Zero-dimensional subschemes of projective spaces related to double points of linear subspaces and to fattening directions

Edoardo Ballico^{i}

Department of Mathematics, University of Trento ballico@science.unitn.it

Received: 18.1.2016; accepted: 8.4.2016.

Abstract. Fix a linear subspace V ⊆ P^{n}and a linearly independent set S ⊂ V . Let ZS,V ⊂ V
or Zs,r with r := dim(V ) and s = ](S), be the zero-dimensional subscheme of V union of all
double points 2p, p ∈ S, of V (not of P^{n}if n > r). We study the Hilbert function of ZS,V and
of general unions in P^{n}of these schemes. In characteristic 0 we determine the Hilbert function
of general unions of Z2,1(easy), of Z2,2and, if n = 3, general unions of schemes Z3,2and Z2,2.

Keywords:zero-dimensional scheme, Hilbert function, postulation

MSC 2000 classification:primary 14N05

Introduction

Fix P ∈ P^{n}. Look at all possible zero-dimensional schemes Z with Z_{red} =
{P } and invariant for the action of the group G_{P} of all h ∈ Aut(P^{n}) such that
h(P ) = P . In characteristic zero we only get the infinitesimal neighborhoods
mP of P in P^{n}, m > 0, i.e the closed subschemes of P^{n} with (I_{P})^{m} as its
ideal sheaf. If we take two distinct points P, Q ∈ P^{n}, P 6= Q, we also have a
line (the line L spanned by the set {P, Q}) and it is natural to look at the
zero-dimensional schemes Z ⊂ P^{n} such that Z_{red} = {P, Q} and h^{∗}(Z) ∼= Z
for all h ∈ Aut(P^{n}) fixing P and Q (or, if we take non-ordered points fixing
the set {P, Q}) and in particular fixing L. A big restriction (if n > 1) is to
look only to the previous schemes Z which are contained in L, not just with
Z_{red} = {P, Q} ⊂ L. The easiest invariant zero-dimensional scheme (after the
set {P, Q}) is the degree 4 zero-dimensional scheme (2P + 2Q, L), i.e. the zero-
dimensional subscheme Z2,1 of L with 2 connected components, both of degree
2, and with {P, Q} as its support. We call them (2, 1)-schemes. This is a kind of
collinear zero-dimensional schemes and hence the Hilbert function of a general

iThis work is partially supported by MIUR and GNSAGA (INDAM) http://siba-ese.unisalento.it/ c 2016 Universit`a del Salento

union of them is known ([1], [6]). We may generalize this construction in the following way.

For any linear space V ⊆ P^{n} and any P ∈ V let (2P, V ) denote the closed
subscheme of V with (I_{P,V})^{2} as its ideal sheaf. We have (2P, V ) = 2P ∩ V ,
(2P, V )_{red}= {P } and deg(2P, V ) = dim(V ) + 1. Fix integer n ≥ r ≥ s − 1 ≥ 0.

Fix an r-dimensional linear subspace V ⊆ P^{n} and a linearly independent set
S ⊂ V with ](S) = s. Set Z_{S,V} := ∪_{p∈S}(2p, V ). Z_{S,V} is a zero-dimensional
scheme, (ZS,V)red = S, ZS,V ⊂ V and deg(Z_{S,V}) = s(r + 1). Any two schemes
Z_{S}^{0}_{,V}^{0} and Z_{S,V} with dim(V ) = dim(V^{0}) and ](S) = ](S^{0}) are projectively
equivalent. In the case s = r + 1 we may see Z_{S,V} as the first order invariant
of the linearly independent set S inside the projective space V (not the full
projective space P^{n} if r < n) and V is exactly the linear span of S, so that it is
uniquely determined by the set S ⊂ P^{n}. In the case s ≤ r, Z_{S,V} is not uniquely
determined by S. The scheme ZS,V prescribes some infinitesimal directions at
each point of S, so that each connected component of Z_{S,V} spans V . Z_{S,V} is the
minimal zero-dimensional subscheme B ⊂ P^{n} such that B_{red}⊇ S and the linear
span of each connected component A of B spans a linear space VA ⊇ V . In all
cases Z_{S,V} depends only on S and V and not on the projective space containing
V . An (s, r)-scheme of P^{n} or a scheme Z_{s,r} of P^{n} is any scheme Z_{S,V} ⊂ P^{n}
for some S, V with dim(V ) = r and ](S) = s. Set Zs := Zs,s−1. We have
deg(Zs,r) = s(r + 1). If ](S) = 1 we say that Z_{S,V} is a 2-point of V . A scheme
Z_{1,2} is called a planar 2-point.

Let Z ⊂ P^{n} be a zero-dimensional scheme. Consider the exact sequence
0 → IZ(t) → O_{P}^{n}(t) → OZ(t) → 0 (0.1)
The exact sequence (0.1) induces the map r_{Z,t} : H^{0}(O_{P}^{n}(t)) → H^{0}(O_{Z}(t))
(the restriction map). We say that Z has maximal rank if for each t ∈ N the
restriction map r_{Z,t} is either injective (i.e. h^{0}(I_{Z}(t)) = 0) or surjective (i.e.

h^{1}(I_{Z}(t)) = 0). For each t ∈ N let hZ(t) be the rank of the restriction map r_{Z,t}.
We have h_{∅}(t) = 0 for all t ∈ N. If Z 6= ∅, then hZ(0) = 1 and the function hZ(t)
is strictly increasing until it stabilizes to the integer deg(Z). The regularity index
ρ of Z is the first t ∈ N such that hZ(t) = deg(Z), i.e. such that h^{1}(I_{Z}(t)) = 0.

