5 r = 21 ∆ = 70 q = 14 ǫ = 0 r − q − ǫ = 7 n = 6 r = 30 ∆ = 126 q = 21 ǫ = 0 r − q − ǫ = 9 n = 7 r = 41 ∆ = 208 q = 29 ǫ = 5 r − q − ǫ = 7 r = 42 ∆ = 216 q = 30 ǫ = 6 r − q − ǫ = 6 n = 8 r = 55 ∆ = 330 q = 41 ǫ = 2 r − q − ǫ = 12 n = 9 r = 71 ∆ = 490 q = 54 ǫ = 4 r − q − ǫ = 13 r = 72 ∆ = 500 q = 55 ǫ = 5 r − q − ǫ = 12 For d = 5 and we get the following table. n value of r ∆ := r ( n + 1) − (cid:0) n + d − n (cid:1) value of q value of ǫ value of r − q − ǫn = 2 r = 7 ∆ = 6 q = 3 ǫ = 0 r − q − ǫ = 4 n = 3 r = 14 ∆ = 21 q = 7 ǫ = 0 r − q − ǫ = 7 n = 4 r = 25 ∆ = 55 q = 13 ǫ = 3 r − q − ǫ = 9 r = 26 ∆ = 60 q = 15 ǫ = 0 r − q − ǫ = 11 n = 5 r = 42 ∆ = 126 q = 25 ǫ = 1 r − q − ǫ = 16 n = 6 r = 66 ∆ = 252 q = 42 ǫ = 0 r − q − ǫ = 24 n = 7 r = 99 ∆ = 462 q = 66 ǫ = 0 r − q − ǫ = 33 We now prove a few basic numeric facts that will be employed later.
Lemma 2.7.
For fixed integers n ≥ , d ≥ and ≤ r ≤ l n +1 (cid:0) n + dn (cid:1)m , let q ∈ Z and ≤ ǫ < n be suchthat nq + ǫ = r ( n + 1) − (cid:0) n + d − n (cid:1) . Then, nǫ + q ≤ (cid:0) n + d − n − (cid:1) ,(2) (cid:0) n + d − n (cid:1) ≤ ( r − q − ǫ )( n + 1) ,(3) r − q − ǫ ≥ n + 1 , for d = 4 and n ≥ .(4) q ≥ ǫ .Proof. (1) We prove the equivalent statement that n ( nǫ + q ) ≤ n (cid:0) n + d − n − (cid:1) . Clearly, nq ≤ r ( n +1) − (cid:0) n + d − n (cid:1) .Since r ≤ l n +1 (cid:0) n + dn (cid:1)m , we have ( n + 1) r ≤ (cid:0) n + dn (cid:1) + n , and so(2.2) n ǫ + nq ≤ n ( n −
1) + (cid:18) n + dn (cid:19) + n − (cid:18) n + d − n (cid:19) = n ( n −
1) + (cid:18) n + d − n − (cid:19) + n. The right-hand side is at most n · (cid:0) n + d − n − (cid:1) except when d = 4 and ≤ n ≤ . In these three caseshowever the inequality still holds, as one can check directly with given values of q and ǫ in the above tables.(2) Since ( r − q − ǫ )( n + 1) = r ( n + 1) − ( nq + ǫ ) − ( nǫ + q ) , we have ( r − q − ǫ )( n + 1) = (cid:18) n + d − n (cid:19) − ( nǫ + q ) ≥ (cid:18) n + d − n (cid:19) − (cid:18) n + d − n − (cid:19) = (cid:18) n + d − n (cid:19) , whence the inequality holds by (1).(3) We prove the equivalent statement that ( r − q − ǫ )( n + 1) ≥ ( n + 1) for d = 4 and n ≥ . By thework in (2) we see that ( r − q − ǫ )( n + 1) ≥ ( n + 1) holds if and only (cid:0) n +3 n (cid:1) − ( nǫ + q ) ≥ ( n + 1) . Thisholds if and only if ( n + 1) (cid:18) ( n + 3)( n + 2)6 − ( n + 1) (cid:19) ≥ nǫ + q ⇐⇒ (cid:18) n + 13 (cid:19) ≥ nǫ + q. By equation (6.2), nǫ + q ≤ n ( n −
1) + (cid:0) n +34 (cid:1) + n . The right-hand side is at most (cid:0) n +13 (cid:1) if and only if n − n + 3 n − ≥ . This equality holds for all n ≥ . The cases n = 8 , are easily checked byusing the above tables.(4) Assume by contradiction q < ǫ then r ( n + 1) − (cid:0) n + d − n (cid:1) = nq + ǫ < ( n + 1) ǫ ≤ ( n + 1)( n − . Bydefinition of r one also sees that r ( n + 1) > (cid:0) n + dn (cid:1) − ( n + 1) , so r ( n + 1) ≥ (cid:0) n + dn (cid:1) − n . Combining the abovetogether we obtain (cid:0) n + dn (cid:1) − n ≤ r ( n + 1) < ( n + 1)( n −
1) + (cid:0) n + d − n (cid:1) , thus (cid:0) n + d − n − (cid:1) < ( n + 1)( n −
1) + n .Since ( n + 1)( n −
1) + n − n + 2)( n − , then (cid:0) n + d − n − (cid:1) ≤ ( n + 2)( n − . It is well-known that (cid:0) n + d − n − (cid:1) increases as d increases, so the left-hand side is at least (cid:0) n +3 n − (cid:1) = (cid:0) n +34 (cid:1) . Inparticular ( n +3)( n +2)( n +1) n ≤ ( n + 2)( n − , thus f ( n ) ≤ , where f ( n ) := ( n + 3)( n + 1) n − n − . It is easily seen that f ( n ) is increasing for n ≥ and f (2) = 6 > , thus f ( n ) > for every n ≥ ,yielding a contradiction. (cid:3) We are ready to present the core inductive argument for Theorem 1.1. The proof follows the four stepswe outlined when we illustrated the m´ethode diff´erentielle . Theorem 2.8.
For fixed n ≥ , d ≥ and j n + 1 (cid:18) n + dn (cid:19)k ≤ r ≤ l n + 1 (cid:18) n + dn (cid:19)m , let q ∈ Z and ≤ ǫ < n be such that nq + ǫ = r ( n + 1) − (cid:0) n + d − n (cid:1) . Suppose that(i) q general double points are AH n − ( d ) .(ii) r − q general double points are AH n ( d − (iii) r − q − ǫ general double points are AH n ( d − Then, r general double points are AH n ( d ) . roof. By Remark 2.1, it suffices to construct a set of r double points in P n which is AH n ( d ) . This set of r double points arises in the form ∪ ∪ t , for some family of parameters t , where the sets Ψ , Λ and Γ t are constructed as in the outlined steps. To understand the construction better, we shall use 21 double pointsin P and degree 6 as our running example; in this particular situation, r = 21 , d = 6 , q = 9 and ǫ = 1 . Step 1.
We first fix a hyperplane L ≃ P n − in P n , with defining equation ℓ = 0 . We take a set of q + ǫ general points in L , let Γ = { γ , . . . , γ ǫ } be a subset of ǫ of these points, and let Λ be the set consistingof the remaining q points. Finally, we take a set Ψ of r − q − ǫ general points in P n outside of L . (In ourrunning example, Γ consists of a single point in L , Λ of 9 general points in L , and Ψ of 11 general pointsoutside of L .) Step 2.
By (ii), we have H R/ ( I (2)Ψ ∩ I (2)Γ ) ( d −
1) = min (cid:26) ( n + 1)( r − q ) , (cid:18) n + d − n (cid:19)(cid:27) = ( n + 1)( r − q ) , where the rightmost equality holds because Lemma 2.7(4) yields (cid:0) n + d − n (cid:1) = ( n + 1)( r − q ) − ǫ + q ≥ ( n + 1)( r − q ) . Now, if we consider Γ | L instead of Γ , then the linear system associated to [ I (2)Ψ ∩ I (2)Γ | L ] d − is obtained by removing ǫ equations from the linear system of equations defined by [ I (2)Ψ ∩ I (2)Γ | L ] d − (moreprecisely, the ones corresponding to setting the partial derivative with respect to ℓ equal to ). One thenobtains H R/ ( I (2)Ψ ∩ I (2)Γ | L ) ( d −
1) = min (cid:26) ( n + 1)( r − q ) − ǫ, (cid:18) n + d − n (cid:19)(cid:27) = ( n + 1)( r − q ) − ǫ (= 47 for the running example), and then H I (2)Ψ ∩ I (2)Γ | L ( d −
1) = (cid:0) n + d − n (cid:1) − H R/ ( I (2)Ψ ∩ I (2)Γ | L ) ( d −
1) = q ( = 9 in the running example). Claim 1. H R/ ( I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ) ( d −
1) = e ( R/ ( I (2)Ψ ∩ I (2)Γ | L ∩ I Λ )) = (cid:0) n + d − d − (cid:1) .(that is, H R/ ( I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ) (5) = (cid:0) − (cid:1) = 56 for the running example.)Notice that Claim 1 implies that ∪ | L ∪ Λ is multiplicity ( d − -independent. The rightmost equalityin the claim holds because (cid:0) n + d − d − (cid:1) = (cid:0) n + d − d − (cid:1) − ǫ + q = e ( R/ ( I (2)Ψ ∩ I (2)Γ | L ∩ I Λ )) . To conclude the proof itthen suffices to show that H R/ ( I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ) ( d −
1) = (cid:0) n + d − d − (cid:1) = H R ( d − , i.e. [ I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ] d − = 0 .If we restate the paragraph before the Claim in terms of linear algebra, we see that the solution set ofthe linear system defined by [2Ψ ∪ | L ] d − is a q -dimensional vector space. Now, for each simple generalpoint in P n that we are adding to ∪ | L , we are adding a general linear equation to this system, sowe are reducing the dimension of the solution set by 1. So, if we add q general simple points in P n to ∪ | L , then the corresponding ideal contains no forms of degree d − . It follows that if we add q pointsto ∪ | L , and these q additional points lie on L , then the defining equation ℓ of L divides the equationof any hypersurface of degree d − passing through ∪ | L and these q points. In particular, any form F ∈ [ I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ] d − is divisible by ℓ , and we can write F = F ℓ .Since ℓ ∈ I Λ (because Λ ⊆ L ) and ℓ is regular on R/I Ψ (because none of the points of Ψ lies on L ), then F is a degree ( d − form in ( I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ) : ℓ = I (2)Ψ ∩ ( I (2)Γ | L : ℓ ) ⊆ I (2)Ψ . owever, by (iii), we know that H I (2)Ψ ( d −
2) = max n , (cid:0) n + d − n (cid:1) − ( r − q − ǫ )( n + 1) o = 0 , thus F = 0 ,and so F = 0 . Then [ I (2)Ψ ∩ I (2)Γ | L ∩ I Λ ] d − = 0 and, therefore, Claim 1 is proved.To prove the theorem we need to prove the following equality H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I (2)Γ ) ( d ) = min (cid:26) ( n + 1) r, (cid:18) n + dn (cid:19)(cid:27) . We now proceed by considering two different cases depending on which of the two possible values takes theright-hand side. Since, by assumption, j n + 1 (cid:18) n + dn (cid:19)k ≤ r ≤ l n + 1 (cid:18) n + dn (cid:19)m , it can be easily seenthat • min n ( n + 1) r, (cid:0) n + dn (cid:1)o = ( n + 1) r holds precisely if r = j n + 1 (cid:18) n + dn (cid:19)k , • min n ( n + 1) r, (cid:0) n + dn (cid:1)o = (cid:0) n + dn (cid:1) > r ( n + 1) holds if r = l n + 1 (cid:18) n + dn (cid:19)m > j n + 1 (cid:18) n + dn (cid:19)k .The running example of 21 double points in P falls in the first possibility.C ASE r = j n + 1 (cid:18) n + dn (cid:19)k . In this case r ( n + 1) ≤ (cid:18) n + dn (cid:19) , nq + ǫ ≤ (cid:0) n + d − n − (cid:1) , and by the abovewe need to show H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I (2)Γ ) ( d ) = ( n + 1) r. Claim 2. H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) = e ( R/ ( I (2)Λ ∩ I (2)Ψ )) = ( n + 1)( r − ǫ ) . Castelnuovo’s inequality (2.1) gives H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) ≥ H R/ ( I Λ ∩ I (2)Ψ ) ( d −
1) + H R/I (2)Λ | L ( d ) . By Claim1 and Lemma C.12(1) H R/ ( I (2)Ψ ∩ I Λ ) ( d −
1) = ( n + 1)( r − q − ǫ ) + q . By assumption (ii) and the inequality nq ≤ (cid:0) n + d − n − (cid:1) we have H R/I (2)Λ | L ( d ) = min (cid:26) nq, (cid:18) n − dn − (cid:19)(cid:27) = nq = e ( R/I (2)Λ | L ) . Thus H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) ≥ ( n + 1)( r − q − ǫ ) + q + nq = ( n + 1)( r − ǫ ) . Since the other inequality alwaysholds by Corollary C.5, then Claim 2 is proved.To finish this case it suffices to prove that I (2)Γ is multiplicity [ I (2)Λ ∩ I (2)Ψ ] d -independent, because then H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I (2)Γ ) ( d ) = H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) + e ( R/I (2)Γ ) = ( n + 1)( r − ǫ ) + ( n + 1) ǫ = ( n + 1) r. Instead of proving it directly, we will prove it using a family of general points Γ t having Γ as a limit (as weshall explain in the upcoming Step 3).For now, we observe the following fact. As above, if we add ǫ general points of L to | L we are adding ǫ general equations to the linear system determined by [ I (2)Λ | L ] d , and then H R/ ( I (2)Λ | L ∩ I Γ ) ( d ) = min (cid:26) nq + ǫ, (cid:18) n − dn − (cid:19)(cid:27) = nq + ǫ. (2.3)(In our running example, H R/I (2)Λ | L ∩ I Γ ( d ) = (3)(9) + 1 = 28 .) Step 3.
