THE INTERPOLATION PROBLEM FOR A SET OF THREE FAT POINTS IN P1×P1 7 1 GIUSEPPEFAVACCHIOANDELENAGUARDO 0 2 n Abstract. ThisisanappendixtotherecentpaperofFavacchioandGuardo. Inthesenoteswedescribe a explicitly a minimal bigraded free resolution and the bigraded Hilbert function of a set of 3 fat points J whosesupportisanalmostcompleteintersection(ACI)inP1×P1.Thissolvetheinterpolationproblem 3 forthreepointswithanACIsupport. 1 ] G A 1. Introduction . h The interpolation problem is an interesting problem which involve many different fields of research t a included algebraic geometry and commutative algebra. It asks to determine the dimension of the vector m space of homogeneous polynomials of each given degree that vanish on a set of points under some con- [ dition. Let R:=k[Pn]=k[x0,...,xn] be the standard polynomial ring over a infinite field, consider the homogeneous ideal 1 s v I := Imi 7 Z Pi 7 i=1 \ 6 of R, where I is the homogeneous ideal associated to P and m are non negative integers. The Pi i i 3 interpolation problem then asks what can be the Hilbert function of R/I , where the Hilbert function Z 0 . H (t):=dim (R/I ) , for all t>0 1 Z k Z t 0 computes the dimension of the homogeneous component (R/I ) of degree t of R/I for all t ∈ N. 7 Z t Z Geramita-Maroscia-Roberts [GeMaRo] and Geramita-Gregory-Roberts [GeGrRo] give an answer for set 1 : of reduced points, i.e. mi = 1 for all i. When some of the multiplicities are greater than one, the v interpolation problem remains open. i X AvariationoftheinterpolationproblemwasintroducedbyGiuffrida-Maggioni-Ragusa[GiMR]chang- r ing the ambient space from a single projective space Pn to Pn1 ×···×Pnr a multiprojective space. One a ofthemaindifferenceisthatthecoordinateringofacollectionofpointsinamultiprojectivespaceisnot always Cohen-Macaulay. The multigraded version of the interpolation problem is still open for reduced set of points in P1×P1. A partialanswer was given in P1×P1, was givenby Giuffrida-Maggioni-Ragusa [GiMR], for arithmetically Cohen-Macaulay reduced sets of points, and by Guardo-Van Tuyl [GVT] for arithmetically Cohen-Macaulay sets of fat points. Recently, the first author (see [F]) gave a description of the Hilbert functions for bigraded algebras in k[P1×P1]. Without the hypothesis of Cohen-Macaulayness,we are still far away from a complete understanding of the problem and any result in this direction could be interesting. In [FG] the authors recursively computed the minimal free resolution of a (non homogeneous) set Z of fat points whose support is an 2000 Mathematics Subject Classification. 13F20,13A15,13D40,14M05. Keywords and phrases. interpolationproblem;multiprojectivespaces;Hilbertfunctions;fatpoints. January 13,2017. 