POSTULATION OF ZERO-DIMENSIONAL SCHEMES ON A SMOOTH QUADRIC SURFACE

Let Q be a smooth quadric surface and Z ⊂ Q a zero-dimensional scheme. We study the postulation of a general union of Z and prescribed numbers of fat points with multiplicity 2 and 3. AMS Subject Classification: 14N05, 14H99


Proofs for 2-Points
Remark 1.It is easy to check that a 3-point of any smooth surface X is a flat limit of a family of disjoint unions of pairs of 2-points of X ( [6]).
Proof.Since a 3-point is a flat limit of a family of disjoint unions of two 2-points (Remark 1), this lemma is a particular case of Lemma 8.
independent conditions to the linear system W .In arbitrary characteristic S imposes independent conditions to any linear system V on L with dimension ≥ ♯(S).This is also true for S ∪ E if either char(K) = 0 or char(K) > c, because E is a general tangent vector of L and our assumption on char(K) implies that the rational map induced by V is separable.
The case e ≤ u is easier.Indeed, we do not use Differential Horace and in the residual we only have Z ∪ S with ♯(S) = e, instead of Z ∪ S ∪ E. From now on we assume γ ≤ c.Since each connected component of U contains a general tangent vector of Q at its support and U red is general, it is sufficient to prove that f := ♯(U red ) ≥ γ/2.Set f ′ := f + e ′ − e.Let g ∈ {0, 1, 2} the congruence class of 2c + 2 + γ modulo 3. First assume that c is odd.In this case we get In particular we assume e ≥ ⌈(c + 1)(2⌉.Fix a line L ⊂ Q of type (0, 1) such that L ∩ Z = ∅.First assume that c is odd.Let A ′ ⊂ Q be a general union of e − (c + 1)/2 2-points.Take S ⊂ L such that ♯(S) = (c + 1)/2.It is sufficient to prove that h )/2 and to prove the lemma for c odd it is sufficient to prove that S imposes ♯(S) independent conditions to H 0 (I Z∪A ′ (c, d + 1)).Let W ⊂ H 0 (O L (c)) be the image of the restriction map ρ : Now assume that c is even.Let A ′′ ⊂ Q be a general union of e − c/2 − 1 2-points of Q.Take a general S ′ ∪{o} ⊂ L such that ♯(S ′ ) = c/2 and o / ∈ S ′ .Let E ⊂ L be the 2-point of L with o as its support.Since deg(2S ∩L)+deg({o}) = c + 1, the Differential Horace Lemma for 2-points shows that to prove that a general union Y of Z ∪ 2S ∪ A ′′ and a 2-point satisfies h 1 (I Y (c, d + 2)) = 0 (and hence to prove the lemma in the case c even), it is sufficient to prove that h 1 (I Z∪A ′′ ∪S∪E (c, d + 1)) = 0. Lemma 3 gives h 1 (I Z∪A ′′ (c, d + 1)) = 0. Let V ⊂ H 0 (O L (c)) be the image of the restriction map ρ ′ : H 0 (I Z∪A ′′ (c, d + 1)) → H 0 (O L (c)).Since h 0 (I Z∪A ′′ (c, d)) = 0 (Lemma 4), ρ is injective.Hence dim(V ) ≥ ♯(S) + deg(E).Hence it is sufficient to use that S ∪ {o} is general in L and hence that E is a general tangent vector of L \ S.
Proof.Adapt either the proof of Lemma 1 or the one of Lemma 5 quoting Lemmas 2 and 1 for the residual with respect to L instead of Lemmas 4 and 3.
Proof.Set x := ⌊(c + 1)/4⌋.Since any 3-point of Q is a flat limit of a family of disjoint unions of pairs of 2-points of Q (Remark 1), the semicontinuity theorem for cohomology shows that it is sufficient to do the case 1 ∈ S ′ and set S ′ 1 := S ′ \ {o ′ 1 }.First assume c ≡ 3 (mod 4).In this case we apply 3 times the Horace Differential Method for 3-points to each point of S ′ with respect to the integers 3 > 2 so that (1, 2, 3) are the degrees of the intersection with L of the first, second and third trace ( [5], [7], [8], [9]), while we specialize x of the 3-points of A and A ′ to 3-points with a point of S as their support.At each step in L we have a scheme of degree c + 1.Now assume c ≡ 1 (mod 4).In the first step we also add 