ARTICLE IN JUNE 2014

The polynomial interpolation problem on the plane is to determine if general multiple points impose independent conditions on plane algebraic curves with given restrictions.. This restri[r]

GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF

PRESCRIBED MONOMIALS IN TWO VARIABLES

DARREN J FINNIGAN, ALAA HAJ ALI, KYUNGYONG LEE, CHRIS M LOCRICCHIO, THE MINH

TRAN, RAFAL URBANIUK

Abstract The polynomial interpolation problem on the plane is to determine if general

multiple points impose independent conditions on plane algebraic curves with given

restric-tion(s) This restriction is classically given by a degree In this paper we use a different

restriction, namely we fix a set of monomials and consider their linear combinations We

study whether singular points in general position impose independent conditions on linear

combinations of monomials satisfying certain parity properties.

1 Introduction Let K be an infinite field It is an important problem to determine if general multiple points on P2

K impose independent conditions on plane curves of a given degree This problem has a very long history (for a very small fraction of references, see [1, 2, 3, 4, 5, 6, 8, 9])

In this paper we continue the characteristic 2 reduction approach following the spirit of [7] The following is

Before we state the main theorem, we need some notation For integers i, j, a, b, we say that (i, j) ≡ (a, b)(mod 2) if i ≡ a(mod 2) and j ≡ b(mod 2) For a, b ∈ {0, 1}, let

Ma,b:= {(i, j) ∈ (Z≥0)2 | (i, j) ≡ (a, b)(mod 2)}

For L ⊂ (Z≥0)2, define L1 = M0,0∩ L, L2 = M0,1 ∩ L, L3 = M1,0 ∩ L and L4 = M1,1 ∩ L Also, define xL= {xiyj | (i, j) ∈ L} The following is our main theorem

Theorem 1.1 Fix a positive integer n Let L ⊂ (Z≥0)2 be a set of lattice points with

|L| ≤ 3n Suppose that |Li| ≤ n for all 1 ≤ i ≤ 4 Then no curve in Span(xL) passes through n general points with multiplicity≥ 2

Applying this to plane curves with a given degree, we obtain several corollaries

Corollary 1.2 Fix positive integers d ≥ 2 and n Let n0 = n + d2d+5

3 e + 1, and let (mi)1≤i≤n0

be a sequence consisting of positive integers with mi = 2 for i ∈ {n + 1, , n0} Assume that

no curve of degree d passes through general points p1, , pn with multiplicity≥ mi at pi Then

no curve of degree d + 2 passes through general points p1, , pn0 with multiplicity≥ mi at pi Corollary 1.3 Let t, d and n be integers with t ≥ 3 and d ≥ t + 5 If (d + 1)(d + 2)/2 ≤ 3n + t(t + 1)/2 then there is no polynomial of degree d that passes through n + 1 general points with multiplicity ≥ 2 at the first n points and with multiplicity≥ t at the last point

1

As another corollary, we give a new proof of the following theorem which is obtained more than a century ago by Campbell [4], Palatini [8] and Terracini [9]

Corollary 1.4 Fix two positive integers n and d with (n, d) 6= (2, 2) and (n, d) 6= (5, 4) Then (d + 1)(d + 2)/2 ≤ 3n if and only if no polynomial of degree d vanishes at n general points with multiplicity ≥ 2

2 Proof of the main theorem

In this section we prove Theorem 1.1

Let L ⊂ (Z≥0)2 be a set of lattice points and (x1, y1), , (xn, yn) be n general points on

K2 A nonzero element f in Span(xL) is of the form

X

(i,j)∈L

ci,jxiyj,

and f vanishes at a point (xk, yk) with multiplicity ≥ 2 if and only if f (xk, yk) = ∂f∂x(xk, yk) =

∂f

∂y(xk, yk) = 0

To prove Theorem 1.1, it is enough to prove it for |L| = 3n If |L| < 3n we can always find a set of lattice points S ⊂ (Z≥0)2 \ L where |S| = 3n − |L| and L0 := L ∪ S satisfies

|L0i| ≤ n for every i ∈ [1, 4] For the rest of this section, we will assume that |L| = 3n and

|Li| ≤ n for every i ∈ [1, 4] Let L = {(i1, j1), (i2, j2), , (i3n, j3n)} If the system of equations























xi1

1yj1

1yj2

1 · · · xi3n

1 yj3n 1

i1xi1 −1

1 yj1

1 i2xi2 −1

1 yj2

1 · · · i3nxi3n −1

1 yj3n 1

j1xi1

1yj1 −1

1 j2xi2

1yj2 −1

1 · · · j3nxi3n

1 yj3n −1 1

xi 1

nyj 1

nyj 2

n · · · xi 3n

n yj 3n n

i1xi1 −1

n yj1

n i2xi2 −1

n yj2

n · · · i3nxi3n −1

n yj3n n

j1xi1

nyj1 −1

n j2xi2

nyj2 −1

n · · · j3nxi3n

n yj3n −1 n































ci1,j1

ci2,j2

ci3n,j3n









= 0

has no nontrivial solution, then no curve in Span(xL) passes through n general points with multiplicity≥ 2

