On the minimal free resolution for fat point schemes of multiplicity at most 3 in P^2
aa r X i v : . [ m a t h . AG ] O c t On the minimal free resolution for fat point schemesof multiplicity at most in P . Edoardo Ballico, Monica Id`a
Abstract.
Let Z be a fat point scheme in P supported on general points. Here we prove that if the multiplicitiesare at most 3 and the length of Z is sufficiently high then the number of generators of the homogeneous ideal I Z ineach degree is as small as numerically possible. Since it is known that Z has maximal Hilbert function, this impliesthat Z has the expected minimal free resolution. msc: 14N05keywords: minimal free resolution; homogeneous ideal; zero-dimensional scheme; fat double point; fat triple point;projective plane.
1. Introduction.
What is the minimal free resolution of a “ general ” zero-dimensional scheme Z ⊂ P ? In this paper“ general ” means that ♯ ( Z red ) is fixed and Z red is general in P . We are interested in the minimal freeresolution of general fat point schemes of P . A fat point mP ⊂ P is the zero-dimensional subscheme of P with support in the point P and ( I P ) m as its ideal sheaf. A general fat point scheme Z := m P + . . . + m r P r of P , m ≥ . . . m r ≥
0, is a general zero-dimensional scheme such that for each P i ∈ Z red the connectedcomponent of Z with support in P i is the fat point m i P i . If m = 1 (resp. m = 2, resp. m = 3) we willsay that mP is a simple (resp. double, resp. triple) point. We recall that length( mP ) = m ( m + 1) / m > r if m = m = . . . = m r ≤
20 ([ Hi], [ C-C-M-O]) and in many other cases([ H-H-F], [ H-R], [ E2], [ R]); it is also known if r ≤ m i ≤ Z = m P + . . . + m r P r are known if r ≤ m ≤
3; i.e., it is known that the homogeneous ideal I ( Z ) of Z is minimally generated for m = 1([ G-M]), m = 2 ([ I]) and m = 3 ([ G-I]).If Z is a fat point scheme of multiplicity at most 3 in P , Z has maximal Hilbert function by [ M], i.e. h ( P , I Z ( k )) · h ( P , I Z ( k )) = 0 for all k ≥
0, provided that k ≥ m + m + m . In this paper we show that, ifthe length of Z is sufficiently high, the multiplication maps µ k ( Z ) : H ( I Z ( k )) ⊗ H ( O P (1)) → H ( I Z ( k +1))are of maximal rank for any k . The following result hence follows:The authors were partially supported by MUR and GNSAGA of INdAM (Italy).1 heorem 1.1. Fix non-negative integers a, b, c such that a + 3 b + 6 c ≥ , and let Z ⊂ P be a generalunion of a simple points, b double points and c triple points in P . Let v be the minimal integer such that a + 3 b + 6 c ≤ ( v + 2)( v + 1) / . Then Z has the expected minimal free resolution, i.e. the homogeneous ideal of Z is minimally generated by ( v +2)( v +1) / − a − b − c forms of degree v and max { , a +6 b +12 c − v − v } forms of degree v + 1 . In the case in which there are only double points, we can handle a few low integers v and prove thefollowing complete result: Theorem 1.2.
Fix integers a, b such that a ≥ , b ≥ , and let Z ⊂ P be a general union of a simplepoints and b double points of P . Let v be the minimal integer such that a + 3 b ≤ ( v + 2)( v + 1) / . Then, forany ( a, b ) = (0 , , (0 , , (1 , , (1 , , Z has the expected minimal free resolution, i.e. the homogeneous idealof Z is minimally generated by ( v + 2)( v + 1) / − a − b forms of degree v and max { , a + 6 b − v − v } forms of degree v + 1 . We raise the following conjecture.
Conjecture 1.3.
Fix integers m > n ≥
2. Then there is an integer α n,m such that for all integers r > α n,m the minimal free resolution of a general union Z ⊂ P n of r multiple points with multiplicity atmost m is the expected one.For related conjectures and discussions, see also [ Ha1] and [ H-H-F]. Also recall the Minimal ResolutionConjecture, which is about the generic union of r simple points in P n , and has been proved for n = 2 (assaid above), for n = 3 ([ B], [ B-G]), in particular cases for n ≥
4, and for any n and r >> Z is lower and c = 0, itis essentially a question of patience; for example, in order to go down to a + 3 b + 6 c ≥
56 we only need toconsider a few more cases, thanks to Lemma 3.1.5 and Lemma 3.1.6; if the fat point scheme is supportedon ≤ a, b, c are such that a + 3 b ≡ b ≤ a and ⌊ a +3 b ⌋ 6 = 2 , ,
5, then [ G-I] and Remark 2.2 gives the result through semicontinuity. There are low lengthcases where the minimal free resolution is not the expected one, for example the union of 2 or 3 or 5 triplepoints.In the following we use the Horace method, introduced by A.Hirschowitz to prove this interpolationtype problems, and the differential Horace method. For the first one, we send the reader to [ Hi]. The secondone has been introduced in [ A-H] for invertible sheaves, and then extended to vector bundles (see [ G-I]Proposition 2.6).We work over an algebraically closed field K such that either char( K ) = 0 or char( K ) >
3. In [ I] and[ G-I] there is the char( K ) = 0 assumption, but in fact this assumption is not necessary in [ I] , while in[ G-I] it is used only in the proof of Lemma 2.9 where a map t i t r i i between formal power series ringsis considered. To get the injectivity of the differential of this map at (0 , . . . ,
0) it is sufficient to assumechar( K ) > r i for all i . Since in our set up (fat point with multiplicity at most 3) we have r i ≤ i , wehence assume char( K ) = 0 or char( K ) >
3. 2 cknowledgements.
The authors wish to thank the referee for her/his accurate report and usefulremarks.
2. Preliminaries and proof of Theorem 1.2.2.1.
Let X be a 0-dimensional scheme in P , and let length X = l ; we say that X has maximal Hilbertfunction in degree k if h ( I X ( k )) = max { , (cid:0) k +22 (cid:1) − l } ; X has maximal Hilbert function if this is true forany k .Now assume that X has maximal Hilbert function. Then X has the expected minimal free resolution ifthe natural multiplication maps µ k : H ( I X ( k )) ⊗ H ( O P (1)) → H ( I X ( k + 1))have maximal rank for each k . Set Ω := Ω P . Taking the cohomology sequence of the Euler sequence in P tensored by the ideal sheaf I X ( k + 1), we see that ker µ k = H (Ω( k + 1) ⊗ I X ).Set v = v ( X ) = min { k ≥ | (cid:0) k +22 (cid:1) − l ≥ } ; then, µ k is trivially injective for k < v because h ( I X ( k )) = 0, and µ k is surjective for k > v by the Castelnuovo-Mumford Lemma because h ( I X ( v )) = 0.Hence X is minimally generated if and only if µ v is of maximal rank.Now let w = w ( X ) = min { k ≥ | k ( k + 2) − l ≥ } ( w is the smallest integer for which the restrictionmap ρ k : H (Ω( k + 1)) → H (Ω( k + 1) | X ) can be surjective). Then 2 l ≤ w ( w + 2) gives l < (cid:0) w +22 (cid:1) , and byassumption X has maximal Hilbert function, hence h ( I X ( w )) = 0 and h ( I X ( w )) = (cid:0) w +22 (cid:1) − l >
0. Now if h ( I X ( k )) = 0 then h ( I X ( k + 1)) = 0 and we get 3 h ( I X ( k )) − h ( I X ( k + 1)) = 3( (cid:0) k +22 (cid:1) − l ) − ( (cid:0) k +32 (cid:1) − l ) = k ( k + 2) − l . Hence w is the smallest integer k for which the map µ k can be surjective, without beingthe 0-map. From what is said above we have v ≤ w ≤ v + 1, so that X is minimally generated if andonly if µ w − is injective and µ w is surjective, and this happens if and only if h (Ω( w ) ⊗ I X ) = 0 and h (Ω( w + 1) ⊗ I X ) = w ( w + 2) − l .If we assume only that the Hilbert function of X is maximal in degree w , without any assumptionson the other degrees, then the same considerations as above show that µ w − is injective if and only if h (Ω( w ) ⊗ I X ) = 0, and µ w is surjective if and only if h (Ω( w + 1) ⊗ I X ) = w ( w + 2) − l .What we do in practice is to look for suitable schemes for which the map ρ k is bijective, and from thesededuce the injectivity or surjectivity for the schemes we are interested in. For arithmetical reasons (think of k odd) it is better to work in the projectivized bundle P (Ω) with the canonical projection π : P (Ω) → P .We set E k := O P (Ω) (1) ⊗ π ∗ O P ( k ). One has (e.g., see [ I2] Lemma 2.1 ) H (Ω( k + 1) ⊗ I X ) ∼ = H ( E k +1 ⊗ I π − X ) , H (Ω( k + 1) | X ) ∼ = H ( E k +1 | π − X ) . If X ⊂ P (Ω) is a 0-dimensional scheme such that lengthX = H ( E k +1 ), and H ( E k +1 ⊗ I X ) = 0, we saythat X is k -settled. Remark 2.2.