2o.In this step we have a scheme whose intersection with L has degree c + 1, while in the second step the scheme on L has only degree c, because Res L (2o) = {o}.Hence at the second step we also use the Horace Differential lemma for double points with respect to o ′ , so that in L we have a degree c + 1 scheme, while at the third step we have a scheme sitting of degree c + 1 inside L, with no connected component with o as its reduction and with a degree 2 component with o ′ as its reduction.Now assume c ≡ 0 (mod 4).We take 3S and apply the Differential Horace Method for 3-points with respect to the integers 3 > 2, i.e. with traces of degrees (1, 2, 3), for each point of S ′ 1 and with respect to the sequence 2 (i.e. with traces of degrees (2, 1, 3)) with respect to o ′ 1 ; hence the intersection of L with this virtual scheme has degree 3x + (x − 1) + 2 = c + 1. (x components of degree 3, x − 1 components of degree 1 and one of degree 2).At the second step we also add the scheme 2o.The new virtual scheme intersects L in a scheme with degree c + 1 (2x + 1 of its connected components have degree 2, the one supported by o ′ 1 has degree 1).In the last step, after taking the virtual residual scheme, we get a scheme of degree c + 1 (x + 1 components of degree 1, x components of degree 3).Now assume c ≡ 2 (mod 4).We first add 3S, apply the Differential Horace lemma for 3-points with respect to the integers 3 > 2, i.e. with traces of degrees (1, 2, 3), at each point of S ′ 1 , the Differential Horace points for 3-points with intersections with L of degree (3, 1, 2) (it is the example done in the introduction of [5]) at o ′ 1 and the differential Horace lemma for double points with respect to o ′ .The intersection of the virtual residual scheme with L has only degree 4x + 1 = c − 1 (2x connected components of degree 2 and one of degree 1).Therefore in the second step we also apply the Differential Horace lemma for 2-points at o ′ and o ′′ ; in this way the restriction of this scheme to L has degree c + 1 (2x components of degree 2 and 3 components of degree 1).The virtual residue scheme with respect to L is contained in L and it has degree c + 1 (it has x components of degree 1 with S as the union of their reductions, x − 1 components of degree x − 1 with S ′ 1 as the union of their reductions and 3 components of degree 2 with o ′ 1 , o ′ and o ′′ as their reduction).
Proof of Theorem 1.We have h Remark 2. The quoted parts of [8] and [9] and the Differential Horace lemma for double points ( [3]) are characteristic free.We only use the condition char(K) to say that a general tangent vector gives two conditions to any linear system of dimension > 1 ([4], [2], Lemma 1.4).In a few cases we state the minimal assumptions for our proofs.In all theorem it is sufficient to assume char(K) > max{a, b, 2c + d} (the only problem is to check Lemma 4 when char(K) > 2c + d).

Proof of Theorem 3 .Proposition 1 .
If a = c and b = d + 2, then we quote Lemma 5. Now assume a = c and b ≥ d + 3.In this case we copy the proof of Lemma 5 and induction on d.If a = c + 1 and b = d + 2, then we quote Lemma 7. Then for a fixed a we get all cases with b > d + 2 using induction on b and the Horace Differential lemma for double points with respect to a general L ∈ |O Q (1, 0)|.Proof of Theorem 2. Taking a union of Z and γ := h 0 (I Z (a, b)) general points of Q we reduce to the case γ = 0. Then we apply Theorem 3. Fix integers m ≥ 3, e > 0, a > m, b > m such that (a, b) = (m + 1, m + 1).Let A ⊂ Q be a general union of two m-points and e 2-points.Then either h 1 (I A (a, b)) = 0 or h 0 (I A (a, b)) = 0.