1 2 GIUSEPPEFAVACCHIOANDELENAGUARDO almost complete intersection (ACI), that in general are not arithmetically Cohen-Macaulay(ACM) even if the support is ACM. The aim of this work is to describe the Hilbert function, and the bigraded Betti numbers, for a set Z ofthreefatpointsinP1×P1 onanACIsupportX i.e. I isgeneratedbythreebihomogeneousforms. In X Section2werecallsomeresultsandnotationfrom[FG]. InSection3weintroduceandstudyanumerical function which will be related with the homologicalinvariantof such sets of points. Finally, in Section 4 first we give a formula to compute the graded Betti numbers and the Hilbert Function of a set of three fatpointsonanACIsupport,thenwegiveananswertotheinterpolationproblemforthosesetofpoints. 2. Notation and preliminary result Throughout this paper R := k[x ,x ,x ,x ] is the coordinate ring of P1×P1 over an infinite field of 0 1 2 3 characteristic 0, with the bigrading given by degx = degx = (1,0) and degx = degx = (0,1). Let 0 1 2 3 H be horizontal lines of type (1,0) and V vertical lines of type (0,1), then points in P1 ×P1 can be i j denoted by P :=H ×V . With an abuse of notation we denote the ideal I =(H ,V ). ij i j Pij i j Given a set of distinct points X and positive integers m , we call Z = m P a set of fat ij Pij∈X ij ij points supported at X. The associated ideal to Z is I := Imij. Z Pij∈X Pij P In these note we use the following notation as in [FG]. T Notation 2.1. We set Z :=m P +m P +m P , where m ≥0 and, without loss of generality, 11 11 12 12 21 21 ij m ≥m . We denote by Z =(m −1) P +m P +(m −1) P , where (n) :=max{n,0}. 12 21 1 11 + 11 12 12 21 + 21 + The following results were provenin[FG]in a more generalsetting. For the convenienceofreader,we recall them in a version which is useful to our focus. Lemma 2.2 (Lemma 2.2, [FG]). Let Z = m P +m P be a set of two collinear fat points. Set 11 11 12 12 M :=max{m ,m }, then a minimal free resolution of I is 11 12 Z M M 0→ R(−t,−(m11−t+1)+−(m12−t+1)+)→ R(t,−(m11−t)+−(m12−t)+)→IZ →0 Mt=1 Mt=0 Lemma 2.3 (Lemma 3.4, [FG]). Let Z := m P +m P be a set of two non collinear fat points. 12 12 21 21 Then a minimal free resolution of I is Z 0→ R(−a−b,−c−d)→ R(−a−b,−c−d)→ (a,b,Mc,d)∈D2 (a,b,Mc,d)∈D1 → R(−a−b,−c−d)→I →0 Z (a,b,Mc,d)∈D0 where: D :={(a,b,c,d)| 0≤a,d≤m , 0≤b,c≤m , a+d=m , b+c=m } 0 12 21 12 21 D :={(a,b,c,d)| 0≤a,d≤m , 0≤b,c≤m , (a+d=m +1, b+c=m )∨(a+d =m , b+c= 1 12 21 12 21 12 m +1)} 21 D :={(a,b,c,d)| 0≤a,d≤m , 0≤b,c≤m , a+d=m +1, b+c=m +1}. 2 12 21 12 21 These two lemmas describe the resolutionof Z in the degeneratingcase when one of the multiplicities is 0. The next result allows to recursively compute the resolution of Z in the remaining cases. Lemma 2.4 (Remark 2.10,Theorem 2.12, [FG]). Using Notation 2.1, let 0 → L → L → L be a 2 1 0 minimal free resolution of I , then a minimal free resolution for I is Z1 Z 0→ R(−a,−b)⊕L2(0,−1)→ (a,b)M∈A2(Z) THE INTERPOLATION PROBLEM FOR A SET OF THREE FAT POINTS IN P1×P1 3 → R(−a,−b)2⊕R(−m11−m21,−(m12−m11)+−1)⊕L1(0,−1)→ (a,b)M∈A1(Z) (2.1) → R(−a,−b)⊕L0(0,−1)→IZ →0 (a,b)M∈A0(Z) A0(Z):={(a,b) | a+b=m11+m21+(m12−m11)+ and 0≤b≤(m12−m11)+} where A1(Z):={(a,b) | a+b=1+m11+m21+(m12−m11)+ and 1≤b≤(m12−m11)+} A2(Z):={(a,b) | a+b=2+m11+m21+(m12−m11)+ and 2≤b≤(m12−m11)++1}. In Section 4 we