By the Castelnuovo - Mumford lemma the homogeneous ideal of Z is generated
in degree ≤ ρ(Z) + 1 and h^{1}(I_{Z}(t)) = 0 for all t ≥ ρ(Z). If Z is contained in a
proper linear subspace W ⊂ P^{n}, then each h_{Z}(t) does not depend on whether
one sees Z as a subscheme of P^{n} or of W and a minimal set of generators of
the homogeneous ideal of Z in P^{n} is obtained lifting to P^{n} a minimal set of
generators of the homogeneous ideal of Z in W and adding n − dim(W ) linear
equations. The integer hZS,V(t) only depends on r, s and t (see Remark 2). See
Proposition 1 for the Hilbert function of each Z_{S,V}.

We study the Hilbert function of general unions of these schemes. There are
obvious cases with non-maximal rank for O_{P}^{n}(2), but we compute the exact
values of hZ(2) (see Propositions 2 and 3). For Z3,2 exceptional cases arise also
with respect to O_{P}^{n}(d) if d = 3 and n ≤ 4 (Proposition 4) and one case with
d = 4 and n = 3 (as expected by the Alexander-Hirschowitz theorem [1], [4])
(see Theorem 3).

For the schemes Z2,2 we prove the following results (only in characteristic zero).

Theorem 1. Fix integers n ≥ 2, d ≥ 3 and k ≥ 2. Let Z ⊂ P^{n} be a general
union of k schemes Z_{2,2}. Then either h^{0}(I_{Z}(d)) = 0 or h^{1}(I_{Z}(d)) = 0.

Theorem 2. Fix integers n ≥ 2, d ≥ 3 and k ≥ 2. Let Z ⊂ P^{n} be a general
union of k schemes Z_{2,2} and one planar 2-point. Then either h^{0}(I_{Z}(d)) = 0
or h^{1}(IZ(d)) = 0, except in the case (n, k, d) = (2, 2, 4) in which h^{0}(IZ(4)) =
h^{1}(IZ(4)) = 1.

For general unions of an arbitrary number of schemes Z3,2and Z2,2 we prove
the case n = 3 (see Theorem 3 for O_{P}^{3}(d), d ≥ 3).

We also explore the Hilbert function of general unions of zero-dimensional schemes and general lines (see Lemma 6 for a non-expected easy case with non maximal rank).

We work over an algebraically closed field K with characteristic 0. We heavily use this assumption to apply several times Remark 3.

1 Preliminaries

For any closed subscheme Z ⊂ P^{n}and every hyperplane H ⊂ P^{n}let ResH(Z)
be the closed subscheme of P^{n} with I_{Z} : I_{H} as its ideal sheaf. For each t ∈ Z
we have a residual exact sequence

0 → I_{Res}_{H}_{(Z)}(t − 1) → IZ(t) → IZ∩H,H(t) → 0 (1.1)
We say that (1.1) is the residual exact sequence of Z and H. We have Res_{H}(Z) ⊆
Z. If Z is zero-dimensional, then deg(Z) = deg(Z ∩ H) + deg(ResH(Z)).

For any scheme Z ⊂ P^{n}let hZ : N → N denote the Hilbert function of Z, i.e.

for each t ∈ N let hZ(t) denote the rank of the restriction map H^{0}(O_{P}^{n}(t)) →
H^{0}(O_{Z}(t)).

Remark 1. Let V ⊆ P^{n} be an r-dimensional linear subspace. Assume Z ⊂
V . Since for each t ∈ N the restriction map H^{0}(O_{P}^{n}(t)) → H^{0}(V, O_{V}(t)) is
surjective, the Hilbert function of Z is the same if we see Z as a subscheme of
P^{n}or if we see it as a subscheme of the r-dimensional projective space V .

Remark 2. Fix integers n ≥ r > 0 and s with 1 ≤ s ≤ r + 1. Fix linear
spaces V_{i} ⊆ P^{n}, i = 1, 2, and sets S_{i} ⊂ V_{i}, i = 1, 2, such that dim(V_{i}) = r,
](Si) = s and each Si is linearly independent. Since there is h ∈ Aut(P^{n}) with
h(V_{1}) = V_{2} and h(S_{1}) = S_{2}, Z_{S}_{1}_{,V}_{1} and Z_{S}_{2}_{,V}_{2} have the same Hilbert function.

Proposition 1. Let Z := Z_{S,V} be an (s, r)-scheme.

(i) If s = 1, then h_{Z}(t) = r + 1 for all t ≥ 1 and the homogeneous ideal
of ZS,V is generated by forms of degree ≤ 2.

(ii) If s ≥ 2, then h_{Z}(0) = 1, h_{Z}(1) = r + 1, h_{Z}(2) = (r + 1)s − s(s − 1)/2,
h_{Z}(t) = s(r + 1) for all t ≥ 3 and the homogeneous ideal of Z_{S,V} is generated
by forms of degree ≤ 4, but not of degree ≤ 3. Outside S the scheme-theoretic
base locus of |I_{Z}(3)| is the union of all lines spanned by 2 of the points of S.

Outside S the scheme-theoretic base locus of |I_{Z}(2)| is the linear span of S.

Proof. Since Z ⊂ V , we may assume n = r, i.e. V = P^{r} (Remark 1).

Part (i) is well-known. The case r = 1, s ≥ 2 is also obvious, by the cohomol-
ogy of line bundles on P^{1}. Hence we may assume r ≥ 2, s ≥ 2 and use induction
on the integer r. We assume that Proposition 1 is true for all pairs (s^{0}, r^{0}) with
1 ≤ r^{0} < r and 1 ≤ s^{0} ≤ r^{0}+ 1. For any (s, r)-scheme Z we have h_{Z}(0) = 1 and
h_{Z}(1) = r + 1. Since S is linearly independent, we have h^{1}(I_{S}(t)) = 0 for all
t > 0.