Give a total order T on the set of all lattice points as follows: (0, 0) < (0, 1) < (1, 0) < (0, 2) < (1, 1) < (2, 0) < (0, 3) < (1, 2) < (2, 1) < (3, 0) < i.e (i, j) < (i0, j0)

if i + j < i0+ j0 or

if i + j = i0+ j0 and if i < i0

GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES 3

Notation Let

DL= det























xi1

1yj1

1yj2

1 · · · xi3n

1 yj3n 1

i1xi1 −1

1 yj1

1 i2xi2 −1

1 yj2

1 · · · i3nxi3n −1

1 yj3n 1

j1xi1

1yj1 −1

1 j2xi2

1yj2 −1

1 · · · j3nxi3n

1 yj3n −1 1

xi 1

nyj 1

nyj 2

n · · · xi 3n

n yj 3n n

i1xi1 −1

n yj1

n i2xi2 −1

n yj2

n · · · i3nxi3n −1

n yj3n n

j1xi1

nyj1 −1

n j2xi2

nyj2 −1

n · · · j3nxi3n

n yj3n −1 n





















 ,

and for every W ⊂ L with |W | = 3 , if W = {(u1, v1), (u2, v2), (u3, v3)} and i ∈ {1, 2, , n}

we let

DWi = det





xu1

i yv1

i yv2

i yv3 i

u1xu1 −1

i yv1

i u2xu2 −1

i yv2

i u3xu3 −1

i yv3 i

v1xu1

i yv1 −1

i v2xu2

i yv2 −1

i v3xu3

i yv3 −1 i





Definition 2.1 A triangle decomposition of L is a sequence T = (W(1), , W(n)) such that

∪1≤i≤nW(i) = L, W(i)∩ W(j) = ∅ for all i 6= j and |W(i)| = 3 for every i Let TL be the

collection of all triangle decompositions of L For every T = {W(1), , W(n)} ∈ TL, let

P(T ) =

n

Q

i=1

DWi (i)

Lemma 2.2 We can always obtain a triangular decomposition T0 = {W(1), W(2), W(n)}

using the following recursive algorithm Let S be a set of lattice points and j and k positive

integers Set S to L and j = 1;

(1) Step 1: Choose k such that |Sk| = M in{|Si|, i ∈ [1, 4]}

(2) Step 2: Let W(j) = ∪i6=k,i∈[1,4]{smallest element in Si}

(3) Step 3: remove W(j) from S and increment j by 1

(4) Step 4: if S 6= ∅ go back to step 1; otherwise the construction is complete

Proof Prooving the Lemma by induction on n

Base Case:

For n = 1, |L| = 3 and |Li| ≤ 1 for every i ∈ [1, 4] So we excute the steps 1,2,3,4 ones and

we exit because S becomes empty

inductive case:

Suppose that the algorithm works successfully for n = m

For n = m + 1, |L| = 3m + 3 and |Li| ≤ m + 1 for every i ∈ [1, 4] But there exists i0 ∈ [1, 4]

such that |Li0| ≤ m, because, if not, we would have |L| = 4(m + 1) > 3(m + 1) which is

a contradiction Thus, after implementing the first three steps of the algorithm, we have

|S| = 3m and |Si| ≤ m for every i ∈ [1, 4]; and by the induction assumption the rest of the

We want to show that P(T0) 6= 0 and P(T0) cannot be cancelled by P

T ∈T L ,T 6=T 0

P(T ) so we can conclude that DL 6= 0

Lemma 2.3 Let D =det





xayb xcyd xeyf

axa−1yb cxc−1yd exe−1yf

bxayb−1 dxcyd−1 f xeyf −1





Then D is of the form D = Cxsytwith C 6= 0 (mod 2) if and only if (a, b) 6≡ (c, d) 6≡ (e, f )

Proof D = [(cf − de) − (af − be) + (ad − bc)]xa+c+e−1+yb+d+f −1 Then C = (cf − de) − (af − be) + (ad − bc)