give a closed formula to compute the bigraded Betti numbers of Z. 3. Numerical facts In order to determine the graded Betti numbers of Z we need to introduce a numerical function depending on a parameter t∈Z. We define inductively the function ϕ :Z→Z as follows: t 1 if n=0 ϕ (n)= 1 (0 otherwise and, for t>1 ϕ (n) if 0≤n0, by Lemma 2.4 we get 11 β (Z )+1 if a+b=m +m and b≤m −m 0,(a,b−1) 1 12 21 12 11 β (Z)= β (Z ) if a+b=m +m and b>m −m 0,(a,b)  0,(a,b−1) 1 12 21 12 11 0 otherwise Thus, set S :=m +m and B :=m −m , we have 12 21 12 11 (min{a,b−m11,m21−m11}+1)++ϕm12,B+1(b−1)+1 ifa+b=S andb≤B β0,(a,b)(Z)=(min{a,b−m11,m21−m11}+1)++ϕm12,B+1(b−1) ifa+b=S andb>B 0 otherwise  THE INTERPOLATION PROBLEM FOR A SET OF THREE FAT POINTS IN P1×P1 5 and by using Proposition 3.3 it is (min{a,b−m11,m21−m11}+1)++ϕm12,B(b) ifa+b=S andb≤B β0,(a,b)(Z)=(min{a,b−m11,m21−m11}+1)++ϕm12,B(b) ifa+b=S andb>B 0 otherwise.  The computation of β (Z) also follows by induction and Lemma 2.4. 1,(a,b) β (Z )+2 if a+b−1=m +m and 1≤b≤m −m 1,(a,b−1) 1 21 12 12 11 β (Z)= β (Z)+1 if (a,b)=(m +m ,m −m +1) 1,(a,b)  1,(a,b) 11 21 12 11 0 otherwise β (Z )+β (Z )+1 if a+b−1=m +m and 1≤b≤m −m 0,(a,b−2)1 0,(a−1,b−1) 1 21 12 12 11 = β (Z )+β (Z ) if (a,b)=(m −m ,m −m +1)  0,(a,b−2) 1 0,(a−1,b−1) 1 11 21 12 11 0 otherwise β (Z)+β (Z)−1 if a+b−1=m +m and 1≤b≤m −m  0,(a,b−1) 0,(a−1,b) 21 12 12 11 = β (Z)+β (Z)−1 if (a,b)=(m +m ,m −m +1)  0,(a,b−1) 0,(a−1,b) 11 21 12 11 0 otherwise Thecomputationofβ2,(a,b)(Z)requiresthesameprocedureasabovebyusingtheinductivehypotheses and Lemma 2.4. (cid:3) We show in the following example how to compute the graded Betti numbers of a set of three fat points using Theorem 4.1. Example 4.2. Consider Z = 2P +5P +4P , to compute β (Z) we first need to compute the 11 12 21 0,(a,b) bigraded Betti numbers of Z′ :=(2−2)P +5P +(4−2)P =5P +2P . By Lemma 2.3, the non 11 12 21 12 21 zero bigraded Betti numbers of R/IZ′ are: β0,(7,0)(IZ′)=1 β1,(7,1)(IZ′)=2 β2,(7,2)(IZ′)=1 β0,(6,1)(IZ′)=2 β1,(6,2)(IZ′)=4 β2,(6,3)(IZ′)=2 β0,(5,2)(IZ′)=3 β1,(5,3)(IZ′)=5 β2,(5,4)(IZ′)=2 β0,(4,3)(IZ′)=3 β1,(4,4)(IZ′)=5 β2,(4,5)(IZ′)=2 β0,(3,4)(IZ′)=3 β1,(3,5)(IZ′)=5 β2,(3,6)(IZ′)=2 β0,(2,5)(IZ′)=3 β1,(2,6)(IZ′)=4 β2,(2,7)(IZ′)=1 β0,(1,6)(IZ′)=2 β1,(1,7)(IZ′)=2 β0,(0,7)(IZ′)=1 Moreover we have ϕ =(1,2,2,2,1,1,0...). Hence, if a+b=9 we have 5,3 β (Z)=β (Z′)+ϕ (b). 0,(a,b) 0,(a,b−2) 5,3 Then the non zero bigraded Betti numbers of R/I are: Z β (Z)=1 β (Z)=2 β (Z)=1 0,(9,0) 1,(9,1) 2,(9,2) β (Z)=2 β (Z)=4 β (Z)=2 0,(8,1) 1,(8,2) 2,(8,3) β (Z)=3 β (Z)=6 β (Z)=3 0,(7,2) 1,(7,3) 2,(7,4) β (Z)=4 β (Z)=7 β (Z)=3 0,(6,3) 1,(6,4) 2,(6,5) β (Z)=4 β (Z)=7 β (Z)=3 0,(5,4) 1,(5,5) 2,(5,6) β (Z)=4 β (Z)=6 β (Z)=2 0,(4,5) 1,(4,6) 2,(4,7) β (Z)=3 β (Z)=5 β (Z)=2 0,(3,6) 1,(3,7) 2,(3,8) β (Z)=3 β (Z)=4 β (Z)=1 0,(2,7) 1,(2,8) 2,(2,9) β (Z)=2 β (Z)=2 0,(1,8) 1,(1,9) β (Z)=1. 