(a) Assume s ≤ r. Let H ⊂ P^{r} be a hyperplane containing S. We have
Z ∩ H = Z_{S,H} and Res_{H}(Z) = S. From the residual exact sequence (1.1) and
the inductive assumption we get h^{1}(IZ(t)) = 0 for all t ≥ 3, h^{1}(IZ(2)) =
h^{1}(H, I_{Z∩H,H}(2)) and h^{0}(I_{Z}(2)) = h^{0}(I_{Z∩H,V ∩H}(2)) + r + 1 − s. The inductive
assumption gives h^{1}(H, I_{Z∩H,H}(2)) = s(s − 1)/2. Hence h_{Z}(2) = deg(Z) − s(s −
1)/2 = s(r + 1) − s(s − 1)/2. We also get that outside S the base locus of |IZ(t)|,
t = 2, 3, and of |IZ∩H(t)| are the same. By the Castelnuovo-Mumford’s lemma
the homogeneous ideal of Z is generated in degree ≤ 4. It is not generated in
degree ≤ 3, because |IZ(3)| has a one-dimensional base locus.

(b) Assume s = r + 1 ≥ 3. Since V = P^{r}, S spans P^{r} and every quadric
hypersurface of P^{r} has as its singular locus a proper linear subspace of P^{r}, we
have h^{0}(IZ(2)) = 0 and hence hZ(2) = ^{r+2}_{2}

and h^{1}(IZ(2)) = (r + 1)^{2} −

r+2 2

= (r + 1)r/2 = s(s − 1)/2. Fix p ∈ S and set S^{0} := S \ {p}. Let H
be the hyperplane spanned by S^{0}. We have Z ∩ H = Z_{S}^{0}_{,H} and Res_{H}(Z) =
2p ∪ S^{0}. We have h^{0}(I_{2p∪S}^{0}(1)) = 0 and so h^{1}(I_{2p∪S}^{0}(1)) = s − 1. We have
ResH(2p ∪ S^{0}) = 2p. Since h^{1}(I2p(x)) = 0 for all x > 0 and S^{0} ⊂ H is linearly
independent, the residual exact sequence of 2p ∪ S^{0} with respect to H gives
h^{1}(I_{2p∪S}^{0}(t)) = 0 for all t ≥ 2. Hence (1.1) and the inductive assumption gives
h^{1}(IZ(t)) = 0 for all t ≥ 3. The scheme-theoretic base locus of |IZ(2)| is V =
P^{r}. The scheme-theoretic base locus E of |I_{Z}(3)| contains the union T of all

lines spanned by two of the points of S, which in turn contains Z. Since Z
is zero-dimensional, we have h^{1}(Z, I_{Z∩H,Z}(3)) = 0 and so the restriction map
H^{0}(OZ(3)) → H^{0}(OZ∩H(3)) is surjective. Since h^{1}(IZ(3)) = 0, the restriction
map H^{0}(O_{P}^{r}(3)) → H^{0}(O_{Z∩H}(3)) is surjective. Therefore E ∩ H is the scheme-
theoretic base locus of |I_{Z∩H,H}(3)| in H. By the inductive assumption we get
E ∩ H = T^{0} outside S^{0}.

Now we check that E_{red} = T . If r = 2, then this is true (even scheme-
theoretically), because h^{0}(IZ(3)) = 1 and so |IZ(3)| = {T }. Now assume r ≥ 3
and that this assertion is true for lower dimensional projective spaces. Fix q ∈
P^{r}\ T . Let S^{00}⊆ S be a minimal subset of S whose linear span contains q. Since
S is linearly independent, if S1 ⊆ S and the linear span of S_{1} contains q, then
S1 ⊇ S^{00}. Since q /∈ T , we have ](S^{00}) ≥ 3. Take 3 distinct points p1, p2, p3 of
S^{00} and let H_{i}, i = 1, 2, 3, be the linear span of S \ {p_{i}}. Since any two points
of S are contained in at least one of the hyperplanes H1, H2 or H3, we have
T ⊂ H1 ∪ H_{2}∪ H_{3}. Since each point of S is contained in at least two of the
hyperplanes H_{i}, we have Z ⊂ H_{1}∪ H_{2}∪ H_{3}. Since p_{i} ∈ S^{00}, i = 1, 2, 3, we have
q /∈ H_{i} and so q /∈ (H_{1}∪ H_{2}∪ H_{3}). Thus q /∈ E_{red}. Hence Ered= T .

To conclude the proof of (ii) it is sufficient to prove that E is reduced outside
S. For any set B ⊂ P^{r} let hBi denote its linear span. Fix p ∈ T \ S and call
p1 and p2 the points of S such that p is contained in the line ` spanned by
{p_{1}, p2}. Since ` ⊂ T , it is sufficient to prove that ` is the Zariski tangent space
T_{p}E of E at p. Assume the existence of a line R ⊂ T_{p}E such that p ∈ R and
R 6= `. Let SR ⊆ S be a minimal subset of S whose linear span contains R.

Since S is linearly independent, any B ⊆ S with R ⊂ hBi contains S_{R}. Set
α := ](S_{R}). Since R * T , we have α ≥ 3. Let G be the set of all h ∈ Aut(P^{r})
such that h(u) = u for all u ∈ S. Note that g(ZS,V) = ZS,V for all g ∈ G.

Set Gp := {g ∈ G : g(p) = p}. Gp acts transitively on the set ∆_{R} of all lines
L ⊂ hS_{R}i such that p ∈ L and S_{R} is the minimal subset of S whose linear span
contains L. Since Gp acts transitively on ∆R, each L ∈ ∆Ris contained in TpE.