It is immediate to see that a 2-fat point of P is the flat limit of a family whose general fiberis the general union of 3 simple points. 3 3-fat point of P is the flat limit of a family whose general fiber is the general union of one 2-fat pointand 3 simple points (see [ E], Proposition 4).Hence, a 3-fat point of P is the flat limit of a family whose general fiber is the general union of 6 simplepoints. Notations 2.3.
We denote by R p , with p = 0 , , , , , ,
11, a certain 0-dimensional scheme of length p in P (Ω) which we now define. Let U be an open subset in P and Ω | U ∼ = E ⊕ E a local trivialization for Ω;then, if p = 0 , R p = η ∪ η , where η , η have support on two distinct points A , A in the same fiber π − ( P ) where A i = P ( E i ) ∩ π − ( P ) and η i ⊂ P ( E i ), i = 1 ,
2, so that length( η i ∩ π − ( P )) = 1. If we consideraffine coordinates { x, y, z } in an affine chart of P (Ω) containing R p , we may suppose π − ( P ) = { x = y = 0 } , A = (0 , ,
0) and A = (0 , , R p is defined as follows: R = ∅ ; R = { A } is just a point in P (Ω); R = { A , A } ; R is made of the point η = A and a length 2 structure η on A , given by an ideal of type ( x, y , z − R is made of a length 2 structure η on A , given by an ideal of type ( x, y , z ) and the first infinitesimalneighbourhood η on A , given by an ideal of type ( x , xy, y , z − R is made of two 4-ple structures η , η of the same type, given by ideals of type ( x , xy, y , z ), ( x , xy, y , z − R is such that η is a 5-ple structure on A given by an ideal of type ( x , x y, y , z ), and η is given by anideal of type ( x , x y, y x, y , z − k even R p is two copies of a nilpotent η ⊂ P with η ∼ = η i , while for k odd R p is the same thing butwith a nilpotent transversal to one of the components of this scheme added. Since we are interested in theschemes R p only for the vanishing of global sections of O P (Ω) (1) ⊗ π ∗ O P ( t ), we can consider (see Lemma2.2 in [ G-I]) that R p is the pull back of η ⊂ P for k even, and for k odd the pull back of η with a nilpotenttransversal to this scheme added. For the same reason, if B ⊂ U is any 0-dimensional subscheme of P andwe set: B ′ := π − ( B ) ∩ P ( E ) , B ′′ := π − ( B ) ∩ P ( E ) , b B := B ′ ∪ B ′′ ,as long as we are concerned only with the vanishing of the global section of E k along π − ( B ), we cansubstitute the last one with b B . Notations 2.4.
With Y ( a, b ) we denote in the following the generic union of a points and b double pointsin P ; we also set ˜ Y ( a, b ) := π − ( Y ( a, b )).For any k ≥
0, let q = q ( k ), r = r ( k ) be positive integers such that k ( k + 2) = 6 q ( k ) + r ( k ), with 0 ≤ r ≤ r are 0 , , , Z ( s, d, p ) will denote the generic union in P (Ω) of ˜ Y ( s, d ) with R p , where 2 s +6 d + p = k ( k +2)and 0 ≤ p ≤ r . Notice that p ≡ r (mod 2), hence p = 4.We set ∆ k = { ( s, d, p ) ∈ N | s + 6 d + p = k ( k + 2) , p = 0 , , , , } .In [ I] the assertion:“ Z (0 , q ( k ) , r ( k )) is k-settled”, denoted by “ A(k) ”, is proved for any k = 2 , Lemma 2.5.
If A(k) in [ I] is true and if ( s, d, p ) ∈ ∆ k , then H ( E k +1 ⊗ I Z ( s,d,p ) ) = 0 . Proof.
We write q = q ( k ), r = r ( k ); by assumption 2 s + 6 d + p = 6 q + r with 0 ≤ p ≤ r ≤
5. Writing s = 3 l + j , 0 ≤ j ≤
2, we find 4 + r ≥ j + p ≡ r (mod 6), hence 2 j + p = r .4ow observe that a double point is specialization of 3 points in the plane; R is the pull back of a pointof P ; R is specialization of the union of the pull back of a point of P with R , which is a point in P (Ω),and if 2 j + p = 5, then R is specialization of the general union of the pull back of j points of P with R p ( j = 1 , Z ( s, d, p ) specializes to Z (0 , q, r ) and we conclude by semicontinuity since by assumption H ( E k +1 ⊗ I Z (0 ,q,r ) ) = 0. ⊓⊔ Lemma 2.6.
Let a, b, k be nonnegative integers such that ( k − k + 1) < a + 6 b ≤ k ( k + 2) . Then,if A(k-1) is true, µ k − is injective for Y ( a, b ) ;if A(k) is true, and if Y ( a, b ) has maximal Hilbert function in degree k , then µ k is surjective for Y ( a, b ) . Proof.
By assumption, w ( Y ( a, b )) = k . Then, in order to prove the first and respectively the secondstatement, it is enough to prove (see preliminaries 2.1) h ( E k ⊗ I ˜ Y ( a,b ) ) = 0 and respectively h ( E k +1 ⊗I ˜ Y ( a,b ) ) = k ( k + 2) − a + 3 b ).Assume A(k-1) holds. We now show that there exists ( s, d, p ) ∈ ∆ k − such that Z ( s, d, p ) is contained in˜ Y ( a, b ). We write q = q ( k − r = r ( k − s, d, p such that 2 s + 6 d + p = ( k − k + 1) =6 q + r < a + 6 b , with 0 ≤ r ≤
5. Set r = 2 l + ǫ , ǫ = 0 ,
1. If q < b , we set d = q , s = 0, p = r ; since R r is contained in the pull back of a double point, we have Z (0 , q, r ) ⊂ ˜ Y ( a, b ). If q ≥ b , we set d = b , s = 3( q − b ) + l , p = ǫ . Since 2(3( q − b ) + l ) + ǫ < a , we have s + 1 ≤ a and moreover if ǫ = 1 R ǫ is containedin the pull back of a simple point, hence Z (3( q − b ) + l, b, ǫ ) ⊂ ˜ Y ( a, b ) both for ǫ = 0 and for ǫ = 1.By the previous lemma H ( E k ⊗ I Z ( s,d,p ) ) = 0; we conclude taking cohomology of the exact sequence:0 → E k ⊗ I ˜ Y ( a,b ) → E k ⊗ I Z ( s,d,p ) → ... .Now assume A(k) holds. We now show that there exists ( s, d, p ) ∈ ∆ k such that Z ( s, d, p ) contains ˜ Y ( a, b ).We write q = q ( k ), r = r ( k ); we are looking for s, d, p such that 2 a + 6 b ≤ s + 6 d + p = k ( k + 2) = 6 q + r ,with 0 ≤ r ≤
5. Set r = 2 l + ǫ , ǫ = 0 ,
1. We have 6( b − q ) ≤ r − a ≤
5, hence b ≤ q , so we set d = b , s = 3( q − b ) + l , p = ǫ ; since 2 a ≤ q − b ) + l ) + ǫ , we have a ≤ s , so Z (3( q − b ) + l, b, ǫ ) ⊃ ˜ Y ( a, b ).By the previous lemma H ( E k +1 ⊗ I Z ( s,d,p ) ) = 0; the cohomology of the exact sequence:0 → E k +1 ⊗ I Z ( s,d,p ) → E k +1 ⊗ I ˜ Y ( a,b ) → E k +1 ⊗ I ˜ Y ( a,b ) ,Z ( s,d,p ) → h ( E k +1 ⊗ I ˜ Y ( a,b ) ) ≤ h ( E k +1 ⊗ I ˜ Y ( a,b ) ,Z ( s,d,p ) ) = 2( s − a ) + 6( d − b ) + p = k ( k + 2) − (2 a + 6 b ). On theother hand h ( E k +1 ⊗ I ˜ Y ( a,b ) ) ≥ h ( E k +1 ) − h ( E k +1 | ˜ Y ( a,b ) ) = k ( k + 2) − (2 a + 6 b ), so we have equality. ⊓⊔ Proof of Theorem 1.2.