0,(0,9) 6 GIUSEPPEFAVACCHIOANDELENAGUARDO 4.2. Second Case m > m . To conclude the description of the graded Betti numbers of R/I we 11 21 Z need some preliminaries. Definition 4.3. Let Z = m P + m P + m P be a set of three fat points in P1 × P1, set 11 11 12 12 21 21 B :=m −m , we define the following sets of integers associated to Z: Z 12 11 D (Z)={(a,b)∈N2 | 0≤b<(−B ) −(−B −m ) and a+2b=m +m } 1 Z + Z 21 + 11 21 D (Z)={(a,b)∈N2 | (−B ) −(−B −m ) ≤bm +|B +m | and 2a+b=m +m }. 4 21 Z 21 11 12 Some immediate remarks follows from Definition 4.3. We recall that Z is an arithmetically Cohen- Macaulay (ACM for short) set of points if R/I is Cohen Macaulay. See [GVT] for more details about Z ACM set of fat points. Remark 4.4. It is a matter of computation to show that i) B =B +1; Z1 Z ii) D (Z)=∅ if and only if B ≥0; 1 Z iii) D (Z)=∅ if and only if B+m ≤1 (in this case Z is ACM by Theorem 6.21 [GVT] ); 2 21 iv) If B <0 and B+m =1 then (a,−B)∈/ D (Z) for any a; 21 3 v) If (a,b−1)∈D (Z ) then (a,b)∈D (Z), for i=1,2,3,4; i 1 i vi) If (a,b)∈D (Z) for some i then (a,¯b),(a¯,b)∈/ D for any a¯6=a, ¯b6=b and j =1,2,3,4. i j vii) If (a−1,b)∈D (Z) then (a,b−1)∈/ D (Z). 4 3 Theorem 4.5. If m >m then the bigraded Betti numbers of R/I are: 11 21 Z 1 if (a,b)∈D1(Z) ϕm21+B,B(b) if (a,b)∈D2(Z) β0,(a,b)(Z)=11+ϕm21+B,B(b) iiff ((aa,,bb))∈∈DD34((ZZ)) 0 otherwise 1 if (a,b−1)∈D1(Z) β1,(a,b)(Z)=β10,(a,b−1)(Z)+β0,(a−1,b)(Z)−1 iiff ((aa,−b−1,1b))∈∈DD42((ZZ))∪D3(Z) 0 otherwise  β2,(a,b)(Z)=(β00,(a−1,b−1)(Z)−1 oifth(ear−wi1s,eb−1)∈D2(Z)∪D3(Z) where the D (Z) are defined in Definition 4.3. i Proof. Weproceedbyinductiononm .Ifm =0thenZ isasetof2collinear(fat)points,D =D =∅ 21 21 1 2 and ϕ (b) = 0. Therefore the statement follows by Lemma 2.2. Assume now m >0, by Lemma 2.4 B,B 21 we have β (Z )+1 if a+b=m +m +(m −m ) 0,(a,b−1) 1 11 21 12 11 + β (Z)= and b≤(m −m ) 0,(a,b)  12 11 + β0,(a,b−1)(Z1) otherwise. If m ≤m , i.e. B <0 we get 12 11  Z 1 if (a,b)=(m +m ,0) 11 21 β (Z)= 0,(a,b) (β0,(a,b−1)(Z1) otherwise. THE INTERPOLATION PROBLEM FOR A SET OF THREE FAT POINTS IN P1×P1 7 So, by the inductive hypothesis and using Remark 4.4, we have 1 if (a,b)=(m +m ,0) 11 21 1 if (a,b−1)∈D (Z )  1 1 β0,(a,b)(Z)=1ϕm+21ϕ+mB2Z1+,BBZZ+,B1(Zb+−1(1b)−1) iiff ((aa,,bb−−11))∈∈DD32((ZZ11)) 1 if (a−1,b)∈D (Z ) 4 1 By using Propos0ition 2.4 we are done. Consoidtherernwoiwse.m12 >m11 i.e. BZ >0. We get β (Z )+1 if a+b=m +m and b≤B β (Z)= 0,(a,b−1) 1 21 12 0,(a,b) (β0,(a,b−1)(Z1) otherwise where, by inductive hypothesis, the bigraded Betti numbers of degree zero of R/I are Z1 ϕm21+BZ,BZ+1(b−1) if(a,b−1)∈D2(Z1) β0,(a,b−1)(Z1)=11+ϕm21+BZ,BZ+1(b−1) iiff((aa,−b−1,1b))∈∈DD34((ZZ11)) 0 otherwise.  Therefore, by using Lemma 2.4 and Remark 4.4, we are done. Finally the computation of β (Z) 1,(a,b) and β (Z) requires the same procedure as above by using the inductive hypothesis and Lemma 2,(a,b) 2.4. (cid:3) When the points have the same multiplicity Z is called a homogeneous set of fat points. The graded Bettinumbersfor ahomogeneoussetofpoints Z aregiveninTheorem4.1. Inthis case,itbecame easier to write them, as the next corollary shows. Corollary 4.6. Let Z = mP +mP +mP