Maximal families of nodal varieties with defect

In this paper we prove that a nodal hypersurface in P^4 with defect has at least (d-1)^2 nodes, and if it has at most 2(d-2)(d-1) nodes and d>6 then it contains either a plane or a quadric surface. Furthermore, we prove that a nodal double cover of P^3 ramified along a surface of degree 2d with defect has at least d(2d-1) nodes. We construct the largest dimensional family of nodal degree d hypersurfaces in P^(2n+2) with defect for d sufficiently large.


Introduction
Let n ≥ 3 be an odd integer and c be a positive integer. Let 1 ≤ w 0 ≤ · · · ≤ w n+c and 2 ≤ d 1 ≤ · · · ≤ d c be integers. For a nodal complete intersection X ⊂ P(w 0 , . . . , w n+c ) =: P of multidegree d 1 , . . . , d c we define the defect of X to be h n+1 (X) − h n−1 (X). In this paper we consider the following problem of determining the minimal number of nodes to have positive defect. In this generality the problem is too hard. In the sequel we concentrate on the following three special cases: hypersurfaces in P n+1 , double solids, i.e., hypersurfaces of degree 2k in P(k, 1, 1, 1, 1) and threedimensional complete intersections in P 3+c . In a subsequent paper we will discuss the case of elliptic threefolds over P 2 , i.e., hypersurfaces of degree 6k in P(2k, 3k, 1, 1, 1).
We start by recalling some previous results on this problem. For hypersurfaces in P 4 , Cheltsov showed that the minimal number of nodes to have defect is (d − 1) 2 [2] and that if X has defect and (d − 1) 2 nodes then X contains a plane [3]. This improves a previously known bound by Ciliberto and Di Gennaro [5]. Ciliberto and Di Gennaro showed that if a hypersurface with defect has at most 2(d − 2)(d − 1) nodes and the defect is caused by a smooth surface, then X contains either a plane or a quadric surface.
In the case of complete intersections of degree (d 1 , d 2 ) in P 5 , the best known lower bound for the number of nodes is by Kosta [19]: If d 1 ≤ d 2 and the complete intersection is nondegenerate (see Section 7) then it has at least (d 1 + d 2 − 2) 2 − (d 1 + d 2 − 2)(d 1 − 1) nodes.
Cynk and Rams [7] considered complete intersection threefolds that are CR-nondegenerate in codimension 3 (see Section 7). They showed that a nodal CR-nondegenerate complete intersection threefold with defect, such that the defect is caused by a smooth complete intersection surface, has at least 1≤i≤j≤c (d i − 1)(d j − 1) nodes, except if c ≤ 4 and d 1 = · · · = d c = 2 holds. In the latter case the weaker bound #X sing ≥ 2 c−1 holds. Moreover, they show that this bound is sharp, i.e., for each choice of d 1 , . . . , d c they give an example of a complete intersection with either 1≤i≤j≤c (d i −1)(d j −1) or 2 c−1 nodes and they conjecture that one can drop the two conditions on the surface causing the defect (smoothness and being a complete intersection).
Polizzi, Rapagnetta and Sabatino [26] obtained a lower bound for the number of singular points for complete intersection threefolds with only isolated ordinary multiple points, provided that the complete intersection is nondegenerate in codimension four. A similar bound for the case of nodal complete intersection threefolds in P 5 was obtained by Ciliberto and Di Gennaro [4].
To illustrate our methods we start by giving a new proof of Cheltsov's theorem Theorem 1.1 (Cheltsov [2,3]). Let X ⊂ P 4 be a nodal hypersurface of degree at least 3. Assume that h 4 (X) ≥ 2. Then X has at least (d − 1) 2 nodes. Moreover, if equality holds then X contains a plane.
We included this proof because it is a good illustration of our techniques and is significantly different from the one Cheltsov gave.
For fixed integers n and d let DEF d ⊂ ⊕C[x 0 , . . . , x 2n+2 ] d the locus of polynomials f such that V (f ) is a nodal hypersurface with defect. Otwinowska showed that [22,Conjecture 1] is implied by the Conjecture of Eisenbud, Green and Harris on the Hilbert functions of an ideal containing a complete intersection ideal.
We also consider the case of double covers of P 3 . In this case we recover a result by Cheltsov: Theorem 1.3 (Cheltsov [1]). Let f ∈ C[x 0 , x 1 , x 2 , x 3 ] be a homogeneous polynomial of degree 2d such that V (f ) is a nodal surface. Let X : y 2 = f be the double cover branched along V (f ). Suppose h 4 (X) > 1. Then X has at least d(2d + 1) nodes.
In the case of hypersurfaces in P 4 we prove the Ciliberto-Di Gennaro conjecture for d ≥ 7: Theorem 1.4. Let X ⊂ P 4 be a nodal hypersurface of degree at least 7. Suppose that X is non-factorial and that X has at most 2(d − 2)(d − 1) nodes then either X contains a plane and has (d − 1) 2 nodes or X contains a quadric surface and has at least 2(d − 1)(d − 2) nodes.
The complete intersection case is more complicated. We say that a complete intersection X ⊂ P 3+c of multidegree (d 1 , . . . , d c ), with d 1 ≤ · · · ≤ d c has induced defect if there exists a four-dimensional complete intersection Y ⊂ P 3+c and a hypersurface H ⊂ P 3+c of degree d c , such that X = Y ∩ H and such that for a general hyperplane H ′ we have that h 4 (Y ∩ H ′ ) > h 2 (Y ∩ H ′ ). If this is the case then the singular locus of Y is one-dimensional. If d c−1 < d c holds then induced defect implies that X is degenerate in codimension three in the sense of [7].
We give a sharp lower bound for the number of nodes for a complete intersection threefolds in P 5 provided that they do not have induced defect. For complete intersections in P 3+c with c ≥ 3 we consider only the case where the complete intersection has no induced defect and d c > We prove the following theorem: (This is a combination of Theorem 8. 16 and Theorem 8.21.) We will now briefly discuss the strategy of proof. To reprove Cheltsov's result we use the following strategy: Let I be the ideal of the nodes of X, where X is a nodal hypersurface with defect. Let H = V (ℓ) be a general hyperplane then X ∩ H is smooth. In particular the ideal I H := (I, ℓ) defines an empty scheme. Since X has defect this implies that the Hilbert polynomial of I and the Hilbert function of I are different in degree 2d − 5. From this it follows that h I H (2d − 4) = 0 holds. Since the partials of the defining equation for X are contained in I d−1 , it follows that I d−1 has finitely many base points. Therefore I d−1,H is base point free in degree d − 1. A combination of results from Macaulay and Gotzmann on which functions occur as Hilbert functions of ideals yields that h I H (k) ≥ k + 1 for k ≤ d − 2 and that h I H (k) ≥ 2d − 3 − k, for d − 1 ≤ k ≤ 2d − 3. Now using that p I ≥ h I (2d − 4) ≥ 2d−4 k=0 h I ′ (k) finishes the proof. The ideal I H is very similar to the ideal used by Green [15] to determine the largest component of the Noether-Lefschetz locus of surfaces in P 3 . The other proofs are variations of this idea. In the case of an hypersurface in P 2n+2 with n ≥ 2, Macaulay's result is not strong enough to obtain the desired lower bound. In this case we use a result by Otwinowska [22] instead. However this result bounds h I H (k) only in a certain interval. This is sufficient to detect the largest dimensional component, but not to establish the minimal number of nodes. If one assumes a conjecture from [22], then one obtains the desired lower bound for the number of nodes. The double solid case is very similar to the case of hypersurfaces in P 4 and we will not comment on this.