First let us check that Y ( a, b ) has maximal Hilbert function (mHf for short)for ( a, b ) = (0 , , (0 , Y ( a, b ) has mHf for any a if b = 0, or for any b = 2 , a = 0 ([ Hi]). Now let L = 0 be a linear system in P ; if P is a point outside of the base locus of L , forexample a generic point of the plane, and L ( P ) is the linear system obtained by L imposing the passagethrough P , then dim L (P) = dim L −
1. Hence if we add (generic) simple points to a scheme Y ( a, b ) withmHf we get a scheme which again has mHf. We conclude that all schemes Y ( a, b ) with b = 2 , a ≥
3, and b = 2 ,
5, the scheme Y ( a, b ) specializes to Y ( a − , b + 1) which has mHf, hence bysemicontinuity Y ( a, b ) has mHf too. It is immediate to check by hand that also in the remaining cases, i.e.( a, b ) = (1 , , (2 , , (1 , , (2 ,
5) the Hilbert function is maximal.5ow we study the maps µ k . Recall that assertion A ( k ) is proved in [ I] for k = 1 and k ≥
4. If l = l ( Y ( a, b )) = a + 3 b >
12, and w is the integer such that ( w − w + 1) < a + 6 b ≤ w ( w + 2), then w = w ( Y ( a, b )) ≥ Y ( a, b ) is minimally generated, provided that ( a, b ) = (0 , w = 4 (i.e. 15 < l ≤
24) and v = v ( Y ( a, b )) = 4 (i.e. 10 < l ≤ µ is surjective, and this is true by Lemma 2.6; hence Y ( a, b ) is minimally generatedalso for 11 ≤ l ≤ l ≤
10. We know that Y ( a, b ) is minimally generated for any a if b = 0 ([ G-M]),or for any b = 2 if a = 0 ([ I]); the remaining cases are { ( a, } a =1 ,..., , { ( a, } a =1 ,..., and (1 , , , (4 , , (1 ,
3) we have h ( I Y ( a,b ) (3)) = h ( I Y ( a,b ) (3)) = 0 so by the Castelnuovo-Mumford Lemma thescheme is minimally generated. The scheme Y (6 ,
1) specializes to Y (3 ,
2) which specializes to Y (0 , H (Ω( k + 1) ⊗ I ) is useful).The few cases left can be recovered from [ Ca]; anyway, we give their explicit description in what follows.For (3 , , (2 ,
1) we have v = 2 and w = 3, so it is enough to prove that µ is injective, and this is true because h ( I Y (3 , (2)) = 0 and h ( I Y (2 , (2)) = 1. For (2 ,
2) we have v = 3 and w = 4, so it is enough to provethat H (Ω(4) ⊗ I Y (2 , ) = 0. Let C be a conic through the four points, and L the line through the twodouble points; since Ω | C ∼ = O P ( − ⊕ , we have H (Ω(4) | C ⊗ I Y (2 , ∩ C,C ) = 0, hence h (Ω(4) ⊗ I Y (2 , ) = h (Ω(2) ⊗ I Y (2 , ) and the last one is zero because H (Ω(1)) = 0 and H (Ω(2) | L ⊗ I Y (2 , ,L ) = 0. Thescheme Y (5 ,
1) specializes to Y (2 , Y (5 ,
1) is minimally generated.For (4 ,
1) we have v = 3 and w = 3, so we want to prove that µ is surjective, or equivalently (see 2.1)that h ( E ⊗ I ˜ Y (4 , ) = h (Ω(4) ⊗ I Y (4 , ) = 15 − l = 1; this is in turn equivalent to H ( E ⊗ I Z (4 , , ) = 0,where Z (4 , ,
1) = ˜ Y (4 , ∪ R . Now we use the Horace method in P (Ω), as we do, for example, in the proofof the forthcoming Lemma 3.1.4. Let C be the conic through the five points in the support of Y (4 , H ( E | π − C ⊗ I Z (4 , , ∩ π − C,π − C ) ) = 0, and H ( E ⊗ I Z (1 , , ) = 0 (the last one is assertion A (1)), so weconclude that H ( E ⊗ I Z (4 , , ) = 0. Y (1 ,
1) is not minimally generated. In fact, let L be the line through the two points. Here v = 2 and w = 2, but µ cannot be surjective since L is in the base locus of H ( I Y (1 , (2)), so there must be a generatorof degree 3. Y (1 ,
2) is not minimally generated. Here v = 3 and w = 3, but µ cannot be surjective sincethe line through the two double points is in the base locus of H ( I Y (1 , (3)), so there must be a generatorof degree 4. ⊓⊔
3. Proof of Theorem 1.1.
In the following we use, beyond the Horacemethod, also the differential Horace method for vector bundles, for which we refer to [ G-I] Section 2 and inparticular Proposition 2.6.; moreover, we’ll use notations 3.1 and the ones estabilished at the beginning ofthe proof of 3.3 in [ G-I], so we recall them briefly here.For the definition of vertically graded subscheme with base a fixed smooth divisor see [ A-H] and [G-I] 2.3. We introduce now some notations that will allow us to express ourselves as if we were workingin P , while our environment is actually P (Ω). Let B be a 0-dimensional scheme of P with support ata point P , vertically graded with base a smooth conic C with local equation y = 0, and let x, y be local6oordinates at P . Notice that B ′ and B ′′ (see Notations 2.3) are vertically graded with base H = π − ( C ).Consider the integers a j where, if I T r jC ( B ) := ( I B : I jC ) ⊗ O C , then T r jC ( B ) = ( x a j , y ). We will denote b B = B ′ ∪ B ′′ by a s ... a . So for example b B is denoted by ( h ) if I B = ( x h , y ); by (cid:18) (cid:19) if I B = ( x, y ) ; by (cid:18) (cid:19) if I B = ( x , xy, y ); by (cid:18) (cid:19) if I B = ( x , x y, y ); by (cid:18) (cid:19) if I B = ( x , y ); by if I B = ( x, y ) .If b B is a (cid:18) (cid:19) , we say that b B is “a (cid:18) (cid:19) scheme over C ” and write (cid:18) (cid:19) overC . We write, for example,“ h (cid:18) (cid:19) schemes general over P ” to mean a union of h schemes of type (cid:18) (cid:19) , whose projection in P isgeneral. Moreover, if h, l ∈ N , we will use, for example, the notation “ h (cid:18) (cid:19) + l (cid:18) (cid:19) ” to denote the unionof h schemes of type (cid:18) (cid:19) and l schemes of type (cid:18) (cid:19) .If a i and b i , i = 1 , ..., m are positive integers, with an abuse of notation we write ( P mi =1 a i b i ) = P mi =1 a i ( b i ), since for our vanishing problem only the length of the scheme over C matters.When we apply [ G-I] Proposition 2.6, i.e. when we do a differential Horace, or HD, step, and we say forexample that we “add over C ” (cid:18) (cid:19) + , this means that we are using the “ground slide” of thevertical scheme (cid:18) (cid:19) and the “first floor slide” of the triple point, so that the HD trace ( 5 ) C is given by thenumbers in square brackets, while the HD residue (( 1 ) + (cid:18) (cid:19) ) C is obtained by the eliminating the part inthe square brackets. Notations 3.1.2.