be a homogeneous set of points in P1 ×P1. Then the 11 12 21 bigraded Betti numbers of I are: Z ϕ (b) if a+b=2m m+1 β (Z)= 0,(a,b) (0 otherwise β (Z)= β (Z)+β (Z)−1 1,(a,b) 0,(a,b−1) 0,(a−1,b) + β (Z)= β (Z)−1 2,(a,b) (cid:0) 0,(a−1,b−1) + (cid:1) (cid:0) (cid:1) Proof. The proof is an immediate consequence of Theorem 4.1 and Proposition 3.3 since we have 1+ϕ (b) if b≥m and a+b=2m m β (Z)= ϕ (b) if bm12+m21. Proof. By Theorem 4.1, we have β (Z)=β (Z)+β (Z)−1 for i+j =m +m +1 1,(i,j) 0,(i−1,j) 0,(i,j−1) 12 21 and zero elsewhere, and β (Z)=β (Z)−1 if and only if i+j =m +m +2 (otherwise 2,(i,j) 0,(i−1,j−1) 12 21 zero). So, by applying Corollary 5.2 and a machinery computation we are done. (cid:3) Proposition 5.4. Let Z =m P +m P +m P . If m >m then 11 11 12 12 21 21 11 21 1 if (a,b)<(i,j) for some (i,j)∈∪D (Z) i ∆H (a,b) 1−β (Z) if (a,b)∈∪D (Z) Z  0,(a,b) i 0 otherwise. Proof. The proof use the same argument as in Proposition 5.3. (cid:3) In the last part of these notes we give a criterion to say if an admissible function, H : N2 → N, as introduced in [GiMR] Definition 2.2, is the Hilbert function of a set of, at most, three (fat) points on an ACI support. THE INTERPOLATION PROBLEM FOR A SET OF THREE FAT POINTS IN P1×P1 9 Theorem 5.5. Let H : N2 → N be an admissible function, and let H(i,j) = γ for (i,j) ≫ (0,0). We denote by h :=∆H(i,j), moreover we set ij d−1 d−1 A(d) := h , B(d) := h i ij j ij j=0 i=0 X X and α:=max{i | A(d) 6=0}+1, and β :=max{j | B(d) 6=0}+1. Then we have the following cases: i j Case 1) There exist d,d ,d ∈ N such that if i < d and j < d then h = 1 iff i+j < d. If H is the 1 2 1 2 ij Hilbert function of a set of points Z := m P +m P +m P then (m ,m ,m ) is the solution 11 11 12 12 21 21 11 12 21 of one of the following systems: x+y =α x+y =α z =α z =α x+z =β y =β x+z =β y =β     y+z =d y+z =d. y+z =d x+1 + y+1 + z+1 =γ.     2 2 2 Case 2)Assume the first case does not occur.Then H is not the H(cid:0)ilbe(cid:1)rt fu(cid:0)ncti(cid:1)on (cid:0)of an(cid:1)y set of points m P +m P +m P . 11 11 12 12 21 21 Proof. Case (1). The condition h = 1 iff i+j < d (i < d , j < d ) is always verified for sets of three ij 1 2 points on an ACI support (see Theorem 4.1 and Theorem 4.5). In both cases we have m +m = d. 12 21 Moreover, from Theorem 2.12 in [GiMR], α and β respectively count the maximum number of point on a line of type (0,1) and (1,0) that are respectively max{m +m ,m } and max{m +m ,m 1}. 11 21 12 11 12 2 These conditions give arise to four linear systems: x+y =α x+y =α z =α z =α x+z =β y =β x+z =β y =β     y+z =d y+z =d. y+z =d y+z =d.     