The complete intersection case is a little different. The main difference is that the smallest ideal I ′ containing I H and such that S/I ′ is Gorenstein, is too big to obtain the desired lower bound for the nodes. Instead of working directly with the complete intersection X in P 3+c we work with an associated hypersurface Y in a P c−1 -bundle over P 3+c . If X is nodal then so is Y and the nodes of Y are in one-to-one correspondence with those of X. We analyse the ideal of the nodes of Y . The advantage of working with Y is that one can rephrase the various nondegeneracy properties easily in terms of the ideal of the nodes of Y . However, it turns out that this case is technically much more involved than the previous ones.
The paper is organized as follows. In Section 2 we recall several standard results on the Hilbert functions of ideals. In Section 3 we recall some standard results on the cohomology of nodal complete intersections. In particular, we present a formula to calculate the defect of a nodal hypersurface. In Section 4 we prove the results for the hypersurfaces and in the double covers, except for the Ciliberto-Di Gennaro conjecture, which is proven in Section 5. In Section 6 we discuss the above mentioned construction of Y and give a formula to calculate the defect of a nodal complete intersection. In Section 7 we discuss various notions of nondegenerate complete intersections and compare them. Finally in Section 8 we prove our result on complete intersections.

Macaulay's and Green's result
Let S = C[x 0 , . . . , x n ] and let I ⊂ S be an ideal. Let d ≥ 1 be an integer. Let c := h I (d). We can write c uniquely as Using the Macaulay expansion of c we define the following numbers: Note that c → c * d , c → c d and c → c d are increasing functions in c. Recall the following theorem by Macaulay: [20]). Let V ⊂ S d be a linear system and c = codim V . Then the codimension of We apply this result mostly in the case where V is the degree-d part of an ideal I. In this case we can also obtain information on h I (d − 1).
For small c we have the following Macaulay expansions in base d: Applying the previous corollary repeatedly yields The following result will be used to detect the Hilbert polynomial of the ideal generated by I d : Theorem 2.4 (Gotzmann [13]). Let V ⊂ S d be a linear system and let We use this result mostly in the case where c ≤ d:

Nodal complete intersections
Notation 3.1. Let n = 2k + 1 be a positive odd integer, c be a positive integer, and (w 0 , . . . , w n+c ) a sequence of positive integers. Let us denote with P := P(w 0 , . . . , w n+c ) the associated weighted projective space. Let S = C[x 0 , . . . , x n+c ] be the graded polynomial ring such that deg Definition 3.2. We say that a codimension c complete intersection X ⊂ P is a nodal complete intersection of codimension c, if (1) for all p ∈ P sing ∩ X we have that X is quasi-smooth at p and (2) for all p ∈ X sing \(P sing ∩X) we have that (X, p) is an A 1 -singularity. Let Σ denote the set X sing \ (P sing ∩ X).
Proof. The first equality follows from the Lefschetz hyperplane theorem [11,Theorem 4.2.6]. To prove the second equality we consider a partial resolution of singularities of X: Since X is quasismooth outside Σ we have that for all i = 2n and for all p ∈ X \ Σ the group H i p (X) vanishes. LetX be the blow up of X along Σ. ThenX is smooth along the exceptional divisor. In particular, for all p ∈X we have that H i p (X) = 0 if i = 2n. This implies thatX is Q-homology manifold and satisfies Poincaré duality.
Consider the Mayer-Vietoris sequence associated with the discriminant square [25, Corollary-Definition 5.37] This is an exact sequence of mixed Hodge structures.
The exceptional divisor E is the disjoint union of #Σ smooth quadrics in P n . Thus its cohomology can be nonzero only in even degree between 0 and 2n − 2. Let E j and E k be distinct irreducible components of E. For i even between 2 and 2n − 2 consider c 1 (E j ) i/2 ∈ H i (X). Then c 1 (E j ) i/2 is mapped to zero in H i (E k ) and to a nonzero element of H i (E j ). If i = n − 1 then H i (E j ) is one-dimensional and therefore the map H i (X) → H i (E) is surjective for even i, different from 0 and n − 1. From this it follows that the above long exact sequence splits in the following exact sequences: Since E is a disjoint union of smooth quadrics it follows that h i (E) = #Σ = h 2n−i (E) for i = 0, n − 1, n + 1, 2n. From Poincaré duality it follows that h i (X) = h 2n−i (X) for all i. Combining this yields that that h i (X) = h 2n−i (X) for i = 0, n − 1, n + 1, 2n.
The proof of the above result suggests that h n+1 (X, Q) may be strictly larger than h n−1 (X, Q).
Remark 3.5. If n = 3 then δ equals the rank of the group CH 1 (X)/ Pic(X). Since this group is free, δ measures the failure of Weil divisors to be Cartier. Lemma 3.6. Let X be a nodal complete intersection. Let D be the equisingular deformation space of X. Then the locus is a Zariski open subset of D.
Proof. Let (X t ) t∈U be an equisingular deformation of X. Possibly after shrinking U , we have that X t has the same number of nodes for all t ∈ U . Blowing up these nodes simultaneously yields a flat familyX t of smooth projective varieties. Hence H n+1 (X t ) is independent of t. Let E be the exceptional divisor of the blow-up of a node, and let s the number of nodes of X. As in the proof of Proposition 3.3 we can consider the Mayer-Vietoris sequence associated with the discriminant square. This time we take also into account the Hodge structures. We obtain the following exact sequence . Since X t is a nodal hypersurface and n is odd we have that H n+2 (X t ) = 0. Since all singularities of X t are nodes or induced by the ambient space it follows that H n+1 (X t ) has a Hodge structure of pure weight n + 1. This yields h n+1 (X t ) = h n+1 (X t ) − s · h n+1 (E). Both terms on the right hand side are independent of t, hence so is h n+1 (X t ). By Proposition 3.3 we have h n−1 (X t ) = 1 for all t and hence δ( One can express δ in terms of the Hilbert function of the ideal of the nodes. We will discuss the case of complete intersections of codimension at least 2 in Section 6. Suppose now that c = 1, i.e., X is a hypersurface. Set m := n+1 2 . The following result is [10, Proposition 3.2]: Proposition 3.7. Let X ⊂ P be a nodal hypersurface. Let Σ ⊂ P be the locus of the nodes of X. Then δ(X) = #Σ − dim(S/I(Σ)) md− w i .

Hypersurfaces with defect
We will use the results from the previous section to reprove the following result by Cheltsov on the minimal number of nodes to have defect: [2,3]). Let X ⊂ P 4 be a nodal hypersurface of degree at least 3. Assume that h 4 (X) ≥ 2, i.e., that X has defect. Then X has at least (d− 1) 2 nodes. If X has precisely (d− 1) 2 nodes then X contains a plane.
Proof. Without loss of generality we may assume that Let I ⊂ R be the ideal of the nodes of X. Since X has defect it follows from Proposition 3.7 that h I (2d − 5) < p I (2d − 5).