For t = ( t , . . . , t ǫ ) ∈ K ǫ , consider a flat family of general points Γ t = { γ ,t , . . . , γ ǫ,t ǫ } in P n anda family of hyperplanes { L t , . . . , L t ǫ } such that(1) for every t i , the points γ i,t i lie in L t i for all i = 1 , . . . , ǫ , γ i,t i L for any t i = 0 and any i = 1 , . . . , ǫ ,(3) L = L and γ i, = γ i ∈ L for any i = 1 , . . . , ǫ. (For the running example, we have a family of general points Γ t = { γ t } ⊆ P and a family of hyperplanes L t , for t ∈ K .) Step 4.
To prove I (2)Γ is multiplicity [ I (2)Λ ∩ I (2)Ψ ] d -independent, by Corollary C.5 and Theorem D.9 it sufficesto prove that there exists t = ( t , . . . , t ǫ ) ∈ K ǫ such that I (2)Γ t is multiplicity [ I (2)Λ ∩ I (2)Ψ ] d -independent,because then ( n + 1) r ≥ H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I (2)Γ ) ( d ) ≥ H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I (2)Γ t ) ( d ) = ( n + 1) r. Suppose, by contradiction, such a t does not exist. Then, by Lemma 2.6, for each t = ( t , . . . , t ǫ ) , thereexist curvilinear ideals J i,t i such that I (2) γ i,ti ⊆ J i,t i ⊆ I γ i,ti and, by letting J t = T ǫi =1 J i,t i , we then have H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J t ) ( d ) < H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) + e ( R/J t ) = ( n + 1)( r − ǫ ) + e ( R/J t ) . (2.4)(For the running example, there is a single point γ t , so I γ t is a linear prime and J t is a curvilinear ideal J t with I (2) γ t ⊆ J t ⊆ I γ t and H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J t ) (6) < (3 + 1)(20) + e ( R/J t ) = 80 + e ( R/J t ) .)Since J t is a curvilinear ideal, by Proposition E.7, for every i = 1 , . . . , ǫ the family { J i,t i } has a limit J i, . Let J = T ǫi =1 J i, .Let A := { i (cid:12)(cid:12) ℓ / ∈ J i, } , B := { i (cid:12)(cid:12) ℓ ∈ J i, } and A ′ := { i ∈ A (cid:12)(cid:12) ℓ ∈ p J i, } . We set a = | A | , a ′ = | A ′ | ,and b := | B | . For each t ∈ K ǫ , set J A t = T i ∈ A J i,t i , and J B t = T i ∈ A J i,t i , in particular J t = J A t ∩ J B t .We also set I A ′ Γ = T i ∈ A ′ I γ i .By the semicontinuity of Hilbert function and (2.4), there exists an open neighborhood U of such thatfor any t ∈ U , we have the equalities e ( R/J t ) = e ( R/J ) , H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J A ∩ J B t ) ( d ) = H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J A t ∩ J B t ) ( d ) < ( n + 1)( r − ǫ ) + e ( R/J t ) , (2.5)(for the running example, we have H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J A ∩ J B t ) ( d ) < ), and H R/ ( I (2)Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B t ) ( d −
1) = H R/ ( I (2)Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B ) ( d − . We want to show H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J A ∩ J B t ) ( d ) ≥ ( n + 1)( r − ǫ ) + e ( R/J t ) , which would then contradict(2.5). For any t ∈ U , Castelnuovo inequality gives H R/I (2)Λ ∩ I (2)Ψ ∩ J A ∩ J B t ( d ) ≥ H R/I Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B t ( d −
1) + H R/ ( I (2)Λ | L ∩ I A ′ Γ | L ) ( d ) , where R ∼ = R/ ( ℓ ) and I A ′ Γ | L is the defining ideal of { γ i | i ∈ A ′ } in R .We examine the first summand which, by the above, equals H R/ ( I (2)Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B ) ( d − for every t ∈ U . By Claim 1, the ideal I Λ ∩ I (2)Ψ ∩ I (2)Γ | L is multiplicity ( d − -independent, so the larger ideal I Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B is multiplicity ( d − -independent too. Then H R/ ( I Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B ) ( d −
1) = e ( R/ ( I Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B )) (2.6) = e ( R/ ( I Λ ∩ I (2)Ψ )) + e ( R/ ( J A : ℓ ) ∩ J B )= q + ( n + 1)( r − q − ǫ ) + e ( R/ ( J A : ℓ ) ∩ J B )= q + ( n + 1)( r − q − ǫ ) + e ( R/J ) − a ′ , here a ′ is the cardinality of { i ∈ A | ℓ is not regular on R/J i, } . The last equality holds because e ( R/J ) = e ( R/J : ℓ ) + e ( R/ ( J , ℓ )) = e ( R/J A : ℓ ) + e ( R/ ( J , ℓ )) = e ( R/ ( J A : ℓ )) + e ( R/J B ) + a ′ .Now, the inclusion | L ∪ { γ i (cid:12)(cid:12) i ∈ F } ⊆ | L ∪ Γ , and (2.3) allow the use of Lemma C.12(1) to deduce H R/ ( I Λ | (2) L ∩ I A ′ Γ | L ) ( d ) ≥ e (cid:16) R/ (cid:16) I Λ | (2) L ∩ I A ′ Γ | L (cid:17)(cid:17) = nq + a ′ . Combining all the above, we obtain for any t ∈ UH R/I (2)Λ ∩ I (2)Ψ ∩ J A ∩ J B t ( d ) ≥ H R/I Λ ∩ I (2)Ψ ∩ ( J A : ℓ ) ∩ J B t ( d −
1) + H R/I (2)Λ | L ∩ I A ′ Γ ( d ) ≥ q + ( n + 1)( r − q − ǫ ) + e ( R/J ) − a ′ + ( nq + a ′ )= ( n + 1)( r − ǫ ) + e ( R/J ) = ( n + 1)( r − ǫ ) + e ( R/J t ) . (For the running example, this implies one of the following inequalities H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J ) (6) ≥ e ( R/J t ) or H R/ ( I (2)Λ ∩ I (2)Ψ ∩ J t ) (6) ≥
80 + e ( R/J t ) .) This is a contradiction to (2.4), and we are done.C ASE r > j n + 1 (cid:18) n + dn (cid:19)k . In this case, r = l n + 1 (cid:18) n + dn (cid:19)m , and we have r ( n + 1) > (cid:18) n + dn (cid:19) and nq + ǫ > (cid:18) n + d − n − (cid:19) . First, one considers the case where nq ≥ (cid:0) n + d − n − (cid:1) . Then by (i), we have H R/I (2)Λ | L ( d ) = min (cid:26) nq, (cid:18) n + d − n − (cid:19)(cid:27) = (cid:18) n + d − n − (cid:19) . On the other hand, by (ii), we have H R/ ( I (2)Γ ∩ I (2)Ψ ) ( d −
1) = min (cid:26) ( n + 1)( r − q ) , (cid:18) n + d − n (cid:19)(cid:27) . For any set ∆ of q − ǫ general points in P n one has H R/ ( I ∆ ∩ I (2)Γ ∩ I (2)Ψ ) ( d −
1) = (cid:0) n + d − n (cid:1) = e ( R/ ( I ∆ ∩ I (2)Γ ∩ I (2)Ψ )) . As in the proof of Claim 1, this yields that for any subset Λ ′ ⊆ Λ consisting of q − ǫ points one has H R/ ( I Γ ′ ∩ I (2)Γ ∩ I (2)Ψ ) ( d −
1) = (cid:0) n + d − n (cid:1) . By Lemma C.12(2), one obtains H R/ ( I Λ ∩ I (2)Γ ∩ I (2)Ψ ) ( d −
1) = (cid:0) n + d − n (cid:1) . Thus, by the Castelnuovo inequality, we get H R/ ( I (2)Λ ∩ I (2)Γ ∩ I (2)Ψ ) ( d ) ≥ H R/ ( I Λ ∩ I (2)Γ ∩ I (2)Ψ ) ( d −
1) + H R/I (2)Λ | L ( d )= (cid:18) n + d − n (cid:19) + (cid:18) n + d − n − (cid:19) = (cid:18) n + dn (cid:19) . Hence, equality holds and ∪ ∪ satisfies AH n,d .We may now assume that < ν := (cid:18) n + d − n − (cid:19) − nq < ǫ , and let Γ ′ = { γ , . . . , γ ν } ⊆ Γ . An argu-ment similar to the above (or the proof of Claim 1) shows that H R/ ( I (2)Λ | L ∩ I Γ ′| L ) ( d ) = (cid:18) n + d − n − (cid:19) = nq + ν. Then, by Lemma C.12(2), one has H R/ ( I (2)Λ | L ∩ I Γ | L ) ( d ) = min (cid:26) nq + ǫ, (cid:18) n + d − n − (cid:19)(cid:27) = (cid:18) n + d − n − (cid:19) . o prove that ∪ ∪ satisfies AH n,d we need to prove H R/ ( I (2)Λ ∩ I (2)Γ ∩ I (2)Ψ ) ( d ) = (cid:18) n + dn (cid:19) . Let t and Γ t be defined as in Case 1. By the semi-continuity of the Hilbert function, there exists a neighbor-hood U of such that for t ∈ U we have H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I Γ ) ( d ) = H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I Γ t ) ( d ) . Claim 3.
To finish the proof it suffices to find an ideal K ⊇ I (2)Γ t such that K is multiplicity [ I (2)Λ ∩ I (2)Ψ ] d -independent, and e ( R/K ) = nǫ + ν .Assume such an ideal K exists. The first assumption gives H R/ ( I (2)Λ ∩ I (2)Ψ ∩ K ) ( d ) = H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) + e ( R/K ) . Also, by Claim 2, H R/ ( I (2)Λ ∩ I (2)Ψ ) ( d ) = ( n + 1)( r − ǫ ) . Finally, e ( R/K ) is precisely the amountneeded to ensure that H R/ ( I (2)Λ ∩ I (2)Ψ ∩ K ) ( d ) = (cid:0) n + dn (cid:1) , because one has H R/ ( I (2)Λ ∩ I (2)Ψ ∩ K ) ( d ) = ( n + 1)( r − ǫ ) + e ( R/K ) = ( n + 1)( r − ǫ ) + nǫ + ν = ( n + 1) r − ǫ + ν = ( n + 1) r − ( nq + ǫ ) + ( nq + ν )= (cid:0) n + d − n (cid:1) + (cid:0) n + d − n − (cid:1) = (cid:0) n + dn (cid:1) . Now, H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I Γ ) ( d ) = H R/ ( I (2)Λ ∩ I (2)Ψ ∩ I Γ t ) ( d ) = (cid:18) n + dn (cid:19) , where the rightmost equality follows from Lemma C.12(2). This proves Claim 3.Finally, it is easily seen that K := I (2)Γ ′ ∩ I (Γ − Γ ′ ) | L satisfies the above properties. This concludes the proofof the theorem. (cid:3) Besides that the exceptional cases are proved in the following sections, we now give a complete proof forthe Alexander–Hirschowitz theorem.
