On the containment problem for fat points
Iman Bahmani Jafarloo Giuseppe Zito


Given an ideal I, the containment problem is concerned with finding the values m and r such that the m-th symbolic power of I is contained in its r-th ordinary power. A central issue related to this is determining the resurgence for ideals I of fat points in projective space. In this paper we obtain complete results for the resurgence of fat point schemes m_0P_0+m_1P_1+m_2P_2 in ℙ^N for any distinct points P_0,P_1,P_2, and, when the points P_0,…,P_n are collinear, we extend this result to fat point schemes m_0P_0+…+m_nP_n. As a by-product of our determining the resurgence for all three points fat point ideals, we give new examples of ideals with symbolic defect zero. In case the points are noncollinear, a three  fat points ideal can be regarded as a monomial ideal, but it is typically not square-free.

Optical conductivity of  overdoped cuprate superconductors: application to LSCO
    P. J. Hirschfeld
    December 30, 2023
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§ INTRODUCTION



 §.§ Background

Let us consider the polynomial ring R=𝕂[ℙ^N]=𝕂[x_0,…,x_N], where 𝕂 is an algebraically closed field of any characteristic and N ≥ 2.
In general, if I is a homogeneous ideal of R, the m-th symbolic power of I is I^(m)=R ∩ (⋂_p∈ASS(I)(I^mR_p)).
However in this paper we will always deal with ideals of fat points that are ideals of the form I=⋂_iI(P_i)^m_i, where P_i are distinct points in ℙ^N, I(P_i) is the ideal of all the forms that vanish at P_i and the multiplicity m_i is a non-negative integer.
For ideals of this type, the m-th symbolic power can be simply defined as I^(m)=⋂_iI(P_i)^mm_i. During the last decades, there has been a lot of interest  comparing powers of ideals with symbolic
powers in various ways; see for example, <cit.>, <cit.>, <cit.>, <cit.>, <cit.>, 
and <cit.>.
It is easy to see that I^r ⊆ I^(r)⊆ I^(m) if and only if r ≥ m.
Furthermore I^(m)⊆ I^r implies m ≥ r but the converse is not true in general. Therefore it makes sense to ask the containment question: given an ideal I, for which  m and r is the symbolic power I^(m) contained in the ordinary power I^r?
In <cit.> and <cit.>, Bocci and Harbourne introduced and studied an asymptotic quantity, known as the resurgence, whose computation is clearly linked to the containment problem.

Given a non-zero proper homogeneous ideal I in R, the resurgence of I, denoted by ρ(I), is defined as the quantity: 
    ρ(I)=sup{m/r: I^(m)⊈I^r }.


From the results of <cit.>, it follows that I^(m)⊆ I^r whenever m ≥ Nr. Thus, we can conclude that ρ(I) ≤ N for any homogeneous ideal in R.
In general, directly computing ρ(I) is quite difficult and it has been determined only in very special cases. For example, it is known that ρ(I)=1 when I is generated by a regular sequence <cit.>. The resurgence is also known for certain cases of ideals of the following kinds: monomial ideals
<cit.>, ideals of a projective cone <cit.> and ideals of points on a reducible conic in ℙ^2 <cit.>.

Another situation where the resurgence is known is for certain ideals I defining
zero-dimensional subschemes of projective space. For example, if α(I) = reg(I), where reg(I) is the Castelnuovo-Mumford
regularity of I and α(I) is the degree of a non-zero element of I of least degree, then the resurgence can be completely described in terms of numerical invariants of I (<cit.>
and <cit.>).
One of these invariants is called the Waldschmidt constant of I.

Let I be a non-zero proper homogeneous ideal of R. The Waldschmidt constant of I, denoted by α(I), is defined as: 

    α(I)=inf_m> 0{α(I^(m))/m}=lim_m →∞α(I^(m))/m.


In particular, when I defines a 0-dimensional subscheme, the authors in <cit.> proved that α(I)/α(I)≤ρ(I) ≤reg(I)/α(I), so ρ(I)= α(I)/α(I) when α(I)=reg(I). 


 §.§ Preliminaries

Hereafter Z is a fat point scheme of ℙ^N.


Let P_1,…,P_n be distinct points in ℙ^N and m_1,…,m_n be non-negative integers. The ideal I=⋂_i=1^nI(P_i)^m_i defines a subscheme of ℙ^N and we will denote it by Z=∑_i=1^nm_iP_i where by definition we set I(Z) = I.



Consider a fat point subscheme Z=∑m_tP_t  where all the points P_t lie on a plane  Π (hence a ℙ^2, unique if and only if the points are not collinear). The subscheme Π∩ Z is a fat point subscheme of Π=ℙ^2. We denote the ideal of Π∩ Z in R_Π=𝕂[Π] by I_Π(Z), or more simply by 
I_ℙ^2(Z)⊆ R_ℙ^2. Thus I_ℙ^2(Z)=⋂I_ℙ^2(P_t)^m_t,
and for emphasis we may denote I(Z)⊆ R=𝕂[ℙ^N] by
I_ℙ^N(Z)⊆ R_ℙ^N=𝕂[ℙ^N].	


In this paper we compute the resurgence of I(Z) for two classes of fat point subschemes.

In Section <ref>, we study the subscheme Z=∑_i=1^nm_iP_i in ℙ^N, where the points P_i are collinear and we determine the resurgence in Theorem <ref>.
 
	Let Z=∑_i=1^nm_iP_i be a fat point scheme, where P_1,…,P_n are distinct collinear points in ℙ^N. Then I(Z)^(m)=I(Z)^m  for all  m ∈ℕ, thus ρ(I(Z))=1.

The proof of Theorem <ref> is a direct consequence of Lemma <ref> and Lemma <ref>. 

The property that I(Z)^(m)=I(Z)^m, presented in the statement of Theorem <ref>, gives us more information than the exact value of the resurgence.
In particular, if we define, as in <cit.>, the m-symbolic defect of a homogeneous ideal I of R as the minimal number sdefect(I,m)  of generators of the R-module I^(m)/I^m, Theorem <ref> tells us that sdefect(I(Z),m)=0 for all m, if Z is a fat point scheme whose support consists of collinear points.
Notice that sdefect(I(Z),m)=0 for all m ≥ 1 implies ρ(I(Z))=1. It is still not known if there exists a scheme Z with ρ(I(Z))=1 and I(Z)^m ≠ I(Z)^(m) for some m, but our results show that any such Z must be supported at more than three points, not all of which can be collinear.

In Section <ref> we start considering the fat point subschemes Z consisting of three noncollinear points, initially focusing on the ℙ^2 case. In particular, we show how  the invariants α(I_ℙ^2(Z)) and α(I_ℙ^2(Z)) depend on the values assigned to the multiplicities and how to relate the value of the resurgence of I_ℙ^2(Z) to
ρ(I_ℙ^N(Z)) (hereafter, we will always use I(Z) to mean I_ℙ^N(Z)).

In Section <ref>, we consider  the subscheme Z=m_0P_0+m_1P_1+m_2P_2, where the P_i's are noncollinear points in ℙ^N and m_0≤ m_1 ≤ m_2 are nonnegative integers. In Theorem <ref> we classify fat point ideals
in ℙ^N supported at three noncollinear points which have m-symbolic defect zero for all m (and hence such that ρ(I(Z))=1). 
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2 ≥max(m_0,m_1). Consider the fat point scheme Z=m_0P_0+m_1P_1+m_2P_2.
Then sdefect(I(Z),m)=0 for all m∈ℕ if and only if one of the following conditions holds:


  (a)
m_0+m_1 ≤ m_2;

  (b)
m_0+m_1> m_2 and m_0+m_1+m_2 is even.