Since TpE is closed, it contains all lines L1 ⊂ hS_{R}i such that p ∈ L_{1}. Hence
(changing if necessary R) we reduce to the case α = 3. Assume for the moment
{p_{1}, p2} ⊂ S_{R} and write SR = {p1, p2, p3}. Set Π := hS_{R}i. E ∩ Π contains
T ∩ Π, i.e. the 3 distinct lines of the plane Π spanned by 2 of the points of SR.
E ∩ Π ) T ∩ Π, because TpE contains the line h{p, p_{3}}i and so the scheme E ∩ Π
contains the tangent vector of h{p, p3}i at p. Since T ∩ Π is a cubic curve, we get
Π ⊂ E. Hence E_{red}6= T , a contradiction. Now assume S_{R}∩ {p_{1}, p2} = ∅. Since
p ∈ `, we have hS_{R}i ∩ ` 6= ∅ and so S_{R}∪ {p_{1}, p_{2}} is not linearly independent, a
contradiction. Now assume ](S_{R}∩ {p_{1}, p_{2}}) = 1. Since p ∈ `, we get ` ⊂ hS_{R}i
and hence {p1, p2} ⊂ S_{R}, a contradiction. ^{QED}
Remark 3. Let X be an integral projective variety with dim(X) > 0, L a

line bundle on X and V ⊆ H^{0}(L) any linear subspace. Take a general p ∈ X_{reg}
and a general tangent vector A of X at p. We have dim(H^{0}(I_{A}⊗ L) ∩ V ) =
max{0, dim(V ) − 2}, because (in characteristic zero) any non-constant rational
map X 99K P^{r}, r ≥ 1, has non-zero differential at a general p ∈ X_{reg}.

Lemma 1. Let V ⊆ H^{0}(O_{P}^{n}(2)), n ≥ 2, be any linear subspace such that
dim(V ) ≥ n + 2. Let L ⊂ P^{n} be a general line. Then dim(V ∩ H^{0}(IL(2))) =
dim(V ) − 3.

Proof. Since two general points of P^{n}are contained in a line, we have dim(V ∩
H^{0}(I_{L}(2))) ≤ max{0, dim(V ) − 2}, without any assumption on dim(V ). Let B
denote the scheme-theoretic base locus. Since dim(V ) ≥ n + 2 > h^{0}(O_{P}^{n}(1)), we
have dim(B) ≤ n − 2. Hence B ∩ L = ∅ for a general line L. Let f : P^{n}\ B → P^{r},
r = dim(V )−1, be the morphism induced by V . We have dim(V ∩H^{0}(I_{L}(2))) =
dim(V ) − 2 if and only if f (L) is a line. Assume that this is the case for a general
L. Since any two points of P^{n} are contained in a line, we get that the closure
Γ of f (P^{n}\ B) in P^{r} is a linear space. Since Γ spans P^{r}, we get Γ = P^{r}. Hence

dim(V ) = r + 1 ≤ n + 1, a contradiction. ^{QED}

Remark 4. Let Z ⊂ P^{n}, n ≥ 2, be a general union of k schemes Z2,1. We
have k general lines L_{i}, 1 ≤ i ≤ k, of P^{n}and on each L_{i}a general subscheme of L_{i}
with 2 connected components, each of them with degree 2. Set T := L_{1}∪· · ·∪L_{k}.
We have h^{0}(IZ(2)) = h^{0}(IT(2)), where T ⊂ P^{n} is a general union of k lines. If
n = 2, then h^{0}(I_{T}(2)) = 0 if k ≥ 3 and h^{0}(I_{T}(2)) = ^{4−k}_{2} if k = 1, 2. If n ≥ 3,
then h^{0}(I_{T}(2)) = max{0, ^{n+2}_{2} − 3k} ([7]).

Fix an integer d ≥ 3. If n = 2 assume 4s ≤ ds + 1 − ^{s−1}_{2} for all s with
2 ≤ s ≤ min{k, d + 1}. Note that the family of all schemes Z has a degeneration
Z^{0} in which Z^{0} has k connected components Wi, 1 ≤ i ≤ k, with Wi ⊂ L_{i} and
(W_{i})_{red}a general point of L_{i}. In the terminology of [2] each W_{i} is a collinear jet.

By semicontinuity we have h^{0}(I_{Z}(d)) ≤ h^{0}(I_{Z}^{0}(d)) and h^{1}(I_{Z}(d)) ≤ h^{1}(I_{Z}^{0}(d)).

Hence either h^{0}(IZ(d)) = 0 or h^{1}(IZ(d)) = 0 ([2]).

Notation Let A ⊆ P^{n}be a plane. Fix 3 non-collinear points p_{1}, p_{2}, p_{3} ∈ A.

Let L, R ⊂ A be lines with L ∩ {p_{1}, p_{2}, p_{3}} = {p_{3}} and R ∩ {p_{1}, p_{2}, p_{3}} =
{p_{2}}. Let Z[8] ⊂ P^{n} denote any scheme projectively equivalent to (2p1, A) ∪
(2p_{2}, A) ∪ (2p_{3}, L). Let Z[7] ⊂ P^{n} denote any scheme projectively equivalent to
(2p_{1}, A) ∪ (2p_{2}, A) ∪ {p_{3}}. In both cases p_{3} is called the vertex of Z[8] or of Z[7]

and L is called the vertex line of Z[8]. Let Z[5] ⊂ P^{n} (resp. Z[4] ⊂ P^{n}, resp.

