The Maximal Ran Conjecture Eric Larson Abstract Let C be a general curve of genus g, embedded in P r via a general linear series of degree d. In this paper, we prove the Maximal Ran Conjecture, which determines the Hilbert function of C P r. 1 Introduction A central object of study in algebraic geometry in the past couple of centuries has been algebraic curves in complex projective space. (In this paper, we wor exclusively over C.) These can be described in two basic ways: Parametric Coordinates: We tae an abstract curve C of genus g, pic a line bundle L on C of degree d, and r + 1 linearly independent sections of H 0 (L) whose span is basepoint free. This gives rise to a degree d map C P r. Cartesian Coordinates: We tae a saturated homogeneous ideal I C[x 0, x 1,..., x r ] of height r 1. This gives rise to a curve V (I) P r. A natural question is: How do these points of view relate to each other? The Brill Noether theorem, proven by Griffiths and Harris [6], Gieseer [5], Kleiman and Lasov [11], and others, describes the space of parametric curves with general source: If C is a general curve of genus g, it states that there exists a nondegenerate degree d map C P r if and only if the Brill Noether number ρ(d, g, r) is nonnegative, where ρ(d, g, r) := (r + 1)d rg r(r + 1); and moreover, in this case, there exists a unique component of the Kontsevich space of stable maps M g (P r, d) that both dominates the moduli space of curves M g, and whose general member is nondegenerate. We term stable maps corresponding to points in this component Brill Noether curves (BN-curves). For r 3, it is nown that a general BN-curve is an embedding of a smooth curve, and so we may identify it with its image. There is a unique component of the Hilbert scheme containing the images of general BN-curves of degree d and genus g; by abuse of notation, we also term curves C P r corresponding to points in this component of the Hilbert scheme BN-curves. A specific instance of the above natural question is: What does the homogeneous ideal ( Cartesian equations ) of a general BN-curve loo lie, in terms of d, g, and r? 1
As a first step, one might as for the dimension of the graded pieces of the ideal, i.e. for space of polynomials of each degree that vanish on C, which can be described as the ernel of the restriction map H 0 (O P r()) H 0 (O C ()). The dimensions of these spaces are nown: ( ) r + dim H 0 (O P r()) = and dim H 0 (O C ()) = d + 1 g. The natural conjecture made originally by Severi in 1915 [7] when the Brill-Noether theorem was still a conjecture is: Conjecture 1.1 (Maximal Ran Conjecture). If C P r is a general BN-curve (r 3), the restriction maps H 0 (O P r()) H 0 (O C ()) are of maximal ran (i.e. either injective or surjective). Or equivalently, the dimension of the space of polynomials of degree which vanish on C is max ( 0, ( ) ) r+ (d + 1 g). Previously, many special cases of the maximal ran conjecture have been studied in the literature including the case of rational curves in P 3 by Hirschowitz [8]; the cases of nonspecial curves (i.e. the case d g + r) [4] and space curves [3] by Ballico and Ellia; the case of quadrics (i.e. for = 2) independently by Ballico [2], and by Jensen and Payne [9]; and many others. However, until now, a proof of the full conjecture was elusive. In this paper, we give the first proof in full generality: Theorem 1.2. Conjecture 1.1 (the Maximal Ran Conjecture) holds. More generally, we say a subscheme T P r satisfies maximal ran for polynomials of degree if the restriction map H 0 (O P r()) H 0 (O T ()) is of maximal ran. Since H 1 (O P r()) = 0, the long exact sequence in cohomolgy attached to the short exact sequence of sheaves 0 I C () O P r() O C () 0 implies that T P r satisfies maximal ran for polynomials of degree if and only if H 0 (I C ()) = 0 or H 1 (I C ()) = 0; the vanishing of H 0 (I C ()) being equivalent to the injectivity of the restriction map, and the vanishing of H 1 (I C ()) being equivalent to the surjectivity. In particular, the condition of satisfying maximal ran is open, and can therefore be approached via degeneration. Most cases of the Maximal Ran Conjecture that have proven thus far in the literature have used a specific degenerative approach due originally to Hirschowitz: 2