The proof that Theorem <ref>
(a) implies sdefect(I(Z),m) = 0 for all m ≥ 1 is Proposition <ref>. The proof that Theorem
<ref> (b) also implies sdefect(I(Z),m) = 0 for all m ≥ 1 is Proposition <ref>. To complete
the proof of Theorem <ref>, it remains to show that sdefect(I(Z),m) > 0 for some m > 0
whenever m_0 + m_1 > m_2 and m_0 + m_1 + m_2 is odd. This follows from Theorem <ref>.

 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and max( m_0,m_1) ≤ m_2. Consider the fat point scheme Z=m_0P_0+m_1P_1+m_2P_2. If  m_0+m_1 > m_2 and m_0+m_1+m_2 is odd, then 
    ρ(I(Z))= m_0+m_1+m_2+1/m_0+m_1+m_2.


The proof follows at once from Corollary <ref> and Proposition <ref>.


§ FAT POINTS ON A LINE IN ℙ^N
 
Let L be a line in ℙ^N and let P_1,…,P_n be distinct points which lie on L. Consider the scheme Z= ∑_i=1^nm_iP_i,  where the multiplicities m_1≤ m_2≤…≤ m_n are nonnegative integers. In this section, we determine the resurgence and we prove Theorem <ref>.
To do so requires some lemmas.

The following lemma plays a significant role throughout this section. 
 
Let F ∈ R be a homogeneous form of degree d.
Then there are uniquely determined forms g_d,i_2,…,i_N∈𝕂[x_0,x_1] of degree d-(i_2+…+i_N) such that

    F=∑_k=0^d ∑_i_2+…+i_N=kg_d,i_2,…,i_N· x_2^i_2⋯ x_N^i_N.

Moreover, given any homogeneous linear form G=bx_0+ax_1 (a,b ∈𝕂 ), let I be the ideal ⟨ G,x_2,…,x_N ⟩^m.
Then F ∈ I if and only if G^m-(i_2+…+i_N) divides g_d,i_2,…,i_N whenever m>i_2+…+i_N.


The claim about F=∑_k=0^d ∑_i_2+…+i_N=kg_d,i_2,…,i_N· x_2^i_2⋯ x_N^i_N follows from thinking of R as R=𝕂[x_0,x_1][x_2,…,x_N]. The second claim, regarding F ∈ I, is clear when a=0 or b=0, taking into account that I is a monomial ideal in these cases. If G=bx_0+ax_1, a,b≠0, consider the 𝕂-algebra automorphism f:R → R defined by f(x_i)=x_i for all i≠1 with f(x_1)=G. Then f(⟨ x_1,…,x_N⟩^m)=I. Taking ϕ to be the inverse automorphism, we have 

    ϕ(F)=∑_k=0^d ∑_i_2+…+i_N=kϕ(g_d,i_2,…,i_N)· x_2^i_2⋯ x_N^i_N∈⟨ x_1,…,x_N⟩^m,

so x_1^m-(i_2+…+i_N) divides ϕ(g_d,i_2,…,i_N) whenever m>i_2+…+i_N, hence G^m-(i_2+…+i_N) divides g_d,i_2,…,i_Nwhenever m>i_2+…+i_N.


Considering the previous proof, since the ideal of the point P=[-a:b:0:…:0] is G=⟨ bx_0+ax_1,x_2,…,x_N⟩, indeed, we showed that
F∈ I(mP) if and only if  G^m-(i_2+…+i_N)|g_d,i_2,…,i_N  whenever m>i_2+…+i_N.

Using unique factorization for homogeneous polynomials in 𝕂[x_0, x_1], the following
corollary is an immediate consequence of the previous lemma.
 
Given distinct points P_i=[-d_i:c_i:0:…:0], i=0,…,n, on the line x_2=x_3=…=x_N=0, let F be a form as (<ref>). Then
F∈ I(∑_i=0^nmm_iP_i) if and only if (c_ix_0+d_ix_1)^mm_i-(i_2+…+i_N)|g_d,i_2,…,i_N  whenever i_2+…+i_N<mm_i for all i=1,…,n.
In other words, we have shown that the homogeneous ideal I(∑_i=0^nmm_iP_i) is generated by “monomials”  of the type G_1^a_1⋯ G_n^a_n· x_2^b_2⋯ x_N^b_N, where G_j=c_jx_0+d_jx_1 and  a_j= max_j(0,mm_j-(b_2+…+b_N)),    j=1,…,n and b_2+…+b_N ≤max(mm_0,…,mm_n).

The following general lemma gives us a simple criterion for an ideal I(Z) of a fat point scheme to be such that I(Z)^(m)=I(Z)^m for all m ∈ℕ.
 
Let Z=Z_1+…+Z_r where Z_1,…,Z_r ⊂ℙ^N are fat point subschemes such that

    I(kZ)=∏_i=1^rI(kZ_i)     ∀ k ∈ℕ,
  
where Z_i is a fat point scheme satisfying the condition I(Z_i)^(m)=I(Z_i)^m, for all m ∈ℕ.
Then we have also 
    I(Z)^m=I(Z)^(m)    ∀ m ∈ℕ.



Considering (<ref>) when k=1, we obtain I(Z)=∏_i=1^rI(Z_i), thus

    I(Z)^m=∏_i=1^rI(Z_i)^m=∏_i=1^rI(Z_i)^(m)=∏_i=1^rI(mZ_i)=I(mZ)=I(Z)^(m).
 	
So the proof is complete.

Taking  into account Lemma <ref>, in order to prove Theorem <ref>, it suffices to exhibit a suitable splitting for an ideal I(Z) of a collinear fat point scheme. The following lemma gives us a precise answer to this problem.
 
	Let Z=∑_i=1^nm_iP_i be a fat point scheme, where the P_i's are collinear points in ℙ^N. 	We can assume that the points lie on the line x_2=x_3=…=x_N=0  and  0=m_0≤ m_1≤…≤ m_n.
	Then 
	
    I(mZ)=∏_i=1^nI((mm_i-mm_i-1)Z_i),

	where Z_i=P_i+…+P_n for i=1,…,n.
 

		Notice that the ideal I(Z_i) defined in the previous lemma satisfies 
	
    I(Z_i)^(m)=I(Z_i)^m  for all  m.

	In fact,   I(Z_i) is a complete intersection scheme (a set of simple points on a line), and by <cit.>, its symbolic powers and ordinary powers are always equal. 
	
	Therefore it is enough to show  I(mZ)=∏_i=1^nI(Z_i)^mm_i-mm_i-1.
 We denote by G_i the linear form in 𝕂[x_0,x_1] such that we have I(P_i)=⟨ G_i,x_2,…,x_N ⟩ for all i=1,…,n. The inclusion “⊇" is immediately concluded from the definition of I(Z). For proving the other inclusion “⊆", it suffices to consider Corollary <ref> and show that a monomial  ℳ=G_1^a_1⋯ G_n^a_n· x_2^b_2⋯ x_N^b_N where a_j=max_j(0,mm_j-∑_i=2^Nb_i) and ∑_i=2^Nb_i≤ mm_n, for all 1≤ j≤ n is contained in ∏_i=1^nI((mm_i-mm_i-1)Z_i).
Regard H=x_2^b_2⋯ x_N^b_N as a product of b_2+⋯+b_N linear forms. Let H_1 be the product of the first mm_1 forms in H, H_2 be the product of the next mm_2-mm_1 linear forms in H, etc., until, for some j, H_j is the product of the remaining forms in H. Since b_2+⋯+b_N ≤ mm_n, we know j ≤ n. If j<n, set H_i=1 for i>j (in particular, if b_2+⋯+b_N<mm_1, then H_1=H and H_i=1 for 1<i≤ n). 
Define ℳ_i=G_i⋯ G_n for i=1,…,n and then we can write

    ℳ=(ℳ_1^a_1H_1)(ℳ_2^a_2-a_1H_2)⋯ (ℳ_n^a_n-a_n-1H_n),

and it is easy to check that ℳ_i^a_i-a_i-1H_i ∈ I(Z_i)^(mm_i-mm_i-1) for each i.