Z^{0}[5] ⊂ P^{n}) denote any scheme projectively equivalent to (2p_{1}, A) ∪ (2p_{2}, R)
(resp. (2p_{1}, A) ∪ {p_{2}}, resp. (2p_{1}, A) ∪ {p_{2}, p_{3}}).

Lemma 2. Fix integers n ≥ 2, d ≥ 3, x ≥ 0 and c ≥ 0 and a zero-
dimensional scheme Γ ⊂ P^{n}such that either h^{0}(I_{Γ∪W}(d)) = 0 or h^{1}(I_{Γ∪W}(d)) =

0, where W ⊂ P^{n} is a general union of x + c schemes Z_{3,2}. Let Z ⊂ P^{n} be a
general union of x schemes Z_{3,2} and c schemes Z[8]. Then either h^{0}(I_{Γ∪Z}(d)) =
0 or h^{1}(IΓ∪Z(d)) = 0.

Proof. We use induction on c, the case c = 0 being true for all x by assumption.

Assume c > 0 and set e := h^{0}(IΓ(d)) − 9x − 8c. First assume e > 0. Let Z^{0} ⊂ P^{n}
be a general union of x + 1 schemes Z_{3,2} and c − 1 schemes Z[8]. The inductive
assumption gives h^{0}(I_{Γ∪Z}^{0}(d)) = e − 1 and h^{1}(I_{Γ∪Z}^{0}(d)) = 0. Since Z is general,
we may find Z^{0} with Z^{0} ⊃ Z and h^{1}(I_{Γ∪Z}^{0}(d)) = 0. Thus h^{1}(IΓ∪Z(d)) = 0. Now
assume e ≤ 0. We need to prove that h^{0}(I_{Γ∪Z}(d)) = 0. Decreasing if necessary
c we may assume e ≥ −7. Let E ⊂ P^{n}be a general union of x schemes Z_{3,2} and
c−1 schemes Z[8]. Let A ⊂ P^{n}be a general plane. Let U ⊂ A be a general scheme
Z_{3,2}. Note that (Γ ∪ E) ∩ U = ∅ even if n = 2. The inductive assumption gives
h^{1}(I_{Γ∪E}(d)) = 0, h^{0}(I_{Γ∪E}(d)) = 8 + e and h^{0}(I_{Γ∪E∪U}(d)) = 0, i.e. U imposes
8 + e independent conditions to H^{0}(IΓ∪E(d)). Let U^{0} be a minimal subscheme
of U with h^{0}(I_{Γ∪E∪U}^{0}(d)) = 0. If U^{0} ( U , then we may find Z[8] ⊇ U^{0} and
so h^{0}(I_{Γ∪Z}(d)) = 0. Now assume U^{0} = U . We need to find a contradiction.

Write U = U1 ∪ U_{2} ∪ U_{3} with Ui = (2pi, U ) and p1, p2, p3 general in A. If
8 + e ≤ 6 we use Remark 3 and that U contains 3 general tangent vectors.

Assume 8 + e = 7. We get h^{0}(I_{Γ∪E∪U}_{i}(d)) < h^{0}(I_{Γ∪E}(d)) − 2 for at least one
index i, say h^{0}(IΓ∪E∪U1(d)) = h^{0}(IΓ∪E(d)) − 3; then we use Remark 3 and that
we may find Z[8] ⊂ A containing U1 and 2 general tangent vectors of A. Now
assume 8 + e = 8. In this case we first get h^{0}(I_{Γ∪E∪U}_{i}∪U_{j}(d)) = h^{0}(I_{Γ∪E}(d)) − 6
for some i 6= j and then apply once Remark 3. ^{QED}

2 Proof of Theorems 1 and 2

Unless otherwise stated from now on a 2-point means a planar 2-point.

For all positive integers n, d set u_{d,n}:= b ^{n+d}_{n} /6c and v_{d,n}= ^{n+d}_{n} − 6u_{d,n}.
We have

6ud,n+ vd,n=n + d d

, 0 ≤ vd,n≤ 5 (2.1)

Note that if Z ⊂ P^{n} is a disjoint union of x schemes Z2,2 we have h^{0}(OZ(d)) ≤
h^{0}(O_{P}^{n}(d)) if and only if x ≤ u_{d,n}. If d ≥ 2 and n ≥ 2 from (2.1) for the integers
d and d − 1 we get

6(u_{d,n}− u_{d−1,n}) + v_{d,n}− v_{d−1,n}=n + d − 1
n − 1

(2.2)
From (2.2) we get that u_{d,n−1} = u_{d,n}− u_{d−1,n} and v_{d,n−1} = v_{d,n} − v_{d−1,n} if
vd,n≥ v_{d−1,n}, while ud,n−1= ud,n− u_{d−1,n}− 1 and v_{d,n−1}= 6 + vd,n− v_{d−1,n} if
v_{d,n}< v_{d−1,n}.

Proposition 2. Fix integers n ≥ 2 and k > 0. Let Z ⊂ P^{n} be a general
union of k schemes Z_{2,2}.

(a) Assume n = 2. If k = 1, then h^{0}(I_{Z}(2)) = 1. If k ≥ 2, then
h^{0}(I_{Z}(2)) = 0.

(b) Assume n = 3. If k = 1, then h^{0}(IZ(2)) = 5. If k = 2, then
h^{0}(I_{Z}(2)) = 1. If k ≥ 3, then h^{0}(I_{Z}(2)) = 0.

(c) If n ≥ 4, then h^{0}(I_{Z}(2)) = max{0, ^{n+2}_{2} − 5k}.