Let I H ⊂ S be the ideal obtained by substituting x 4 = 0 in I. From the fact that none of the nodes of X is contained in V (x 4 ) it follows that the following sequence is exact: If h I H (2d−4) vanishes then we have h I H (k) = 0 for k ≥ 2d−4. In particular, h I (k) = h I (k + 1) for k ≥ 2d − 5. Since we know that h I (2d − 5) < p I (2d − 5) this cannot be the case and hence h I d (2d − 4) > 0 holds. Fix now a codimension one subspace W of S 2d−4 containing (I H ) 2d−4 . Define I ′ ⊂ S by I ′ e = {f | f S 2d−4−e ⊂ W } if e ≤ 2d − 4 and I ′ e = S e for e ≥ 2d − 3. Then I ′ is an ideal, containing I H . Moreover S/I ′ is a Gorenstein ring with socle degree 2d − 4. In particular, Combining this information we obtain If p I = (d − 1) 2 then we have that h I equals the Hilbert function of a complete intersection of degree (1, 1, d − 1, d − 1). Since I contains the partials of f it follows that the linear system |I d−1 | has finitely many base points.
Now h I has two generators in degree 1 and two further generators in degree d − 1. In particular these four generators defines a codimension four scheme and hence these four generators form a regular sequence. The Hilbert of the ideal generated by these four forms equals the Hilbert function of I hence I is a complete intersection ideal, generated by At each node of X the polynomials f 1 , f 2 , f 3 , f 4 induce a local system of coordinates. Since at each singular point of X the polynomial f vanishes up to order two it follows that f is an element of the ideal generated by the f i f j with i ≤ j. These forms have degree at most d if and only if i ≤ 2. In particular f is in the ideal generated by f 1 and f 2 and therefore contains the plane f 1 = f 2 = 0.
Recall that I d is the tangent space to the equisingular deformation space of X [16]. Hence it follows that any family of degree d nodal hypersurfaces with defect has codimension at least 1 2 (d 2 + 3d − 10) in S d . Moreover, if equality holds then the above proof shows that X contains a plane.
Consider now hypersurfaces containing a fixed plane P . They form a family of codimension 1 2 (d + 1)(d + 2). Since the Grassmannian of planes in P 4 has codimension 6 it follows that the total family has codimension 1 2 (d 2 +3d−10). A general element of this family is of the form ℓ 1 f 1 +ℓ 2 f 2 with deg(ℓ i ) = 1 and deg(f i ) = d − 1. In particular, a general element is a nodal hypersurface. Hence the largest-dimensional family of nodal hypersurfaces with defect consists of hypersurfaces containing a plane.

The bound we obtained for
is also used in some of the proofs for the explicit Noether-Lefschetz theorem for surfaces in P 3 (e.g., see [15]). However, if n > 3 then Corollary 2.5 is insufficient to deduce the explicit Noether-Lefschetz theorem. Similarly, we were not able to deduce a good lower bound for the number of nodes to have defect from this Corollary. To obtain an explicit Noether-Lefschetz theorem in higher (even) dimension Otwinowska [22] proved a result on the Hilbert function of ideals containing the ideal of a certain complete intersection. This results seems still to be insufficient to obtain a sharp lower bound for the number of nodes to have defect. However, Otwinowska's result is strong enough to determine the largest component of the locus of nodal hypersurfaces with defect. Moreover, if the famous conjecture [12, Conjecture V m ] of Eisenbud, Green and Harris on the Hilbert function of ideals containing a complete intersection holds true, then the result of Otwinowska is strong enough to deduce the minimal number of nodes.
If the f i are chosen sufficiently general then the singular locus is The tangent space to the equisingular deformation space has codimension p n,d and an easy calculation shows that this space is nonreduced, i.e., the actual deformation space has the same codimension.
Moreover, if Conjecture 1 of [22] holds then we may take D = 2 and any hypersurface in DEF d has at least (d − 1) n+1 nodes.
Proof. Let X ∈ L. From Lemma 3.6 it follows that a general equisingular deformation of X also has defect, i.e., L is also an irreducible component of the equisingular deformation space of X.
Fix a general hyperplane H. Since X has defect there is a class γ in H 2n+2 (X, Q) which is not the multiple of the intersection of classes of hyperplanes. The intersection product of γ with H yields a nonzero Hodge class in H 2n (X H , Q) prim . The Noether-Lefschetz locus of hypersurface of degree d in P 2n+1 parametrizes hypersurfaces having a nonzero Hodge class in H 2n (X H , Q) prim . In particular, we have a morphism from an open subset of L to an irreducible component NL(γ H ) of this Noether-Lefschetz locus. The differential of this map defines a map d H from the tangent space T X L to the tangent space of NL(γ H ) at X H .
The tangent space T X L can be identified with the degree d part of the saturation of the Jacobian ideal of F . Without loss of generality we may assume that Hence we can apply [22,Théorème 1]. From this it follows that there is a constant D depending on n such that for d ≥ D we have codim I d ≥ d+n n − (n + 1) 2 . Let J be the ideal of the nodes of X. Then If codim I d = d+n n − (n + 1) 2 holds then we have by [22, Théorème 1] that I up to degree d coincides with a complete intersection ideal of multidegree If the codimension of I d is larger then d+n n − (n + 1) 2 then NL(γ H ) is different from the component of NL parametrizing hypersurfaces containing an n-dimensional linear space. From [23] it follows that for d sufficiently large, the largest component of this type consists of hypersurfaces containing a quadric of dimension n. This locus has codimension If n ≥ 16 then the Macaulay expansion of c 0 equals For n ≤ 15 we have that the the Macaulay expansion of c 0 equals Suppose now that n ≥ 16. Since c → c <d> increases with c and In particular, there exists a constant C n depending only on n such that the right hand side equals d+n+1 d + d+n d − C n . Therefore we have that for d sufficiently large h J (d) > p n,d holds. If n < 16 then a similar argument will yield the proof for large d.
Remark 4.5. Otwinowska shows in [22] that [22,Conjecture 1] is implied by the Eisenbud-Green-Harris conjecture on the Hilbert function of ideals containing a complete intersection.
We switch now to the case of double covers.
If dgeq2 holds and X has precisely d(2d + 1) nodes then there exist forms ℓ, g, h of degree 1, d and 2d − 1 respectively such that f = ℓg + h 2 .
Proof. Without loss of generality we may assume that Note that the nodes of X correspond one-to-one with the nodes of V (f ). Moreover, since y is in the Jacobian ideal of X we have that the Jacobian rings of X and of V (f ) are isomorphic. Since As in the proof of Theorem 4.1 we obtain that Then I ′ is an ideal, containing I H . Moreover S/I ′ is a Gorenstein ideal with socle degree 3d − 3 and hence h I ′ (k) = h I ′ (3d − 3 − k).
The linear system I ′ 2d−1 contains the partials of f specialized at x 3 = 0 and since X H is smooth this linear system must be base point free. From Corollary 2.5 we obtain that Combining everything we obtain Suppose now that p I is exactly d(2d − 1). Then we have h I (1) = 3. In particular there is a linear form ℓ that vanishes at all the nodes. If ℓ is a factor of f then we can write f = ℓf 1 . All the nodes of V (f ) are contained in V (ℓ, f 1 ) which consists of 2d − 1 points. Since we know that X has at least d(2d − 1) nodes this cannot happen and therefore ℓ is not a factor of f .