With Y ( a, b, c ) we denote in the following the generic union of a points, b double pointsand c triple points in P ; we also set ˜ Y ( a, b, c ) := π − ( Y ( a, b, c )).For any k ≥
0, let u = u ( k ), ρ = ρ ( k ) be the positive integers such that k ( k + 2) = 12 u + ρ , with 0 ≤ ρ ≤ k modulo 6, we get (see [ G-I] 1.2; but notice that there u ( k ), resp. ρ ( k ), are denoted by q ( k ),resp. r ( k )):for k=6l u ( k ) = 3 l + l ρ ( k ) = 0for k=6l+1 u ( k ) = 3 l + 2 l ρ ( k ) = 3for k=6l+2 u ( k ) = 3 l + 3 l ρ ( k ) = 8for k=6l+3 u ( k ) = 3 l + 4 l + 1 ρ ( k ) = 3for k=6l+4 u ( k ) = 3 l + 5 l + 2 ρ ( k ) = 0for k=6l+5 u ( k ) = 3 l + 6 l + 2 ρ ( k ) = 11.For a fixed k , let Z ( s, d, t, p ) denote the generic union in P (Ω) of ˜ Y ( s, d, t ) with R p , where 2 s + 6 d +12 t + p = k ( k + 2) and 0 ≤ p ≤ ρ , p = 0 , , , , , , k = { ( s, d, t, p ) ∈ N | s + 6 d + 12 t + p = k ( k + 2) , ≤ p ≤ ρ, p = 0 , , , , , , } .In the following, “ H ( s, d, t, p, k )” denotes the statement:“If ( s, d, t, p ) ∈ Λ k , then H ( E k +1 ⊗ I Z ( s,d,t,p ) ) = 0 . ”7e want to prove that H ( s, d, t, p, k ) holds for k ≥
12, and this will be done through some lemmas. Noticethat if t = 0 the statement H ( s, d, t, p, k ) reduces to Lemma 2.5, so we can assume t ≥ Remark 3.1.3.
Recall that B ( k ) in [ G-I] is nothing else but H (0 , , u ( k ) , ρ ( k ) , k ), and B ( k ) is true for k ≥
10 ([ G-I] Proposition 3.9). To be punctual about that, notice that for the remainder scheme T definedin [ G-I] 1.3, which is the analogous of our scheme R , one has the choice between two 4-ple structures η , η of the same type, given in local coordinates x, y by an ideal of type ( x , xy, y ) or by an ideal of type ( x , y );in the languages of vertical schemes, T is (cid:18) (cid:19) or (cid:18) (cid:19) . Anyway, the unique point of the proof of Theorem1.1 in [ G-I] where one chooses to use T as a (cid:18) (cid:19) is the next to last step in the proof of Proposition 4.1,where we specialize it on (the pull back of ) a smooth conic C . It is possible to choose T = (cid:18) (cid:19) alsohere, specializing it as a (in other words, we consider it as vertical scheme with respect to the y -axisinstead that to the x -axis); in the last step, the residue (cid:18) (cid:19) can be specialized as a (2) on C . Hence wecan always assume that the remainder scheme T in [ G-I] is our scheme R .The following lemma allows to construct a lot of well generated schemes containing 2-fat points; theproof goes exactly as the proof of Lemma 2.1 in [ I], but we repeat it here for the reader’s sake. Lemma 3.1.4.
Let k be an integer ≥ and R be a 0-dimensional scheme in P (Ω) such that h ( E k − ⊗ I R ) =0 . Let A be the union in P (Ω) of R with π − ( Y ) where Y denotes the union of k − P supported at general points; then, h ( E k +1 ⊗ I A ) = 0 . Proof.
Let C be a smooth conic; we denote by Z the scheme obtained by A specializing k among the 2-fatpoints of Y on C and consider the exact sequence:0 → E k − ⊗ I Res π − C Z → E k +1 ⊗ I Z → E k +1 ⊗ I Z ∩ π − C,π − C → C there is a scheme of length 2 k , H ( E k +1 ⊗ I Z ∩ π − C,π − C ) = 0 so that h ( E k − ⊗ I Res π − C Z ) = h ( E k +1 ⊗ I Z ) ≥ h ( E k +1 ⊗ I A ).The scheme Res π − C Z is the union of R and of the pull-back of k − k simple pointson C . Now let C ′ be another smooth conic; we denote by B the scheme obtained by Res π − C Z specializingthe k − C ′ and 4 among the simple points on C ∩ C ′ ; consider the exact sequence:0 → E k − ⊗ I Res π − C ′ B → E k − ⊗ I B → E k − ⊗ I B ∩ π − C ′ ,π − C ′ → C ′ there is a scheme of length 2 k −
4, the third H is 0 so that h ( E k − ⊗ I Res π − C ′ B ) = h ( E k − ⊗ I B ) ≥ h ( E k − ⊗ I Res π − C Z ).The scheme Res π − C ′ B is the union of R and of the pull-back of k − C and k − C ′ . We denote by D the scheme obtained by Res π − C ′ B specializing the k − C ′ -points on C (fordetails, see [ I]); now consider the exact sequence:0 → E k − ⊗ I Res π − C D → E k − ⊗ I D → E k − ⊗ I D ∩ π − C,π − C → C there is a scheme of length 2 k −
8, the third H is 0. Since Res π − C D is R , we finally get0 = h ( E k − ⊗ I Res π − C D ) = h ( E k − ⊗ I D ) ≥ h ( E k − ⊗ I Res π − C ′ B ), that is, h ( E k +1 ⊗ I A ) = 0. ⊓⊔ Lemma 3.1.5. If d ≤ s and k ≥ , then H ( s, d, t, p, k ) is true. Proof.
Since d ≤ s , by Remark 2.2 the scheme Z ( s, d, t, p ) specializes to Z ( s ′ , , t ′ , p ) where s ′ = s − d − (cid:2) s − d (cid:3) , t ′ = t + d + (cid:2) s − d (cid:3) .Since s ′ ≤ p ≤ ρ , we have 10 + ρ ≥ s ′ + p ≡ ρ (12) so that 2 s ′ + p = ρ , hence it is easy to see that thescheme the union of ˜ Y ( s ′ , ,
0) and of R p specializes to R ρ (see description of schemes R p in Notations 2.3).So finally the scheme Z ( s ′ , , t ′ , p ) specializes to Z (0 , , u ( k ) , r ( k )) and we conclude by semicontinuity that H ( s, d, t, p, k ) holds. ⊓⊔ Lemma 3.1.6.
The statements H ( s, d, t, p, k ) with d > s are true if both the statements H (0 , d, t, ρ ( k ) , k ) with any d and the statements H (0 , d, t, , k ) with k ≡ and with any d are true. Proof.
Since d > s , Remark 2.2 allows us to say that the scheme Z ( s, d, t, p ) specializes to Z ( σ, δ, τ, p )where σ = s − (cid:2) s (cid:3) , δ = d − (cid:2) s (cid:3) , τ = t + (cid:2) s (cid:3) ; moreover, 0 ≤ σ ≤ δ ≥
1. Let u = u ( k ), ρ = ρ ( k ),and set δ = 2 e + j , 0 ≤ j ≤
1; we have: 2 σ + 6 δ + 12 τ + p = 12( τ + e ) + 6 j + 2 σ + p = k ( k + 2) = 12 u + ρ ,0 ≤ p ≤ ρ . Since 10 + ρ ≥ j + 2 σ + p ≡ ρ (mod 12), we get 6 j + 2 σ + p = ρ . Hence it is easy to seethat if ( σ, j, p ) = (0 , ,
5) the union of ˜ Y ( σ, j,
0) and of R p specializes to R ρ , so that finally the scheme Z ( s, d, t, p ) specializes to Z (0 , δ − j, τ, ρ ), where δ − j ≡ σ, j, p ) = (0 , ,
5) then ρ = 11 so that k ≡ Y ( σ, j, R p specializes to R , essentially because two double points in the plane do not specialize to a triplepoint. ⊓⊔ Remark and notations 3.1.7.