Theorem 2.9. (Alexander–Hirschowitz) For every n ≥ and d ≥ , a set X of r general double points in P n C is AH n ( d ) , with the following exceptions:(1) d = 2 and ≤ r ≤ n ;(2) d = 3 , n = 4 and r = 7 ;(3) d = 4 , ≤ n ≤ and r = (cid:0) n +22 (cid:1) − .Proof. By Remarks C.9 and C.10, we may assume that r ≥ and d ≥ . The statement for n = 1 is provedin Proposition C.11. The case where n = 2 is treated in Section 4. Thus, we may also assume that n ≥ .The exceptional cases are discussed in Sections 3,4 and 5. Furthermore, it will be shown that for fixed d and n , the given value of r is the only exceptional case of r general double points not being AH n ( d ) . Finally, for n and d not in the list of exceptional cases, by Lemma 2.1, we only need to consider values of r such that (cid:22) n + 1 (cid:18) n + dd (cid:19)(cid:23) ≤ r ≤ (cid:24) n + 1 (cid:18) n + dd (cid:19)(cid:25) . Our argument proceeds by considering small values of d and then using induction together with Theorem2.8. The statement for d = 2 is proved in Lemma 3.1. The statement d = 3 is examined in Section 5.Therefore, we may assume now that n ≥ and d ≥ .We will use induction on n to prove the assertion for d = 4 . Note that the statement for d = 4 and ≤ n ≤ is proved in Lemma 3.2. On the other hand, if the statement has been shown for ≤ n ≤ , thenTheorem 2.8 applies to prove the desired assertion for all n ≥ . This is because condition (i) holds by the nduction hypothesis on n , condition (ii) holds as shown in Section 5, and condition (iii) holds because, for n ≥ , by Lemma 2.7(3) we have r − q − ǫ ≥ n + 1 , and thus AH r − q − ǫ (2) holds as shown in Lemma 3.1.It remains to consider d = 4 and ≤ n ≤ . We shall leave this case until later in the proof.In general, for d ≥ , the proof proceeds by a double induction on d and n . Observe that if the statementhas been proved for d = 5 , and n = 3 , , then Theorem 2.8 applies to prove the statement for all d ≥ and n ≥ . Therefore, we only need to establish the desired assertion for d = 5 , and n = 3 , .We conclude the proof by analyzing the needed cases, i.e. when d = 4 and ≤ n ≤ , or when d = 5 , and n = 3 , . Most cases are proved via Theorem 2.8. Case 1: d = 4 , n = 5 . In this case, we need to consider r = 21 double points in P , q = 14 and ǫ = 0 .Direct Macaulay 2 [28] computation can be used to verify that the assertion holds. Case 2: d = 4 , n = 6 . In this case, we need to consider r = 30 general double points in P , q = 21 and ǫ = 0 . Theorem 2.8 applies because 21 general double points are AH (4) by Case 1, and 9 general doublepoints are AH (3) (as shown in Section 5) and AH (2) (by Lemma 3.1). Case 3: d = 4 , n = 7 . In this case, we need to consider r = 41 or general double points in P , q = 29 or and ǫ = 5 or . Direct Macaulay 2 [28] computation shows that 41 and 42 general double points in P are indeed AH (4) . Case 4: d = 5 , n = 3 . In this case, we need to consider r = 14 general double points in P , q = 7 and ǫ = 0 . Theorem 2.8 applies because 7 general double points are AH (5) (by Theorem 4.1), AH (4) (byLemma 3.2), and AH (3) (as shown in Section 5). Case 5: d = 5 , n = 4 . In this case, we need to consider r = 25 or general double points in P , q = 13 or q = 15 , and ǫ = 3 or . For r = 25 , q = 13 and ǫ = 3 , Theorem 2.8 applies because 13 general doublepoints are AH (5) by Case 4, 12 general double points are AH (4) (by Lemma 3.2), and 9 general doublepoints are AH (3) (as shown in Section 5). For r = 26 , q = 15 and ǫ = 0 , Theorem 2.8 applies because general double points are AH (5) by Case 4, 11 general double points are AH (4) (by Lemma 3.2) andAH (3) (as shown in Section 5). Case 6: d = 6 , n = 3 . In this case, we need to consider r = 21 general double points in P , q = 9 and ǫ = 1 . Theorem 2.8 applies because general double points are AH (6) (by Theorem 4.1), 12 generaldouble points are AH (5) by Case 4, and general double points are AH (4) (by Lemma 3.2). Case 7: d = 6 , n = 4 . In this case, we need to consider r = 42 general double points in P , q = 21 and ǫ = 0 . Theorem 2.8 applies because general double points are AH (6) by Case 6, and 21 points areAH (5) by Case 5 and AH (4) (by Lemma 3.2). (cid:3)
3. T
HE EXCEPTIONAL CASES
In this section, we consider the exceptional cases listed in Theorem 1.1 and show that they are indeed theonly exceptional cases for given n and d . We begin by considering the case where d = 2 . Lemma 3.1.
A set of r ≥ general double points in P n is not AH n (2) if and only if ≤ r ≤ n .Proof. A single double point is AH n (2) (e.g. by Remark C.9), so we may assume r ≥ . First we prove thata set of r ≥ n + 1 general double points in P n is AH n (2) . Let Y = { P , . . . , P r } denote a set of r ≥ n + 1 general points in P n and let X = 2 Y . It is easily seen that X is AH n (2) if and only if I X contains noquadrics.Then, by Lemma C.12, it suffices to show that I X contains no quadrics when r = n + 1 . When r = n + 1 ,by a change of variables, we can assume that P i is the i -th coordinate point, for i = 1 , . . . , n + 1 . That is, P i = [0 : · · · : 0 : 1 : 0 : · · · : 0] , where the value 1 appears at the i -th position. In this case, I Y is the quarefree monomial ideal I Y = p ∩ . . . ∩ p n = ( x i x j | ≤ i < j ≤ n ) where p i = ( x j | ≤ j ≤ n, j = i ) for every i = 0 , . . . , n . It is well-known that I X = I (2) Y = ( x i x j x h | ≤ i < j < h ≤ n ) (e.g. [23, Cor. 3.8], or [40, Cor. 4.15(a)]), thus I X contains no quadric.To conclude the proof we need to show that any set X of ≤ r ≤ n general double points in P n isnot AH n (2) . Since r ≤ n , we may assume that P i is the i -th coordinate point for ≤ i ≤ r . We firstclaim that I X contains precisely (cid:0) n − r +22 (cid:1) linearly independent quadrics. Indeed, again, let p i be the definingideal of P i , for i = 1 , . . . , r . It is easy to see that ( x r , x r +1 , . . . , x n ) ⊆ p i for all i = 1 , . . . , r . Thus, ( x r , . . . , x n ) ⊆ T ri =1 p i = I X . By modularity law, it follows that I X = ( I r − ,r ) (2) + ( x r , . . . , x n ) where ( I r − ,r ) (2) = T ≤ j
Suppose that ≤ n ≤ . Then, a set of r general double points in P n is not AH n (4) if andonly if r = (cid:0) n +22 (cid:1) − . Proof.
Let Y = { P , . . . , P r } be a set of r general points in P n and let X = 2 Y . We shall first show thatfor r = (cid:0) n +22 (cid:1) − , X is not AH n (4) . Indeed, since r < (cid:0) n +22 (cid:1) , I Y contains a nonzero quadric, say Q . Then, Q is a nonzero quartic in I Y ⊆ I (2) Y = I X . This implies that H R/I X (4) ≤ (cid:0) n +44 (cid:1) − . It is easy to checkthat for ≤ n ≤ , (cid:0) n +44 (cid:1) − < (cid:2)(cid:0) n +22 (cid:1) − (cid:3) ( n + 1) = r ( n + 1) . Therefore, X is not AH n (4) .We shall now show that r = (cid:0) n +22 (cid:1) − is indeed the only exceptional case. The statement for n = 2 isproved in Theorem 4.1. Suppose that ≤ n ≤ .For n = 3 , by Corollary D.5, it suffices to prove that a set of 8 general double points and a set of 10general double points in P are both AH (4) . Similarly, for n = 4 , it suffices to show AH (4) property fora set of 13 general double points and a set of 15 general double points in P . n = 3 and r = 8 . Observe that j (cid:0) (cid:1)k = 8 = r , so Theorem 2.8 applies if its hypotheses are satisfied.In this case, we have q = 4 and ǫ = 0 . Thus, condition (i) holds because general double points in P areAH (4) (by Theorem 4.1), and condition (iii) holds because general double points are AH (2) (by Lemma3.1). To prove that condition (ii) holds, we need to show that general double points are AH (3) . Thisfollows from Section 5.We can also prove this statement directly by considering the 4 coordinate points in P . Let I be thedefining ideal of these coordinate points. Then, I = ( x i x j | ≤ i < j ≤ , and it can be checked that I (2) is minimally generated by the four squarefree monomials of degree 3. In particular, H R/I (2) (3) = 16 which is the expected dimension, so condition (ii) of Theorem 2.8 holds. = 3 and r = 10 . We shall apply Theorem 2.3 for q = 6 . Clearly, general double points is AH (4) (byTheorem 4.1). Thus, it remains to show that the union of 4 general double points and 6 general simple pointson a hyperplane is AH (3) .Let Y be the set of the four coordinate points in P . As shown above, we have H R/I (2) Y (2) = 10 and H R/I (2) Y (3) = 16 . Let L be a hyperplane not containing any point of Y . By taking I = I (2) Y , Proposition C.13(2) holds for any u satisfying H R/I (3) + u ≤ H R/I (2) + (cid:18) − − (cid:19) , i.e. whenever
16 + u ≤
10 + 10 , i.e. u ≤ . Therefore, if we let Y be a set of u = 4 general points on L , then I (2) Y ∩ I Y does not contain any cubic. Now, let Y be obtained by adding two points to Y , then I (2) Y ∩ I Y ⊆ I (2) Y ∩ I Y contains no cubics. That is, Y ∪ Y is AH (3) . n = 4 and r = 13 . We shall apply Theorem 2.3 for q = 8 . So one may take Y to be the set of the 5coordinate points of P and L to be a hyperplane not containing any of these points. Then I Y is againgenerated by all squarefree monomials of degree 2 in R , and I (2) Y by the squarefree monomials of degree 3.It follows that Y is AH (3) , and in particular H R/I (2) Y (3) = 25 . Then inequality (2) of Proposition C.13then becomes
25 + q ≤
15 + 20 , so if we add 10 general simple points in L to Y we obtain a schemecontaining no cubics.In particular, if we take Y to be a set of q = 8 general points on L , then assumption (2) of Theorem 2.3is satisfied, so Y ∪ Y is a set of 13 points in P which is AH (4) . By Lemma D.4 any set of 13 generalpoints is AH (4) . n = 4 and r = 15 . We shall apply Theorem 2.3 for q = 10 . Clearly, a set of q = 10 general double points isAH (4) as shown above. Thus, it suffices to show that the union of 5 general double points and 10 generalsimple points in a hyperplane is AH (3) . This follows by the same argument of the previous case. (cid:3) We conclude this section with the case where d = 3 and n = 4 . Lemma 3.3.
A set of r general double points in P is AH (3) if and only if r = 7 .Proof. We first prove that a set of 7 general double points in P is not AH (3) . Let Y = { P , . . . , P } ⊆ P be a set of 7 general points, a simple computation shows that Y is AH (3) if and only if I (2) X contains nonon-zero cubic.By a result of Castelnuovo (e.g. [20, Thm 1]), given any set of t + 3 points in general position in P t , thereexists a unique rational normal curve C t passing through all of them, whose equations are given by the × minors of a matrix. In particular, there is a (unique) rational normal curve C passing through our7 points in P , whose equation, in an appropriate coordinate system, is I := I (cid:18) x x x x x x x x (cid:19) . One can check directly that I (2) contains (precisely) one cubic, namely x − x x x + x x + x x − x x x . Thus, a set of 7 general double points in P is not AH (3) . lternatively, it is also known that I = I x x x x x x x x x and it can be seen that f = det x x x x x x x x x is singular at all points of C .By Corollary D.5, the assertion completes by showing that sets of r = 6 and r = 8 general double pointsin P are AH (3) . We invoke Theorem 2.3 in both cases. First, observe that by Lemma 3.1, sets of 5 generaldouble points in P are AH (2) . If r = 6 , to apply Theorem 2.3 we need q with ≤ q ≤ , thus q = 4 .Then, assumption (1) holds for the reasons stated in the proof of Lemma 3.2 (the case where n = 3 and r = 8 ), and (2) holds because r − q = 2 general double points are AH (2) (because 5 double coordinatepoints are, and because of Lemma C.12(1)) and by Proposition C.13 (we need to add u = q = 4 generalpoints to the two double points).The case r = 8 is proved similarly. In this case one may take q satisfying ≤ q ≤ , so if we take q = 4 , then as above assumption (1) of Theorem 2.3 is satisfied, and assumption (2) is satisfied as above,by taking 4 coordinate double points (which are AH (2) because 5 of them are) and invoking PropositionC.13 (we add u = q = 4 simple general points lying on a hyperplane which avoids the four coordinatepoints). (cid:3) We end this section by noting that the case of cubics, i.e., when d = 3 , for an arbitrary value of n is muchmore subtle. Section 5 is devoted to handle this case.4. T HE CASE OF P ( n = 2 )This section focuses on the double points in P . Particularly, we shall identify all exceptional cases when n = 2 . While one could prove this case with more elementary arguments, we have chosen to employTheorem 2.8 to provide the reader with a further illustration of its application. Theorem 4.1.