If p is a node of V (f ) then g vanishes at p and p is a double point of f 0 = 0. If f 0 contains a component with multiplicity at least three then X contains a singularity which is not a node, in particular, we can write f 0 = f 2 1 f 2 , such that f 1 and f 2 are coprime and both are squarefree. Hence the locus of the nodes of V (f ) consists of points p such that Denote with e i the degree of f i . Then there are precisely e 1 (2d − 1) points of the former type and at most 1 2 (e 2 − 1)e 2 points of the second type. Their sum is strictly less than d(2d − 1) if e 1 = 0, 2d. If e 1 were 2d then the set of nodes of V (f ) is also the set of nodes of a (reducible) plane curve of degree 2d. From [18,Proposition 3.6] it follows that then h Example 4.7. In order to show that the bound d(2d − 1) for the number of nodes is sharp, consider y 2 = h 2 + ℓg with deg(f ) = d and deg(g) = 2d − 1. Then for general f, g, ℓ the singular locus is a complete intersection of multidegree (1, d, 2d − 1), i.e., it consists of d(2d − 1) points. Moreover ℓ = y − f = 0 defines a Weil divisor that is not Q-Cartier and hence the double cover has defect.

The Ciliberto-Di Gennaro conjecture
In this section we prove the following conjecture for d ≥ 7: Conjecture 5.1. Let X ⊂ P 4 be a non-factorial nodal threefold of degree d with at most 2(d − 2)(d − 1) nodes then either X contains a plane or a quadric surface and if X contains a quadric surface then X has precisely 2(d − 2)(d − 1) nodes.
Note that in [5] Ciliberto-Di Gennaro proved a weaker form of this conjecture, namely they showed that if X is non-factorial, has at most 2(d−2)(d−1) nodes then X contains a plane, a quadric surface or a singular surface.
. , x n ] = be an ideal such that S/I is Artinian Gorenstein of socle degree N. Let d k be the smallest integer t such that the dimension of the base locus of I t is at most k. Then Proof. The ideal I contains a complete intersection ideal I ′ of multidegree (d n−1 , . . . , d −1 ). In particular, Proof. From [27, Section 1] it follows that I d−4 is the degree d − 4-part of the ideal of either a line or a conic.
Suppose first that I d−4 is the degree d − 4-part of the ideal of a line ℓ.
. Since both S/I and S/I ′ are Artinian Gorenstein rings of socle degree 2d − 4 we have From this it follows that for k ≤ d − 2 the base locus of J k contains a plane P. Without loss of generality we may assume that P is given by x 0 = x 1 = 0. Since the base locus J d−1 is finite it follows that we can find two further polynomials h 1 , h 2 of degree d − 1, contained in J such that x 0 , x 1 , h 1 , h 2 form a regular sequence. For a general linear form ℓ we have ℓ, x 0 , x 1 , h 1 , h 2 form a regular sequence, and from the previous paragraph it follows that (J, ℓ) ⊂ (ℓ, x 0 , x 1 , h 1 , h 2 ). This implies that J ⊂ (x 0 , x 1 , h 1 , h 2 ) and therefore that X sing contains a complete intersection of multidegree Suppose we are now in the case that I d−4 is the degree d − 4 part of the ideal of a conic. Without loss of generality we may assume that the conic is defined by Then I contains a complete intersection ideal I ′ of (1, 1, 2, d − 2, d − 1). From I 2d−4 = I ′ 2d−4 it follows that I = I ′ . From this it follows that J is contained in a complete intersection ideal of multidegree (1, 2, d − 2, d − 1). Using that p J ≤ 2(d − 2)(d − 1) and that p I ′ = 2(d − 2)(d − 1) it follows that J = I ′ and that X sing is a complete intersection of multidegree ( consists of a conic. If the base locus of I d−2 is also a conic then we have that h I (k) = 2k + 1 for k ≤ d − 2. Using Gorenstein duality we get that Hence the base locus of I d−2 is not a conic. Since the base locus of I d−4 is a conic and the base locus of I d−2 is one dimensional we have that the conic is a union of two line lines h 1 h 2 = 0 and one of the lines, say h 1 = 0 is contained in the base locus of I d−2 and the linear form h = h 2 . Recall that I = (I ′ : h 2 ). Since h 1 h 2 ∈ I ′ it follows that h 1 ∈ I and therefore that the base locus of I 1 is contained in a line, a contradiction, hence d 0 = d − 1 is impossible. If Then h is the Hilbert function of a complete intersection ideal of multidegree (1, 2, d − 2, d − 1).
Suppose that for some k we have that h I (k) > 2k + 1. Then from Theorem 2.1 it follows that h I (j) > 2j + 1 = h(j) for 2 ≤ j ≤ k.
In our case we have that h I (d − 4) > 2d − 7. Hence h I (k) > h(k) for 2 ≤ k ≤ d − 4. Using Gorenstein duality we get h I (k) > h(k) for d ≤ k ≤ 2d − 6. In particular, . This finishes the proof in the case d > 7.
If d = 7 then . Suppose now that equality holds. Then h I takes the following values 1, 3, 6, 8, 8, 8, 8, 8, 6, 3, 1. This implies that the base locus in degree 1 and 2 is a plane. The base locus of I 3 has dimension at most one. From Lemma 5.2 it follows that the dimension of the base locus is at least one. Since h I (3) = 8 it follows that the base locus of I 3 is the intersection of two plane cubics having either a line or a conic as a common component.
In the first two case we would have that the socle degree of S/I ′ is 10. Since S/I has also socle degree 10 this implies that I = I ′ . However I and I ′ have different Hilbert functions, hence this is not the case.
If I ′ is of multidegree (1, 3, 5, 6) then there exists a linear form h such that I = (I ′ : h). From this it follows that the base locus of I ′ k is contained in the base locus of I ′ k−1 union with V (h). Note that the base locus of I ′ 4 is a cubic curve C and the base locus of I 3 is the intersection of two cubics. Hence the base locus of I 3 is a conic Q = 0 together with a point and V (h) is a component of C. Moreover we have that Qh ∈ I ′ 3 . This implies that Q ∈ (I ′ : h) = I, contradicting that h I (2) = 6, hence we can exclude this case.
If I ′ is of multidegree (1, 3, 6, 6) then there exists a quadratic form h such that I = (I ′ : h). From this it follows that the base locus of I ′ k is contained in the base locus of I ′ k−2 union with V (h). Note that the base locus of I ′ 5 is a cubic curve C and the base locus of I 3 is the intersection of two cubics. Hence the base locus of I 3 is contains either a line L = 0 or a conic Q = 0, and this curve is a component of C. In the first case we have Lh ∈ I ′ 3 and by construction that L ∈ I 1 , a contradiction. In the second case we have that h = h 1 h 2 and Qh 1 ∈ I ′ 3 . This implies that Qh 1 h 2 ∈ I ′ 4 and that Q ∈ (I ′ : h) = I. A contradiction. (1) X is factorial.
(2) X contains a plane and X has at least (d − 1) 2 nodes.
Proof. Suppose X is not factorial. It follows directly from Lemma 5.3 and 5.4 that X sing contains a complete intersection Σ either of multidegree (1, . In the first case Σ consists of (d − 1) 2 points in the second case of 2(d − 2)(d − 1) points.