§ THREE NONCOLLINEAR POINTS: ℙ^N VERSUS ℙ^2
 

Let Z be a three non-collinear fat points scheme. If I_ℙ^N(mZ)⊆ I_ℙ^N(Z)^r, then I_ℙ^2(mZ)⊆ I_ℙ^2(Z)^r. 


We have the canonical ring quotient q:R_ℙ^N→ R_ℙ^2. The key fact is that q(I_ℙ^N(Z))=I_ℙ^2(Z). Hence, if I_ℙ^N(mZ)⊆ I_ℙ^N(Z)^r, then I_ℙ^2(mZ)=q(I_ℙ^N(mZ))⊆ q(I_ℙ^N(Z)^r)=I_ℙ^2(Z)^r.



    ρ(I_ℙ^2(Z))≤ρ(I_ℙ^N(Z))



By the previous lemma it follows that

    {m/r: I_ℙ^2(Z)^(m)⊈I_ℙ^2(Z)^r }⊆{m/r: I_ℙ^N(Z)^(m)⊈I_ℙ^N(Z)^r },

so the desired result easily follows from the definition of resurgence and from the properties of the supremum.

 
Let Z = m_0P_0 + m_1P_1 + m_2P_2⊂ℙ^N, assuming max(m_0,m_1) ≤ m_2 and that the points are noncollinear. Then α(I_ℙ^2(Z)) is as follows:


  (a) m_2 if m_2≥ m_1+m_0

  (b) (m_0 + m_1 + m_2)/2 if m_2≤ m_0 + m_1 and m_0 + m_1 + m_2 is even

  (c) (m_0 +m_1 +m_2 +1)/2 if m_2≤ m_0 +m_1 and m_0 +m_1 +m_2 is odd.



We may choose coordinates so that the points P_0, P_1, P_2 are the coordinate vertices of ℙ^2. Namely  we assume that P_0=[1:0:0], P_1=[0:1:0] and P_2=[0:0:1]. 
The proof in case (a) is: x_0^m_2-m_0x_1^m_0∈ I_ℙ^2(Z) hence α(I_ℙ^2(Z))≤ m_2, but no non-zero form of degree less than m_2 can vanish to order m_2 at a point, hence
α(I_ℙ^2(Z))≥ m_2 too. 
The proof in case (b) is: 

    x_0^(m_2+m_1-m_0)/2x_1^(m_2+m_0-m_1)/2x_2^(m_1+m_0-m_2)/2∈ I_ℙ^2(Z)

 so α(I_ℙ^2(Z))≤(m_0+m_1+m_2)/2. But I_ℙ^2(Z) is monomial and
 there are irreducible conics through the three points. Thus 2α(I_ℙ^2(Z)) ≥ m_0+m_1+m_2 by Bezout's Theorem. Thus, α(I_ℙ^2(Z))=(m_0+m_1+m_2)/2. The proof in the last case is:  all three of m_2+m_1-m_0, m_2 +m_0-m_1 and m_1 +m_0-m_2 are odd and nonnegative, hence at least one.  Then 

    x_0^(m_2+m_1-m_0+1)/2x_1^(m_2+m_0-m_1+1)/2x_2^(m_1+ m_0-m_2-1)/2∈ I_ℙ^2(Z),

 so α (I_ℙ^2(Z))≤(m_0+m_1+m_2+1)/2. But as before there are irreducible conics through the three points. Thus 2α (I_ℙ^2(Z))≥ m_0+m_1+m_2 by Bézout's Theorem (as before), and thus 2α (I_ℙ^2(Z))≥ m_0+m_1+m_2+1 (since m_0+m_1+m_2 is odd). Thus α (I_ℙ^2(Z))= (m_0+m_1+m_2+1)/2.

 
	Let P_0,P_1 and P_2 be three noncollinear  points in ℙ^N and consider  Z= m_0P_0+m_1P_1+ ⁠ m_2P_2. Suppose m_0 ≤ m_1 ≤ m_2. If m_0+m_1 > m_2 and ∑_i=0^2m_i  is odd, then  ρ(I_ℙ^N(Z)) ≥1+∑_i=0^2 m_i/∑_i=0^2 m_i.


	The points P_0,P_1, P_2 span a plane  ℙ^2 ⊂ℙ^N. Without loss of generality we assume in this ℙ^2 that P_0=[1:0:0], P_1=[0:1:0] and P_2=[0:0:1]. We want to use the following inequality 
	
    ρ(I_ℙ^2(Z)) ≥α(I_ℙ^2(Z))/ α(I_ℙ^2(Z)),

	which was proved in <cit.>.
	From the part (c) of the last proposition we have α (I_ℙ^2(Z))= (m_0+m_1+m_2+1)/2.
	Now, consider m ∈ℕ and the 2m-th symbolic power I_ℙ^2(Z)^(2m). Considering the definition of symbolic powers, 
	
    I_ℙ^2(Z)^(2m)= ⟨ x_1,x_2⟩^2m · m_0∩⟨ x_0,x_2⟩^2m · m_1∩⟨ x_0,x_1⟩^2m · m_2.

Since I_ℙ^2(Z)^(2m)=I_ℙ^2(2mZ) we have α(I_ℙ^2(Z)^(2m))=m(m_0+m_1+m_2) by Proposition <ref> (b). 	 Thus we obtain the Waldschmidt constant of I_ℙ^2(Z) as follows:
	 
	 α(I_ℙ^2(Z))   = lim_m →∞α(I_ℙ^2(Z)^(m))/m=lim_m →∞α(I_ℙ^2(Z)^(2m))/2m
   = lim_m →∞m(∑_i=0^2m_i)/2m=∑_i=0^2m_i/2.
	  
	Hence by (<ref>) and (<ref>), 
	
    1+∑_i=0^2m_i/∑_i=0^2m_i≤α(I_ℙ^2(Z))/α(I_ℙ^2(Z))≤ρ(I_ℙ^2(Z)),

	and by Corollary <ref> the desired result is obtained.



§ THREE NONCOLLINEAR POINTS IN ℙ^N
 
In this section we obtain additional results for the fat point scheme Z=m_0P_0+m_1P_1+m_2P_2 in ℙ^N, 
where P_0, P_1 and P_2 are noncollinear and each m_i is a nonnegative integer. We can assume P_0=[1:0:0:… : 0], P_1=[0:1:0: …:0] and P_2=[0:0:1:0: …:0] and max(m_0,m_1) ≤ m_2. Notice that the ideal I(P_i) is a square-free monomial ideal, and hence I(Z) is a monomial ideal. We are interested in computing the resurgence ρ(I(Z)) of the ideal I(Z). In particular, we want to understand how the resurgence of the scheme Z depends on the values of the multiplicities m_i.

The following lemma gives some conditions for a monomial to belong to I(Z).