Now our purpose is to prove the statements H (0 , d, t, ρ ( k ) , k ) with any d and the statements H (0 , d, t, , k ) with k ≡ d , for k ≥ d of double points in the statement H (0 , d, t, ρ ( k ) , k ) is necessarily even, since byassumption 6 d + 12 t = k ( k + 2) − ρ ( k ) ≡ d of double points in the statement H (0 , d, t, , k ) with k ≡ k ≡ ρ ( k ) = 11, and by assumption6( d −
1) + 12 t = k ( k + 2) − ≡ X ( d, t, k ) := Z (0 , d, t, ρ ) where 6 d + 12 t + ρ = k ( k + 2) and k ( k + 2) = 12 u + ρ ,with 0 ≤ ρ ≤
11; notice that t and k , as well as d and k , determine X ( d, t, k ), and d is always even.With ¯ R we denote in the following the generic union of the inverse image of a double point with R .If k ≡ X ( d, t, k ) the scheme obtained by X ( d, t, k ) substituting to R the scheme¯ R .So finally what we want to prove is that X ( d, t, k ) is k-settled for any k ≥
12 and that ¯ X ( d, t, l + 5) is(6 l + 5)-settled for any l ≥ In the following C denotes a smooth conic. 9et R = (cid:0) x (cid:18) (cid:19) + y (cid:18) (cid:19) + z (cid:18) (cid:19) + ( e ) (cid:1) over C + v + u (cid:18) (cid:19) + R ρ be a 0-dimensional scheme withlength R = h ( E k +1 ) and length( R ∩ π − C ) ≤ k , and assume we want to prove that R is k -settled. Wenow define what a standard step is; the idea is that we specialize (in the sense of a differential Horace step)the maximum possible of triple points on C , and if no more triple points are available, we specialize doublepoints on C , so that to get a scheme ¯ R with exactly 2 k conditions on C ; at this point in order to prove that R is k -settled it is enough to prove that the residual scheme is k − standard step k → k − :we “add” over C g + n + p [1]23 + r (cid:18) (cid:19) + s (cid:18) [1]2 (cid:19) where: v ≥ g + n + p , u ≥ r + s , 2 x + 3 y + 3 z + e + 3 g + 2 n + p + 2 r + s = 2 k , 0 ≤ n + p ≤
1, 0 ≤ s ≤
1, and finally r = s = 0 if it is possible to find g, n, p such that 2 x + 3 y + 3 z + e + 3 g + 2 n + p = 2 k .The residue is: (cid:0) ( x + y + 2 z + r + 2 s ) + g (cid:18) (cid:19) + n (cid:18) (cid:19) + p (cid:18) (cid:19) (cid:1) over C + ( v − g − n − p ) + ( u − r − s ) (cid:18) (cid:19) + R ρ .Notice that this construction is possible if 2 x + 3 y + 3 z + e ≤ k and if v ≥ g + n + p , u ≥ r + s . Notation 3.2.2.
We recall here Definition 3.2 given in [ G-I]:Let b, c, d, e, f, ρ, k be integers ≥
0; with Z ( b, c, d, e, f, ρ, k ) we denote a 0-dimensional subscheme of X,union of: b (cid:18) (cid:19) + c (1) + d (cid:18) (cid:19) + e (cid:0) (cid:1) over C, and f + T r general over P , with the following assumptions:( ) k b + c + 3 d + 3 e ≤ k , 0 ≤ d + e ≤ ) k b + c + 4 d + 5 e + 6 f ) + r = k ( k + 2) (i.e., length ( Z ( b, c, d, e, f, r, k )) = h ( E k +1 ))( ) k ρ = 0 or ρ = 8 if k is even; ρ = 3 or ρ = 11 if k is odd.In [ G-I] it is proved that Z ( b, c, d, e, f, ρ, k ) is k -settled for k ≥
12: see Definition 3.2 and proof of Proposition3.9 there.If k = 6 l + 5, i.e. if ρ = 11, we denote here by ¯ Z ( b, c, d, e, f, , k ) the scheme obtained by Z ( b, c, d, e, f, , k )substituting to R the scheme ¯ R . Lemma 3.2.3.
Let k = 6 l + 5 . The schemes ¯ Z ( b, c, d, e, f, , k ) are k-settled for k ≥ . Proof.
Use the proof of the fact that Z ( b, c, d, e, f, ρ, k ) is k -settled given in [ G-I], substituting to R thescheme ¯ R ; it is hence enough to prove the initial cases with this substitution. So it is enough to prove ananalogous of Lemma 5.2 in [ G-I], where 11 schemes of type Z ( b, c, d, e, f, ,
7) are proved to be 7-settled.We’ll do the same here substituting to R the scheme ¯ R ; so we want to prove that the schemes:¯ Z (0 , , , , , , Z (1 , , , , , , Z (1 , , , , , , Z (1 , , , , , , Z (2 , , , , , , Z (2 , , , , , , Z (2 , , , , , , Z (3 , , , , , , Z (4 , , , , , , Z (4 , , , , , , Z (1 , , , , , , (cid:0) ( 2 ) + (cid:18) (cid:19) (cid:1) over C + (cid:18) / (cid:19) or ( 5 ) over C + (cid:18) / (cid:19) (where (cid:18) / (cid:19) denotes the scheme R ). Now thesecond scheme specializes to the first one, since 5 points on a conic are general, so it is enough to prove thatthe first one is 1-settled.We do a step 3 →
1: we “add” over C (cid:18) / (cid:19) and the residue is ( 1 ) + ( 1 / / over C is just a point of P (Ω)). ⊓⊔ Lemma 3.2.4. If d ≤ k , X ( d, t, k ) is k-settled for any k ≥ and ¯ X ( d, t, k ) is k-settled for any k ≥ and k ≡ (mod 6). Proof.
Let u = u ( k ), ρ = ρ ( k ). In [ G-I] it is proved that H ( E k +1 ⊗ I Z ( b, , , ,f,ρ,k ) ) = 0 for k ≥
12, andLemma 3.2.3 says that for k = 6 l + 5, the schemes ¯ Z ( b, , , , f, , k ) are k-settled for k ≥
12. Now thescheme Z ( b, , , , f, ρ, k ) is union in P (Ω) of R ρ with the inverse image of f general triple points and of b double points whoose suports lie on a smooth conic, with b ≤ k (this is condition (0) k ), so that we concludeby semicontinuity that X ( d, t, k ) is k-settled for any k ≥
12. Analougously for ¯ X ( d, t, l + 5). ⊓⊔ Lemma 3.2.5. If k ≥ and k < d ≤ k + (cid:2) k − (cid:3) , or if k ≥ and k + (cid:2) k − (cid:3) < d < k − , then X ( d, t, k ) is k-settled and if k = 6 l + 5 ¯ X ( d, t, k ) is k -settled. Proof.
We first prove the statement about X = X ( d, t, k ). Let C be a smooth conic. We do an Horacestep. We specialize on C k double points; now on C there is a scheme of length 2 k , so that H ( E k +1 ⊗I X ∩ π − C,π − C ) = 0, hence h ( E k − ⊗ I Res π − C X ) = h ( E k +1 ⊗ I X ). The residual scheme Res π − C X is thegeneric union of ˜ Y (0 , d − k, t ) with the pull back of k points on C and with R ρ .Now we do an HD step; this time we need to have 2 k − P in total on C , so we still need k − case 1: d − k ≤ (cid:2) k − (cid:3) :we add on C ( d − k ) (cid:18) (cid:19) + g + h + i [1]23 where k + 2( d − k ) + 3 g + 2 h + i = 2 k − ≤ h + i ≤
1. The HD residue is the generic union of ˜ Y (0 , , t − g − h − i ) of R ρ and of ( d − k )(1) + g (cid:18) (cid:19) + h (cid:18) (cid:19) + i (cid:18) (cid:19) on C , which in the notation of [ G-I] is a generalization of Z ( g, d − k, h, i, t − g − h − i, ρ, k − k − and (2) k − are automatically verified, while the condition(0) k − is true for k ≥
4. Moreover, t − g − h − i ≥ k ≥ case 2: d − k > (cid:2) k − (cid:3) :we set k − m + l , l = 0 ,
1, and we add on
C m (cid:18) (cid:19) + l (cid:18) [1]2 (cid:19) where k + 2 m + l = 2 k −
4. The HDresidue the generic union of ˜ Y (0 , d − k − m − l, t ), of R ρ and of m (1) + l (2) on C .Now we do another HD step: we need 2 k − P in total on C , so we add on C ( d − k − m − l ) (cid:18) (cid:19) + g + h + i [1]23 where m +2 l +2( d − k − m − l )+3 g +2 h + i = 2 k − ≤ h + i ≤