Let f 1 , f 2 , f 3 , f 4 be the generators of I(Σ), ordered by degree. Since the points of Σ are on X it follows that f ∈ (f 1 , f 2 , f 3 , f 4 ). Write f = h i f i . Since the points of Σ are in X sing and the f i form a system of local coordinates each point of Σ it follows that h i ∈ I(Σ). In particular f is in the ideal generated by f i f j . Such a product is of degree at most d only if one of i, j is at most 2. Hence f ∈ (f 1 , f 2 ) and therefore X contains either a plane or a quadric surface, depending on the multidegree of the complete intersection.
Remark 5.6. This result also implies Theorem 4.1. However, in the above proof we used results from [27], which we can avoid in the proof of Theorem 4.1.

Lemma 6.1 (Cayley trick). For 0 ≤ i ≤ 2n we have a natural isomorphism
Proof. Note that we have the following chain of isomorphisms In the first and last line we used the Gysin exact sequence for cohomology with compact support. In the second and second to last line we used Poincaré duality (the complement of a hypersurface in weighted projective space is a Q-homology manifold) and in the middle we used that P \ Y is a C c−1 -bundle over P(w) \ X.
For a subvariety of a toric variety one has the notion of quasismoothness. It follows easily that X ⊂ P(w) is not quasismooth at p if and only if the rank of M (p) = ∂f i ∂x j (p) is strictly less than c. Similarly, we have that Y is not quasismooth at (p, q) ∈ P(E) if all the partial derivatives of c i=1 y i f i vanish simultaneously. Definition 6.2. Let T be a toric variety and let X ⊂ T be a subvariety, then with X sing we denote the set of points p ∈ X such that X is not quasi-smooth at p. Proposition 6.3. The projection P(E) → P(w) restricts to a surjective morphism ψ : Y sing → X sing . For p ∈ X sing the fiber ψ −1 (p) is a linear space of dimension c − rank M (p) − 1. In particular, if X is quasi-smooth then so is Y .
Proof. A point (p, q) on Y is not quasismooth if and only if all the ∂F y i and all the ∂F x j vanish simultaneously at (p, q). The partials with respect to y i are precisely the f i . Hence f i (p) = 0 holds for i = 1, . . . c, which yields ψ(Y sing ) ⊂ X.
The partials of F with respect to x i are j y j ∂f j x i . Hence, if (p, q) is a singular point of Y , then rank M (p) < c and therefore ψ(Y sing ) ⊂ X sing .
Let p ∈ X sing . Then (p, q) is a singular point of Y if and only if M (p)q T = 0. In particular, the fiber of ψ is a linear space of dimension c − rank M (p) − 1.

Proposition 6.4. Suppose p is an isolated hypersurface singularity of X.
Then ψ −1 (p) is a point.
Proof. Locally (X, p) is defined by c equations g 1 = g 2 = · · · = g c = 0 in C n+c . Using the chain rule it follows that the rank the of the Jacobian matrix of (g 1 , . . . , g c ) is independent of the choice of local coordinates and equations. Since (X, p) is a hypersurface singularity we can find local coordinates x 1 , . . . , x n+c on C n+c such that g i = x i for i = 1, . . . , c − 1 and g c defines the hypersurface singularity, i.e., all its partials vanish at p. In particular, the rank of the Jacobian matrix is precisely c − 1.
The previous proposition implies that ψ −1 (p) is a zero-dimensional linear space, i.e., a point. Lemma 6.5. If X is a nodal complete intersection, then Y is a nodal hypersurface.
Proof. Let p be a singular point of X, and (p, q) be the corresponding point on Y . Then we can find a coordinate change in a neighbourhood of p such that (X, p) is locally given by n+1 i=1 (x 2 i ) = x n+2 = · · · = x n+c = 0. Similarly, after a local coordinate change in a neighbourhood of (p, q) we have that the equation for Y is given by The corresponding singular point on Y is (0, . . . , 0) × (1 : 0 : · · · : 0). This point is clearly a node.
For a finite subscheme ∆ ⊂ P(w) we say that the linear system of degree k polynomials through ∆ has defect if for I = I(∆) we have that h I (k) < p I holds.
The following bound for the defect of complete intersections follows from the main result of Cynk [6]: Proposition 6.6. Let X = V (f 1 , . . . , f c ) be a three-dimensional nodal complete intersection in P 3+c such that X c := V (f 1 , . . . , . . . , f c−1 ) is smooth. Then the defect of X is at most the defect of the linear system of degree d c + c i=1 d i − w i polynomials through the points, where X has a node. Since the defect of X and the defect of Y are equal we can compute the defect on Y : Proof. The proof essentially follows the proof of [10, Proposition 3.2] for hypersurfaces in weighted projective space. We refer to that paper for further details.
The defect of Y equals the dimension of the cokernel of Since Y is a nodal hypersurface, the dimension of H n+2c−1 Y sing (Y ) equals the number of nodes of Y and its Hodge structure is of pure (m+c−1, m+c−1) type. There is a residue map which is a morphism of Hodge structures of degree (−1, −1) and surjective for n + 2c − 2 ≥ 3.
Hence the defect of Y equals the dimension of the cokernel On H n+2c−1 (P(E) \ Y ) we have the filtration by the pole order as defined in [8]. The main result of [8] shows that the filtration by the pole order is contained in the Hodge filtration. In particular, we have a surjective map The composed map H 0 (O(D − w, m − 1)) → ⊕ p∈Y sing C is the evaluation map, as in the case of hypersurfaces in P(w). Example 6.8. In the case n = 3 we have that the defect of Y equals the cokernel of ⊕ c i=1 y i S D+d i → ⊕ p∈Y sing C.

Degenerate complete intersections and induced defect
There is a very simple construction of complete intersection with defect, which is often excluded if one attempts to determine the minimal number of nodes on a complete intersection with defect. Let c be an integer and n be an odd integer. Fix an integers d 1 ≤ · · · ≤ d c−1 and polynomials f i such that the X c = V (f 1 , . . . , f c−1 ) is a complete intersection fourfold in P 3+c satisfying h n+3 (X c ) = h n−1 (X c ) = 1. Then X c has a one-dimensional singular locus, say of degree s. Let γ ∈ H n+3 (X c , Z) be such that γ prim = 0. For a general polynomial g of degree d c the complete intersection X(g) := X c ∩ V (g) has defect and this defect is caused by the cycle γ ∩ V (g) ∈ H n+1 (X(g), Z). The number of singular points of X(g) equals sd c . In particular, the number of singular points grows linearly in d c . With some effort one can produce examples of this type where the transversal types of all non-isolated singularities of X c are A 1 , and therefore X(g) is a nodal hypersurface.
The lower bound on the number of nodes we prove later on is quadratic in each of the d i and hence we have to exclude this example. In the literature there are various notions of nondegenerate complete intersections. Each of these notions are attempts to exclude examples as above.
Definition 7.1. Let n be an odd integer and d 1 , . . . , d c integers such that d 1 ≤ d 2 ≤ · · · ≤ d c . Let X ⊂ P n+c be an n-dimensional complete intersection of multidegree (d 1 , . . . , d c ) with isolated singularities.