It is easy to show that for any (α, β), (γ, δ) ∈ (Z≥0)2, αδ − βγ = 1 (mod 2) iff (α, β) 6≡ (γ, δ) 6≡ (0, 0) (mod 2)

If (a, b) ≡ (0, 0), C 6= 0 (mod 2) iff cf − de = 1(mod 2) i.e (c, d) 6≡ (e, f ) 6≡ (0, 0)

If not, and (c, d) ≡ (0, 0), C 6= 0 (mod 2) iff af − be ≡ 1 i.e (a, b) 6≡ (e, f ) 6≡ (0, 0)

Otherwise, that is (a, b) 6≡ (0, 0) and (c, d) 6≡ (0, 0),

if (a, b) ≡ (c, d) then ad − bc = 0 (mod 2) and (cf − de) − (af − be) = 0 (mod 2) Thus

C = 0 (mod 2)

if (a, b) 6≡ (c, d) then ad − bc = 1 (mod 2) Thus C 6= 0 (mod 2) iff (cf − de) − (af − be) = 0 (mod 2) i.e [(e, f ) ≡ (0, 0)] or [(cf − de) = 1 (mod 2) and(af − be) = 1 (mod 2)] Thus,

Lemma 2.4 Let W and W0 two subsets of L with |W | = |W0| = 3 and let DW

i = CW

i xα

iyiβ and DW0

i = CW0

i xγiyδ

i If CW

i 6= 0 (mod 2) and CW0

i 6= 0 (mod 2) then there exist (a, b) and (c, d) ∈ (Z≥0)2 such that W ∩ Ma,b= ∅ and W0 ∩ Mc,d= ∅

Lemma 2.5 Let W = {(u1, v1), (u2, v2), (u3, v3)} and W0 = {(u01, v10), (u02, v20), (u03, v30)} be two subsets of L with |W | = |W0| = 3 and let DW

i = CiWxαiyiβ and DW

0

0

i xγiyiδ Assume that CW

i 6= 0 (mod 2) and CW0

i 6= 0 (mod 2) and let (a, b) and (c, d) ∈ (Z≥0)2 such that

W ∩ Ma,b= ∅ and W0 ∩ Mc,d= ∅

(1) if (a, b) 6= (c, d) then (α 6= γ)or (β 6= δ)

(2) if (a, b) = (c, d) and (ui, vi) ≤ (u0i, vi0) for every i ∈ {1, 2, 3} then (α 6= γ)or (β 6= δ)

Proof for (1)

α = u1+ u2+ u3− 1, β = v1+ v2+ v3 − 1, γ = u01+ u02+ u03− 1, and δ = v01+ v02+ v30 − 1 Thus, α = a − 1 (mod 2), β = b − 1 (mod 2) and γ = c − 1 (mod 2), δ = d − 1 (mod 2) Therefore, if (a, b) 6= (c, d), (α 6= γ)or (β 6= δ)

for (2),

If ui+ vi < u0i+ vi0 for some i then

3

P

i=1

(ui+ vi) <

3

P

i=1

(u0i+ v0i) which implies that

3

P

i=1

ui 6=P3

i=1

u0i or

3

P

i=1

vi 6=

3

P

i=1

vi0 Thus (α 6= γ)or (β 6= δ)

GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES 5

If ui+ vi = u0i+ v0i for all i then

3

P

i=1

ui <

3

P

i=1

u0i and (α 6= γ)or (β 6= δ)



Proof of theorem 1.1 Let T0 = {W(1), W(2), W(n)} the triangular decomposition found

in lemma 2.3 It is sufficient to show that P(T0) 6= 0 and P(T0) cannot be cancelled by

P

T ∈TL,T 6=T 0

P(T )

Let P(T0) =

n

Q

i=1

DW (i)

n

Q

i=1

CW (i)

i xαi

i yβi i

By construction, |W(i)∩ Mα,β| ≤ 1 for any (α, β) ∈ {0, 1}2 and for any i ∈ {1, 2, , n}.Thus,

by lemma 2.4, CW (i)

i 6= 0 (mod 2) for any i ∈ {1, 2, , n} and the coefficient of P(T0) is odd

In particular, P(T0)6= 0

For any T ∈ TL, T 6= T0, if T = {V(1), V(2), V(n)} let P(T ) =

n

Q

i=1

DV (i)

n

Q

i=1

CV (i)

i xγi

i yδi

i and

j the first integer such that W(j) 6= V(j) Also, let W(j) = {(u1, v1), (u2, v2), (u3, v3)} and