Proof. Let W be any scheme of type Z_{2,2} and let A be the plane containing
W . We have h^{0}(A, IW,A(2)) = 1 (the only conic of A containing W is a double
line) and hence h^{1}(A, I_{W,A}(2)) = 1. Hence h^{0}(I_{Z}(2)) ≥ max{0, ^{n+2}_{2} − 5k}

and h^{1}(I_{Z}(2)) ≥ k. Let J be the line spanned by {p_{1}, p_{2}} := W_{red}. We have
h^{0}(IW(2)) = H^{0}(IB(2)), where B is the union of (2p1, A) and any degree 2
scheme with p2as its reduction, contained in A and not contained in J . Therefore
it is sufficient to prove that a general union Z^{0}of k degree 5 schemes projectively
equivalent to B satisfies h^{0}(IZ^{0}(2)) = max{0, ^{n+2}_{2} − 5k}, except in the case
(n, k) = (3, 2). B is a scheme Z[5].

If n = 2, then the result is obvious. Now assume n = 3. If k = 1, then Z
is contained in a plane and from the case n = 2, k = 1 we get h^{1}(IZ(2)) = 1.

If k = 2, then h^{0}(I_{Z}(2)) ≥ 1, because Z is contained in a reducible quadric.

Let A_{1}, A_{2} be the two planes containing the two schemes Z_{2,2} of Z and call
Ni ⊂ A_{i} the scheme Z2,2 contained in Ai and let Li be the line spanned by
(Ni)_{red}. Fix p ∈ A1\ L_{1}. By the case n = 2 we have h^{0}(A1, I_{N}_{1}_{∪{p},A}_{1}(2)) = 0.

Taking p = A_{1}∩ L_{2} we get that f_{|A}_{1} ≡ 0 for each f ∈ H^{0}(I_{Z}(2)) vanishes on
A1. Similarly f|A_{2} ≡ 0. Hence |I_{Z}(2)| = {A1∪ A_{2}}. The case k = 2 obviously
implies the case k ≥ 3.

Now assume n ≥ 4 and that Proposition 2 is true in P^{n−1}. It is sufficient to
do the cases k = b ^{n+2}_{2} /5c and k = d ^{n+2}_{2} /5e and in particular we may assume
that k ≥ d(n + 1)/3e. Fix a hyperplane H ⊂ P^{n}.

(i) First assume n+1 ≡ 0, 2 (mod 3). Write n+1 = 3a+2b with a ∈ N and
0 ≤ b ≤ 1. Let Ai, 1 ≤ i ≤ a + b, be general planes. If 1 ≤ i ≤ a, let Li ⊂ A_{i} be
a general line and let p_{i1} be a general point of L_{i}; set {p_{i2}} := L_{i}∩ H and let v_{i}
be the connected zero-dimensional scheme with p_{i2}as its support and contained
in the line H ∩ Ai. For i = 1, . . . , a set Bi := (2pi1, Ai) ∪ vi. Each Bi is a degree
5 subscheme of A_{i} projectively equivalent to the scheme B described in the first
paragraph of the proof. If b = 1 set L_{a+1} := A_{a+1}∩ H, fix two general points
pa+11 and pa+12 of La+1, and set Ba+1:= (2pa+1, Aa+1) ∪ va+12, where va+12 is
a degree 2 zero-dimensional scheme contained in A_{a+1}, not contained L_{a+1} and
with p_{a+12} as its support. Let E ⊂ H be a general union of k − a − b schemes
Z[5]. Set F := E ∪Sa+b

i=1Bi. We have ResH(F ) = ∪^{a}_{i=1}(2pi1, Ai) ∪ G with G = ∅
if b = 0 and G = {p_{a+11}∪ p_{a+12}} if b = 1. Thus h^{i}(I_{Res}_{H}_{(F )}(1)) = 0, i = 0, 1. Set

G^{0} := ∅ if b = 0 and G^{0} = (2p_{a+11}, L_{a+1}) ∪ {p_{a+12}} if b = 1. The scheme F ∩ H
is the union of E, G^{0} and a general tangent vectors. The inductive assumption
gives that either h^{0}(H, IE,H(2)) = 0 or h^{1}(IE,H(2)) = 0. If h^{0}(IE,H(2)) = 0,
then we get h^{0}(I_{F}(2)) = 0, proving Proposition 2 in this case. Now assume
h^{1}(H, I_{E,H}(2)) = 0. If b = 0, then it is sufficient to use Remark 3. Now assume
b = 1. By Remark 3 we have h^{0}(H, I_{E∪G}^{0}_{,H}(2)) ≤ max{0, h^{0}(IE,H(2)) − 2}

and to prove the proposition in this case it is sufficient to exclude the case
h^{0}(H, I_{E∪G}^{0}_{,H}(2)) = h^{0}(I_{E,H}(2)) − 2 > 0, i.e. the case in which a general line
of H imposes only 2 conditions to |IE,H(2)|.

First assume n = 4. We have a = b = 1 and it is sufficient to prove that
h^{0}(IZ(2)) = 0 when k = 3. Since k − a − b = 1, we have h^{0}(H, IE,H(2)) = 5 and
so h^{0}(H, I_{E∪L,H}(2)) = 2 for a general line L ⊂ H by the case n = 3 of Lemma
1.

Now assume n ≥ 5 and b = 1. We have h^{1}(H, I_{E∪U,H}(2)) = 0 for a general
degree 5 scheme U ⊂ H projectively equivalent to B by the inductive assump-
tion. Since the base locus of |IU(2)| contains the line spanned by Ured, we get
h^{0}(H, I_{E∪G}^{0}(2)) = h^{0}(H, I_{E,H}(2)) − 3. Apply a times Remark 3.