Degeneration to a reducible curve C C with C contained in a hyperplane H and C transverse to H. In this case, from the long exact sequence in cohomology attached to the short exact sequence of sheaves 0 I C P r( 1) I C C P r() I C (C H) H() 0, we conclude that to show H i (I C C Pr()) = 0 as desired, it suffices to show H i (I C P r( 1)) = Hi (I C (C H) H()) = 0. One can thus try to argue by induction on r and, reducing the desired result for (r, ) inductively to (r 1, ) and (r, 1). However, there are several critical issues with this approach, that have limited previous attempts to prove the maximal ran conjecture to special cases: 1. We need a uniform way to construct the reducible curves C C (previous methods were more or less ad-hoc). 2. We want: (a) Either H 0 (I C P r( 1)) = H0 (I C (C H) H()) = 0 H 1 (I C P r( 1)) = H1 (I C (C H) H()) = 0. (H 0 (I C P r( 1)) = H1 (I C (C H) H()) = 0, for example, proves nothing). (b) The fiber dimension of the map from the Hilbert scheme to M g at [C C ] to be ρ(d, g, r) + dim Aut P r, which forces C C to be a BN-curve. (c) The points C C to be general in H, so we may tae C and C each general. But each of these imply inequalities on the degree and genus of C, which do not in general have a solution. 3. This method relates maximal ran for C C to maximal ran for C and maximal ran for C (C H). Note that C (C H) is not a curve, so we need a stronger inductive hypothesis. But even worse, C and C must satisfy incidence conditions, so C and C H are not independently general and there is no nice description of C (C H) that doesn t reference the entire reducible curve C C. The ey innovations introduced some here and some elsewhere by the author to get around these difficulties are as follows: or 3
First Difficulty (uniformity): We leverage results on the interpolation problem for normal bundles, which determine the number of general points a BN-curve of given degree and genus can pass through (as well as variants which wor when some of the points are constrained to lie in a transverse hyperplane). If we find two curves C 1 and C 2 both passing through a finite set of points Γ, their union produces a reducible curve. Results on the interpolation problem thus let us build desired curves uniformly by showing certain systems of inequalities have integer solutions. The interpolation problem for normal bundles was studied in a sequence of papers by the author and others [1, 14, 16, 18, 19], which contains all the results on this topic that we shall need here. Second Difficulty (desiderata conflict): We relax our second desideratum, and show that certain reducible curves constructed as above are BN-curves, even when the fiber dimension of the map from the Hilbert scheme to M g is too large. This is done by first leveraging results on the interpolation problem for restricted tangent bundles to calculate this fiber dimension at certain other reducible curves of this form, and showing that for these other curves it is correct. Then we use iterative specialization and deformation to construct a broen arc in the Kontsevich space (which may not mae sense in the Hilbert scheme compactification) between the desired reducible curves and the other ones. Provided that the specializations are to smooth points of the Kontsevich space, this shows the desired curves are in the same component of the Kontsevich space, and thus the same component of the Hilbert scheme, as these other curves. This program is carried out in a sequence of papers by the author [12, 13, 15], which contains all the results on this topic that we shall need here. Third Difficulty (C (C H)): The ey idea to get around this difficulty is to degenerate to a 3-component curve C 1 C 2 C, with C a general BN-curve in a hyperplane H, and C 1 C 2 a BN-curve transverse to H such that [C 2 H] [C 1 C 2] Sym deg C 2 P r Sym #(C 1 C 2 ) H is general. We then smooth C 1 C 2 to a general BN-curve C, while preserving the incidence conditions with C. The method of Hirschowitz relates maximal ran for C C P r to maximal ran for C P r, which we can apply induction to; and to maximal ran for C (C H), which can be further specialized to C (C 1 H) (C 2 H). But now C 1 H is a set of general