Let P_0, P_1 and P_2 be noncollinear points in ℙ^N as above and m_i ≥ 0. We define the fat point scheme Z=m_0P_0+m_1P_1+m_2 P_2. Then the monomial 𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(Z) if and only if (a_0,…,a_N) satisfies the following system of inequalities

    Cond(Z):=
    a_1+a_2+a_3+⋯+a_N ≥ m_0 
     a_0+a_2+a_3+⋯+a_N ≥ m_1 
     a_0+a_1+a_3+⋯+a_N ≥ m_2.



The result easily follows from the fact that the ideal I(Z) is the monomial ideal
⋂_i=0^2 I(P_i)^m_i with I(P_0) = (x_1,x_2,x_3,...,x_N), I(P_1) = (x_0,x_2,x_3,...,x_N) and I(P_2) = (x_0,x_1,x_3,...,x_N).

Notice that in the previous lemma, in order to simplify the notation, we made implicit the dependence of Cond(Z) on m_0, m_1 and m_2.

We divide this section into two subsections where we study distinct configurations for the multiplicities m_i.


 §.§ Case m_0+m_1≤ m_2

The aim of this subsection is to prove the following result.
 
	Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2 ≥max(m_0,m_1). Let Z=∑_i=0^2m_iP_i be a fat point scheme. If m_0+m_1≤ m_2, then I(Z)^(m)=I(Z)^m  for all  m ∈ℕ and consequently ρ(I(Z))=1.	

  Proposition <ref> follows at once by accordingly using  Lemma <ref> if we can find a suitable splitting for the ideal I(Z). As explained in Remark <ref> (following the proof of Lemma <ref> ), the following lemma gives a suitable splitting.
 
	Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2≥max(m_0,m_1). 
Consider the fat point scheme Z=m_0P_0+m_1 P_1+m_2P_2.  If m_0+m_1≤ m_2, then 
    I(Z)=I(m_0(P_0+P_2)) · I(m_1(P_1+P_2))· I((m_2-m_0-m_1)P_2).



	Notice that, if m_2=m_0+m_1, then I((m_2-m_0-m_1)P_2)=R thus the desired splitting  in this case is I(Z)=I(m_0(P_0+P_2)) · I(m_1(P_1+P_2)).
Set Z_1=m_0(P_0+P_2), Z_2=m_1(P_1+P_2) and Z_3=(m_2-m_0-m_1)P_2. The inclusion I(Z_1) · I(Z_2)· I(Z_3) ⊆ I(Z) is trivial since Z=Z_1+Z_2+Z_3. Now, we show the other inclusion holds.
Thus, let us consider a  monomial 𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(Z) where the a_i's satisfy the system Cond(Z), and set b=∑_i=3^Na_i. We have the following cases:

(a) Assume a_1+b < m_0. By Cond(Z) it follows that a_2 ≥ m_0-a_1-b >0 and a_0 ≥ m_2-a_1-b= (m_0-a_1-b)+m_1+(m_2-m_0-m_1). Then the monomial

    (x_0^m_0-a_1-bx_1^a_1x_2^m_0-a_1-bx_3^a_3⋯ x_N^a_N) · (x_0^m_1)· (x_0^m_2-m_0-m_1)
 divides 𝒩 and belongs to I(Z_1)· I(Z_2)· I(Z_3) because the j-th part of the above product is in I(Z_j) by Cond(Z_j) for j=1,2,3.

(b) Assume that a_0+b<m_1. The proof is similar to the previous case by using a_0+b<m_1.

(c) Consider a_1+b ≥ m_0 and a_0+b ≥ m_1 and the following four cases:


  (1) Assume a_1 ≥ m_0 and a_0≥ m_1. We can write 𝒩 as

    (x_1^m_0) · (x_0^m_1)· (x_0^a_0-m_1x_1^a_1-m_0x_2^a_2x_3^a_3⋯ x_N^a_N).

The first two factors belong respectively to I(Z_1) and I(Z_2) while the third one is in I(Z_3) because Cond(Z) implies a_0+a_1+a_3+⋯+a_N-m_1-m_0 ≥ m_2-m_1-m_0.
Hence, Cond(Z_3) is satisfied.


  (2) Assume a_1<m_0 and a_0 ≥ m_1. By a_1+b ≥ m_0, we deduce 
∑_i=3^Na_i≥ m_0-a_1.
Then, for each i=3,…,N, we can choose 0≤ b_i ≤ a_i such that ∑_i=3^Nb_i=⁠ m_0-a_1.
It can be written
𝒩=(x_1^a_1x_3^b_3⋯ x_N^b_N) · (x_0^m_1)· (x_0^a_0-m_1x_2^a_2x_3^a_3-b_3⋯ x_N^a_N-b_N), where it is easy to check that the first two factors belong respectively to I(Z_1) and I(Z_2) while
the third term is in I(Z_3) because Cond(Z) implies that
a_0+a_3+⋯+a_N-m_1-∑_i=3^Nb_i=a_0+a_1-m_1-m_0+∑_i=3^Na_i≥ m_2-m_1-m_0. So, Cond(Z_3) is satisfied.

  (3)Assume a_1 ≥ m_0 and a_0 < m_1. The proof of this case is similar to the proof of the previous one.

  (4)
Assume a_1 < m_0 and a_0 < m_1. Cond(Z) implies that

   ∑_i=3^Na_i=b

   =(a_0+a_1+b-m_0-m_1)+(m_0-a_1)+(m_1-a_0)

   ≥(m_2-m_1-m_0)+(m_0-a_1)+(m_1-a_0).

Because the  last three summands are all positive we can choose for all i=3,…,N, some integers 0≤ c_i,d_i,e_i ≤ a_i such that 
c_i+d_i+e_i ≤ a_i for all i=3,…,N, ∑_i=3^Nd_i=m_1-a_0, ∑_i=3^Nc_i=m_0-a_1, and ∑_i=3^Ne_i=m_2-m_1-m_0. So, the monomial ℳ= (x_1^a_1x_3^c_3⋯ x_N^c_N) · (x_0^a_0x_3^d_3⋯ x_N^d_N)· (x_3^e_i⋯ x_N^e_N) divides 𝒩 and belongs to I(Z_1)· I(Z_2)· I(Z_3) because its j-th factor belongs to I(Z_j) by Cond(Z_j) for j=1,2,3.


 
Notice that the splitting presented in the previous lemma satisfies the condition of Lemma <ref>. In fact the ideals involved in the product are ideals of   fat point schemes whose support consists of collinear points, and by means of Theorem <ref> we  have 

    I(Z_i)^(m)=I(Z_i)^m  for all  m.

Furthermore,  Lemma <ref>  can be applied to the fat point scheme kZ where k ∈ℕ, deducing that

    I(kZ)=∏_i=1^3I(kZ_i).




 §.§ Case m_0+m_1 > m_2

In this subsection, we deal with the case m_0+m_1 > m_2 showing how the value of the resurgence depends on the parity of the sum ∑_i=0^2m_i. 
Using the same approach as in  the previous subsection, we want to split the ideal I(Z) in a convenient way as a product of ideals I(Z_i).
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2 ≥max(m_0,m_1). We consider the scheme Z=m_0P_0+m_1 P_1+m_2P_2. If m_0+m_1 > m_2, then I(Z)= I(Z_1) · I(Z_2)· I(Z_3) where,

	
  * Z_1=(m_0+m_1-m_2)(P_0+P_1+P_2)
	
  * Z_2=(m_2-m_1)(P_0+P_2)
	
  * Z_3=(m_2-m_0)(P_1+P_2).