(ii) Now assume n ≡ 0 (mod 3). Write a := n/3 − 1. Let Ai, 1 ≤ i ≤ a + 2,
be general planes. If 1 ≤ i ≤ a, let L_{i} ⊂ A_{i} be a general line and p_{i1} a general
point of L_{i}; set {p_{i2}} := L_{i}∩ H and let v_{i} be the connected zero-dimensional
scheme with pi2as its support and contained in the line H ∩ Ai. For i = 1, . . . , a
set B_{i} := (2p_{i1}, A_{i}) ∪ v_{i}. For i = a + 1, a + 2 set L_{i} := A_{i}∩ H, fix two general
points p_{i1} and p_{i2} of L_{i}, and set B_{i} := (2p_{i1}, A_{i}) ∪ v_{i2}, where v_{i2} is a degree
2 zero-dimensional scheme contained in Ai, not contained Li and with pi2 as
its support. Let E ⊂ H be a general union of k − a − 2 schemes Z[5]. Set
F := E ∪Sa+2

i=1B_{i}. We conclude as above using Remark 3 and twice Lemma

1. ^{QED}

Proposition 3. Fix integers n ≥ 2 and k > 0. Let Z ⊂ P^{n} be a general
union of k schemes Z2,2 and one planar 2-point.

(a) If n = 2, then h^{0}(IZ(2)) = 0.

(b) Assume n = 3. If k = 1, then h^{0}(I_{Z}(2)) = 2. If k ≥ 2, then
h^{0}(IZ(2)) = 0.

(c) Assume n ≥ 4. Then h^{0}(I_{Z}(2)) = max{0, ^{n+2}_{2} − 5k − 3}.

Proof. Part (a) and the second half of part (b) follow from Proposition 2. As-
sume n = 3 and k = 1. Write Z = U t M with U a (2, 2)-scheme and M a
planar 2-point. Since h^{0}(I_{U}(2)) = 5 (Proposition 2), we have h^{0}(I_{Z}(2)) ≥ 2.

Let N be the plane spanned by M and let L be the line spanned by U_{red}. Fix
a general p ∈ N . For a general Z we have U ∩ N = ∅ and L ∩ N is a gen-
eral point of N . Since L is in the base locus B of |I_{Z∪{p}}(2)|, we have N ⊂ B.

Since h^{0}(I_{U}(1)) = 1, we get h^{0}(I_{Z∪{p}}(2)) ≤ 1 and so h^{0}(I_{Z}(2)) ≤ 2. Now
assume n ≥ 4. Proposition 2 gives h^{0}(I_{Z}(2)) ≥ max{0, ^{n+2}_{2} − 5k − 3}. By
Proposition 2 and Remark 3 it is sufficient to do the case k = b ^{n+2}_{2} /5c and
only for the integers n ≥ 4 such that ^{n+2}_{2} ≡ 3, 4 (mod 5). Fix a hyperplane
H ⊂ P^{n}. If n + 1 ≡ 0, 2 (mod 3) we use part (i) of the proof of Proposition 2
taking as E ⊂ H a general union of one planar 2-point and k − a − b scheme
Z[5]; if n ≡ 0 (mod 3) we use part (ii) of the proof of Proposition 2 with as
E ⊂ H a general union of a planar 2-point and k − a − 2 schemes Z[5]. We
explain now why this construction works. In the proof of Proposition 2 we con-
structed a zero-dimensional scheme W ⊂ P^{n} for which we proved that either
h^{0}(H, IW∩H,H(2)) = 0 or h^{1}(IW∩H,H(2)) = 0, deg(ResH(W)) = n + 1 and
ResH(W) is linearly independent. Thus h^{i}(I_{Res}_{H}_{(W)}(1)) = 0, i = 0, 1. ResH(W)
does not depend on the scheme E ⊂ H and so it is the same as in Proposition
2. Assume n = 5c + 1 with c a positive integer. We have ^{n+2}_{2} = 5k + 3 and
so we need to prove that h^{i}(IZ(2)) = 0, i = 0, 1. So we need to prove that
h^{i}(H, IW∩H,H(2)) = 0, i = 0, 1. We have the inductive assumption in H to han-
dle E and then we continue as in steps (i) and (ii) of the proof of Proposition

2. ^{QED}

Lemma 3. Let G ⊂ P^{3} be a general union of 3 planar 2-points. Then
h^{1}(IG(2)) = 0 and h^{0}(IG(2)) = 1.

Proof. Let A be the plane spanned by G_{red}. Keeping A fixed and moving G
among the union of 3 planar 2-points with support on A we see that for a general
G the scheme G ∩ A is a general union of 3 tangent vectors of A. Remark 3 gives
h^{i}(A, I_{G∩A}(2)) = 0. Since Res_{A}(G) = G_{red}, we have h^{0}(I_{Res}_{A}_{(G)}(1)) = 1 and

h^{1}(I_{Res}_{A}_{(G)}(1)) = 0. ^{QED}

Proofs of Theorems 1 and 2. First assume n = 2. Since any two points of P^{n}
are collinear, in the case n = 2 of Theorem 1 (resp. Theorem 2) Z is a general
union of 2k (resp. 2k + 1) general 2-points of P^{2}. The Alexander-Hirschowitz
list ([1], [4]) gives Theorems 1 and 2. We assume n ≥ 3 and that Theorems
1 and 2 are true in P^{n−1}. To prove Theorem 1 is sufficient to do the cases
k = b ^{n+d}_{n} /6c = u_{d,n} and k = d ^{n+d}_{n} /6e. Let H ⊂ P^{n} be a hyperplane.