Remark 7.2. A more natural definition of without induced defect would be to require h n+1+2(c−j) (X j ) = h n−1 (X j ) for every choice of (f 1 , . . . , f c ) ∈ S d 1 ⊕ · · · ⊕ S dc such that X = V (f 1 , . . . , f c ) and for every j ∈ {1, . . . , c − 1} we have. However this definition seems insufficient for the proofs in the next section.
Hence X has positive defect if and only if Y has at least three nodes, or Y has two nodes with the same (y 1 : y 2 )-coordinate.
Suppose we are in the latter case. Then after choosing different generators (f 1 , f 2 ) for I(X) we may assume y 2 = 0. This implies that V (f 1 ) is singular at the two nodes of Y . Since f 1 is a quadric it follows that the singular locus of f 1 is one-dimensional. In particular we may assume that Hence if X has defect and has two nodes then the defect is induced. Cynk and Rams showed that there exist CR-nondegenerate complete intersections of degree (2, 2) with two nodes that have defect. . . . , f c ) ⊂ P 3+c be a nodal complete intersection of multidegree (d 1 , . . . , d c ). Throughout this section we assume that d 1 ≤ · · · ≤ d c . Moreover, if c = 2 and d 1 = d 2 = 2 holds then it follows from Example 7.4 that a complete intersection with defect, but without induced defect has at least 3 nodes. Hence we proved Theorem 8.21 in this case. Throughout this section we will assume that either c > 2 or d 2 > 2 holds. We need this assumption since Proposition 8.1 does not hold true if a general hyperplane section of X is a rational surface.

Complete intersection threefolds
Note that the codimension of W k in ⊕ i R k−dc+d i is at most the number of nodes on Y . From Proposition 6.4 and Lemma 6.5 it follows that X and Y have the same number of nodes.
Let H ⊂ P 3+c a general hyperplane. Then X H is a smooth complete intersection surface in P 2+c . Denote g i := f i | H . Without loss of generality we may assume that H is given by x 3+c = 0. With W ′ we denote the S := C[x 0 , . . . , x 2+c ]-module W | x 3+c =0 . Then for all k we have In the sequel we provide a lower bound for h W ′ (k). We provide first a lower bound for h W ′ (k) for k ≥ d c and here we exploit that X has no induced defect.
To obtain a good bound for lower degrees turns out to be more difficult: In the cases we considered before we could use Gorenstein duality, but this does not seem to work in the complete intersection case. We provide a lower bound for h W (k) for k < d c − 1 in the case that c = 2, or if d c is large compared with c−1 j=1 d j . Note that the module W depends on the choice of generators for I(X), but that h W does not depend on it. For a subspace V ⊂ ⊕S k+d i −dc and a point p ∈ P 2+c we denote with V (p) ⊂ C c the vector space obtained by evaluating all elements of V at p.
Then for all p ∈ P 2+c we have that V dc (p) = C c .
Proof. We have h W (k) = #Y sing for k sufficiently large. Since X has defect, we have that In particular, we can find a subspace V D+dc−c−3 satisfying the first condition. Note that by construction X has a Weil divisor P that is not Q-Cartier. Hence X H contains a divisor which is not the multiple of a hyperplane section. The elements of W are tangent vectors to the equisingular deformation space of X and these equisingular deformations preserve the defect of X (Lemma 3.6). Hence W ′ dc is contained in the tangent space to some component L of the Noether-Lefschetz locus of complete intersections of multidegree (d 1 , . . . , d c ) in P 2+c .
Consider now X ′ c = V (f 1 , . . . , f c−1 ) ∩ H, which is either smooth or has isolated singularities. Since H is a general hyperplane and the transversal type of the non-isolated singularities of V (f 1 , . . . , f c−1 ) is A 1 it follows that X ′ c is a nodal threefold. Define the Noether-Lefschetz locus inside |O X ′ c (d c )| as the locus of smooth surfaces with Picard number at least one. If the Noether-Lefschetz locus is Zariski open in |O X ′ c (d c )| then it follows from [9, Theorem 2.1(c)] that h 4 (X ′ c ) ≥ 2. Since X does not have induced defect we can exclude this. Since the closure of each irreducible component of the Noether-Lefschetz locus is a proper subset of |O X ′ c (d c )| it follows from the same result that Hence we can choose V D+dc−c−3 satisfying both (1) and (2). Denote with e i ∈ C c the i-th standard basis vector. By construction the partials ((g 1 ) x i , . . . , (g c ) x i ) and the elements g j e i are contained in W . If p ∈ P 2+c is a point where the Jacobian matrix of (g 1 , . . . , g c ) has full rank, then dim W ′ dc−1 (p) = c and hence dim V dc−1 (p) = dim V dc (p) = c. Hence we need only to consider points p such that the Jacobian matrix is not of full rank. Since X ′ is smooth there is some i such that f i (p) = 0. Then for any j with d j ≥ d i we have e j ∈ W ′ dc (p). Hence p is a singular point of a partial complete intersection V (f 1 , . . . , f k ) with d k < d i .
Let p be a point such that dim V dc (p) < c. Then there exists a 1 , . . . , a c ∈ C such that a i h i (p) = 0 for all (h 1 , . . . , h c ) ∈ W dc . From e c ∈ V dc (p) it follows that a c vanishes.
Note that the same linear relation holds for (W ′ dc · S D−c−2 )(p). From [8] it follows that there is a surjection ⊕ c i=1 S D+d i −c−2 → H 1,1 (X H , C) prim . Let γ ∈ H 2 (X H , C) prim be a nonzero class mapped to the subspace of W ′ D+dc−c−2 where a i h i (p) = 0 holds. Let Λ ⊂ H 2 (X H , Z) be sub-Hodge structure which contains γ. Then for any infinitesimal deformation of X H in |O X ′ c (d c )| we have that Λ remains a subhodge structure of H 2 . This is ruled out by [9, Theorem 2.1(c)].
Remark 8.2. If we assume that X is nondegenerate in codimension three then X ′ c (as constructed in the above proof) is a smooth threefold. In this case we have that the monodromy representation on H 2 (X H , C) prim is irreducible. If c > 2 or d 2 > 2 this implies that the Noether-Lefschetz locus in |O X ′ c (d c )| is not a Zariski open subset. From this we get points (2) and (3) almost directly. Hence in this case we can avoid the application of [9]. A similar reasoning can also avoid the other application of this paper in the case that X is nondegenerate in codimension three.
If X has induced defect then it is easy to see that the conclusion of the above Proposition may not hold.
For the rest of this section we will choose a V dc+D−c−3 satisfying the conclusion of the proposition and we set Then for all integers k we have Note that by the construction of V we have that the rows of the Jacobian matrix of the complete intersection V (g 1 , . . . , g c ) are contained in V and that g i e j ∈ V . . . . , f c ) be a complete intersection with defect, but without induced defect. Then for Proof. Since X has no induced defect we have for all p ∈ P 2+c that the dimension of V dc (p) equals c. It follows from [17, We want to bound h V in degree at most d c − 1. In order to do this we will define a filtration on V .
Let us define P i V j by pr 1 (F i V )(d c − d j ), i.e., the projection of F i onto the first factor, with an appropriate degree shift.