The inclusion I(Z_1) · I(Z_2)· I(Z_3) ⊆ I(Z) is trivial since Z=Z_1+Z_2+Z_3. We just need to show that if a monomial 𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(Z), then 𝒩∈ I(Z_1)· I(Z_2)· I(Z_3).
Thus, suppose 𝒩∈ I(Z), and set b= ∑_i=3^Na_i. We have the following cases:

(a) Let a_1+b<m_2-m_1. Considering the system Cond(Z),


  * a_2 ≥ m_0-a_1-b=(m_0+m_1-m_2)+(m_2-m_1-a_1-b) 

  * a_0 ≥ m_2-a_1-b=(m_0+m_1-m_2)+(m_2-m_1-a_1-b)+(m_2-m_0)

where all the numbers between parenthesis are nonnegative. So,  the  monomial ℳ=((x_0x_2)^m_0+m_1-m_2)  ·((x_0x_2)^m_2-m_1-a_1-bx_1^a_1x_3^a_3⋯ x_N^a_N)·(x_0^m_2-m_0)
divides 𝒩. Furthermore, ℳ belongs to I(Z_1)· I(Z_2)· I(Z_3) because the j-th factor belongs to I(Z_j) by Cond(Z_j) for j=1,2,3. Thus 𝒩∈ I(Z_1)· I(Z_2)· I(Z_3).

(b)   a_0+b < m_2-m_0: the proof is similar to previous case using a_0+b<m_2-m_0.

(c)  a_1+b ≥ m_2-m_1 and a_0+b ≥ m_2-m_0: we have four subcases,


  (1)a_1 ≥ m_2-m_1 and a_0 ≥ m_2-m_0: we can write
 𝒩=(x_0^a_0-m_2+m_0x_1^a_1-m_2+m_1x_2^a_2x_3^a_3⋯ x_N^a_N) · (x_1^m_2-m_1)· ( x_0^m_2-m_0) where the first factor is in I(Z_1) because Cond(Z) implies
 
 	
  ∙ a_1+a_2+a_3+⋯+a_N-m_2+m_1 ≥ m_0+m_1-m_2
 	
  ∙ a_0+a_2+a_3+⋯+a_N-m_2+m_0 ≥ m_0+m_1-m_2
 	
  ∙ a_0+a_1+a_3+⋯+a_N-2m_2+m_0+m_1 ≥ m_0+m_1-m_2
 
So, Cond(Z_1) is satisfied. Furthermore, it is easy to check that x_1^m_2-m_1∈ I(Z_2) and x_0^m_2-m_0∈⁠ I(Z_3). Thus 𝒩∈  I(Z_1)· I(Z_2)· I(Z_3).

  (2) a_1 ≥ m_2-m_1 and a_0 < m_2-m_0: a_0+b ≥ m_2-m_0 implies that ∑_i=3^Na_i≥ m_2-m_0-a_0>0.
For each i=3,…,N we can choose 0≤ b_i ≤ a_i such that ∑_i=3^Nb_i=m_2-m_0-a_0.
We can write
𝒩=(x_1^a_1-m_2+m_1x_2^a_2x_3^a_3-b_3⋯ x_N^a_N-b_N) · (x_1^m_2-m_1)· (x_0^a_0x_3^b_3⋯ x_N^b_N),
where the first factor is in 
I(Z_1) because by Cond(Z), it follows

	
  ∙ a_1+a_2+b-m_2+m_1-∑_i=3^Nb_i=-2m_2+m_0+m_1+∑_i=0^Na_i≥  m_0+m_1-m_2
	
  ∙ a_2+b-∑_i=3^Nb_i =a_0+m_0-m_2+∑_i=2^Na_i≥ m_0+m_1-m_2
	
  ∙ a_1+b+m_1-m_2-∑_i=3^Nb_i=a_0+a_1 +m_0+m_1-2m_2+ ∑_i=3^Na_i≥ m_0+m_1-m_2.

So, the conditions at Cond(Z_1) are satisfied. As we have seen in the previous subcase, the second factor belongs to I(Z_2). Furthermore, it is easy to check, using Cond(Z_3), that x_0^a_0x_3^b_3⋯ x_N^b_N∈⁠ I(Z_3). Thus 𝒩∈  I(Z_1)· I(Z_2)· I(Z_3).

  (3) a_1 < m_2-m_1 and a_0 ≥ m_2-m_0: the proof  is similar to the previous one.

  (4) a_1 < m_2-m_1 and a_0 < m_2-m_0: by Cond(Z), we have	
b=∑_i=3^Na_i
=(a_1+a_0+b+m_0+m_1-2m_2)+(m_2-m_1-a_1)
+(m_2-m_0-a_0)≥ (m_0+m_1-m_2)+(m_2-m_1-a_1)
+(m_2-m_0-a_0). Because the last three summands are all positive, it is possible to choose for all i=3,…,N some integers 0≤ c_i, d_i, e_i ≤ a_i such that c_i+d_i+e_i ≤ a_i for all i=3,…,N, ∑_i=3^Nc_i=m_0+m_1-m_2, ∑_i=3^Nd_i=m_2-m_1-a_1 and ∑_i=3^Ne_i=m_2-m_0-a_0.
By Cond(Z_i), it  follows that 

    ℳ=(x_3^c_3⋯ x_N^c_N) · (x_1^a_1x_3^d_3⋯ x_N^d_N)· (x_0^a_0x_3^e_i⋯ x_N^e_N)
 belongs to I(Z_1)· I(Z_2)· I(Z_3). Since ℳ divides 𝒩, we deduce 𝒩∈ I(Z_1)· I(Z_2)·⁠ I(Z_3). 

So, in all the possible cases,  𝒩∈  I(Z_1)· I(Z_2)· I(Z_3) and
the proof of the lemma is complete.
The next lemma helps us to deal with subschemes of the type (2q+r)(P_0+P_1+P_2)  that appeared as a factor in the splitting presented in Lemma <ref>.
 
	Let P_0, P_1 and P_2 be three noncollinear points in ℙ^N. If  q,r ∈ℕ with 0≤ r <2, then
	I((2q+r)(P_0+P_1+P_2))=I(2(P_0+P_1+P_2))^q· I(P_0+P_1+P_2)^r.


We give a proof by induction on q. In order to prove the base case q=0, we need to show that
I(r(P_0+P_1+P_2))=I(P_0+P_1+P_2)^r.
But this is trivial for r=0,1. For the induction, we suppose that the lemma is true for q-1 and we prove that it holds for q. We claim that 

    I((2q+r)(P_0+P_1+P_2))=I(2(P_0+P_1+P_2))· I((2(q-1)+r)(P_0+P_1+P_2)).

Proof of the claim. 
Set Z_1=(2q+r)(P_0+P_1+P_2), Z_2=2(P_0+P_1+P_2) and Z_3=(2(q-1)+r)(P_0+P_1+P_2). The inclusion I(Z_2) · I(Z_3) ⊆ I(Z_1) is trivial from the definition. Therefore, we  show that if a monomial 𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(Z_1)
then 𝒩∈ I(Z_2)· I(Z_3).
Thus, consider 𝒩∈ I(Z_1). Set b= ∑_i=3^Na_i. We have the following cases:

(a) Let b ≥ 2, for each i=3,…,N, it can be chosen 0≤ b_i ≤ a_i such that ∑_i=3^Nb_i=2.
If we write
𝒩=(x_3^b_3⋯ x_N^b_N) · (x_0^a_0x_1^a_1x_2^a_2x_3^a_3-b_3⋯ x_N^a_N-b_N), then we can easily deduce by Cond(Z_2) and Cond(Z_3) that 𝒩∈  I(Z_2)· I(Z_3).

(b) Let b=1. We have three subcases.