V(j) = {(u01, v10), (u02, v02), (u03, v03)}

If CjV(j) ≡ 0 (mode 2), then the coefficient of P(T ) is even and it cannot cancel P(T0)

If CV (j)

j 6≡ 0 (mode 2), then by lemma 2.4 |V(j)∩ Ms,t| ≤ 1 for any (s, t) ∈ {0, 1}2

Then there exist (a, b), (c, d) ∈ {0, 1}2 such that W(j)∩ Ma,b= ∅ and V(j)∩ Mc,d = ∅

If (a, b) 6= (c, d) then by lemma 2.6 (1),(αj 6= γj)or (βj 6= δj) Thus P(T )cannot cancel P(T0)

If (a, b) = (c, d) then by the construction of P(T0), (ui, vi) ≤ (u0i, vi0) for every i ∈ {1, 2, 3}

Thus by lemma 2.6 (2), (αj 6= γj)or (βj 6= δj) and P(T ) cannot cancel P(T0)

Therefore, P(T0) cannot be cancelled by P

T ∈T L ,T 6=T 0

3 Proofs of Corollaries

Proof of Corollary 1.2 Let (x1, y1), , (xn, yn), (xn+1, yn+1), , (xn0, yn0) be n0general points

Let n00 = d2d+53 e + 1 and P (x, y) =P

i+j≤d+2ci,jxiyj a polynomial of degree d + 2 We have

a system ofPn

i=1

m i +1

2
+ 3n00 homogeneous linear equations with (d+3)(d+4)/2 unknowns

Let A be the matrix corresponding to this system of linear equations, N = {(i, j), i + j ≤ d}

and L = {(i, j), d + 1 ≤ i + j ≤ d + 2} Also let B be the matrix formed by

choos-ing the rows correspondchoos-ing to the first n points and the columns correspoondchoos-ing to N ;

and C be the submatrix formed by choosing the rows corresponding to the last n00 points

and the columns corresponding to L To show that our system of equations has no

triv-ial solution it is enough to show that both of the systems of homegeneous linear equations

associated with B and C have no trivial solutions The claim is true for B by

assump-tion For C, we try to use theorem 1.1 We have |L| = 2d + 5 On the other hand,

3n00 = 3d2d+53 e + 3 = 2d + 5 − (2d + 5)(mod2) + 3 ≥ 2d + 5 Thus |L| ≤ 3n00 To use theorem

1.1 we only need to show that |Li| ≤ n00 for every i ∈ [1, 4] It is easy to show that

|L1| = (d + 4)/2 if d is even

(d + 3)/2 if d is odd;

|L2| = |L3| = (d + 2)/2 if d is even

(d + 3)/2 if d is odd;

|L4| = (d + 2)/2 if d is even

(d + 1)/2 if d is odd;

Since by assumption we have (d + 4)/2 ≤ (2d + 5)/3; therefore, |Li| ≤ n00 for every i ∈

Proof of Corollary 1.3 Let (x1, y1), , (xn+1, yn+1) be n + 1 general points

As in the previous two corollaries we let P (x, y) =P

i+j≤dci,jxiyj be a polynomial of degree

d We have a system of 3n + t(t + 1)/2 homogeneous linear equations with (d+1)(d+2)/2 unknowns

Let A be the matrix corresponding to this system of linear equations, N = {(i, j), i+j ≤ t−1} and L = {(i, j), t ≤ i + j ≤ d} Also let B be the matrix formed by choosing the rows corresponding to the (n + 1)th point and the columns correspoonding to N ; and C be the matrix formed by choosing the rows corresponding to the first n points and the columns corresponding to L

To proove that our system of equations has no trivial solution it is enough to show that both of the systems of homogeneous linear equations corresponding to the matrices B and

C have no trivial solutions The claim is true for B because there is no curve of degree t − 1 that passes through a point with multiplicity ≥ t

Since |N | = t(t + 1)/2 and |L| = (d + 1)(d + 2)/2 − t(t + 1)/2 ≤ 3n, we try to use theorem 1.1 to proove our claim for the matrix C We need to show that |Li| ≤ n for every i ∈ [1, 4]

As in the first corollary

|L1∪ N1| = (d + 2)(d + 4)/8 if d is even

(d − 1)(d + 1)/8 if dis odd;

|L2∪ N2| = |L3∪ N3| = |L4∪ N4| = (d(d + 2)/8 if d is even

(d + 1)(d + 3)/8 if dis odd;

and

|N1| = (t + 1)(t + 3)/8 if d is even

(t − 2)t/8 if dis odd;

|N2| = |N3| = |N4| = ((t − 1)(t + 1)/8 if d is even

t(t + 2)/8 if dis odd;