(a) Assume d = n = 3. Since u_{3,3} = 3 and v_{3,3} = 2, to prove Theorem
1 it is sufficient to prove that h^{1}(IZ(3)) = 0 if k = 3 and h^{0}(IZ(3)) = 0 if
k = 4. First assume k = 3. Let Y ⊂ P^{3} be a union of 3 schemes Z_{2,2} such that
one of them is contained in H and that each of the other ones have a point in
its support contained in H and that Y is general with these restrictions. The
scheme Y ∩ H is a general union of one scheme Z_{2,2} (call it β) and 2 tangent

vectors. We obviously have h^{1}(H, I_{β,H}(3)) = 0 and hence h^{i}(H, I_{Y ∩H,H}(3)) = 0,
i = 0, 1, by Remark 3. Hence it is sufficient to prove that h^{1}(I_{Res}_{H}_{(Y )}(2)) = 0.

We have ResH(Y ) = U1tU_{2}with Uispanning a general plane Aiand Uiunion of
a general 2-point (2P_{i}, A_{i}) of A_{i}and a general point q_{i}of the line A_{i}∩H. We have
h^{1}(I_{(2P}_{1}_{,A}_{1}_{)∪(2P}_{2}_{,A}_{2}_{)}(2)) = 0 (Lemma 3) and so h^{0}(I_{(2P}_{1}_{,A}_{1}_{)∪(2P}_{2}_{,A}_{2}_{)}(2)) = 4.

The scheme (2P1, A1) ∪ (2P2, A2) does not depend on H. Since any two points
of P^{3} are collinear, moving H we may assume that (q_{1}, q_{2}) is a general element
of A_{1}× A_{2}.

Since h^{0}(I_{A}_{1}_{∪(2P}_{2}_{,A}_{2}_{)}(2)) = h^{0}(I_{(2P}_{2}_{,A}_{2}_{)}(1)) = 1 and q_{1} is general in A_{1}, we
get h^{0}(I_{(2P}_{1}_{,A}_{1}_{)∪(2P}_{2}_{,A}_{2}_{)∪{q}_{1}}(2)) = 3. Since h^{0}(I_{(2P}_{1}_{,A}_{1}_{)∪(2P}_{2}_{,A}_{2}_{)∪{q}_{1}}∪A_{2}(2)) =
h^{0}(I_{(2P}_{2}_{,A}_{2}_{)∪{q}_{1}_{}}(1)) = 1, we obtain that h^{0}(I_{(2P}_{1}_{,A}_{1}_{)∪(2P}_{2}_{,A}_{2}_{)∪{q}_{1}_{}∪{q}_{2}_{}}(2)) = 2,
i.e. we have h^{1}(I_{Res}_{H}_{(Y )}(2)) = 0.

Now assume k = 4. Since two general points of P^{n}are contained in a scheme
Z2,2 and v3,3 = 2, the case k = 4 of Theorem 1 follows from the case k = 3. For
Theorem 2 it is sufficient to do the cases k = 2 (true because Z is contained
in a general disjoint union of 3 schemes Z_{2,2}) and k = 3 (we use that a planar
2-point contains a tangent vector, Remark 3 and that v3,3 = 2).

(b) Assume d = 3 and n ≥ 4. To prove Theorem 1 is sufficient to prove
the cases k = b ^{n+3}_{3} /6c and k = d ^{n+3}_{3} /6e. Fix any k disjoint schemes B_{i}
projectively equivalent to Z_{2,2}. Let A_{i} be the plane containing B_{i} and let L_{i}
be the line spanned by the reduction of Bi. We assume that Li∩ L_{j} = ∅ for
all i, j such that i 6= j. Since ^{6}_{2}

= 15 and ^{n+2}_{2}

≥ 20 for all n ≥ 5, we
may write ^{n+2}_{2}

= 5x + 4a with a, x non-negative integers and 0 ≤ a ≤ 4.

Now we check that k ≥ x + a. Assume k ≤ x + a − 1. Since a ≤ 4, we get
5k ≤ 5x + 5a − 5 ≤ ^{n+2}_{2} − 1, contradicting the inequality 6k ≥ ^{n+3}_{3} − 5.

Let G ⊂ P^{n} be a general union of x schemes of type Z_{2,2} and let S ⊂ H be
the intersection with H of the lines associated to each scheme Z_{2,2}. Since G is
general, these x lines are x general lines of P^{n} and so S is a general subset of
H. Fix a general planes A_{i}, 1 ≤ i ≤ a, and let B_{i} ⊂ A_{i} be a general scheme
of type Z_{2,2} with the restriction that one of the points of (B_{i})_{red} is contained
in H ∩ Ai. Let E ⊂ H be a general union of k − a − x schemes Z2,2 of H.

Set Y := G ∪ E ∪Sa

i=1Bi. For all i = 1, . . . , a the scheme Res_{H}(Bi) is a
general union of a 2-point of A_{i} and a general point of the line A_{i}∩ H. Let M
be the union of the reduced components of ∪^{a}_{i=1}ResH(Bi). Since any Z2,2 is a
degeneration of a family of general planar 2 points, by Proposition 3 we have
h^{1}(I_{Res}_{H}_{(Y )\M}(2)) = 0 and so h^{0}(I_{Res}_{H}_{(Y )\M}(2)) = a. Res_{H}(Y ) \ M does not
depend on H. Since a ≤ 4 ≤ n, any a points of P^{n}are contained in a hyperplane.

Hence, writing M = {q1, . . . , qa} with q_{i} ∈ A_{i}, we may assume that (q1, . . . , qa)
is a general element of ×^{a}_{i=1}A_{i}. As in step (a) we get h^{i}(I_{Res}_{H}_{(Y )}(2)) = 0,
i = 0, 1. Hence it is sufficient to prove that h^{0}(H, I_{S∪(Y ∩H)}(3)) = max{0, ^{n+2}_{3} −