Butthelastsystemisnotdetermined,soweneedtoreplaceoneequationwith x+1 + y+1 + z+1 =γ, 2 2 2 that is the degree of a set of three fat points. (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) Moreover,from Proposition 5.3 and Proposition 5.4 we can see that case (2) does not lead to any set of at most three fat points on an ACI support. (cid:3) GivenanadmissiblenumericalfunctionH,Theorem5.5allowsustofindthemultiplicitiesofthethree points candidate to have as Hilbert function H. Supposed to be in Case (1), we construct sets of points with multiplicity as found by solving the systems. Then, by using Theorem 4.1 and Proposition 5.3 or 4.5 and Proposition 5.4, we compute the Hilbert function of these points and hence we compare them with H. The next example shows this procedure. Example 5.6. Let H :N2 →N be a numerical function such that 0 1 2 3 4 5 6 7 ··· 0 1 1 1 1 1 1 1 0 ··· 1 1 1 1 1 1 0 0 0 ··· 2 1 1 1 1 0 0 0 0 ··· 3 1 1 1 −1 0 0 0 0 ··· ∆H = 4 1 1 −2 0 0 0 0 0 ··· 5 1 −1 0 0 0 0 0 0 ··· 6 0 0 0 0 0 0 0 0 ··· ... ... ... ... ... ... ... ... ... ... 10 GIUSEPPEFAVACCHIOANDELENAGUARDO Note that we are in Case (1) of Theorem 5.5, in particular we have d=6. Thus γ = ∆H(i,j)=18 α=4 and β =7. So we get the following systems to solve P x+y =4 x+y =4 z =4 z =4 (i) x+z =7 (ii) y =7 (iii) x+z =7 (iv) y =7     y+z =6 y+z =6. y+z =6 x2+x+18=0.     Note that (i),(ii),(iv) have not solution in N3, i.e. H is the Hilbert function of a set of fat points     Z = m P +m P +m P if and only if (m ,m ,m ) = (3,2,4), that is the solution of (iii). 11 11 12 12 21 21 11 12 21 But from Proposition 5.4 and Theorem 4.5 we have that Z =3P +2P +4P has the first difference 11 12 21 of the Hilbert function equal to 0 1 2 3 4 5 6 7 ··· 0 1 1 1 1 1 1 1 0 ··· 1 1 1 1 1 1 0 0 0 ··· 2 1 1 1 1 0 0 0 0 ··· 3 1 1 1 −1 0 0 0 0 ··· ∆H = Z 4 1 1 −1 0 0 0 0 0 ··· 5 1 −1 0 0 0 0 0 0 ··· 6 0 0 0 0 0 0 0 0 ··· ... ... ... ... ... ... ... ... ... ... so H 6= H and hence H is not the Hilbert function of any set of at most three fat points on an ACI Z support. References [F] G.Favacchio, The Hilbert function of bigraded algebras in k[P1×P1].Preprintarxiv.org/pdf/1609.06950v2.pdf [FG] G. Favacchio, E. Guardo The minimal free resolution of fat almost complete intersections in P1 ×P1. Canadian Journal ofMathematics.PublishedelectronicallyonDecember 23,2016.http://dx.doi.org/10.4153/CJM-2016-040-4 [GeGrRo] A.V. Geramita, D. Gregory, L. Roberts,Monomial ideals and points in projective space. J. Pure Appl. Algebra 40(1986), no.1,33–62. [GeMaRo] A.V.Geramita, P. Maroscia, L.G. Roberts, The Hilbert function of a reduced k-algebra. J. London Math. Soc. (2)28(1983), no.3,443–452. [GiMR] S.Giuffrida,R.Maggioni,A.Ragusa,Onthepostulationof0-dimensionalsubschemesonasmoothquadric.Pacific J.Math.155(1992), no.2,251–282. [GVT] E. Guardo, A. Van Tuyl, Aritmetically Cohen-Macaulay Sets of points in P1×P1. SpringerBriefs inMathematics (2015) Dipartimentodi Matematica e Informatica, Viale A. Doria, 6 - 95100 - Catania,Italy E-mail address: favacchio@dmi.unict.it URL:/ Dipartimentodi Matematica e Informatica, Viale A. Doria, 6 - 95100 - Catania,Italy E-mail address: guardo@dmi.unict.it URL:http://www.dmi.unict.it/∼guardo/

The interpolation problem for a set of three fat points in $\mathbb{P}^1\times\mathbb{P}^1$

Similar The interpolation problem for a set of three fat points in $\mathbb{P}^1\times\mathbb{P}^1$