	
  (1)
a_0=0: by Cond(Z_1) it follows that 

    a_1 ≥ 2q+r-1 ≥ 1   and  a_2 ≥ 2q+r-1 ≥ 1.

Therefore,
𝒩=(x_1x_2x_3^a_3⋯ x_N^a_N) · (x_1^a_1-1x_2^a_2-1),
where  it is easy to see that the  first factor is in I(Z_2).  So we need to show that the second one is in I(Z_3). By Cond(Z_1) it follows


  ∙  a_j-1 ≥2(q-1)+r for j=1,2 by (<ref>)

  ∙ a_1+a_2-2≥ a_2-1 ≥ 2(q-1)+r by (<ref>).

Thus the conditions  at Cond(Z_3) are satisfied.

  (2) a_1=0 and a_2=0:	these cases are similar to the previous one.

  (3) a_0,a_1,a_2>0:
we can write 

    𝒩=(x_0x_1x_2) · (x_0^a_0-1x_1^a_1-1x_2^a_2-1x_3^a_3⋯ x_N^a_N),
 where it is easy to prove that the two factors belong  to I(Z_2) and I(Z_3) respectively.

(c) If we consider b=0, then it is a known case in ℙ^2 (see the end of section 6 in <cit.>). Using the canonical inclusion R_ℙ^2⊆ R_ℙ^N we get I_ℙ^2(Z_i)⊂ I_ℙ^N(Z_i). Hence, we have 
    𝒩∈ I_ℙ^2(Z_1)=I_ℙ^2(Z_2)· I_ℙ^2(Z_3)⊂ I_ℙ^N(Z_2)· I_ℙ^N(Z_3).

So, the proof of the claim is complete. By the inductive step 

   I((2q+r)(P_0+P_1+P_2))

   =I(2(P_0+P_1+P_2))·I((2(q-1)+r)(P_0+P_1+P_2))

   =I(2(P_0+P_1+P_2))·I(2(P_0+P_1+P_2))^q-1·I(P_0+P_1+P_2)^r,

and the proof  is complete.

Now, we can solve our main problem when  ∑_i=0^2m_i is even. 
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_0 ≤ m_1 ≤ m_2. Denote by Z the corresponding fat point scheme Z=m_0P_0+m_1 P_1+m_2P_2. If m_0+m_1> m_2 and m_0+m_1+m_2 is even, then I(Z)^(m)=I(Z)^m  for all  m ∈ℕ and hence ρ(I(Z))=1.

 Since m_0+m_1+m_2 is even, we can set m_0+m_1-m_2=2q.
By applying Lemma <ref> to the fat point scheme kZ=km_0P_0+km_1P_1+⁠ km_2P_2, we obtain 

I(kZ)=  I(k2q(P_0+P_1+P_2))    ·I(k(m_2-m_1)(P_0+P_2))

   ·I(k(m_2-m_0)(P_1+P_2)).

From Lemma <ref> we can also deduce

    I(2q(P_0+P_1+P_2))^(m)=I(2q(P_0+P_1+P_2))^m    ∀ m∈ℕ.

Thus the conditions of Lemma <ref> are satisfied for Z, and we can deduce the desired result.

Let ∑_i=0^2m_i be odd.
Our aim is proving  Theorem <ref>.
Proposition <ref> gives us a suitable lower bound, so we need to prove that ρ(I(Z))≤1+∑m_i/∑m_i. We  will do it by directly considering the definition of resurgence and using further preliminary lemmas on the splitting of the symbolic powers.
By Lemma  <ref> and Lemma <ref>, we can deduce the following corollary.
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_0 ≤ m_1 ≤ m_2. Denote by Z the corresponding fat point scheme Z=m_0P_0+m_1 P_1+m_2P_2. 
If m_0+m_1>m_2 and m_0+m_1+m_2 is odd, then for all k ∈ℕ,

I(Z)^(k)=I(P_0+P_1+P_2)^(k)· I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^(k)= I(P_0+P_1+P_2)^(k)· I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^k.	


Consider k ∈ℕ. We write k=2q_1+r where 0≤ r<2. Since m_0+m_1+m_2 is odd, it follows
m_0+m_1-m_2 is odd and we can write m_0+m_1-m_2=2q_2+1.
Set Z_1=P_0+P_1+P_2.
Because km_0+km_1=k(m_0+m_1) > km_2, we can apply Lemma <ref> to the scheme kZ and we obtain that I(Z)^(k) is equal to

I(k(m_0+m_1-m_2)Z_1) · I(k(m_2-m_1)(P_0+P_2))· I(k(m_2-m_0)(P_1+P_2))=I((2(2q_1q_2+q_1+rq_2)+r)Z_1)· I(k(m_2-m_1)(P_0+P_2))· I(k(m_2-m_0)(P_1+P_2))
=I(2Z_1)^2q_1q_2+q_1+rq_2· I(Z_1)^r · I(k(m_2-m_1)(P_0+P_2))· I(k(m_2-m_0)(P_1+P_2)),
where the last equality holds by Lemma <ref>.
By Lemma <ref>

    I(Z_1)^(k)=I((2q_1+r)Z_1)= I(2Z_1)^q_1· I(Z_1)^r.

 By applying Lemma <ref>  to the scheme 
Z'=k(m_0-1)P_0+k(m_1-1)P_1+k(m_2-1)P_2 (we can use it because k(m_0-1)+k(m_1-1)=k(m_0+m_1-2) ≥ k(m_2-1) since m_0+m_1 ≥ m_2+1) we have
I(Z')=I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^(k)
= I(k(m_0+m_1-m_2-1)Z_1)· I(k(m_2-m_1)(P_0+P_2))· I(k(m_2-m_0)(P_1+P_2))
=I((2(2q_1q_2+rq_2)Z_1)· I(k(m_2-m_1)(P_0+P_2))· I(k(m_2-m_0)(P_1+P_2))
=I(2Z_1)^2q_1q_2+rq_2· I(k(m_2-m_1)(P_0+P_2))· I(k(m_2-m_0)(P_1+P_2)),
where the last equality holds by Lemma <ref>.
Thus, 
    I(Z_1)^(k)· I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^(k)
       =I(2Z_1)^q_1· I(Z_1)^r· I(Z') 
       =I(2Z_1)^2q_1q_2+rq_2+q_1· I(Z_1)^r· I(k(m_2-m_1)(P_0+P_2))
       · I(k(m_2-m_0)(P_1+P_2))=I(Z)^(k).

Finally notice that by Propositions <ref> and <ref> it follows that 

   I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^(k)

   =I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^k.

Notice that in general the equality  I(Z)^(a+b)=I(Z)^(a)· I(Z)^(b) is not satisfied. 
However, the previous results  imply the following corollary which tells us when this splitting is possible for  I(Z). 
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2 ≥max(m_0,m_1). Denote by Z the fat point scheme Z=m_0P_0+m_1 P_1+m_2P_2. If m_0+m_1>m_2 and m_0+m_1+m_2 is odd, then

    I(Z)^(k)=I(Z)^(2i)·I(Z)^(k-2i)   1≤i ≤k/2-1       k    

    I(Z)^(k)=I(Z)^(i)·I(Z)^(k-i)     1≤i ≤k-1       k   ,

i.e., I(Z)^(k)=I(Z)^(i)· I(Z)^(k-i) as long as i and k-i are not both odd.


(a) Suppose that k=2q. Then I(Z)^(2q)=I(2Z)^(q), where 2Z is a fat point scheme that satisfies the condition of the Proposition <ref>. Therefore 

I(2Z)^(q)=I(2Z)^q=I(2Z)^i· I(2Z)^q-i=I(2Z)^(i)· I(2Z)^(q-i)=I(Z)^(2i)· I(Z)^(k-2i).