points, and C and C 2 H are independently general (and so can in particular be described without reference to the incidence conditions)! Since these hyperplane sections are independently general, they may be studied separately using results of the author in [17]. This construction is studied in Section 2 of the present paper. The upshot is that we can then argue by induction on the the following stronger hypothesis (note that taing n = ɛ = 0 recovers the maximal ran conjecture, so this proves Theorem 1.2 as desired): Theorem 1.3. Fix an inclusion P r P r+1, and let be a positive integer. Let C P r be a general BN-curve (which is alternatively permitted to be a general rational curve of degree d < r 4
if 3); D 1, D 2,..., D n P r+1 be independently general BN-curves which are required to be nonspecial if = 2 and r 4; and p 1, p 2,..., p ɛ P r be a general set of points, where { 2 if n > 0 and 5 and r 4; ɛ ɛ 0 := 0 otherwise. Then any subset of T := C ((D 1 D 2 D n ) P r ) {p 1, p 2,..., p ɛ } P r which contains C {p 1, p 2,..., p ɛ } satisfies maximal ran for polynomials of degree. As explained above, our argument will be by induction on r and ; we shall reduce Theorem 1.3 for (r, ) inductively to Theorem 1.3 for (r 1, ) and (r, 1). But first, for fixed (r, ), we may mae the following reductions: 1. Applying the uniform position principle and Lemma 2.5 of [17], it suffices to prove that T satisfies maximal ran in Theorem 1.3. (Theorem 1.3 is stated for any subset of T containing C {p 1, p 2,..., p ɛ } only because that is a more convenient inductive hypothesis.) 2. If n = 0 or r = 3 or = 2, then again by Lemma 2.5 of [17], we may reduce to the case ɛ = 0. (Note that we may always lower ɛ if desired, since general points impose independent conditions on any subspace of sections of O P r().) Notation: In the proof of Theorem 1.3, we write d and g for the degree and genus of C, and d i and g i for the degrees and genera of the D i. We also define h := ɛ ɛ 0 + n d i. i=1 Organization of the remainder of the paper: We begin, in Section 2, by studying the ey degenerations that we shall use in our inductive argument. Then in Sections 3 and 4, we establish Theorem 1.3 for space curves (r = 3) and quadrics ( = 2), leveraging results of Ballico and Ellia [3] and Ballico [2] which establish Theorem 1.2 in these cases; these will form the base cases for our larger inductive argument. In Sections 5 and 6 we study the easy cases of Theorem 1.3, where the restriction map is far from being an isomorphism (i.e. either the dimension of the source is much larger than the dimension of the target, or vice versa); here we leverage results of Ballico and Ellia which establish Theorem 1.2 for nonspecial curves [4]. Intuitively, these are the easier cases since the codimension in the space of all linear maps of those with non-maximal ran is highest when the dimensions are far apart. Finally, in Section 7, we use the machinery of the preceding sections to reduce Theorem 1.3 to a computation involving the existence of integers satisfying certain systems of inequalities. This computation is taen care of in Appendices A, B, and C. 5
Acnowledgements The author would lie to than Joe Harris for his guidance throughout this research, as well as Atanas Atanasov, Edoardo Ballico, Brian Osserman, Sam Payne, Ravi Vail, Isabel Vogt, David Yang, and other members of the Harvard and MIT mathematics departments, for helpful conversations or comments on this manuscript. The author would also lie to acnowledge the generous support both of the Fannie and John Hertz Foundation, and of the Department of Defense (NDSEG fellowship). 