Let X be any set of r general points in P . Then X is AH ( d ) for every d ≥ , except forthe exceptional cases of r = 2 and d = 2 , and r = 5 and d = 4 .Proof. Let R = C [ x, y, z ] be the homogeneous coordinate ring of P . We shall consider different casesbased on the values of d . Case 1: d = 1 . It suffices to prove the assertion for r = 1 since the degree of a double point in P is H R (1) . This case follows from Remark C.9. Case 2: d = 2 . The assertion is true for r = 1 by Remark C.9. The case where r = 2 is an exceptional casesince a double line going through 2 general points contains 2 general double points. Suppose that r ≥ .Since the degree of 3 double points in P is 9, which is bigger than H R (2) , it suffices to prove that thereis no conic in P with 3 double points. By B´ezout theorem, the equation of every conic with 3 double pointsis divisible by the equations of the three lines connecting 2 of these points – this gives a contradiction. Case 3: d = 3 . The statement is true for r = 1 , again by Remark C.9. When r = 2 we need to showthat H R/I (2) X (3) = 4 . Observe that by B´ezout theorem, a cubic with 2 double points must contain the lineconnecting these points. That is, this cubic factors as a line and a conic going through these 2 points. Sincethe Hilbert function of 2 general points in P is , , , . . . , it follows that the space of conic going throughthese 2 points has dimension 4. Particularly, the space of cubic with 2 double points has dimension 4. Thus,the assertion is true for r = 2 .Observe further that by B´ezout theorem, a cubic with 3 double points must contain 3 lines connecting2 of these points, and so there is a unique such cubic, which is the union of the 3 lines. It follows that H I (2) X (3) = 10 − e ( R/I (2) X ) , therefore, the assertion is true for r = 3 . uppose that r ≥ . Since the degree of 4 double points is >
10 = H R (3) , it suffices to show thatthere is no cubic containing 4 double points. By B´ezout theorem again, if such a cubic existed then it wouldcontain the 6 lines connecting any 2 of these 4 points, a contradiction. Case 4: d ≥ . Note that the degree of r double points in P is r . If r > (cid:0) d +22 (cid:1) then we restrict to asubscheme of r ′ double points, where r ′ = l (cid:0) d +22 (cid:1)m . Thus, we may assume that r ≤ l (cid:0) d +22 (cid:1)m .Furthermore, if r ≤ (cid:0) d +12 (cid:1) then we may consider degree ( d − instead. Therefore, we may also assumethat (cid:0) d +12 (cid:1) < r .Let q ∈ N and ≤ ǫ ≤ be chosen such that
21 + ǫ = 3 r − (cid:0) d +12 (cid:1) . By Lemma 2.7(1), we have also ǫ + 1 ≤ d. It follows from Theorem 2.8 that a set of r general double points in P is AH ( d ) if(1) general double point in P is AH ( d ) (which holds by Proposition C.11),(2) r − general double points in P are AH ( d − , and(3) r − − ǫ general double points in P are AH ( d − .When d = 4 , we get ≤ r ≤ . If r = 4 , then and ǫ = 0 . By the above (1)–(3) are satisfied,proving AH (4) holds. The case r = 5 is an exceptional case, and indeed in this case , ǫ = 1 andcondition (3) is not satisfied, because it is the exceptional case of double points in degree 2.For d = 5 , the induction hypotheses (1)–(3) are satisfied with the only possible exception of (2) when r − (as it reduces to the exceptional case of 5 double points in degree 4). Since
21 + ǫ = 3 r − ,we get ǫ = r − . Since (cid:0) (cid:1) < r and ≤ ǫ ≤ , we deduce that r can only be 6. The degree of 6 doublepoints is 18. Thus, it remains to show that the space of quintics with 6 general double points has dimensionat most 3.Suppose, on the contrary, that the space of quintics with 6 general double points has dimension at least 4.Then, by imposing one of these general points to be a triple point, we impose at most additional 3 conditions(the degree of a triple point is 6 and that of a double point is 3). It follows that the space of quintics with 5general double points and a general triple point is nonempty. Let Q be such a quintic with 5 general doublepoints and a general triple point, and let H be a quintic with 6 general double points. Their intersectioncontains 5 general double points, plus a scheme of multiplicity (3)(2) supported at the last point. Thus Q and H meet at a subscheme of degree > . This violates B´ezout theorem, unless Q and H share acommon factor. Observe further that, by Bertini’s theorem, the general element in the space of quintic with6 general double points is an irreducible curve. This forces Q to coincide with this irreducible curve, whichthen implies that all quintics with 6 double general double points are in fact the same as Q , a contradiction.For d = 6 , the induction hypotheses (1)–(3) are satisfied with the only possible exception of (3) when r − − ǫ = 5 , because in this case (3) reduces to the exceptional case of 5 double points in degree 4.Since
21 + ǫ = 3 r − , we get r − and ǫ = 11 − r . Since r ≤ l ( ) m = 10 and ≤ ǫ ≤ ,we must have r = 10 , ǫ = 1 and . This particularly shows that general double points in degree is not an exceptional case. As a consequence, there is a unique sextic containing 9 general double points(since the degree of 9 double points is 27). On the other hand, the Hilbert function of 9 general points is , , , , , . . . , and so there is only one cubic passing through 9 general points. Thus, the unique sexticwith 9 general double points is the double cubic going through these 9 general points. It can now be seenthat imposing another general point will result in no such sextic (since the Hilbert function of 10 generalpoints is , , , , , . . . , i.e., there is no cubic going through 10 general points). Thus, 10 general doublepoints in degree is not an exceptional case.Since there are no exceptional cases in degree 5 and 6, by Theorem 2.8, we conclude that there is noexceptional cases in any degree d ≥ . This completes the proof. (cid:3) . T HE CASE OF CUBICS ( d = 3 )In this section, we consider the case of cubics for any value of n . The main result in this section extendsLemma 3.3 and completes the case where d = 3 . Theorem 5.1.
Suppose that n ≥ . A set of r general double points in P n is not AH n (3) if and only if n = 4 and r = 7 .Proof. The case where n = 2 was already proved in Section 4. The case of n = 4 has been discussed inLemma 3.3. For n ≥ and n = 4 we proceed by considering two possibilities depending on the congruenceof n modulo 3.C ASE n ≡ , (mod 3). We shall use induction on n to show that the ideal of r general doublepoints in P n contains no cubics. The first base case, when n ≡ (mod 3), is n = 3 . By Remark 2.1, theassertion amounts to showing that a set X of 5 general double points in P is AH (3) , ie. its defining idealcontains no cubics. Without loss of generality we may write X = Y ∪ { Q } where Q = [1 : 1 : 1 : 1] , Y = { P , P , P , P } and P i = [ e i ] = [0 : . . . : 1 : 0 . . . : 0] for ≤ i ≤ . Then Y is a star configurationof 4 points, and a basis of [ I (2) Y ] is x x x , x x x , x x x , x x x (e.g. [23, Cor. 3.8]). So any cubic f in I (2) X ⊆ I (2) Y is a linear combination of these basis elements; It is easily seen that imposing that the partialderivatives ( ∂/∂x i ) f ( Q ) = 0 forces f = 0 .The other base case, when n ≡ (mod 3), is n = 7 . By Remark 2.1, to prove the assertion when n = 7 ,it suffices to show that a set of 15 general double points in P is AH (3) . Macaulay 2 [28] computationsindeed verify this statement.Suppose now that n ≥ and n = 7 . It can be seen, since n ≡ , (mod 3), that ( n + 2)( n + 3) is a multiple of 6. Thus, n +1 (cid:0) n +33 (cid:1) = ( n +2)( n +3)6 ∈ Z . By Remark 2.1, it suffices to show that a set of r = ( n +2)( n +3)6 general double points in P n is AH n (3) .Let r = n ( n − and let L be a codimension 3 linear subspace in P n . By a change of variables if necessary,we can assume that the defining ideal of L is p L = ( x n − , x n − , x n ) . Let X be a set of r general doublepoints in L together with r − r = n + 1 general double points outside of L . By the semicontinuity ofHilbert function (Remark 2.1), it is enough to show that I X contains no cubics. Consider a point Q in thesupport of X that lies in L , and let q be its defining ideal. Clearly, q ⊇ p L . Thus, we can write q = q + p L ,where q is a linear prime in R = C [ x , . . . , x n − ] ≃ R/P L . It follows from [31, Theorem 3.4] that q (2) = q (2) + q · p L + p (2) L . Particularly, it implies that q (2) + p L = q (2) + p L is the defining ideal of the double point Q in L . Thus,by letting X be the set of r general double points of X in L , considered as a subscheme of L ≃ P n − , weobtain I X + p L ⊆ I X + p L . Moreover, by the induction hypothesis applied to X ⊆ L ≃ P n − , we have (cid:2) I X (cid:3) = (0) . Therefore, I X + p L / p L contains no cubics. Hence, by considering the exact sequence −→ I X ∩ p L −→ I X −→ I X + p L / p L −→ , to prove that I X contains no cubics, it remains to show that I X ∩ p L contains no cubics. This is the contentof Claim 5.1.1 below. Claim 5.1.1.
Suppose that n ≥ and n = 4 . Let L be a codimension 3 linear subspace of P n and let X bethe union of r = n ( n − general double points in L and n + 1 general double points outside of L . Then, I X ∩ p L contains no cubics. roof of Claim 5.1.1. We use also induction on n to prove the assertion. The base case n = 3 holds because,by the above, I X contains no cubics. The other base case n = 7 can be verified directly, or by Macaulay 2[28] computations. Assume that n ≥ . For the inductive step, let M be a codimension 3 linear subspace of P n such that L ∩ M has codimension 6 in P n (any general codimension 3 linear subspace would work). Let p M be the defining ideal of M . We specialize to the following situation: • r := ( n − n − − = ( n − n − of the points of X in L are general double points in L ∩ M ; • the r − r = n − remaining points of X in L lie outside M ; • n − of the n + 1 points of X lying outside of L are general double points in M ; • and the last 3 points of X outside of L are general double points outside L ∪ M .By the semicontinuity of Hilbert function, it suffices to show that I X ∩ p L contains no cubics in thisparticular case. From the short exact sequence −→ I X ∩ p L ∩ p M −→ I X ∩ p L −→ ( I X ∩ p L ) + p M / p M −→ , it suffices to prove the other two terms of this exact sequence contain no cubics. As before, observe that ( I X ∩ p L ) + p M ⊆ ( I X ∩ p L ) + p M , where X denotes the set of points of X lying in M ≃ P n − , and L denotes the codimension 3 subspace L ∩ M of M ≃ P n − . As above, it can be seen that, in M , X is the union of r general double pointslying in L and n − general double points outside of L . Thus, by the induction hypothesis, the ideal ( I X ∩ p L ) + p M / p M of R/ p M ≃ C [ y , . . . , y n − ] contains no cubics. Hence, it remains to show that I X ∩ p L ∩ p M contains no cubics. This follows from Claim 5.1.2 below. (cid:3) Claim 5.1.2.
Suppose that n ≥ and n = 4 . Let L, M be two general codimension 3 linear subspaces of P n . Let X ⊆ P n be the union of r = ( n − n − general double points in L ∩ M , n − general doublepoints in L \ M , n − general double points in M \ L , and 3 general double points outside of L ∪ M . Then, I X ∩ p L ∩ p M contains no cubics.Proof of Claim 5.1.2. Let Z be the set of double points obtained by removing the r double points in L ∩ M from X . Clearly, I Z ⊇ I X . We shall prove a stronger statement that I Z ∩ p L ∩ p M contains no cubics. Thestatement for n = 3 , , , and can be verified by direct computations (e.g. using Macaulay 2 [28]). Weshall use induction to prove the statement for n ≥ .Let N be another general codimension 3 linear subspace of P n and let p N be its defining ideal. Wespecify n − of the n − double points of Z lying in L to be in L ∩ N , specify n − of the n − doublepoints of Z lying in M to be in M ∩ N , and specify the 3 general double points of Z outside of L ∪ M to be in N ≃ P n − . By the semicontinuity of Hilbert function, it suffices to show that for this particularconfiguration of Z , I Z contains no cubics.Consider the following short exact sequence −→ I Z ∩ p L ∩ p M ∩ p N −→ I Z ∩ p L ∩ p M −→ I Z ∩ p L ∩ p M + p N / p N −→ . By an argument similar to the proof of Claim 5.1.1, we have I Z ∩ p L ∩ p M + p N ⊆ I Z ∩ p L ∩ p M + p N , where − represents the restrictions to N ≃ P n − . The induction hypothesis applies to Z , so I Z ∩ p L ∩ p M + p N / p N contains no cubics. Therefore, to establish the desired statement, it remains to show that I Z ∩ p L ∩ p M ∩ p N contains no cubics. This follows from Claim 5.1.3 below, noting that n − ≥ . (cid:3) Claim 5.1.3.
Suppose that n ≥ . Let L , M , and N be general codimension linear subspaces of P n . Let X ⊆ P n be the union of 3 general double points in L \ ( M ∪ N ) , 3 general double points in M \ ( L ∪ N ) ,and 3 general double points in N \ ( L ∪ M ) . Then I X ∩ p L ∩ p M ∩ p N contains no cubics. roof of Claim 5.1.3. Direct computations (e.g. via Macaulay 2 [28]) verify the statement for n = 5 and n = 6 . (Notice that in [8, Prop. 5.2] it is incorrectly stated that when n = 6 the ideal p L ∩ p M ∩ p N containsno quadrics.) Assume that n ≥ . Without loss of generality, we may assume that p L = ( x , x , x ) and p M = ( x , x , x ) , so p L ∩ p M = p L p M . Let κ := p L ∩ p M ∩ p N , so we need to show that I X ∩ κ containsno cubics.We shall first show that κ contains no quadrics. Indeed, if n ≥ then we may assume that p N =( x , x , x ) . In this case, κ = p L p M p N is generated in degree at least 3. On the hand, if n = 7 then we mayassume that p N = ( x , x , x − x ) . The short exact sequence −→ R/ p L ∩ p M −→ R/ p L ⊕ R/ p M −→ R/ p L + p M = R/ ( x , . . . , x ) −→ gives H R/ p L ∩ p M (2) = 15 + 15 − . Similarly, consider the short exact sequence −→ R/κ −→ R/ p L ∩ p M ⊕ R/ p N −→ R/ ( p L ∩ p M ) + p N −→ . Since R/ p L ∩ p M + p N = R/ p L p M + p N = R/ ( x , x , x )( x , x , x ) , x − x , x , x ) is isomorphic to B := C [ x , . . . , x ] / ( x , x , x )( x , x , x ) , then we have H R/κ (2) = 27 + 15 − H B (2) . It is easy to see that B contains all the quadrics in C [ x , . . . , x ] except for the 9 generators of ( x , x , x )( x , x , x ) .Thus, H B (2) = 15 − . Therefore, H R/κ (2) = 42 − , showing that [ κ ] = 0 .Now, since dim R ≥ one has depth R/κ ≥ . Let h be a general linear form in R and let H be thehyperplane in P n defined by h ; since depth R/κ ≥ , we may assume h is regular on R/κ . Let R = R/ ( h ) and κ be the image of κ in R . From the standard short exact sequence −→ R/κ −→ R/κ −→ R/κ −→ one obtains that depth R/κ ≥ , i.e. κ is saturated in R .We now specialize the configuration so that all 9 double points of X are on the hyperplane H ≃ P n − and let I = I X ∩ κ for simplicity of notation. Consider the short exact sequence −→ ( I : h )( − −→ I −→ ( I, h ) / ( h ) −→ . Since the points in X are lying on H , we have I : h = κ : h = κ . Thus, this sequence can be rewritten as −→ κ ( − −→ I −→ ( I, h ) / ( h ) −→ . As we have shown, κ has no quadrics, so κ ( − has no cubics. Hence, to show that I contains no cubics,it remains to show that the image I of I in R has no cubics. This is indeed true by induction on n , since I ⊆ ( I ) sat and ( I ) sat is the defining ideal of X in H ≃ P n − . (cid:3) C ASE n ≡ (mod 3). Let r = ( n +2)( n +3)6 − = ( n +1)( n +4)6 and δ = n +13 . By Remark 2.1, to provethe desired statement, it suffices to show that sets of r = r , r + 1 general double points are AH n (3) . Tothis end, it is enough to show that a scheme X ⊆ P n consisting of r general double points and a generalsubscheme η supported at another general point with degree δ is AH n (3) . Indeed, it is easy to see that X has multiplicity exactly (cid:0) n +33 (cid:1) . Thus, by a similar argument as in Lemma C.12, it can be show that if X is AH n (3) then so is a set of r general double points in P n . On the other hand, a set of r + 1 generaldouble points contains X as a subscheme, so its Hilbert function in degree d is at least that of X , which is (cid:0) n +33 (cid:1) , i.e. it is already maximal. Particularly, a set of r + 1 general double points also has maximal Hilbertfunction in degree 3. s in Case 1, we shall use induction on n ≥ to show that X is AH n (3) . The case n = 2 is provedin Theorem 4.1. The induction step proceeds along the same lines as Case 1. The only difference is atClaim 5.1.1, which shall be replaced by the following Claim 5.1.4.