(b) 
Suppose that k=2q+1.
By Proposition <ref>, Lemma  <ref>, Corollary <ref> and the even case, it follows that

   I(Z)^(2q+1)=I(P_0+P_1+P_2)^(2q+1)

   ·I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^(2q+1)=I(P_0+P_1+P_2)^(2q)

   ·I(P_0+P_1+P_2)·I((m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2)^2q+1

   = I(Z) ·I(Z)^(2q) =I(Z)·I(Z)^(2i)·I(Z)^(2q-2i)

   =I(Z)^(2i+1)·I(Z)^(2q-2i)
 
and the desired result follows.

As a consequence of the results which were proved in <cit.>, we can deduce the following corollary for three simple points in ℙ^2.
 
Let P_0=[1:0:0], P_1=[0:1:0], P_2=[0:0:1]. Then 
    ρ(I_ℙ^2(P_0+P_1+P_2))=4/3.
	

From the previous corollary we can deduce the following useful lemma.
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N. Then 
    I(P_0+P_1+P_2)^(r)⊆ I(P_0+P_1+P_2)^r-1 for   1≤ r ≤ 4.



We work by induction on r. It is trivial for r=1 . For the induction suppose that it is true for r-1 and we prove it for r. Consider
𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(P_0+P_1+P_2)^(r)
then

    a_1+a_2+a_3+⋯+a_N ≥ r 
    
    a_0+a_2+a_3+⋯+a_N ≥ r 
     
    a_0+a_1+a_3+⋯+a_N ≥ r.

Set b=∑_i=3^Na_i. We have the following cases:

(a) Assume b=0. We can see the monomial 𝒩 as an element of the ideal
I_ℙ^2(P_0+P_1+P_2)^(r).
By Corollary <ref>,  ρ(I_ℙ^2(P_0+P_1+P_2))=4/3. Furthermore, r < 4 implies 4r-4 < 3r,  so r/(r-1) > 4/3=ρ(I_ℙ^2(P_0+P_1+P_2)). Then, using the definition of resurgence I_ℙ^2(P_0+P_1+P_2)^(r)⊆ I_ℙ^2(P_0+P_1+P_2)^r-1  for r<4, while it is possible to check computationally that I_ℙ^2(P_0+P_1+P_2)^(4)⊆ I_ℙ^2(P_0+P_1+P_2)^3. Thus 𝒩∈ I_ℙ^2(P_0+P_1+P_2)^r-1. Hence, 𝒩∈ I(P_0+P_1+P_2)^r-1.

(b) Assume ∑_i=3^Na_i=b>0.  There exists i ∈{ 3,…,N} such that a_i>0.
We  may assume i= ⁠ 3.
We  write 𝒩= ⁠(x_3)· (x_0^a_0x_1^a_1x_2^a_2x_3^a_3-1⋯ x_N^a_N), where x_3 ∈ I(P_0+P_1+P_2).
By (<ref>) it follows that x_0^a_0x_1^a_1x_2^a_2x_3^a_3-1⋯ x_N^a_N∈ I(P_0+P_1+P_2)^(r-1)⊆ I(P_0+P_1+P_2)^r-2, where the last inclusion holds for the induction.
 Hence, 𝒩∈ I(P_0+P_1+P_2)^r-1.

Now we can prove the following important lemma.
 
Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2 ≥max(m_0,m_1) and suppose that m_0+m_1 > m_2 and m_0+m_1+m_2  is odd. Let Z=m_0P_0+m_1P_1+m_2P_2 be a scheme of fat points.  Then
	

  (a) I(Z)^(q(1+∑m_i))⊆ I(Z)^q(∑m_i) for all q ∈ℕ, 

  (b)
I(Z)^(q(1+∑m_i)+r)⊆ I(Z)^q(∑m_i)+r-1  for all q ∈ℕ and 0<r<1+∑m_i.
	


Let us start with proving (a) by induction on q. First, we let q=1 as the base case. Thus, we need to prove  I(Z)^(1+∑m_i)⊆ I(Z)^∑m_i.
Set Z_1=P_0+P_1+P_2 and Z_2=(m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2, and we define:

    W(n_0,n_1,n_2)=(n_0+∑_i=0^2n_i)P_0+(n_1+∑_i=0^2n_i)P_1+(n_2+∑_i=0^2n_i)P_2,

for n_i ≥⁠ 1. We claim that I(W(n_0,n_1,n_2)) ⊆ I(Z_1)^∑n_i,  for all  n_i ≥ 1.

Proof of the claim.
We prove by induction on the  sum ∑_i=0^2n_i. The base case is ∑_i=0^2n_i=3, with  n_0=n_1=n_2=1 and by  Lemma <ref> it holds.  Now, we suppose the claim  holds for  n
_i' such that ∑_i=0^2n'_i<∑_i=0^2n_i and we prove it for n_i.
Because we have already considered the case  n_0=n_1=n_2=1, there must exist an i such that n_i>1. We can assume that n_2>1.
We consider the monomial 𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(W(n_0,n_1,n_2)).
Set b=∑_i=3^Na_i.

(i) Let b=0. We have the following subcases.


  (1)
Let a_1=0. By Cond(W(n_0,n_1,n_2)), it follows a_0≥ n_0+n_1+2n_2 ≥∑_i=0^2n_i and a_2≥ 2n_0+n_1+n_2 ≥∑_i=0^2n_i, then it can be written 𝒩=(x_0x_2)^∑n_i x_0^a_0-∑n_ix_2^a_2-∑n_i∈ I(Z_1)^∑n_i, because x_0x_2 ∈ I(Z_1).

  (2)
Let a_0=0: similar to the subcase a_1=0.

  (3)
a_1,a_0 >0: we can write, 𝒩= (x_0x_1)x_0^a_0-1x_1^a_1-1 x_2^a_2, where x_0x_1 ∈ I(Z_1).

Using the fact that the a_i's satisfy  Cond(W(n_0,n_1,n_2)), we can check that x_0^a_0-1x_1^a_1-1 x_2^a_2∈ I(W(n_0,n_1,n_2-1)).  So, by induction (n_2-1 ≥ 1) 
x_0^a_0-1x_1^a_1-1 x_2^a_2∈ I(W(n_0,n_1,n_2-1)) ⊆  I(Z_1)^∑n_i-1,
and 𝒩∈ I(Z_1)^∑n_i.

(ii) Let ∑_i=3^Na_i=b >0. Without loss of generality, let a_3 >0. We can write
𝒩= (x_3) · (x_0^a_0x_1^a_1 x_2^a_2x_3^a_3-1⋯ x_N^a_N),
 where x_3 ∈ I(Z_1). By using Cond(W(n_0,n_1,n_2)), the second factor is in I(W(n_0,n_1,n_2-1))⊆ I(Z_1)^∑n_i-1, where the last inclusion holds by induction. Hence 𝒩∈ I(Z_1)^∑n_i. So, the claim is proved. Now, from the definition  I(Z_1)^1+∑m_i· I(Z_2) ⊆ I(W(m_0,m_1,m_2)). By Corollary <ref>  it follows	that

I(Z)^(1+∑m_i)   =I(Z_1)^(1+∑m_i) ·I(Z_2)^1+∑m_i

   =I(Z_1)^(1+∑m_i)·I(Z_2) ·I(Z_2)^∑m_i

   ⊆I(W(m_0,m_1,m_2)) ·I(Z_2)^∑m_i

   ⊆I(Z_1)^∑m_i·I(Z_2)^∑m_i

   =(I(Z_1) ·I(Z_2))^∑m_i=I(Z)^∑m_i,

and the base case is proved.