2 Degenerations In this section, we outline the main degenerations we shall use in the proof of Theorem 1.3. Definition 2.1. We say r,, d, g, d, g, h, h, and ɛ 0 satisfy I(r,, d, g, d, g, h, h, ɛ 0, ɛ 0) (respectively satisfy S(r,, d, g, d, g, h, h, ɛ 0, ɛ 0)) if ( ) ( 1)d + 1 g + h + ɛ r + 0 (respectively ) r + (1) ( ) d g ( 1)d + g + h h + ɛ 0 ɛ r r + 0 (respectively ) r +. (2) Proposition 2.2. Suppose that Theorem 1.3 holds for (r 1, ) and (r, 1), and let H P r be a hyperplane. Assume there exists: 1. A specialization of C to an interior BN-curve C = C C, with C contained in H and C transverse to H, with C C general, and with C of degree d and genus g. 2. Specializations {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} of {p ɛ0 +1, p ɛ0 +2,..., p ɛ }, and D i of each D i, with #({p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H) + #(D i H) = h h. 3. A specialization {p 1, p 2,..., p ɛ 0 } of {p 1, p 2,..., p ɛ0 }, with #({p 1, p 2,..., p ɛ 0 } H) = ɛ 0 ɛ 0. Assume also that for some deformation C of C, A := ( C D 1 D 2 D n {p 1,..., p ɛ} ) H H and B := C (( D 1 D 2 D n {p 1,..., p ɛ} ) (P r H) ) P r satisfy the assumptions of Theorem 1.3 or are the union of subsets hyperplane sections of general BN-curves, which are nonspecial if = 2. If r,, d, g, d, g, h, h, ɛ 0, and ɛ 0 satisfy either then Theorem 1.3 holds for T. I(r,, d, g, d, g, h, h, ɛ 0, ɛ 0) or S(r,, d, g, d, g, h, h, ɛ 0, ɛ 0), 6
Proof. Since C C is general in H, and C is transverse to H, we may arrange for C to pass through C C, so that C C is a deformation of C C. Since C C is an interior BN-curve by assumption, C C is also a specialization of C. Write T = C C ((D 1 D 2 D n) P r ) {p 1, p 2,..., p ɛ}. Then we have an exact sequence 0 I B ( 1) I T () I T H() 0. In particular, to show H i (I T ()) = 0, it suffices to show H i (I B ( 1)) = H i (I T H()) = 0. Since T H can be specialized to A, it thus suffices to show H i (I B ( 1)) = H i (I A ()) = 0, or equivalently that the restriction maps H 0 (O H ()) H 0 (O A ()) and H 0 (O P r( 1)) H 0 (O B ( 1)) are either both injective or both surjective. Since A H and B P r satisfy the assumptions of Theorem 1.3 or Theorem 1.3 of [17], we now by our inductive hypothesis that each of these maps is either injective or surjective. It thus remains to show that either or dim H 0 (O B ()) dim H 0 (O P r( 1)) and dim H 0 (O A ()) dim H 0 (O H ()), dim H 0 (O B ()) dim H 0 (O P r( 1)) and dim H 0 (O A ()) dim H 0 (O H ()). But these are exactly the conditions I(r,, d, g, d, g, h, h, ɛ 0, ɛ 0) and S(r,, d, g, d, g, h, h, ɛ 0, ɛ 0) respectively. Proposition 2.3. Suppose that Theorem 1.3 holds for (r 1, ), and let H P r be a hyperplane. Assume there exists: 1. A specialization of C to a general BN-curve C H. 2. Specializations {p 1, p 2,..., p ɛ} of {p 1, p 2,..., p ɛ }, and D i of each D i, such that B := ( D 1 D 2 D n {p 1,..., p ɛ} ) (P r H) P r is a union of subsets of hyperplane sections ) ( of general BN-curves, which are nonspecial if = 3, and has cardinality = r+ 1 ). If in addition (r+ r+ 1 A := C ( (D 1 D 2 D n {p 1,..., p ɛ}) H ) H satisfies the assumptions of Theorem 1.3, then Theorem 1.3 holds for T. Proof. Write T = C ((D 1 D 2 D n) P r ) {p 1, p 2,..., p ɛ}. Then we have an exact sequence 0 I B ( 1) I T () I A () 0. In particular, to show H i (I T ()) = 0, it suffices to show H i (I B ( 1)) = H i (I A ()) = 0. By Theorem 1.3 of [17] and our second assumption, we have H 0 (I B ( 1)) = H 1 (I B ( 1)) = 0. It thus suffices to note that, by our inductive hypothesis for Theorem 1.3 either H 0 (I A ()) = 0 or H 1 (I A ()) = 0. 7
Definition 2.4. We say that ɛ 0, ɛ 0, and t satisfy E(ɛ 0, ɛ 0, t) if t ɛ 0 ɛ 0. Proposition 2.5. If t 0, there exists a specialization {p 1, p 2,..., p ɛ 0 } of {p 1, p 2,..., p ɛ0 }, with #({p 1, p 2,..., p ɛ 0 } H) = ɛ 0 ɛ 0, so that {p 1, p 2,..., p ɛ 0 } H H and {p 1, p 2,..., p ɛ 0 } P r H P r are sets of general points, the second of cardinality at least t, provided that ɛ 0, ɛ 0, and t satisfy E(ɛ 0, ɛ 0, t). Proof. This is clear since {p 1, p 2,..., p ɛ0 } are general points, so can be specialized to an arbitrary set of points. Definition 2.6. We say that r, h, and h satisfy A(r, h, h ) if 0 h h r + 1. Proposition 2.7. There exist specializations {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} of {p ɛ0 +1, p ɛ0 +2,..., p ɛ }, and Di of each D i, with so that are sets of general points, and #({p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H) + #(D i H) = h h, {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H H and (D 1 D 2 D n {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ}) (P r H) P r D 1 H, D 2 H,..., D n H H is a set of hyperplane sections of independantly general BN-curves, for some integer h satisfying A(r, h, h ). Proof. Since general points in P r can be specialized to general points in either H or P r, we reduce to the case ɛ = ɛ 0, which follows from combining Lemmas 5.3 and 6.1 of [17]. Definition 2.8. We say that r, h, and h satisfy J(r, h, h ) if h r + h h, r + 1 satisfy K(r, h, h ) if h 2r + 1 and 0 h h, 8
