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

0.2 MB·

by  Giuseppe Favacchio

THE INTERPOLATION PROBLEM FOR A SET OF THREE FAT POINTS IN
P1×P1
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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.
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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≤n<t−1
t−1
ϕ (n)= ϕ (n)+1 if t−1≤n<2t−1
t  t−1
0 otherwise.
We will use the convention ϕt(n)=0 if t≤0.
Remark 3.1. One can inductively check that ϕ (n)= min{n, 2t−2−n} +1 . But, for our purposes,
t 2
+
we prefer the use of the recursive definition. (cid:16)j k (cid:17)
Let t,d ∈ Z be two integers such that t ≥ d, we define the function ϕ (n) : Z → Z in the following
t,d
way:
ϕ (n)=ϕ (n+d)−ϕ (n+d).
t,d t d
We give an example in order to clarify the notation.
Example 3.2. To shorten the notation, we represent the functions as tuples, where the first entry is
their value in 0, the second entry is their value in 1 and so on.
ϕ =(1,0,0,...)
1
ϕ =(1,1,1,0,0,...)
2
ϕ =(1,1,2,1,1,0,...)
3
ϕ =(1,1,2,2,2,1,1,0,...)
4
···
ϕ =(1,1,2,2,3,3,4,3,3,2,2,1,1,0,...)
7
ϕ −ϕ =(0,0,0,0,1,2,3,3,3,2,2,1,1,0,...)
7 4
ϕ =(1,2,3,3,3,2,2,1,1,0,...)
7,4
ϕ =(0,0,0,1,1,2,2,2,1,1,0,...)
4,−3
We have the following property.
4 GIUSEPPEFAVACCHIOANDELENAGUARDO
Proposition 3.3. Let t≥d be two integers, then:
ϕ (n)−1 if 0≤n≤d
t,d
ϕ (n−1)=
t,d+1
(ϕt,d(n) otherwise.
Proof. By definition, we have ϕ (n−1)=ϕ (n+d)−ϕ (n+d), hence
t,d+1 t d+1
ϕ (n+d)−ϕ (n+d) if 0≤n+d<d
t d
ϕ (n−1)= ϕ (n+d)−ϕ (n+d)−1 if d≤n+d<2d+1
t,d+1  t d
ϕt(n+d)−0 otherwise
ϕt(n+d)−ϕd(n+d) if 0≤n+d<d
=ϕt(n+d)−ϕd(n+d)−1 if d≤n+d<2d+1
ϕt(n+d)−0 otherwise
=ϕt(n+d)−ϕd(n+d)−1 if 0≤n≤d. (cid:3)
(ϕt(n+d)−ϕd(n+d) otherwise
4. The graded Betti numbers of I
Z
Let X be a set of fat points in P1 ×P1 and let I ⊆ R := k[P1 ×P1] be the bihomogeneous ideal
X
associated to X. Then we can associate to I a minimal bigraded free resolution of the form
X
0→ R(−i,−j)β2,(i,j)(X) → R(−i,−j)β1,(i,j)(X) → R(−i,−j)β0,(i,j)(X) →R→R/IX →0
M M M
where R(−i,j) is the free R-module obtained by shifting the degrees of R by (i,j). The graded Betti
number β (X) of R/I counts the number of a minimal set of generators of degree (i,j) in the u-th
u,(i,j) X
syzygy module of R/I .
X
Using the same strategy as in [FG] we split the description in two cases.
4.1. First case m ≤m .
11 21
Theorem 4.1. With the Notation 2.1, if m ≤ m then the bigraded Betti numbers of I are:
11 21 Z
β0,(a,b)(Z)=((0min{a,b−m11,m21−m11}+1)++ϕm12,m12−m11(b) iofthae+rwbis=e m21+m12
β1,(a,b)(Z)= β0,(a,b−1)(Z)+β0,(a−1,b)(Z)−1 +
β2,(a,b)(Z)=(cid:0)β0,(a−1,b−1)(Z)−1 + (cid:1)
(cid:0) (cid:1)
Proof. We proceed by induction on m . If m = 0 then ϕ (b) = 0 and statement is true by
11 11 m12,m12
Lemma 2.3. Assume m >0, 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 ) ≤b<m and a+b=max{m ,m +m }}
2 Z + Z 21 + 21 11 12 21
D (Z)={(a,b)∈N2 | m ≤b≤m +|B +m | and a+b=max{m ,m +m }
3 21 21 Z 21 11 12 21
D (Z)={(a,b)∈N2 | b>m +|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 b<m and a+b=2m
0,(a,b)  m
0 otherwise.
(cid:3)

5. The Hilbert Function of Z
In this section we explicitly compute the Hilbert function of the set of fat points Z. Recall that the
Hilbert function of Z is a numeric function H :=H :N2 →N, defined by
Z R/IZ
H (a,b)=dim (R/I ) =dim R −dim I (a,b).
Z K Z (a,b) k (a,b) k Z
8 GIUSEPPEFAVACCHIOANDELENAGUARDO
The first difference of the Hilbert function is defined as
∆H (a,b)=H (a,b)+H (a−1,b−1)−H (a−1,b)−H (a,b−1).
Z Z Z Z Z
From now on we will write (i,j)≤(a,b) iff both i≤a and j ≤b.
The following are the multigraded version of well known results for standard graded algebras. Next
lemma shows how we can compute the Hilbert function of Z from a minimal free resolution of R/I .
Z
Lemma 5.1. Let Z be a set of fat points in P1×P1, then
H (a,b)=(a+1)(b+1)− (a−i+1)(b−j+1)(β (Z)−β (Z)+β (Z)).
Z 0,(i,j) 1,(i,j) 2,(i,j)
(i,jX)≤(a,b)
Proof. Since a minimal free resolution of R/I
Z
0→ R(−i,−j)β2,(i,j)(Z) → R(−i,−j)β1,(i,j)(Z) → R(−i,−j)β0,(i,j)(Z) →R→R/IZ →0
has bigrMaded morphisms, we get thMe following exact sequencMe of vector spaces
0→ R(−i,−j)β2,(i,j)(Z) → R(−i,−j)β1,(i,j)(Z) → R(−i,−j)β0,(i,j)(Z) →(R)(a,b) →(R/I)(a,b) →0
(a,b) (a,b) (a,b)
M(cid:16) (cid:17) M(cid:16) (cid:17) M(cid:16) (cid:17)
Moreover
dimk R(−i,−j)βu,(i,j)(Z) = dimk R(−i,−j)βu,(i,j)(Z) = βu,(i,j)(Z)(a−i+1)(b−j+1).
(a,b) (a,b)
(cid:16)M (cid:17) (i,jX)≤(a,b) (cid:16) (cid:17) (i,jX)≤(a,b)
(cid:3)
Corollary 5.2. Let B := β (Z) then
u,(a,b) (i,j)≤(a,b) u,(a,b)
∆PHZ(a,b)=1−B0,(a,b)+B1,(a,b)−B2,(a,b)
Proof. This follows from Lemma 5.1. (cid:3)
In the following two propositions we explicitly compute the first difference of the Hilbert Function of
Z using Corollary 5.2.
Proposition 5.3. Let Z =m P +m P +m P . If m ≤m then
11 11 12 12 21 21 11 21
1 if a+b<m +m
12 21
∆H (a,b)= 1−β (Z) if a+b=m +m
Z  0,(a,b) 12 21
0 if a+b>m12+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)
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