Suppose that n ≥ . Let L be a general codimension 3 linear subspace in P n . Let X ⊆ P n the union of r ′ = ( n − n +1)6 general double points in L , ( n + 1) general double points outside of L , anda general subscheme η supported at a point Q ∈ L and of multiplicity δ such that η ∩ L has multiplicity δ − n − . Then, I X ∩ p L contains no cubics.Proof of Claim 5.1.4. One proceeds by induction exactly as in the proof of Claim 5.1.1. (cid:3)
The proof of Theorem 5.1 is now completed. (cid:3)
6. O
PEN PROBLEMS
In this section we discuss a few open problems. Let us state clearly that there are many more interestingquestions outside the ones we have included in this section. For instance, as indicated by Appendix Abelow, the polynomial interpolation is closely connected to secant varieties and Waring rank. Thus, thereare many other problems and questions that are of interest to researchers working in these areas or studying,for instance, containment problems for ordinary and symbolic powers of ideals, other interpolation problemsand invariants associated to symbolic powers of ideals.However, to keep this section aligned with the other sections, we restrict ourselves to some questions related to the Alexander–Hirschowitz theorem . It is implicit that this small set of questions and conjecturesis far from being comprehensive, and it should be considered a sample – aimed at young researchers – ofthe many problems in this active area of research.We begin by observing that Theorem 1.1 describes the Hilbert function of I (2) Y for every set Y of general points in P n with a finite list of exceptions (the Hilbert functions in these cases can be worked out individu-ally). Characterizing the Hilbert function of I (2) Y for any set of points Y in P n , on the other hand, seems outof reach; see, for instance, the surveys of Gimigliano [27] and Harbourne [32]. Despite the large body ofliterature dedicated to understanding the Hilbert function of double points, this problem continues to eludeus.For any n ≥ and r ≥ H n ( r ) be the set of all Hilbert functions H R/I (2) Y where Y is a set of r points in P n Problem 6.1.
Characterize the numerical functions which are Hilbert functions of I (2) Y for some set Y ofpoints in P n , i.e. for every n ≥ characterize all elements in H n := [ r ≥ H n ( r ) = n H R/I (2) Y | Y is a set of points in P n o . Problem 6.1 is easy for points in P ; see Proposition C.11. To the best of our knowledge, Problem 6.1remains open even in P , so one might attempt to start with this first nontrivial case, i.e., for double pointsin P . See [26] and [10] for some work in this direction. Problem 6.2.
Characterize the numerical functions which are Hilbert functions of I (2) Y for some set Y ofpoints in P , i.e. characterize all elements in H := n H R/I (2) Y | Y is a set of points in P o . n investigating a family of Hilbert functions, it is natural to determine the existence of “minimal” and“maximal” elements in the following sense. Define a partial order on H n ( r ) by setting H R/I (2) Y ≤ H R/I (2) Z if H R/I (2) Y ( d ) ≤ H R/I (2) Z ( d ) for every d ≥ . Notice that every H ∈ H n ( r ) satisfies H ( d ) ≤ min (cid:26)(cid:18) n + dd (cid:19) , r ( n + 1) (cid:27) and, by Theorem 1.1, equality holds for any general set of points (with a few exceptions). Therefore,Theorem 1.1 in particular proves the existence of maximal elements in H n ( r ) (with a few exceptions), andnumerically characterizes what these maximal Hilbert functions are. It is a natural problem to determine thepotential existence and characterization of minimal elements of H n ( r ) . Problem 6.3.
Fix n, r ≥ . ( a ) Prove the existence of a minimal element in H n ( r ) . ( b ) Determine the minimal element in H n ( r ) .A partial answer to Problem 6.3 was given for double points in P in [26, 24]. Even this special caseproves to be difficult. An affirmative answer is only known to exist if the number of points is a binomialcoefficient number or at most 11.Another natural approach in examining the Hilbert function of double points is to specify that the pointsare lying on a given subscheme . For instance, if the points are on a rational normal curve or on a conic. Problem 6.4.
For n ≥ , let C n be the rational normal curve in P n . For any r ≥ , determine the Hilbertfunction of R/I (2) Y where Y is a set of r general points on C n .If the rational normal curve C n is replaced by a conic then Problem 6.4 has a satisfactory answer, givenby Geramita, Harbourne and Migliore [25].Another problem in the lines of the Alexander–Hirschowitz theorem is to determine the Hilbert functionsof sets of general double points in multiprojective spaces . In general, however, points in multiprojectivespaces are harder to understand then points in projective spaces. (e.g. a set of points in P n × · · · × P n k doesnot need to be Cohen–Macaulay.) Much work has been put forward to understand, in general, numericalinvariants and properties of points in the first nontrivial case of a multiprojective space, i.e. P × P , (e.g.[30]).While the Hilbert function for a general set of double points in P × P is known (see [45]), that foran arbitrary set of double points in P × P is not yet completely classified, except when the support isCohen–Macaulay [29]. Problem 6.5.
Let R = C [ x , . . . , x ] and fix any r ≥ . Determine the possible Hilbert functions of R/I (2) Y where Y is any set of r points in P × P .We observe, in passing, that similarly to how the Alexander-Hirschowitz theorem is closely related to thestudy of secant varieties of Veronese embeddings of P n , so Problem 6.5 is intimately connected to the studyof secant varieties of Segre-Veronese varieties (cf. [11]).In general, understanding the symbolic square I (2) Y of a set Y of simple points is far from being a com-pleted task. Since for certain questions I Y is more understood than I (2) Y , a possible approach is to compare I (2) Y and I Y , or simply the module I (2) Y /I Y . or instance, Galetto, Geramita, Shin and Van Tuyl [23] defined a first possible measure aimed at quanti-fying the gap between the m -th symbolic power of an ideal and the m -th ordinary power. They dubbed thismeasure the m -th symbolic defect of an ideal J , and they defined it to be sdef( J, m ) := µ ( J ( m ) /J m ) . (Here, µ ( M ) denotes the minimal number of generators of a finitely generated R -module M .) The problemof determining symbolic defects of an ideal is open, even for the defining ideal of a general set of points. Problem 6.6.
Compute sdef( I Y , for any set Y of general simple points in P n .Problem 6.6 seems to be open even in P . Problem 6.7.
Compute sdef( I Y , for any set Y of general simple points in P .A first partial result towards Problem 6.7 is [23, Thm 6.3], where the authors determined the secondsymbolic defect when | Y | ≤ and | Y | 6 = 6 . These are precisely the set of points whose second symbolicdefect is either 0 or 1. They also proved that if | Y | = 6 of | Y | > , then sdef( I Y , > , however, theprecise value is not known.Inspired by studies on symbolic defects of an ideal, we can consider a similar invariant defined by exam-ining the Hilbert function instead of the minimum number of generators. Particularly, for m ∈ N , define the m -th symbolic HF-defect of an ideal J to be the Hilbert function of J ( m ) /J m , i.e. sHFdef( J, m ) := H J ( m ) /J m . Problem 6.8.
Compute sHFdef( I Y , for any set Y of general points in P n . Equivalently, compute theHilbert function H R/I Y for any set Y of general points in P n .The equivalence of the statements given in Problem 6.8 follows because H I (2) Y /I Y = H R/I Y − H R/I (2) Y ,and by Theorem 1.1 we already know H R/I (2) Y .To the best of our knowledge, Problem 6.8 is solved only for the case of n + 1 general points in P n [5].Most of the above problems are aimed at understanding symbolic squares of ideals of points; however, themost natural, important and challenging question raised by Theorem 1.1 is to prove an analogue of Theorem1.1 for any symbolic power of any ideal defining a set of general points in P n . Problem 6.9.
Let n ≥ and R = C [ x , . . . , x n ] . For every fixed m ≥ , determine the Hilbert function of R/I ( m ) Y for a set Y of general points in P n .Problem 6.9 is one the big open problems about interpolation. Even the case where m = 3 is still wideopen. Problem 6.10.
Let n ≥ and R = C [ x , . . . , x n ] . Determine the Hilbert function of R/I (3) Y for a set Y ofgeneral points in P n .As we have seen in Theorem 1.1, one expects to have a finite list of exceptional cases, for which thegeneral statement does not hold. A starting point toward Problem 6.10 is to determine a similar list ofexceptional cases for triple general points. Problem 6.11.
Let n ≥ and R = C [ x , . . . , x n ] . Determine all the potential exceptional cases for Problem6.10, i.e., find a finite list L such that if Y , for a general set of points Y ⊆ P n , is not AH n ( d ) then Y ∈ L .A well-known conjecture, often referred to as the SHGH Conjecture , raised (and refined) over the yearsby Segre, Harbourne, Gimigliano and Hirschowitz provides the first step toward solution to Problem 6.11 by redicting what these exceptional cases are expected to be. We shall state a special case of this conjecture,namely, the uniform points in P . See, for instance, [9] for a more general statement and details on theSHGH Conjecture.An irreducible homogeneous polynomial F ∈ R = C [ x, y, z ] is said to be exceptional for a set Y = { P , . . . , P r } of points in P if deg( F ) − r X i =1 n i = − F ) + r X i =1 n i = − , where n i is the highest vanishing order of F at P i , for i = 1 , . . . , r . Conjecture 6.12. (SHGH Conjecture) Let Y be a general set of points in P and let m ∈ N . Then, mY isnot AH n ( d ) if and only if there exists an irreducible homogeneous polynomial F ∈ R that is exceptionalfor Y such that F s , for some s > , divides every homogeneous polynomial of degree d in I ( m ) Y .The ultimate goal naturally would be to determine the Hilbert function of every non-uniform symbolicpower of any set Y of general points in P n (i.e. the Hilbert function of p m ∩ . . . ∩ p m r r , where Y = { P , . . . , P r } is a set of general points in P n and p i is the defining ideal of P i for every i ).B. Harbourne [34] showed that this problem would be solved if one is able to determine α ( p m ∩ . . . ∩ p m r r ) for every choice of the multiplicities m i ∈ Z + . Here, for any homogeneous ideal J , α ( J ) := min { d ≥ | [ J ] d = 0 } , is the initial degree of J . Hence, the problem of determining the initial degree of symbolic powers of idealsof points would solve the ultimate problem on interpolation. However, as one may expect, determining α is usually very challenging, even in the uniform case and even for points in P . For instance, the followingcelebrated conjecture of Nagata, which arose from his work on Hilbert’s 14-th problem [42], remains open. Conjecture 6.13. (Nagata’s Conjecture) For any set Y of r ≥ general points in P , and any m ≥ onehas α ( I ( m ) Y ) > m √ r. Conjecture 6.13 was proved by Nagata when r is a perfect square. A large body of literature is dedicatedto this conjecture (cf. [33] and references therein and thereafter). Connections have also been made withother problems, for instance symplectic packing problems (e.g. [6, Section 5]). Nevertheless, the conjecturestill seems out of reach at the moment. The interested reader will find in the literature many variations andand different viewpoints on Nagata’s Conjecture, e.g. [12].Given the difficulty in establishing the bound predicted by Nagata’s conjecture, it is natural to ask forweaker bounds. In this direction, we mention that for an arbitrary set of points Y in P n , a weaker bound for α ( I ( m ) Y ) was formulated by G. V. Chudnovsky in [15]. Conjecture 6.14 (Chudnovsky) . Let Y be an arbitrary set of points in P n . For every m ≥ , we have α ( I ( m ) Y ) m ≥ α ( I Y ) + n − n . Chudnovsky’s conjecture has been established for • points in P (see [15, 35]), • general points in P (see [18]), • points on a quadric (see [22]), • very general points in P n (in [19] for large number of points, and in [22] for any number of points), • large number of general points in P n (see [7]). ery recently, in a personal communication with the authors, R. Lazarsfeld suggested a geometric intu-itive evidence for why one may expect the existence of counterexamples to Conjecture 6.14. Thus, insteadof trying to prove Conjecture 6.14, one may look for counterexamples. It should be noted that partial re-sults stated earlier suggest that a potential counterexample should be a special configuration and have highsingularity outside of the given set of points. Problem 6.15.