1. The HD11esidue is the generic union of ˜ Y (0 , , t − g − h − i ), of R ρ and of ( d − k − m − l )(1) + g (cid:18) (cid:19) + h (cid:18) (cid:19) + i (cid:18) (cid:19) on C , which in the notation of [ G-I] is a generalization of Z ( g, d − k − m − l, h, i, t − g − h − i, ρ, k − k − and (2) k − are automatically verified, while (0) k − is true for k ≥
16. Moreover, it iseasy to check that t − g − h − i ≥ k ≥ Z ( b, c, d, e, f, r, k )are k -settled for k ≥ X ( d, t, l + 5) is proved exactly in the same way using Lemma 3.2.3 instead of [ G-I]. ⊓⊔ In the following Lemma 3.2.7 we treat the case of double and triple points with a lot of double points.Hence it is convenient to give some definitions.Let k, n be integers with 1 ≤ n ≤ k and let R be a 0-dimensional scheme in P (Ω) such that h ( E k +1 − n ⊗I R ) = 0. Let A be the union in P (Ω) of R with the inverse image of the union of (2 k −
4) + (2( k − − . . . + (2( k − n − −
4) = n (2 k − n + 2) double points in P supported at general points; then, Lemma3.1.4 applied n times gives h ( E k +1 ⊗ I A ) = 0 (the condition n ≤ k assures that k + 1 − n ≥
1, and alsothat each addend in the sum is positive, since n ≤ k +46 ).For all k ≥ ≤ n ≤ k +46 we set α ( n, k ) := n − X i =0 (cid:0) k − i ) − (cid:1) = n (2 k − n + 2) . If we fix a k ≥ α ( n, k ) is hence increasing as long as it is defined, and strictly increasing if n < k +46 .Now consider a scheme X ( d, t, k ) with k ≥
6; we can set¯ n = n ( d, k ) := max { n, ≤ n ≤ k , d ≥ α ( n, k ) } if d ≥ k − , ¯ n = n ( d, k ) := 0 if d < k − . Let d ≥ k − m be an integer, 1 ≤ m ≤ ¯ n ; we have seen above that H (0 , d, t, ρ, k ) is true if H (0 , d − α ( m, k ) , t, ρ, k − m ) is true. Moreover, the scheme X ( d − α ( m, k ) , t, k − m ) verifies α (¯ n + 1 − m, k − m ) > d − α ( m, k ) ≥ α (¯ n − m, k − m ) ( ∗ )so that n ( d − α ( m, k ) , k − m ) = n ( d, k ) − m ( ∗∗ ) . In fact, one has α (¯ n, k ) − α ( m, k ) = α (¯ n − m, k − m ), so that the second inequality is clear. For the firstone, there are two possibilities:i) α (¯ n + 1 , k ) is defined and > α (¯ n, k ), i.e. ¯ n + 1 < k +46 ; then by definition α (¯ n + 1 , k ) > d so that α (¯ n + 1 − m, k − m ) > d − α ( m, k ).ii) α (¯ n + 1 , k ) is not defined, i.e. ¯ n + 1 > k +46 , or α (¯ n + 1 , k ) = α (¯ n, k ), i.e. ¯ n + 1 = k +46 .Since ¯ n ≤ k we have: ¯ n = k − or ¯ n = k in the first case, ¯ n = k − in the second. We recall that6 d + 12 t + ρ ( k ) = k ( k + 2), hence d ≤ k ( k +2) − ρ = 2 u ( k ).If ¯ n = k − then k ≡ ρ = 8; hence, α (¯ n, k ) = ( k − k +4)6 = 2 u ( k ).12f ¯ n = k − then k ≡ ρ = 3; hence, α (¯ n, k ) = ( k − k +3)6 = 2 u ( k ).If ¯ n = k then k ≡ ρ = 0; hence, α (¯ n, k ) = k ( k +2)6 = 2 u ( k ). Since by definition d ≥ α (¯ n, k ), in each of the three cases we get d = α (¯ n, k ); hence the first inequality in ( ∗ ) becames α (¯ n + 1 − m, k − m ) > α (¯ n − m, k − m ) which is true (notice that ¯ n + 1 − m ≤ ¯ n so that α (¯ n + 1 − m, k − m )is defined). Lemma 3.2.7. If k ≥ then X ( d, t, k ) is k-settled and when k = 6 l + 5 also ¯ X ( d, t, k ) is k-settled. Proof.
We prove the statement about X ( d, t, k ), since the statement about ¯ X ( d, t, k ) can be proved exactlyin the same way.If d < k −
4, then X ( d, t, k ) is k -settled for k ≥
18 by Lemma 3.2.4 and Lemma 3.2.5, and for 12 ≤ k ≤ d ≥ k −
4, write k = 6 h + j , 0 ≤ j ≤
5, and let ¯ n = n ( d, k ). If ¯ n ≤ h −
3, then k − n ≥
18 + j , so that H (0 , d − α (¯ n, k ) , t, ρ, k − n ) is true by Lemma 3.2.4 and Lemma 3.2.5, since by ( ∗ ) d − α (¯ n, k ) < k − n ) − n ≥ h −
2, then we apply h − H (0 , d, t, ρ, k ) is true if H (0 , d − α ( h − , k ) , t, ρ,
12 + j ) is true. So we conclude by Lemma 3.3.1. Lemma 3.3.1. If k = 12 + j with ≤ j ≤ then X ( d, t, k ) is k-settled and when k = 17 also ¯ X ( d, t, k ) isk-settled. Proof. If d ≤ k the statement is true by Lemma 3.2.4. If k = 16 ,
17 and k < d ≤ k + (cid:2) k − (cid:3) , the statementis true by Lemma 3.2.5. We are hence going to prove that the union X ( d, t, k ) of d double and t triple pointsand of R ρ verifies H ( E k +1 ⊗ I X ( d,t,k ) ) = 0, where d + 2 t = k ( k +2) − ρ = 2 u ( k ) so that d is even, and d > k if 12 ≤ k ≤ d > k + (cid:2) k − (cid:3) if k = 16 ,
17. For k = 17 we are also going to prove that the same holdssubstituting to R the scheme ¯ R . When j = 0 , , , t = 0 is proved in [ I], section 2 so thatwe’ll assume t > k = 12 , , ,
16. Recall that at each step l → l −
2, which can be an Horace or adifferential Horace step, the divisor is the pull back of a smoth conic C and the points of P needed on C are 2 l . All the assertions about unions of double points plus a scheme R ρ are proved in [ I] section 2.Assume 0 ≤ t ≤ (cid:2) k (cid:3) . We define the following algorithm (A)(t,k) applying one standard step (see 3.2.1)to X ( d, t, k ), and then another standard step to the residue (in these assumptions the standard steps areparticularly simple): step k → k −
2: We “add” over
C t + b (cid:18) (cid:19) + a (cid:18) [1]2 (cid:19) where 3 t + 2 b + a = 2 k , a = 0 ,
1; sincethe condition 3 t ≤ k is verified by assumption, this is possible if d − a − b ≥ (cid:0) t (cid:18) (cid:19) + ( b + 2 a ) (cid:1) over C + ( d − a − b ) (cid:18) (cid:19) + R ρ . step k − → k −
4: We “add” over
C h (cid:18) (cid:19) + i (cid:18) [1]2 (cid:19) where 2 t + b + 2 a + 2 h + i = 2( k − i = 0 , t + b + 2 a ≤ k −
2) and d − a − b − h − i ≥ t + h + 2 i ) over C + ( d − a − b − h − i ) (cid:18) (cid:19) + R ρ . We set w = w ( t, k ) := t + h + 2 i , and q = q ( t, k ) := d − a − b − h − i ; we are reduced to prove that the scheme ( w ) over C + q (cid:18) (cid:19) + R ρ is( k − − settled .Assume t > (cid:2) k (cid:3) . We define the following algorithm (C)(t,k) applying one standard step to X ( d, t, k ) (againin these assumptions the standard step is very simple):step k → k − C g + n + p [1]23 where 3 g + 2 n + p = 2 k , n + p = 0 , (cid:0) g (cid:18) (cid:19) + n (cid:18) (cid:19) + p (cid:18) (cid:19) (cid:1) over C + ( t − g − n − p ) + d (cid:18) (cid:19) + R ρ .In the following we apply the standard step 3.2.1, (A)(t,k) and (C)(t,k) a number of times; easy calculationsassure that the conditions respectively 2 x + 3 y + 3 z + e ≤ k , v ≥ g + n + p , u ≥ r + s for the standard step,and 2 t + b + 2 a ≤ k − d − a − b − h − i ≥ case k = : here ρ = 0; we have to treat the cases 1 ≤ t ≤ ≤ t ≤ w, q ) = (10 ,
10) for t = 6 ,
7, ( w, q ) = (7 ,
11) for 2 ≤ t ≤
5, ( w, q ) = (4 ,
12) for t = 1.In all three cases we do a standard step 8 →
6, and the residue is ( 3 ) over C + 7 (cid:18) (cid:19) , which specializes to 8double points and it is 6-settled, or ( 6 ) over C + 6 (cid:18) (cid:19) . In the last case, we do another standard step 6 → over C + 3 (cid:18) (cid:19) , which specializes to 4 double points and it is 4-settled. case k = : here ρ = 8; we have to treat the cases 0 ≤ t ≤ t ≤ (cid:0) t (cid:18) (cid:19) +( b + 2 a ) (cid:1) over C +( d − a − b ) (cid:18) (cid:19) + R is 12-settled; recall that R is a scheme (cid:18) (cid:19) .step 12 →