It is easy to show that for all k and all i ∈ {1, . . . , c − 1}. Note that P i V for i = 1, . . . , c and F c V are ideals and that by construction S/F c V is Artinian Gorenstein of socle degree D + d c − c − 3. . . . , f c ) be a complete intersection with defect, but without induced defect. Then for Proof. Note that the rows of the Jacobian matrix of X H are contained in V dc−1 . From this it follows that the determinant of the Jacobian matrix is contained in F c V D−c . Recall that g i = f i | H and that g i ∈ F c V for all i. Hence a base point of F c V D−c is a singular point of X H . Since X H is smooth it follows that F c V D−c is base point free. Using that h F c V (D+d c −c−3) ≥ 1 holds and Corollary 2.5 we obtain that Proof. This follows from Gorenstein duality and the previous lemma.
Recall that we set D = c i=1 d i . In the sequel we will concentrate on the case D < 2d c . If c = 2 this covers every case except d 1 = d 2 , which we will treat separately. However, for c > 2 the condition D < 2d c requires d c to be large compared with d 1 , . . . , d c−1 .
The following result turns out to be very useful.
Lemma 8.7. Suppose that D ≤ 2d c and that at least one of the following holds: Proof. In the first case there is a line ℓ, such that I(ℓ) ⊂ F c V . In the second case we can apply Theorem 2.4 and we obtain that the base locus of F c V dc−1 , resp. F c V dc−2 consists of a line ℓ together with finitely many points.
Let X ′ be the partial complete intersection V (g 1 , . . . , g c−1 ). The space F c V d is contained in the tangent space to a component of the Noether-Lefschetz locus in |O X ′ (d c )|. We want to apply the strategy from [24,Section 7] to conclude that ℓ ⊂ X H . However, in that paper it is assumed that X ′ is smooth and d c is sufficiently large. In the case where X ′ is smooth, i.e., X is nondegenerate in codimension 3, the proof needs little adaptation: We mostly use the results of Section 1 and 3 of [24], which hold for arbitrary d c . To conclude that X H contains a line we need also the results of Section 7 of loc. cit. and this is the place where Otwinowska needs that d c is sufficiently large. In this section one chooses a general linear space L ⊂ P 2+c of codimension three and uses it to define a spaceT (L) ⊂ ⊕H 0 (X ′ , T X ′ (i)). In particular, for any v ∈T (L) and any element f of the coordinate ring of X ′ one can consider the Lie derivative L v f .
If ℓ ′ is a line not contained in X H , then it is shown in the proof of [24, Lemma 10] that there is a v ∈T (L) such that for any point p ∈ X H ∩ ℓ ′ we have L v g c (p) = 0. In particular, the ideal I ′ generated by g 1 , . . . , g c−1 , g c , I(ℓ ′ ) and L v g c defines the empty scheme. We can find a v such that deg L v (g c ) = D − c, (i.e., we could choose v such that L v (g c ) is a maximal minor of the Jacobian matrix of X H ) and using Corollary 2.2 we obtain that Let E 1 := (F c V : g c ) and let R L be a linear form vanishing on the "cone" C(L, ℓ) ∼ = P c . From [24,Lemme 11] it follows that (I(ℓ), g 1 , . . . , g c , L v g c | v ∈T (L)) is contained in (E 1 : R L ) and from [24,Lemme 12] it follows that If X ′ has isolated singularities we have to proceed in a slightly different way. Since X ′ is smooth in a neighbourhood of X H , we can consider X H as a hypersurface in a resolution of singularitiesX ′ of X ′ . This allows us to define the ideals E i as in [24,Section 1]. More problematic is the definition ofT andT (L). Let C * X ′ be the affine cone over X ′ minus the vertex and C * X ′ the fiber product C * X ′ × X ′X ′ . We can now take the global section of the tangent sheaf of C * X ′ and define a grading as in [24,Section 3]. Then Proposition 1 of loc. cit. holds true.
We define nowT (L) similarly as in [24,Section 7] but we have to restrict to elements inT such that their pushforward to X ′ is well defined. If ℓ ′ ⊂ X ′ then we can find again a v ∈ T (L) such that h (I(ℓ ′ ),f 1 ,...,fc,Lvfc) (D − c + d c − 1) = 0. Similarly we can show that h (E 1 :R L ) )(3d c − 2) = 0 and we obtain again that ℓ ⊂ X H . Lemma 8.8. Suppose that X has defect, but no induced defect. Moreover, suppose that the intersection of a general member of the equisingular deformation space of X with a general hyperplane is a surface containing a line. Then the number of nodes of X is at least Proof. The map (X t , H) → X t,H defines a map from the product of the equisingular deformation space of X and (P 3+c ) ∨ to a component NL(γ) of the Noether-Lefschetz locus of complete intersections of degree (d 1 , . . . , d c ) in P 2+c . Since (X, H) are chosen general and X H contains a line we may assume that the image of this map lands in the component parametrizing surfaces containing a line, i.e, we can take γ to be the primitive part of the class of the line. Then W D+dc−c−3 is contained in the lift to ⊕S D+d i −c−3 of the orthogonal complement of γ in H 1,1 (X, C) prim and we may take V D+dc−c−3 to be precisely this orthogonal complement.
It follows directly from Noether-Lefschetz theory that ⊕I(ℓ) D+d i −c−3 ⊂ V D+d i −c−3 and therefore that L := ⊕I(ℓ)(d i − d c ) is contained in V . If p ∈ ℓ then dim L dc−1 (p) = c. If p ∈ ℓ then p ∈ X H since X H is smooth at p and the rows of the Jacobian matrix of X H are contained in V dc−1 it follows that dim V dc−1 (p) = c.
In particular dim V dc−1 (p) = c for all p ∈ P 2+c . From Since h V (d c − 1) − h L (d c − 1) ≤ c + 1 holds, we obtain that each of these generators in c + 1 and that L k = V k for k ≤ d c − 2. In particular, We denote with h V L the Hilbert function of the S-algebra constructed in the previous proof. Then We will next focus on the case where X H does not contain a line. We consider first the case where h F c V (d c − 1) is sufficiently large:

The latter is at least
for all k and there is at least one k for which the inequality is strict, which implies
and that if equality holds then the base locus of F c V D−c−1 contains a line ℓ. Suppose first that equality holds. Since g 1 , . . . , g c are contained in F c V and D − c − 1 ≥ d c holds, we have that g 1 , . . . , g c ∈ I(ℓ). In particular, X H contains a line, contradicting our assumptions. Hence In particular, by Theorem 2.4 we have that there is a zero-dimensional scheme ∆ of length l such that F c V D−c−1 = I(∆) D−c−1 holds. By the construction of F c V we have then F c V k = I(∆) k for k ≤ D − c − 1. From d c ≤ D − c − 1 and g 1 , . . . , g c ∈ F c V it follows that ∆ is contained in X H . In particular, for any p ∈ ∆ we have dim V dc−1 (p) = c.
Let p ∈ ∆, let v 1 ∈ V be an element of minimal degree k such that v 1 (p) is nonzero. Since dim V dc−1 (p) = c there exist c − 1 elements v 2 , . . . , v c ∈ V dc−1 such that the determinant of v 1 , . . . , v c does not vanish at p. This determinant is contained in F c V and its degree equals k + D − d c − c + 1. Since p is a base point of F c V D−c−1 it follows that k ≥ d c − 1. Hence for k ≤ d c − 2 we have that any element of V k vanishes along ∆ and hence Proof. Since D − 2 = d 2 holds, we have that F d 2 is base point free. The socle degree of F 2 V equals D + d c − c − 3 = 2d 2 − 3 and therefore we have Hence we are done if we can show h V (k) > k+1 for some 1 ≤ k ≤ d 2 −2. Suppose that this is not the case then h F 2 V (k) = k + 1 for k ≤ d 2 − 2. By Lemma 8.7 we have that X H contains a line and therefore X has at least (d 2 − 1) 2 + (d 2 − 1) + 1 nodes.
nodes. Moreover, if equality holds then X H contains a line.