Fix n ∈ Z + . Either prove Conjecture 6.14 for any set of points in P n or find a counterexam-ple for the conjecture.A. A PPENDIX : S
ECANT VARIETIES AND THE W ARING PROBLEM
In this section, we briefly describe the connection between the polynomial interpolation problem, partic-ularly the Alexander–Hirschowitz theorem, and studies on secant varieties and Waring problem for forms.Throughout this section, let V be a vector space of dimension ( n + 1) over C . Then P n can also be viewedas P ( V ) , the projective space of lines going through the origin in V . For f ∈ V \ { } , let [ f ] denote the linespanned by f in V and, at the same time, the corresponding point in P ( V ) .Let S be the symmetric algebra of V . Then S is naturally a graded algebra, given by S = L d ≥ S d V ,where the d -th symmetric tensor S d V is a C -vector space of dimension (cid:0) n + dn (cid:1) . Note that the dual R = S ∗ is the polynomial ring C [ x , . . . , x n ] identified as the coordinate ring of P ( V ) . Definition A.1.
Let V be a ( n + 1) -dimensional vector space over C .(1) The d -th Veronese embedding of P ( V ) is the map ν d : P ( V ) → P ( S d V ) , given by [ v ] [ v d ] = [ v ⊗ · · · ⊗ v | {z } d times ] . Equivalently, ν d is the map P n → P N , where N = (cid:0) n + dd (cid:1) − , defined by [ a : · · · : a n ] [ a d : a d − a : · · · : a dn ] , where the coordinates on the right are given by all monomials of degree d in the a i ’s.(2) The d -th Veronese variety of P ( V ) , denoted by V nd , is defined to be the image ν d ( P ( V )) . Lemma A.2.
Let f ∈ V \ { } . The tangent space T [ f d ] ( V nd ) of V nd at the point [ f d ] is spanned by { [ f d − g ] ∈ P ( S d V ) (cid:12)(cid:12) g ∈ V } . Proof.
Let g ∈ V \ { } . The line ℓ passing through [ f ] ∈ P ( V ) , whose tangent vector at [ f ] is given by [ g ] , is parameterized by ǫ [ f + ǫg ] . The image of this line via the Veronese embedding ν d is given by [( f + ǫg ) d ] . Thus, the tangent vector of ν d ( ℓ ) at ν d ([ f ]) is (cid:2) ddǫ (cid:12)(cid:12)(cid:12) ǫ =0 ( f + ǫg ) d (cid:3) = [ df d − g ] = [ f d − g ] . The statement then follows. (cid:3)
Lemma A.3.
Let f ∈ V \ { } .(1) There is a one-to-one correspondence between hyperplanes in P ( S d V ) containing [ f d ] and hyper-surfaces of degree d in P ( V ) containing [ f ] .(2) There is a one-to-one correspondence between hyperplanes in P ( S d V ) containing T [ f d ] ( V nd ) andhypersurfaces of degree d in P ( V ) singular at [ f ] . roof. (1) Let z , . . . , z N , where N = (cid:0) n + dd (cid:1) − be the homogeneous coordinate of P ( S d V ) . The equationfor a hyperplane H in P ( S d V ) has the form a z + · · · + a N z N = 0 . By replacing z i with the corresponding monomial of degree d in the x i ’s, this equation gives a degree d equation that describes a degree d hypersurface in P ( V ) . Clearly, this hypersurface is the preimage ν − d ( H ) of H . Furthermore, since ν − d ([ f d ]) = [ f ] , if H passes through [ f d ] then ν − d ( H ) contains [ f ] .(2) Let { e , . . . , e n } be a basis of V whose dual basis in R is { x , . . . , x n } . By a linear change ofvariables, we may assume that f = e . That is, [ f ] = [1 : 0 : · · · : 0] ∈ P nK . Then, the defining ideal of [ f ] is ( x , . . . , x n ) .It follows from Lemma A.2 that T [ f d ] ( V nd ) is spanned by { [ e d ] , [ e d − e ] , . . . , [ e d − e n ] } . As before, the equation for a hyperplane H in P ( S d V ) has the form a z + · · · + a N z N = 0 . By using lexi-cographic order, we may assume that z , . . . , z n are variables corresponding to monomials x d , x d − x , . . . , x d − x n . Then, H contains T [ f d ] ( V nd ) if and only if a = · · · = a n = 0 . It follows that the equation for ν − d ( H ) is a linear combination of monomials of degree d not in the set { x d , x d − x , . . . , x d − x n } . Par-ticularly, these monomials have degree at least 2 in the variables x , . . . , x n . Hence, ν − d ( H ) is singular at [ f ] . (cid:3) We obtain an immediate corollary.
Corollary A.4.
Let X = { [ f ] , . . . , [ f k ] } ⊆ P ( V ) be a set of points. Let m , . . . , m k be the defining idealof these points. Then, there is a bijection between the vector space of degree d elements in T ki =1 m i and thevector space of hyperplanes in P ( S d V ) containing the linear span of T [ f d ] ( V nd ) , . . . , T [ f dk ] ( V nd ) .Proof. The conclusion follows from Lemma A.3, noticing that m i is the ideal of polynomials in R singularat [ f i ] ∈ P ( V ) for all i = 1 , . . . , k . (cid:3) Corollary A.5.
Let X = { [ f ] , . . . , [ f k ] } ⊆ P ( V ) be a set of points. Let m , . . . , m k be the defining idealsof the points in X . Then, dim C (cid:2) k \ i =1 m i (cid:3) d = N − dim h T [ f d ] ( V nd ) , . . . , T [ f dk ] ( V nd ) i . Proof.
The conclusion is an immediate consequence of Corollary A.4 and basic linear algebra facts. (cid:3)
Definition A.6.
Let X be a projective variety. For any nonnegative integer r , the r -secant variety of X ,denoted by σ r ( X ) , is defined to be σ r ( X ) = [ P ,...,P r ∈ X h P , . . . , P r i Zariski closure . Note that σ r ( V nd ) is an irreducible variety for all r . Remark A.7.
Let X ⊆ P N be a projective scheme of dimension n . Then, dim σ r ( X ) ≤ min { rn + r − , N } = min { ( n + 1) r − , N } . When the equality holds we say that σ r ( X ) has expected dimension . emma A.8 (First Terracini Lemma) . Let Y ⊆ P n be a projective scheme. Let p , . . . , p r be general pointsin Y . Let z ∈ h p , . . . , p r i be a general point in the linear span of p , . . . , p r . Then, T z ( σ r ( Y )) = h T p ( Y ) , . . . , T p r ( Y ) i . Proof.
Let Y ( τ ) = Y ( τ , . . . , τ n ) be a local parametrization of Y . Let Y j ( τ ) represent the partial deriv-ative with respect to τ j , for j = 1 , . . . , n . Suppose that p i corresponds to τ i = ( τ i , . . . , τ in ) in this localparametrization.By definition, T p i ( Y ) is spanned by the tangent vectors Y ( τ i ) + ǫY j ( τ i ) , for j = 1 , . . . , n . Thus, h T p ( Y ) , . . . , T p r ( Y ) i is the affine span of { Y ( τ i ) , Y j ( τ i ) (cid:12)(cid:12) i = 1 , . . . , r, j = 1 , . . . , n } .On the other hand, a general point z in σ r ( Y ) is parametrized by Y ( τ k ) + P r − i =1 γ i Y ( τ i ) . By consid-ering partial derivatives at z , it can be seen that T z ( σ r ( Y )) is also the affine span of { Y ( τ i ) , Y j ( τ i ) (cid:12)(cid:12) i =1 , . . . , r, j = 1 , . . . , n } . The lemma is proved. (cid:3) The following theorem establishes the equivalence between being AH n ( d ) for double points and havingexpected dimension for secant varieties. Theorem A.9.
A general collection of r double points in P n is AH n ( d ) if and only if σ r ( V nd ) has expecteddimension.Proof. Let X = { q , . . . , q r } be a set of r general simple points in P n , and let m , . . . , m r be their definingideal. Set p i = ν d ( q i ) for i = 1 , . . . , r . By Corollary A.5, we have dim C (cid:2) r \ i =1 m i (cid:3) d = N − dim h T p ( V nd ) , . . . , T p r ( V nd ) i . It follows that h P n (2 X, d ) = (cid:18) n + dd (cid:19) − dim C (cid:2) r \ i =1 m i (cid:3) d = dim h T p ( V nd ) , . . . , T p r ( V nd ) i + 1 . By the genericity assumption of the points and the fact that σ r ( V nd ) is an irreducible variety, Lemma A.8now gives h P n (2 X, d ) = dim σ r ( V nd ) + 1 . The conclusion now follows, noting that h P n (2 X, d ) ≤ min { ( n +1) r, (cid:0) n + dn (cid:1) } and dim σ r ( V nd ) ≤ min { ( n +1) r − , (cid:0) n + dn (cid:1) − } . (cid:3) Via its connection to secant varieties of Veronese varieties, the interpolation problem is also closelyrelated to the Waring problem for forms.
Definition A.10.
Let F ∈ R = be a homogeneous polynomial of degree d . The Waring rank of F , denotedby rk( F ) , is defined to be the minimum s such that F = ℓ d + · · · + ℓ ds for some linear forms ℓ , . . . , ℓ s ∈ R .The Waring problem for forms is to find bounds for the Waring rank of homogeneous polynomials. Definition A.11.
Let n, d be positive integers and R = C [ x , . . . , x n ] .(1) Set G ( n, d ) := min { s ∈ N (cid:12)(cid:12) rk( F ) ≤ s for a general element F ∈ R d } . (2) Set g ( n, d ) := min { s ∈ N (cid:12)(cid:12) rk( F ) ≤ s for any element F ∈ R d } . he Big Waring Problem and
Little Waring Problem , respectively, are to determine G ( n, d ) and g ( n, d ) .It is easy to see that g ( n, d ) = max { rk( F ) (cid:12)(cid:12) F ∈ R d } . The connection between G ( n, d ) and secant varietiescomes from the following result. Lemma A.12. G ( n, d ) = min { r (cid:12)(cid:12) σ r ( V nd ) = P ( S d V ) } . Proof.
Fix a basis { e , . . . , e n } of V whose dual basis in R is { x , . . . , x n } . Let θ : R → S be the naturalisomorphism defined by x i e i . Consider a form F ∈ R d . By definition, rk( F ) ≤ r if and only if thereexist linear forms ℓ , . . . , ℓ r ∈ R such that F = ℓ d + · · · + ℓ dr (A.1)Let h ′ i = θ ( ℓ i ) for i = 1 , . . . , r . Then, (A.1) holds if and only if θ ( F ) = h d + · · · + h dr . By scalar scaling ifnecessary, this is the case if and only if [ θ ( F )] = h [ h d ] , . . . , [ h dr ] i .By the definition of σ r ( V nd ) (being the Zariski closure of the union of secant linear subspaces), it thenfollows that σ r ( V nd ) = P ( S d V ) if and only if S P ,...,P r ∈ V nd h P , . . . , P r i contains a general point of P ( S d V ) .Hence, rk( F ) ≤ r for a general element F ∈ R d if and only if σ r ( V nd ) = P ( S d V ) . (cid:3) B. A
PPENDIX : S
YMBOLIC POWERS
Considering the considerable literature on symbolic powers of ideals, we have included in this appendixonly a minimal amount of definitions and results. We refer the interested reader to the well-written, recent,comprehensive survey on the subject [17].
Definition B.1.
Let R = C [ x , . . . , x n ] and let I be an ideal with no embedded associated primes. Forevery m ∈ Z + , the m -th symbolic power of I is the R -ideal I ( m ) = \ p ∈ Ass(
R/I ) ( I m R p ∩ R ) . Additionally, one sets I (0) = R .For every integer m ≥ and ideal I as above one has I m ⊆ I ( m ) , but in general one has I m ( I ( m ) . Anotable exception is when I is a complete intersection, in which case I m = I ( m ) for every m ≥ .The following result gives a way to compute symbolic powers of ideals of points. Proposition B.2.
Let X = { P , . . . , P r } be a set of points in P n , and I X = p ∩ . . . ∩ p r be its definingideal in R = C [ x , . . . , x n ] , then for every m ≥ I ( m ) X = p m ∩ · · · ∩ p mr . Definition B.3. If P is a point in P n , for every m ≥ we write mP for the subscheme of P n with definingideal p ( m ) . One often calls mP a fat point subscheme of P n .If P , . . . , P r are points in P n , the defining ideal of the fat points scheme X = m P + m P + . . . + m r P r is I X := p ( m )1 ∩ . . . ∩ p ( m r ) r . Example B.4. (The defining ideal of 3 non-collinear double points in P ) Let X = { [1 : 0 : 0] , [0 : 1 :0] , [0 : 0 : 1] } ⊆ P be the three coordinate points in P , and I X = ( x , x ) ∩ ( x , x ) ∩ ( x , x ) =( x x , x x , x x ) ⊆ C [ x , x , x ] be its defining ideal, then the defining ideal of mX is I ( m ) X = ( x , x ) m ∩ ( x , x ) m ∩ ( x , x ) m . For instance, X is defined by I (2) X = ( x , x ) ∩ ( x , x ) ∩ ( x , x ) , and by computing this intersectionone obtains I (2) X = ( x x x ) + I X . n particular, I X = I (2) X . (In fact, more generally, x t x t x t ∈ I (2 t ) X for all t ∈ Z + .)An important theorem proved by Zariski [46] and Nagata [43] (and generalized by Eisenbud and Hochster[21]) provides with a first illustration of the geometric relevance of the symbolic powers of ideals: theyconsist of all hypersurface vanishing with order at least m on the variety defined by I . Theorem B.5 (Zariski-Nagata) . Let I be a radical ideal in R = C [ x , . . . , x n ] and let s ≥ N . Then, I ( s ) = \ m ∈ Max( R ) , I ⊆ m m s . = { f ∈ R | all partial derivatives of f of order ≤ s − lie in I } . Example B.6.