10: We “add” over
C h ′ (cid:18) (cid:19) + i ′ (cid:18) [1]2 (cid:19) + (cid:18) (cid:19) where 2 t + b +2 a +2 h ′ + i ′ +3 = 2( k −
2) = 24, i ′ = 0 ,
1; this is possible since the conditions 2 t + b + 2 a + 3 ≤
24 and d − a − b − h ′ − i ′ ≥ t + h ′ + 2 i ′ + 1 ) over C + ( d − a − b − h ′ − i ′ ) (cid:18) (cid:19) . We set w ′ = w ′ ( t ) := t + h ′ + 2 i ′ + 1, and q ′ = q ′ ( t ) := d − a − b − h ′ − i ′ ; we are reduced to prove that the scheme ( w ′ ) over C + q ′ (cid:18) (cid:19) is 10 − settled ;an easy calculation shows that ( w ′ , q ′ ) = (12 ,
16) for 8 ≤ t ≤
9, ( w ′ , q ′ ) = (9 ,
17) for 4 ≤ t ≤ w ′ , q ′ ) = (6 ,
18) for 0 ≤ t ≤
3. These three configurations specialize to the first residual scheme of (A)(t,12)(i.e. (cid:0) t (cid:18) (cid:19) + ( b + 2 a ) (cid:1) over C + ( d − a − b ) (cid:18) (cid:19) ) obtained respectively in cases t = 1 , ,
4, and we haveproved that they are 10-settled.If t = 10 apply (C)(10,14); it is now enough to prove that (cid:0) (cid:18) (cid:19) + (cid:18) (cid:19) (cid:1) over C +16 (cid:18) (cid:19) + R is 12-settled.14tep 12 →
10: we “add” over C (cid:18) (cid:19) (which is the scheme R ) and the residue is ( 12 ) over C + 16 (cid:18) (cid:19) ,which is the case ( w ′ , q ′ ) = (12 ,
16) previously treated. case k = : here ρ = 0; we have to treat the cases 1 ≤ t ≤ ≤ t ≤
10 (A)(t,16) gives ( w, q ) = (12 ,
24) for 6 ≤ t ≤
10, ( w, q ) = (9 ,
25) for 2 ≤ t ≤
5, ( w, q ) =(6 ,
26) for t = 1. Now we do a standard step 12 →
10, and if w = 12, the residue is ( 6 ) over C + 18 (cid:18) (cid:19) ,while if w = 9 or w = 6, the residue is ( 9 ) over C + 17 (cid:18) (cid:19) , and these are the cases ( w ′ , q ′ ) = (9 ,
17) or(6 ,
18) previously treated in k = 14.For 11 ≤ t ≤
12 apply (C)(t,16); it is now enough to prove that (cid:0) (cid:18) (cid:19) + (cid:18) (cid:19) (cid:1) over C + ( t − + d (cid:18) (cid:19) is 14-settled. We now do two standard steps more, 14 →
12 and 12 →
10, and the last residue is inboth cases ( 6 ) over C + 18 (cid:18) (cid:19) , which we have just recalled is 10-settled. case k = : here ρ = 3; we have to treat the cases 1 ≤ t ≤ ≤ t ≤ w, q ) = (12 ,
12) for t = 8, ( w, q ) = (9 ,
13) for 4 ≤ t ≤
7, ( w, q ) = (6 ,
14) for1 ≤ t ≤ →
7, and if w = 12, the residue is ( 3 ) over C + 9 (cid:18) (cid:19) + R , whichspecializes to 10 double points + R and it is 7-settled.If w = 9 or w = 6, the residue is ( 6 ) over C + 8 (cid:18) (cid:19) + R ; we do another standard step 7 →
5, and theresidue is ( 4 ) over C + 4 (cid:18) (cid:19) + R , which specializes to 5 double points + R and it is 5-settled.For t = 9 apply (C)(9,13); it is now enough to prove that (cid:0) (cid:18) (cid:19) + (cid:18) (cid:19) (cid:1) over C + 14 (cid:18) (cid:19) + R is 11-settled.We do a standard step 11 →
9, and the residue is ( 12 ) over C + 12 (cid:18) (cid:19) + R , which is one of the previouscases. case k = : here ρ = 3; we have to treat the cases 1 ≤ t ≤ ≤ t ≤
10 (A)(t,15) gives ( w, q ) = (13 ,
19) for 8 ≤ t ≤
10, ( w, q ) = (10 ,
20) for 4 ≤ t ≤ w, q ) = (7 ,
21) for 1 ≤ t ≤
3. These three configurations specialize to the first residual scheme of (A)(t,13)(i.e. (cid:0) t (cid:18) (cid:19) + ( b + 2 a ) (cid:1) over C + ( d − a − b ) (cid:18) (cid:19) + R ) obtained respectively in cases t = 1 , ,
4, and wehave proved that they are 11-settled.For 11 ≤ t ≤
13 apply (C)(t,15); it is now enough to prove that (cid:0) (cid:18) (cid:19) (cid:1) over C +( t − + d (cid:18) (cid:19) + R is 13-settled. If 11 ≤ t ≤
12 , this scheme specializes to the scheme used in the first step of (A)(t,13) (i.e. (cid:0) t + b (cid:18) (cid:19) + a (cid:18) (cid:19) (cid:1) over C + ( d − a − b ) (cid:18) (cid:19) + R ) respectively in cases t = 1 (where b = 11, a = 1)and t = 2 (where b = 10, a = 0), and we have proved that they are 13-settled. If t = 13 we do three standard15teps from 13 to 7 and the last residue is ( 6 ) over C + 8 (cid:18) (cid:19) + R ; in k = 13 we have proved that it is7-settled. case k = : here ρ = 11; we have to treat the cases 0 ≤ t ≤ X ( d, t,
17) is 17-settled.For 0 ≤ t ≤
11 we have d ≥
30, and by Lemma 3.1.4 X ( d, t,
17) is 17-settled if X ( d − , t,
11) is 11-settled.If t = 11, d = 30, so that X ( d − , t,
11) is 11-settled by [ G-I].If 0 ≤ t ≤ X ( d − , t,
11) gives ( w, q ) = (8 ,
6) for 4 ≤ t ≤ w, q ) = (5 ,
7) for 0 ≤ t ≤
3. Now we do another standard step, and the residue is ( 3 ) over C + 3 (cid:18) (cid:19) + R ,resp. ( 6 ) over C + 2 (cid:18) (cid:19) + R .step 5 →
3: we “add” over
C r (cid:18) (cid:19) + s (cid:18) [1]2 (cid:19) + / with 0 ≤ s ≤ C ,and in both cases the residue is (cid:0) ( 2 ) + (cid:18) / (cid:19) (cid:1) over C + (cid:18) (cid:19) .step 3 →
1: we “add” over C (cid:18) (cid:19) and the residue is ( 1 ) + ( 1 / / over C is just a point of P (Ω)).For 8 ≤ t ≤
10, we do 3 standard step from 11 to 5, and the residue is ( 3 ) over C + 3 (cid:18) (cid:19) + R if t = 8 , (cid:0) ( 2 ) + (cid:18) (cid:19) (cid:1) over C + 2 (cid:18) (cid:19) + R if t = 10, in which case we proceed asfollows:step 5 →
3: we “add” over C / + (cid:18) (cid:19) and the residue is (cid:0) ( 2 ) + (cid:18) / (cid:19) (cid:1) over C + (cid:18) (cid:19) treatedabove.If 12 ≤ t ≤
14 , we do 3 standard steps 17 →
15, 15 →
13 and 13 →
11, and the residue is ( 6 ) over C +20 (cid:18) (cid:19) + R in all the three cases; we go on with 3 other standard steps 11 →
9, 9 → →
5, andthe residue is ( 6 ) over C + 2 (cid:18) (cid:19) + R , already treated above.Now we want to prove that ¯ X ( d, t,
17) is 17-settled, so we substitute to R the scheme ¯ R , and, if t = 11,we do the same steps from 17 to 5 as above; now we have to prove that the following schemes are 5-settled:( 6 ) over C + 2 (cid:18) (cid:19) + R + (cid:18) (cid:19) ;( 3 ) over C + 3 (cid:18) (cid:19) + R + (cid:18) (cid:19) ; (cid:0) ( 2 ) + (cid:18) (cid:19) (cid:1) over C + 2 (cid:18) (cid:19) + R + (cid:18) (cid:19) . We do a standard step 5 → over C + (cid:18) / (cid:19) + (cid:18) (cid:19) in the first case, ( 5 ) over C + (cid:18) / (cid:19) in the second and third case (here (cid:18) / (cid:19) denotes thescheme R ); in Lemma 3.2.3 we proved that both are 3-settled.Now let t = 11. By Lemma 3.1.4 it is enough to prove that ¯ X (0 , ,
11) is 11-settled. In order to prove this,we do 4 standard steps from 11 to 3 and the residue is (cid:0) ( 2 ) + (cid:18) (cid:19) (cid:1) over C + (cid:18) / (cid:19) which is proved to be3-settled in Lemma 3.2.3. ⊓⊔ orollary 3.3.2. H ( s, d, t, p, k ) is true for k ≥ proof. It follows by Lemma 3.1.5, Lemma 3.1.6 and Lemma 3.2.7. ⊓⊔ Proof of Theorem 1.1.