Proof. If X H contains a line and the same holds for any equisingular deformation of X then the result follows from Lemma 8.8. Assume now that there is an equisingular deformation of X such that a general hyperplane section does not contain a line. Since a small equisingular deformation leaves both the number of nodes and the defect invariant, we may replace X by such an equisingular deformation and hence we may assume that X H does not contain a line.
If D < d c + c + 1 then c = 2 and d 1 = 2. The case d 2 = 2 is covered by Example 7.4, the case d 2 > 2 is covered by Lemma 8.13.
If D ≥ d c + c + 1 then it follows from Lemma 8.7, 8.10 and 8.11 that If equality holds then from Lemma 8.7 it follows that X H contains a line, which we excluded. Hence In particular, it follows that Using Gorenstein duality for F c V and the inequality Using that D < 2d c , c ≥ 2 and d c ≥ 4 we obtain that the right hand side is at least 1 and we are done.
holds for all k, and for one value of k we have a strict inequality. It remains to consider the case where   Proof. Suppose h V (0) = 1. Then after a linear change of variables in y 1 , y 2 (i.e., by choosing a new basis for I(X H )) we may assume (0, 1) ∈ V . This contradicts the fact that 0 ⊕ S ⊂ V from Proposition 8.1. Proof. Suppose d = 2 then h V (0) = 2 by Lemma 8.17. From Lemma 8.5 it follows that h F 2 V (1) = 1 and hence h V (0) + h V (1) ≥ 3.
If h F 2 V (3) ≥ 5 then one gets h F 2 V (2) ≥ 4 and h F 2 V (1) ≥ 3. Then using Gorenstein duality we get that h F 2 V (k) ≥ 26. Since h V (0) = 2 it follows that h V (k) ≥ 27. If h F 2 V (3) = 4 and h F 2 V (2) ≥ 4 holds then h F 2 V (4) = 4 by duality and h F 2 V (5) ≤ 4 by Theorem 2.1. Hence h F 2 V (2) = h F 2 V (5) = 4. This means that the base locus of F 2 V 5 consists of four points and we obtain that V k (p) = 0 for k ≤ 2 and p a base point. In particular h V (k) ≥ 2h F 2 V (k) for k ≤ 2. This implies h V (k) ≥ 32. It remains to consider the case h F 2 V (3) = 4 and h F 2 V (2) = 3. In this case we have by Lemma 8.7 that X H contains a line and we are done. Suppose now d = 5. then the socle degree is 10. Note that h F 2 V (5) ≥ 6. If h F 2 V (4) = 5 then by Theorem 2.4 we know that the base locus of F 2 V 5 contains a line and therefore that X H contains a line. Hence h F 2 V (6) = h F 2 V (4) ≥ 6. Using Corollary 2.2 it follows that h F 2 V (3) ≥ 5 and h F 2 V (2) ≥ 4 and that h F 2 V (1) ≥ 3. Using Gorenstein duality we get that h F 2 V (k) ≥ 44. If we have equality in degree 3 then we have it also in degree 1 and 2 and the base locus of F 2 V 4 consists of a line and a point and the base locus of F 2 V 6 consists of at least 5 collinear points. This implies that V k (p) = 0 for k ≤ 2 and all four base points. In particular we get a contribution of at least 8 from P 1 V , and hence k h V (k) is at least 52. If h F 2 V (3) ≥ 6 then h F 2 V (2) ≥ 4. In this case we obtain h F 2 V (k) ≥ 48 and we are done. Now assume that h F 2 V (4) ≥ 6 then h F 2 V (2) ≥ 5. This implies that the total contribution of F 2 V is at least 48. Since there is a contribution from P 1 V we are done.
Lemma 8.20. Suppose that X H does not contain a line and that d : then by Theorem 2.4 we obtain that the base locus of F c V d contains a line and therefore that X H contains a line. Hence h F 2 V (d) ≤ h F 2 V (d − 1) holds. Using Gorenstein duality we get by the same argument h F 2 V (d) = h F 2 V (2d − 5) ≥ h F 2 V (2d − 4) = h F 2 V (d − 1) and hence h F 2 V (d) = h F 2 V (d − 1). Using Gorenstein duality again we get that To conclude this case if h P 1 V (d − 1) ≥ d − 1 then we obtain h V (d − 1) ≥ 2d − 2. If h P 1 V (d − 1) < d − 1 then we get that h P 1 V (d) < h P 1 V (d − 1) and hence that h V (d − 1) > h V (d) = 2d − 4 holds. In both cases we have that h V (d − 1) ≥ 2d − 3.
We consider next degree d − 2. If h F 2 V (d − 2) = d − 1 and either holds then we have that F 2 V d−1 is the degree d − 1 part of the ideal of a line and hence that X H contains a line by Lemma 8.7. Hence we may exclude this and we have in any case that h F 2 V (k) ≥ k + 2 for k ≤ d − 3 and h F 2 V (d − 2) ≥ d − 1. If h P 1 V (d − 1) ≥ d − 1 then we have h P 1 V (k) ≥ k + 1 for k ≤ d − 2 by Corollary 2.3, hence h V (k) ≥ 2k + 2 for k ≤ 2 and we are done. We assume now that h P 1 V (d − 1) < d − 1. If From Theorem 2.1 it follows that h F 2 V (k) ≥ min(k + a, 2k + 1) and h P 1 V (k) ≥ min(k + 1, b). If h V (k) < 2k + 2 then k + a + b < 2k + 2, i.e. k ∈ {d − 2, d − 3}. The total difference d−1 k=0 h V L (k) − h V (k) is at most 3.
By duality we have that 3d−5 k=d h V (k) − h V L (k) at least d − 1 and since d > 5 we are done.
The final case is h P 1 V (d − 1) = 0. In this case h F 2 V (d) = h V (d) ≥ 2d − 4 and by Theorem 2.1 we have that Then h F 2 V (k) ≥ min(k + a, 2k + 1). The total miss d−1 k=0 h V L (k) − h V (k) is at most d + 6. If d ≥ 6 then a ≥ 2 and by duality we have that k≥d h V (k) − h V L (k) is at least 2d + 1, which is again at least d + 6.
Theorem 8.21. Suppose X is a nodal complete intersection of bidegree (d, d) with defect and without induced defect. Then X has at least 3(d − 1) 2 nodes.
Proof. If d = 2 then by Lemma 8.17 we have that h V (0) = 2. Since X has defect we have that h V (1) ≥ 1 and we are done.
Suppose now that d > 2 holds. If X H contains a line and the same holds for any equisingular deformation then the result follows from Lemma 8.8. If X H contains a line but this property does not hold for any equisingular deformation then we may replace X by this equisingular deformation and therefore assume that X H does not contain a line. Depending on d and h F c V (d − 1) this is covered by one of Lemma 8.18, Lemma 8.19 or Lemma 8.20.