Let X , I X and R be as in Example B.4. We have claimed that x x x ∈ I (2) X . One can checkthis easily using Zariski–Nagata theorem, because each of the partial derivatives of x x x with respect toone of the variables is a minimal generator of I X , thus all partial derivatives of order at most of x x x lie in I X , thus by Zariski–Nagata theorem x x x ∈ I (2) X .More geometrically, V ( x x x ) is the union of the three lines V ( x ) ∪ V ( x ) ∪ V ( x ) . Each of thepoints at the intersection of two of the three lines are singular points. Since these three intersections are thepoints of X , it follows by Zariski and Eisenbud–Hochster theorem that x x x ∈ I (2) X .C. A PPENDIX : H
ILBERT FUNCTION
In this section, we give basic properties of Hilbert functions, especially those of set of points. We startwith a simple lemma allowing the use of Linear Algebra to investigate interpolation problems.
Lemma C.1.
Let P ∈ P n be a point with defining ideal p ⊆ R . Then [ p m ] d consists of all solutions of ahomogeneous linear system of (cid:0) n + m − n (cid:1) equations in (cid:0) n + dn (cid:1) variables. In particular, the rank of this linearsystem is H R/ p m ( d ) .Proof. Let F be a generic homogeneous equation of degree d in n + 1 variables, i.e. F = X M ∈ T d c M M ∈ C [ { c M } , x , . . . , x n ] where T d consists of all the (cid:0) n + dd (cid:1) monomials of degree d in C [ x , . . . , x n ] .By Zariski–Nagata’s theorem, the equation F vanishes with multiplicity m at a point P ∈ P n ⇐⇒ all the ( m − -order (divided power) derivatives of F vanish at P .Now, consider any ( m − -th partial order derivative of F with respect to the x i ’s and substitute thecoordinates of P in for the variables. We obtain is a linear combination of the coefficients c M ’s; this linearcombination is zero if and only if that partial order of F vanish at P . Therefore, the elements of the desiredhomogeneous linear system are obtained from solutions to all these (cid:0) n + m − n (cid:1) linear equations of unknownsbeing the coefficients c M ’s. The assertion then follows. (cid:3) The Hilbert function of a graded ring counts the number of linearly independent forms in a given degree.
Definition C.2.
Let R = C [ x , . . . , x n ] , and let M = L i ≥ M i be a graded R -module (e.g. M = R/I where I is a homogeneous ideal). Then, M i is a C -vector space for every i ≥ . The Hilbert function of M in degree d is the function H M : Z ≥ → N , given by H M ( d ) := dim C M d . For any a ∈ Z , M ( a ) is defined as the graded R -module whose degree j component is [ M ( a )] j = M a + j . n general, for d large, the Hilbert function of M agrees with a polynomial of degree dim( M ) − , whichis called the Hilbert polynomial of M . Its normalized leading coefficient is an integer e ( M ) called the multiplicity of M . When M is 1-dimensional, this implies that H M ( d ) is non-decreasing and eventuallyequals the multiplicity of M . We recapture this property in the following proposition. Proposition C.3.
Let R = C [ x , . . . , x n ] and let M be a Cohen-Macaulay graded R -module with dim( M ) =1 . Then H M ( d − ≤ H M ( d ) for all d ∈ N , and H M ( d ) = e ( M ) for d ≫ . In particular, H M ( d ) ≤ e ( M ) for every d ∈ N .Proof. Since dim( M ) = 1 , then the Hilbert polynomial of M is just the constant function e ( M ) , so H M ( d ) = e ( M ) for d ≫ .Since M is Cohen-Macaulay there exists a linear form x ∈ R that is regular on M . Let R = R/ ( x ) bethe Artinian reduction of R , and let M = M/ ( x ) M . The short exact sequence −→ M ( − · x −→ M −→ M −→ and the additivity of Hilbert function under short exact sequence yield H M ( d ) = H M ( d ) − H M ( d − ,which is, of course, non-negative for every d ∈ N , and so H M ( d − ≤ H M ( d ) for all d ∈ N . (cid:3) To complement the previous result, using the so-called Associativity Formula for e ( R/I ) one obtains thefollowing statement that works particularly for points in P n . Proposition C.4.
Let Y = { P , . . . , P r } be a set of points in P n , and let X = { m P , . . . , m r P r } . Then e ( R/I X ) = r X i =1 (cid:18) n + m i − n (cid:19) . By counting equations and variables, one immediately obtains an upper bound for the Hilbert function ofany ideal associated to (possibly fat) points.
Corollary C.5.
Let X = { m P , . . . , m P r } be a set of points in P n with multiplicities m , . . . , m r . Then H R/I X ( d ) ≤ min ((cid:18) d + nn (cid:19) , r X i =1 (cid:18) n + m i − n (cid:19)) , or, equivalently, H I X ( d ) ≥ max n , (cid:0) d + nn (cid:1) − P ri =1 (cid:0) n + m i − n (cid:1)o .Proof. By Proposition C.3 or C.4 we have H R/I X ( d ) ≤ e ( R/I X ) = P ri =1 (cid:0) n + m i − n (cid:1) . Since R = C [ x , . . . , x n ] one has H R/I X ( d ) ≤ H R ( d ) = (cid:0) d + nn (cid:1) . (cid:3) When equality is achieved in the above inequality we obtain the definition of AH n ( d ) , or maximal Hilbertfunction in degree d . In other papers, thisproperty is often referred to as X imposes independent conditionson degree d hypersurfaces in P n . Definition C.6.
Let X = { m P , . . . , m r P r } ⊆ P n be a set of r points with given multiplicities m , m , . . . , m r .We say that X is AH n ( d ) (or has maximal Hilbert function in degree d ), if H R/I X ( d ) = min ((cid:18) d + nn (cid:19) , r X i =1 (cid:18) n + m i − n (cid:19)) . The number min n(cid:0) d + nn (cid:1) , P ri =1 (cid:0) n + m i − n (cid:1)o is called the expected codimension in degree d (for r generalpoints in P n with multiplicities m , . . . , m r ). he simplest situation for Definition C.6 is when m = · · · = m r = 1 , i.e., the points in X are all simplepoints. Theorem C.7.
A set X of r general points in P n is AH n ( d ) for every d ≥ .Proof. The statement follows from a simple observation, from Lemma C.1, that the condition H R/I X ( d ) =min { (cid:0) n + dd (cid:1) , r } is an open condition. (cid:3) Example C.8.
Let X = { P , P , P } ⊆ P , then e ( R/I X ) = 25 , and X is AH ( d ) if and only if H R/I X ( d ) = min (cid:26)(cid:18) d + 33 (cid:19) , (cid:27) i.e. if its Hilbert function is H R/I X = (1 , , , , , , , . . . ) .Other simple situations, where the property AH n ( d ) trivially holds, are when X is supported at a singlepoint, when d = 1 , and when n = 1 . Remark C.9.
A single double point X = { P } in P n is AH n ( d ) for all n and d . Proof.
After a change of coordinates we may assume I P = ( x , . . . , x n ) , so I X = I (2) P = I P . If d = 1 then [ I (2) P ] = 0 , so H R/I (2) P (1) = n + 1 = min { n + 1 , n + 1 } . If d ≥ then [ R/I (2) P ] d = h x d , x d − x , . . . , x d − x n i . Thus, H R/I (2) P ( d ) = n + 1 = min (cid:26)(cid:18) n + dd (cid:19) , n + 1 (cid:27) , and the statement follows. (cid:3) Remark C.10.
Any set Y of r double points (not necessarily general) in P n is AH n (1) . Proof.
For any r ≥ , one needs to show that H I (2) Y (1) = max { , (cid:18) n + 1 n (cid:19) − r ( n + 1) } = 0 . Let P ∈ Y be any point, then [ I (2) Y ] ⊆ [ I P ] = (0) . Thus, I (2) Y contains no linear forms. (cid:3) Proposition C.11.
Let Y be a set of r distinct simple points in P . Then mY is AH ( d ) for every d, m ∈ Z + .Proof. Notice that the defining ideal of any point in P is just a principal prime ideal generated by a linearform, thus if Y = { P , . . . , P r } , then I Y is a principal ideal generated by a form of degree r . It follows that I ( m ) Y = I mY ⊆ R = C [ x , x ] is a principal ideal of degree rm , so I ( m ) Y ∼ = R ( − rm ) . Thus, H I ( m ) Y ( d ) is max (cid:26) , (cid:18) d + 1 − rm (cid:19)(cid:27) = max { , d + 1 − rm } = max (cid:26) , (cid:18) d + 1 d (cid:19) − rm (cid:27) Therefore, mY is AH ( d ) . (cid:3) The following lemma allows us to restrict attention to only a finite number of values of r in proving theAlexander-Hirschowitz theorem. Lemma C.12.
Let X be a set of r points in P n with multiplicities m , . . . , m r which are AH n ( d ) .(1) If H R/I X ( d ) = (cid:0) d + nn (cid:1) , then X ′ is also AH n ( d ) for any larger set X ′ ⊇ X consisting of r ′ ≥ r points of X with multiplicities m ′ ≥ m , . . . , m ′ r ′ ≥ m r ′ .
2) If H R/I X ( d ) = P ri =1 (cid:0) n + m i − n (cid:1) = e ( R/I X ) , then X ′ is also AH n ( d ) for any subset X ′ ⊆ X consisting of r ′ ≤ r points of X with multiplicities m ′ ≤ m , . . . , m ′ r ′ ≤ m r ′ .Proof. (1) Since X ⊆ X ′ then I X ′ ⊆ I X . By assumption [ I X ] d = 0 , thus also [ I X ′ ] d = 0 , which impliesthat H R/I X ′ ( d ) = (cid:0) d + nn (cid:1) .(2) By Lemma C.1 and the numerical assumption, the homogeneous linear system associated to [ I X ] d hasrank P ri =1 (cid:0) n + m i − n (cid:1) , which is the number of rows of the linear system similar to that described in LemmaC.1. It follows that any subset of the rows of this linear system is again linearly independent. This provesimmediately the case where r ′ < r and m ′ i = m i for all i = 1 , . . . , r ′ .The general case follows by the same argument after the following adjustment. For any m ′ < m , thehomogeneous linear system consisting of all the ( m − -th partial order of a generic form of degree d in n + 1 variables is equivalent to the homogeneous system of all the ( m ′ − -th partial order derivatives plussome of the m -th order derivatives. By iteration, it suffices to illustrate the case where m ′ = m − , and byreplacing F by any ( m − -th partial order derivative of F , we may assume m ′ = 1 and m = 2 . In thiscase, we have a system of n + 2 equations, namely F ( P ) = 0 and ∂ x i F ( P ) = 0 i = 0 , . . . , n . This linearsystem has rank n + 1 , and since the linear equation corresponding to F ( P ) = 0 is not identically zero(because of the pure power monomial x di corresponding to any non-zero coordinate of P ), then the system { ∂ x i F ( P ) = 0 } i =0 ,...,n +1 is equivalent to the linear system { F ( P ) = 0 } ∪ { ∂ x i F ( P ) = 0 } i = j obtained byreplacing one of partial derivative equation by F ( P ) = 0 . (cid:3) We conclude this appendix by stating a numerical characterization of when it is possible to find simplepoints to be added to a given scheme in order to change its Hilbert function by a prescribed value. The mainingredient in the proof is that if I is a saturated ideal, then its homogeneous d component I d is the null spaceof a linear system obtained by adding H R/I ( d ) general points of V ( I ) . Proposition C.13 ([13, Lemma 3]) . Let I be a saturated homogeneous ideal in R = C [ x , . . . , x n ] , and let ℓ be a linear form that is regular on R/I . TFAE(1) there exists a set Y of u points in V ( ℓ ) such that H R/ ( I ∩ I Y ) ( t ) = H R/I ( t ) + u, (2) H R/I ( t ) + u ≤ H R/I ( t −
1) + (cid:0) n + t − t (cid:1) . D. A
PPENDIX : S
EMI - CONTINUITY OF THE H ILBERT FUNCTION AND REDUCTION TO SPECIALCONFIGURATIONS
The starting point of the proof of Theorem 1.1 is the observation that to establish the AH n ( d ) propertyfor a general set of double points, in non-exceptional cases, we only need to exhibit a specific collectionof double points with the AH n ( d ) property. This is because Hilbert functions have the so-called lowersemi-continuity property. This is the content of this appendix.We begin by defining generic points and the specialization of points. Throughout this appendix, we shallfix a pair of positive intgers n and r . Recall that R = C [ x , . . . , x n ] is the homogeneous coordinate ring of P n . Let z = { z ij (cid:12)(cid:12) ≤ i ≤ r, ≤ j ≤ n } be a collection of r ( n + 1) indeterminates, and let C ( z ) be thepurely transcendental field extension of C be adjoining the variables in z . Let S = C ( z )[ x , . . . , x n ] be thehomogeneous coordinate ring of P n C ( z ) . Set-up D.1. (1) By the generic set of r points , we mean the set Z = { Q , . . . , Q r } , where Q i = [ z i : · · · : z in ] , for i = 1 , . . . , r , are points with the generic coordinates in P n C ( z ) . Let I Z ⊆ S denote the defining idealof Z .