First let us check that in our assumptions Y ( a, b, c ) has maximal Hilbert function(mHf for short). Notice that Y ( a, b, c ) has mHf if and only if h ( I Y ( a,b,c ) ( v − h ( I Y ( a,b ) ( v )) = (cid:0) v +22 (cid:1) − l where l = l ( Y ( a, b, c )) = a + 3 b + 6 c and v = v ( Y ( a, b, c )) (see 2.1).By [ M], a general fat point scheme Z = m P + . . . + m r P r , 4 ≥ m ≥ . . . ≥ m r ≥
0, has mHf in anydegree k such that k ≥ m + m + m . Since for our schemes Y ( a, b, c ) one always has m + m + m ≤ Y ( a, b, c ) has mHf for v ≥
10, i.e. for l > k be the integer such that ( k − k + 1) < a + 6 b + 12 c ≤ k ( k + 2), so that w ( Y ( a, b, c )) = k .Rephrasing what is done in the proof of Lemma 2.6 for the analogous statements with c = 0, it is easy toshow thati) there exists ( s, d, t, p ) ∈ Λ k − such that Z ( s, d, t, p ) ⊆ ˜ Y ( a, b, c );ii) there exists ( s ′ , d ′ , t ′ , p ′ ) ∈ Λ k such that Z ( s ′ , d ′ , t ′ , p ′ ) ⊇ ˜ Y ( a, b, c ).If 2 l > ·
14, i.e. l >
84, then k ≥
13, hence ( s, d, t, p ) ∈ Λ k − , respectively ( s ′ , d ′ , t ′ , p ′ ) ∈ Λ k , impliesthat H ( s, d, t, p, k − H ( s ′ , d ′ , t ′ , p ′ , k ) is true, i.e. H ( E k ⊗ I Z ( s,d,t,p ) ) = 0, respectively H ( E k +1 ⊗ I Z ( s ′ ,d ′ ,t ′ ,p ′ ) ) = 0 (see Corollary 3.3.2 and 3.1.2).So we see, exactly as in the proof of Lemma 2.6, that µ k − ( Y ( a, b, c )) is injective and µ k ( Y ( a, b, c )) issurjective and we conclude that Y ( a, b, c ) is minimally generated (see 2.1). If k = 12 (i.e. 143 < l ≤ v = 12 (i.e. 78 < l ≤ Y ( a, b, c ) is minimally generated it is enough toprove that µ is surjective (see 2.1), and this is true again by ii) and Corollary 3.3.2; hence Y ( a, b, c ) hasthe expected resolution also for 79 ≤ l ≤ ⊓⊔ References [ A-H] J. Alexander and A. Hirschowitz,
An asymptotic vanishing theorem for generic unions of multiplepoints , Invent. Math. 140 (2000), no. 2, 303–325.[ B] E. Ballico,
Generators for the homogeneous ideal of s general points of P , J. Algebra 106 (1987),no. 1, 46–52.[ B-G] E. Ballico and A. V. Geramita, The minimal free resolution of the ideal of s general points in P , in: Proceedings of the 1984 Vancouver Conference in Algebraic Geometry, CMS Conference ProceedingsVol. 6, 1–11, Canadian Math. Soc. and Amer. Math. Soc, Providence, RI, 1986.[ Ca] M.V. Catalisano, ”Fat” points on a conic , Comm. Algebra 19 (1991), 2153–2168.[ C] K. A. Chandler, A brief proof of a maximal rank theorem for generic double points in projectivespace , Trans. Amer. Math. Soc. 353 (2000), no. 5, 1907–1920.[ C-C-M-O] C. Ciliberto, F. Cioffi, R. Miranda and F. Orecchia.
Bivariate Hermite interpolation andlinear systems of plane curves with base fat points , in: Computer mathematics, 87–102, Lecture Notes Ser.Comput., 10, World Sci. Publishing, River Edge, NJ, 2003.[ E] L. Evain,
Calculs de dimensions de syst`emes lin´eaires de courbes planes par collisions de gros points ,C.R.Acad.Sci.Paris, t.325, S´erie I (1997), 1305–1308.17 E2] L. Evain,
Computing limit linear series with infinitesimal methods , preprint 2004 (arXiv:math.AG/0407143).[ F] S. Fitchett,
Maps of linear systems on blow ups of the Projective Plane , J. Pure Appl. Algebra 156(2001), 1–14.[ F-H-H] S. Fitchett, B. Harbourne and S. Holay,
Resolutions of Fat Point Ideals Involving Eight GeneralPoints of P , J. Algebra 244 (2001), 684–705.[ G-I] A. Gimigliano and M. Id`a, The ideal resolution for generic -fat points in P , J. Pure Appl.Algebra 187 (2004), no. 1-3, 99–128.[ G-M] A. V. Geramita and P. Maroscia, The ideal of forms vanishing at a a finite set of points in P n ,J. Algebra 90 (1984), no. 2, 528–555.[ Ha1] B. Harbourne, The Ideal Generation Problem for Fat Points , J. Pure Appl. Alg. 145(2), 165–182(2000).[ Ha2] B. Harbourne,
Anticanonical rational surfaces , Trans. Amer. Math. Soc. 349, 1191–1208 (1997).[ Ha3] B. Harbourne,
An Algorithm for Fat Points on P , Can. J. Math. 52 (2000), 123–140.[ H-H-F] B. Harbourne, S. Holay and S. Fitchett, Resolutions of ideals of quasiuniform fat point sub-schemes of P , Trans. Amer. Math. Soc. 355 (2003), no. 2, 593–608.[ H-R] B. Harbourne and J. Ro´e. Linear systems with multiple base points in P , Adv. Geom. 4 (2004),41–59.[ Hi] A. Hirschowitz, La m´ethod d’Horace pour l’interpolation `a plusieurs variables , Manuscripta Math.50 (1985), 337–388.[ H-S] A.Hirschowitz - C.Simpson:
La r´esolution minimale de l’id´eal d’un arrangement g´en´eral d’ungrand nombre de points dans P n , Invent. Math. 126 (1996), 467–503.[ I] M. Id`a, The minimal free resolution for the first infinitesimal neighborhood of n general points inthe plane , J. Algebra 216 (1999), no. 2, 741–753.[ I2] M.Id`a: Generators for the generic rational space curve: low degree cases . ”Commutative Algebraand Algebraic Geometry, Proc. of the Ferrara Conference in honour of M.Fiorentini, 1997”, Lecture Notesin Pure and Applied Math. Dekker (1999), 169–210.[ M] T. Mignon,
Syst´emes de courbes plane `a singularit´es impos´ees: le cas des multiplicit´es inf´erieuresou ´egales `a quatre , J. Pure Appl. Algebra 151 (2000), no. 2, 173–195.[ N] M. Nagata,
On rational surfaces, II , Mem. Coll. Sci. Univ. Kyoto, Ser. A Math. 33 (1960), 271–293.[ R] J. Ro´e,
Limit linear systems and applications , e-print arXiv:math.AG/0602213.[ Y] S. Yang,