Since |Li| = |Li∪ Ni| − |Ni| for every i ∈ [1, 4], it is suffices to show that the greatest possible value of |Li∪ Li| mince the smallest possible values of |Li| is ≤ n That is it is suffices to show that (d + 2)(d + 4)/8 − (t − 2)t/8 ≤ n

(d + 1)(d + 2)/2 ≤ t(t + 1)/2 + 3n is equivalent to d ≤ (−32 +p(1

4 + t2+ t + 6n))

To show that (d + 2)(d + 4)/8 − (t − 2)t/8 ≤ n is equivalent to show that d ≤ (−3 +p(1 +

GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES 7

t2− 2t + 8n))

It can be shown that if n ≥ t + 5, d ≤ (−32 +p(1

4+ t2+ t + 6n)) ≤ (−3 +p(1 + t2− 2t + 8n))

But for d ≥ t + 4, (d + 1)(d + 2)/2 ≤ t(t + 1)/2 + 3n implies that t + 5 ≤ 35n + 2 We notice

that in this case, n is necessarly ≥ 5 Because otherwise we would have t + 5 ≤ 5 i.e t ≤ 0

Thus n ≥ 5 and t + 5 ≤ 35n + 2 ≤ n which prooves the claim 

Proof of Corollary 1.4 A polynomial P (x, y) of degree d is of the formP

i+j≤dci,jxiyj Such

a polynomial vanishes at n general points with multiplicity ≥ 2 iff the system of 3n

homo-geneous linear equations

P (xk, yk) = ∂P/∂x(xk, yk) = ∂P/∂y(xk, yk) = 0 for any 1 ≤ k ≤ n

has at least one solution

The number of unknown in this system is the cardinality of the set {(i, j), i + j ≤ d} This

number is equal to (d + 1)(d + 2)/2

If (d + 1)(d + 2)/2 > 3n, the number of unknown is strictly greater than the number of

equations; therefore, there are an infinitely many curves of degree d vanishing at n general

points with multiplicity ≥ 2

If (d + 1)(d + 2)/2 ≤ 3n, Let L = {(i, j), i + j ≤ d} Then |L| = (d + 1)(d + 2)/2 ≤ 3n To

apply theorem 1.1 it suffices to show that |Li| ≤ n for every i ∈ [1, 4] It is easy to show that

|L1| = (d + 2)(d + 4)/8 if d is even

(d − 1)(d + 1)/8 if d is odd;

|L2| = |L3| = |L4| = (d(d + 2)/8 if d is even

(d + 1)(d + 3)/8 if d is odd;

If d is even, |Li| ≤ |L1| for every i ∈ [1, 4] So it suffices to proove that |L1| ≤ n

(d + 1)(d + 2)/2 ≤ 3n is equivalent to d ≤ (−32 +p(1

4 + 6n))

To show that (d + 2)(d + 4)/8 ≤ n is equivalent to show that d ≤ (−3 +p(1 + 8n))

It can be shown that if n ≥ 15 then (−32 +p(1

4+ 6n)) ≤ (−3 +p(1+8n)) therefore |L1| ≤ n

in this case

If n < 15, the only possible even values of d that satisfies d ≤ (−32 +p(1

4 + 6n)) are 2 and 4; so we still need to study these two cases

For d = 2 , (d + 1)(d + 2)/2 ≤ 3n with n 6= 2 implies that n ≥ 3 But (d + 2)(d + 4)/8 = 3;

therefore |L1| ≤ n in this case

For d = 4, (d + 1)(d + 2)/2 ≤ 3n with n 6= 5 implies that n ≥ 6 But (d + 2)(d + 4)/8 = 6;

therefore |L1| ≤ n in this case

For d is odd, |Li| ≤ |L2| for every i ∈ [1, 4] So it is sufficient to show that |L2| ≤ n

This is true for n ≥ 15 because (d + 1)(d + 3)/8 < (d + 2)(d + 4)/8 So, we only need to

show that the claim is true for n < 15

If n < 15, the only possible odd values of d that satisfy d ≤ (−32 +p(1

4 + 6n)) are 1 and 3

For d = 3, (d + 1)(d + 2)/2 ≤ 3n implies that n ≥ 4 But (d + 1)(d + 3)/8 = 3 ≤ n; therefore

|L2| ≤ n in this case

For d = 1, (d + 1)(d + 3)/8 = 1 ≤ n for any n ≥ 1 therefore |L2| ≤ n in this case



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Department of Mathematics, Wayne State University, Detroit, MI 48202