2) Let λ = ( λ ij ) ∈ A r ( n +1) C be such that for each i = 1 , . . . , r , λ ij = 0 for some j . Define the set Z ( λ ) = { Q ( λ ) , . . . , Q r ( λ ) } of points in P n , with Q i ( λ ) = [ λ i : · · · : λ in ] . Let I λ ⊆ R be thedefining ideal of Z ( λ ) .We call Z ( λ ) the specialization of the generic points at λ , and call I λ the specialization of the ideal I Z at λ .To define precisely the notions of general points and very general points one often employs Chow vari-eties. However, one can also use dense Zariski-open subsets of A r ( n +1) (see e.g. [22, Lemma 2.3]) for thesepurposes, and this is the point of view we take. Definition D.2.
One says that a property P• holds for a general set of r points of P n C if there is a dense Zariski-open subset U ⊆ A r ( n +1) C suchthat P holds for Z ( λ ) for all λ ∈ U ; • holds for a very general set of r points of P n C if P holds for Z ( λ ) for all λ ∈ U where U is an anintersection of countably many dense Zariski-open subsets of A r ( n +1) C .The lower semi-continuity of Hilbert functions that we shall use is stated in the following theorem. Theorem D.3 (Lower-semi-continuity of the Hilbert function) . Assume Set-up D.1. Then, for any m, d ∈ N ,we have H I ( m ) Z ( d ) ≤ H I ( m ) Y ( d ) for any set Y of r points in P n . Moreover,(1) for fixed m ≥ and d ≥ , the equality H I ( m ) Z ( d ) = H I ( m ) Y ( d ) holds for a general set of points Y ⊆ P n ; and(2) for fixed m ≥ , the equality of Hilbert functions H I ( m ) Z = H I ( m ) Y holds for a very general set ofpoints Y ⊆ P n .Proof. Note that every set Y of r points in P n can be viewed as a specialization Z ( λ ) of the generic set of r points. The proof is similar to the proof of [22, Thm 2.4]. For every s ≥ , set W s := { λ ∈ A r ( n +1) C | H I ( m )( λ ) Z ( d ) ≥ s } . We claim that W s is a Zariski-closed subset of A r ( n +1) C for any s ≥ .To see it, let f = P | α | = d C α x α ∈ R [ C α ] be a generic homogeneous polynomial of degree d , where x α are the monomials of degree d in R . Let ∂ β x α denote the divided power partial derivative of x α with respectto β .Now, let D m,d be the matrix with as many columns are monomials of degree d in R , its rows are indexedby partial derivatives β with | β | ≤ m − , and its rows are h ∂ β x d . . . ∂ β z iα . . . ∂ β x dn i . Let [ B m,d ] λ be the r by block matrix B m,d = D m,d ( P ) D m,d ( P ) ... D m,d ( P k ) where D m,d ( P ) is the specialization of the matrix D m,d at the point P i , i.e. we replace x , . . . , x n by λ i, , . . . , λ i,n , respectively. hen the forms f = P | α | = d C α x α of degree d in I λ ( m ) are in a bijective correspondence with the non-trivial solutions to the system of equations (in the variables C α ) [ B m,d ] λ · (cid:2) C ( d,..., . . . C α . . . C (0 ,...,d ) (cid:3) T = 0 . It follows that λ ∈ W s if and only if the null-space of this linear system has dimension at least s , whichis holds if and only if the number r (cid:0) m + nm − (cid:1) of rows of [ B m,d ] λ is less than (cid:0) d + nn (cid:1) − ( s − or r (cid:0) m + nm − (cid:1) ≥ (cid:0) d + nn (cid:1) − ( s − and rk[ B m,d ] λ < (cid:0) d + nn (cid:1) − ( s − . In either case we have a closed condition in A r ( n +1) .This proves the claim.To prove the inequality in the statement we prove that when one takes s := H I ( m ) Z ( d ) , then W s is alsocontains a dense Zariski-open subset, thus showing that W s is the entire space.Indeed, let f , . . . , f s be linearly independent forms of degree d in I ( m ) Z . We may assume that each f i ∈ C ( z )[ x , . . . , x n ] . Let M be the matrix whose i -th row consists of the coefficients of each monomial x α in f i .By assumption M has maximal rank, i.e. s , so at least one of the minors of size s of M does not vanish. Itfollows that there exists a dense Zariski-open subset f U t of specializations z λ ensuring that the special-ization does not make this minor vanish, thus for any λ ∈ e U we have that ( f ) z λ , ( f ) z λ , . . . , ( f s ) z λ are s linearly independently forms of degree d in I ( m ) λ . This concludes the proof of the first part.The equality in (1) follows from this last paragraph, as it is shown in there that for any λ ∈ f U d one hasthat s := H I ( m ) Z ( d ) = H I ( m ) λ ( d ) .The equality in (2) follows by part (1), since for any λ ∈ e U := T d ≥ f U d one has the equality of the entireHilbert functions. (cid:3) We obtain the following immediate consequences of Theorem D.3.
Corollary D.4.
Fix positive integers n , r , d and m . TFAE:(1) There exists a set Y of r points in P n such that mY is AH n ( d ) .(2) For every set Y of r general points in P n one has that mY is AH n ( d ) . Corollary D.5.
Fix n, d ∈ Z + . Then every set Y of r general double points in P n is AH n ( d ) if and only ifthere exist sets of r double points in P n which are AH n ( d ) for j n +1 (cid:0) d + nn (cid:1)k ≤ r ≤ l n +1 (cid:0) d + nn (cid:1)m .Similarly, if a set of r general points is not AH n ( d ) , then any set of r = r general double points in P n is AH n ( d ) if and only if there exist sets of r − and r + 1 double points in P n that are AH n ( d ) .Proof. The desired statements are direct consequences of Corollary D.4 and Lemma C.12. (cid:3)
In the last part of this section we prove a semi-continuity results in the more general setting of flat familiesof projective schemes.
Definition D.6.
Let f : X → Y be a morphism of schemes, and let F be a sheaf of O X -modules. We saythat F is f -flat at x ∈ X if the stalk F x , seen as an O Y,f ( x ) -module, is flat. We say that F is f -flat if it is f -flat at every point in X . Definition D.7. A family of (closed) projective schemes f : X → Y is a morphism f of (locally) Noetherianschemes which factors through a closed embedding X ⊆ P r × Y = P , for some r . The family is flat if O X if f -flat.Let p be a point in Y . Let C ( p ) be the residue field of the local ring O Y, p . Let X p = X × Y Spec( O Y, p ) andlet P p = P × Y Spec( O Y, p ) . For example, if Y = Spec( A ) and X = Proj( R/I ) , where R = A [ x , . . . , x r ] and I ⊂ R is a homogeneous ideal, then X p = Proj(( R/I ) ⊗ A K ( p )) and P p = Proj( R ⊗ A K ( p )) . Note hat, in general, the defining ideal of X p in P p may not be the same as I ⊗ A K ( p ) ; rather, it is the image ofthe canonical map ( I ⊗ A K ( p ) → R ⊗ A K ( p )) .The following result is well-known; see, for example, [37, Theorem III.12.8]. Theorem D.8.
Let f : X → Y be a family of projective schemes and let F be a coherent sheaf over X which is also f -flat. Then, for each i ≥ , the function Y → Z p dim K ( p ) ( H i ( X p , F p )) is upper semicontinuous on Y . Theorem D.9.
Let f : X → Y be a flat family of projective schemes. Then, for any degree d ≥ , thefunction Y → Z p h P p ( X p , d ) is lower semicontinuous on Y .Proof. Let I be its ideal sheaf of the embedding X ⊆ P . Let p ∈ Y be any point and let A = O Y, p . Wehave a short exact sequence → I → O P → O X → . By tensoring with C ( p ) , we obtain the followingshort exact sequence → I ⊗ A K ( p ) → O P p → O X p → . Particularly, this shows that
I ⊗ A K ( p ) is the ideal sheaf of the embedding X p ⊆ P p . Set I p = I ⊗ A K ( p ) .We then have h P p ( X p , n ) = h ( O P p ( n )) − h ( I p ( n )) . Observe that O X is f -flat, and so I is also f -flat. Therefore, by Theorem D.8, the function p h ( I p ( n ) is an upper semicontinuous function on Y . The conclusion now follows, since h ( O P p ( n )) is constant on Y . (cid:3) E. A
PPENDIX : H
ILBERT SCHEMES OF POINTS AND CURVILINEAR SUBSCHEMES
We end the paper with our last appendix giving basic definitions and properties of curvilinear subschemesthat allow the deformation argument in the m´ethode diff´erentielle to work.
Definition E.1.
A finite zero-dimensional scheme Z is said to be curvilinear if Z locally can be embeddedin a smooth curve. That is, for every point P in Z , the dimention T P ( Z ) of the tangent space is at most 1. Lemma E.2.
Let Z be a zero-dimensional scheme supported at one point P . Then Z is curvilinear if andonly if Z ≃ Spec C [ t ] / ( t l ) , where l is the degree of Z .Proof. Without loss of generality, assume that ( x , . . . , x n ) are local parameters at P . Let C be a smoothcurve to which Z can be embedded in. Clearly, P ∈ C . Let I C = ( f , . . . , f s ) be the defining ideal of C in O P = C [ x , . . . , x n ] (particularly, s ≥ n − ). Since C is smooth at P , the Jacobian matrix of C at P has rank n − . Thus, by a change of variables and a re-indexing, if necessary, we may further assume that f i = x i + g i , for i = 1 , . . . , n − , and g , . . . , g n − ∈ O P .Let I Z be the defining ideal of Z in O P . Since Z can be embedded in C , we have I C ⊆ I Z . Therefore,locally at P , O Z is a quotient ring of C [ x n ] . It follows that, locally at P , O Z ∼ = C [ x n ] / ( x ln ) for some l .The converse is clear by the same arguments. Observe further that localizing at P (a minimal prime in O Z ) does not change the multiplicity of O Z , or equivalently, the degree of Z . Hence, deg( Z ) = l . (cid:3) Corollary E.3.
Let Z be a curvilinear subscheme of a double point. Then the degree of Z is either 1 or 2.Proof. By Lemma E.2, we have Z ∼ = Spec C [ t ] / ( t l ) . Since Z is contained in a double point, we must have l is equal to 1 or 2. Hence, deg( Z ) is either 1 or 2. (cid:3) emark E.4. Let Z be a zero-dimensional scheme with irreducible components Z , . . . , Z r . Then, Z iscurvilinear if and only if Z , . . . , Z r are curvilinear.The next lemma gives another way of seeing curvilinear schemes. Lemma E.5.
Let Z be a zero-dimension scheme supported at one point P . Then, Z is curvilinear if andonly if, locally at P , the O Z is generated by one element, that is, O Z = C [ f ] for some f ∈ O Z .Proof. By Lemma E.2, if Z is curvilinear then, clearly, O Z is generated by one element. Suppose, con-versely, that O Z = C [ f ] for some f ∈ O Z . Since Z is zero-dimensional, we must have f l = 0 for some l .By taking the smallest such l , we then have O Z ∼ = C [ t ] / ( t l ) , and so Z is curvilinear by Lemma E.2. (cid:3) The main result about curvilinear subschemes that we shall use is that they form an open dense subset inthe Hilbert scheme of zero-dimensional subscheme of a given degree in P n . Particularly, this allows us totake the limit of a family of curvilinear subschemes. For this, we shall need the following lemma. Lemma E.6.
Let R be a free A -algebra of rank n . Show that the set of p ∈ Spec A such that the K ( p ) -algebra R ⊗ A K ( p ) is generated by one element is an open subset U of Spec A . Here, K ( p ) is the residuefield A p / p A p at p .Proof. Let U := { p ∈ Spec( A ) | the K ( p ) -algebra R ⊗ A K ( p ) is generated by one element } . Clearly, if p ∈ U and q ⊆ p then q ∈ U . Thus, by Nagata’s topological criterion (e.g. [41, Thm 24.2]), to prove that U is open it suffices to show that if p ∈ U , then there exists a non-empty open subset of V ( p ) contained in U .Write R = A [ T , . . . , T r ] /J , since p ∈ U then (after possibly relabelling) we may assume that there exist a , . . . , a r − ∈ A \ p and g , . . . , g r − ∈ A [ T , . . . , T r ] with T i / ∈ supp( g i ) for any i = 1 , . . . , r − suchthat ( a T + g , . . . , a r − T r − + g r − ) ⊆ J. Clearly, for any q ∈ V ( p ) \ [ V ( a ) ∪ V ( a ) ∪ . . . ∪ V ( a r )] we have q ∈ U ; this concludes the proof. (cid:3) We are now ready to state and prove the density result of curvilinear subschemes.
Proposition E.7.
Let H l denote the Hilbert scheme of zero-dimensional subscheme of degree l in P n andlet H curv l denote the subset of H l consisting of curvilinear subschemes of degree l in P n . Then H curv l is anopen dense subset of H l .Proof. The statement follows from Lemmas E.5 and E.6. (cid:3) R EFERENCES [1] J. Alexander,
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