 We suppose that (a) is true for  q-1, then we prove it for q.
By induction and Corollary <ref>, using the fact that 1+∑m_i is even,

I(Z)^(q(1+∑m_i))   =I(Z)^((q-1)(1+∑m_i))·I(Z)^(1+∑m_i) 

   ⊆I(Z)^(q-1)(∑m_i)·I(Z)^∑m_i=I(Z)^q(∑m_i).

For proving (b), we work by induction on q as before.
First of all, we need to prove the base case of q=0. Hence, we need to show I(Z)^(r)⊆ I(Z)^r-1 for  1<r<1+∑m_i. Set Z_1=P_0+P_1+P_2 and Z_2=(m_0-1)P_0+(m_1-1)P_1+(m_2-1)P_2 . We define:

    V(n_0,n_1,n_2,r)=(r+n_0-1)P_0+(r+n_1-1)P_1+(r+n_2-1)P_2,

for n_i ≥ 1 and 1<r<1+∑n_i.	We claim that I(V(n_0,n_1,n_2,r)) ⊆ I(Z_1)^r-1 always holds.
	
Proof of the claim.
We work by induction on the sum ∑n_i. The base case is   n_i=1. Then we have to prove 
I(P_0+P_1+P_2)^(r)⊆ I(P_0+P_1+P_2)^r-1,  for  1<r<4
and this is true by Lemma <ref>.
We suppose that the claim is true for assignment n
_i' such that ∑n_i'<∑n_i, then we prove it for n_i.
Because we have already considered the case  n_0=n_1=n_2=1, there must exist an i such that n_i>1. We can assume that n_2>1.
We consider 𝒩=x_0^a_0x_1^a_1⋯ x_N^a_N∈ I(V(n_0,n_1,n_2,r)).
Set b=∑_i=3^Na_i, and we consider cases depending upon b.

(i)
let b=0. We have the following subcases.


  (1)
a_1=0: by Cond(V(n_0,n_1,n_2,r)), it follows	that

    a_0≥ r+n_2-1≥ r-1	  and  a_2≥ r+n_0-1 ≥ r-1.
			
So we can write	
𝒩=(x_0x_2)^r-1 x_0^a_0-r+1x_2^a_2-r+1∈ I(Z_1)^r-1, because x_0x_2 ∈ I(Z_1).

  (2)
a_0=0: similar to the case a_1=0.			

  (3)
a_1,a_0 >0: we write,
	𝒩= (x_0x_1) x_0^a_0-1x_1^a_1-1 x_2^a_2, where x_0x_1 ∈ I(Z_1).
 By  Cond(V(n_0,n_1,n_2,r)), we deduce x_0^a_0-1x_1^a_1-1 x_2^a_2∈ I(V(n_0,n_1,n_2-1,r-1)). 
So for the inductive step (n_2-1 ≥ 1 and r-1<(n_0+n_1+n_2-1)+1) we conclude
x_0^a_0-1x_1^a_1-1 x_2^a_2∈ I(V(n_0,n_1,n_2-1,r-1)) ⊆ I(Z_1)^r-2, and 𝒩∈ I(Z_1)^r-1.

(ii) Let ∑_i=3^Na_i=b >0. We can assume that a_3 >0. We can write
𝒩= (x_3) (x_0^a_0x_1^a_1 x_2^a_2x_3^a_3-1⋯ x_N^a_N), where x_3 ∈ I(Z_1). By Cond(V(n_0,n_1,n_2,r)), we have that


    x_0^a_0x_1^a_1 x_2^a_2x_3^a_3-1⋯ x_N^a_N∈ I(V(n_0,n_1,n_2-1,r-1)).

So by induction (n_2-1 ≥ 1 and r-1<(n_0+n_1+n_2-1)+1) we see

x_0^a_0x_1^a_1 x_2^a_2x_3^a_3-1⋯ x_N^a_N∈ I(V(n_0,n_1,n_2-⁠1,r-1)) ⊆ I(Z_1)^r-2, and 𝒩∈ I(Z_1)^r-1. So the claim is true. From the definition  I(Z_1)^r· I(Z_2) ⊆ I(V(m_0,m_1,m_2,r)). By Corollary <ref>, 

I(Z)^(r)   =I(Z_1)^(r)·I(Z_2)^r=	I(Z_1)^(r) ·I(Z_2) ·I(Z_2)^r-1
 
   ⊆I(V(m_0,m_1,m_2,r))·I(Z_2)^r-1 ⊆I(Z_1)^r-1·I(Z_2)^r-1

    = (I(Z_1)·I(Z_2))^r-1=I(Z)^r-1,

and the base case is proved. Now we can proceed with the inductive step. We suppose that (b) is true for q-1, then we prove it for q. By induction and  Corollary <ref> we can write, using the fact that 1+∑m_i is even,

	I(Z)^(q(1+∑m_i)+r)   =I(Z)^((q-1)(1+∑m_i)+r)·I(Z)^(1+∑m_i)

     ⊆   I(Z)^(q-1)(∑m_i)+r-1·I(Z)^∑m_i=I(Z)^q(∑m_i)+r-1.

Thus the proof  is  complete.

By Lemma <ref> we can deduce the following crucial corollary.
 
	Let P_0, P_1 and P_2 be noncollinear points in ℙ^N and m_2 ≥max(m_0,m_1), and suppose that m_0+m_1 > m_2 and ∑_i=0^2m_i is odd. If Z=m_0P_0+m_1P_1+m_2P_2, then
	ρ(I(Z)) ≤ (1+∑_i=0^2m_i)/(∑_i=0^2m_i).


It is enough to show that if m/n ≥ (1+∑m_i)/(∑m_i) then  I(Z)^(m)⊆⁠ I(Z)^n. Suppose that m and n are such that
m/n ≥ (1+∑m_i)/(∑m_i). Then we can deduce m ≥⌈1+∑m_i/∑m_in ⌉. Now, n can be written as n=q∑m_i+r with 0≤ r<∑m_i. Thus 

    m ≥⌈  q(1+∑m_i)+r+r/∑m_i⌉= 
    q(1+∑m_i)  if  r = 0 
    
    q(1+∑m_i)+r+1  if  r ≠ 0.

If r=0, by Lemma <ref> 

    I(Z)^(m)⊆ I(Z)^(q(1+∑m_i))⊆ I(Z)^q∑m_i=I(Z)^n.

If r ≠ 0, then r'=r+1<1+∑m_i.   By Lemma <ref>

    I(Z)^(m)⊆ I(Z)^(q(1+∑m_i)+r')⊆ I(Z)^q∑m_i+r'-1=I(Z)^n.

Then  ρ(I(Z)) ≤ (1+∑_i=0^2m_i)/ (∑_i=0^2m_i).

 From Corollary <ref> and Proposition <ref> we can immediately deduce  Theorem <ref>, therefore we have a complete description for the resurgence of a fat point scheme consisting of three noncollinear points of ℙ^N. 


§ ACKNOWLEDGEMENT

This project started during the summer school PRAGMATIC 2017. The authors would like to thank Enrico Carlini,  Tai Huy Ha, Brian Harbourne and Adam Van Tuyl,  for giving very interesting lectures and for sharing open problems. They wish to thank all the organizers of PRAGMATIC 2017 for giving them the opportunity to attend the school. The authors owe also special thanks to the anonymous referee for the interesting and extensive comments on an earlier version of this paper.


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