satisfy L(r, h, h ) if satisfy M(r, h, h ) if and satisfy N(r, h, h ) if h 3r + 2 and h = 2, h 3r + 2 and r + 2 h h h r + h h r r + 1 h 2r + 2 r 2,. Proposition 2.9. There exist specializations {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} of {p ɛ0 +1, p ɛ0 +2,..., p ɛ }, and Di of each D i, with so that are sets of general points, and #({p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H) + #(D i H) = h h, {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H H and {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} P r H P r D 1 H, D 2 H,..., D n H H and D 1 P r H, D 2 P r H,..., D n P r H P r are sets of subsets of hyperplane sections of independantly general BN-curves, provided that r, h, and h satisfy J(r, h, h ). Moreover, such specializations exist so that the second of these sets can be specialized to a set of subsets of hyperplane sections of independantly general nonspecial BN-curves if r, h, and h satisfy one of K(r, h, h ), L(r, h, h ), M(r, h, h ), or N(r, h, h ). Proof. Since general points in P r can be specialized to general points in either H or P r, we reduce to the case ɛ = ɛ 0. By combining Lemmas 5.4 and 6.1 of [17], we see such a specialization exists if r, h, and h satisfy J(r, h, h ); moreover, we can tae the second of these sets to be a set of subsets of hyperplane sections of independantly general nonspecial BN-curves if N(r, h, h ) is satisfied. It thus remains to consider the case when one of K(r, h, h ), L(r, h, h ), or M(r, h, h ) is satisfied. Note that d i d i ρ(d i, g i, r + 1) = (r + 1) (g i + r + 1 d i + 1). In particular, D i can only be special if its degree is at least 2r + 2. If K(r, h, h ) is satisfied, we thus conclude all D i are nonspecial. Similarly, if L(r, h, h ) or M(r, h, h ) is satisfied, then since a BN-curve in P r+1 has degree at least r +1, we thus conclude that either all D i are nonspecial or n = 1 and g 1 = d 1 r. Since the the desired result when all D i are nonspecial follows from Corollary 4.2 of [17], it remains to consider the case when L(r, h, h ) or M(r, h, h ) is satisfied, n = 1, and g 1 = d 1 r. 9
If L(r, h, h ) is satisfied, the result now follows from Lemma 5.3 of [17]. It thus remains to consider the case when r, h, and h satisfy L(r, h, h ) and not N(r, h, h ), i.e. when r h 3r + 2 and h r + 1 h h. 2 By the uniform position principle, the points of D 1 P r are in linear general position. We may therefore apply an automorphism of P r so that exactly h h r 1 of these points lie in H. Since the automorphism group of P r acts transitively on sets of h h r 1 points in linear general position, D 1 H can be assumed to be general; in particular, it is a subset of the hyperplane section of a general rational normal curve. It remains to see D 1 P r H, which is a subset of a hyperplane section of a general special BN-curve, can be specialized to a subset of a hyperplane section of a general nonspecial BN-curve. For this, we apply Theorem 1.8 of [12] to degenerate D 1 P r to a stable map f : D Γ P 1 P r, with f D a general BN-curve of degree d 1 r 2 and genus d 1 r 2 r 1 (which is in particular nonspecial), and f P 1 of degree r 2, and #Γ = r 2 +2. Since h h r 2 by assumption, we can arrange for D 1 P r H f(d) H. Definition 2.10. We say that r, d, g, d, and n satisfy X(r, d, g, d, n) if g n + 1 0 (3) r(d d ) (r 1)g + (r 1)n r 2 + 1 0 (4) n 1 0 (5) d n 0 (6) r + 2 n 0 (7) 2n + d d g r 1 0. (8) Proposition 2.11. Let r 4 and d, g, and d be integers which satisfy (r + 1)d rg r(r + 1) 0, and suppose there exists an integer n which satisfies X(r,, d, g, d, n). Then there exists a BNcurve C C P r of degree d and genus g, with C a general BN-curve in a hyperplane H, and C a general rational curve of degree d transverse to H. Proof. Let n be some such integer. Write g = g n + 1 and d = d d. Upon rearrangement, (3) gives g 0, and (4) gives ρ(d, g, r 1) 0. We may therefore let C H be a general BN-curve of degree d and genus g. Note that C passes through r + 2 general points in H, by Corollary 1.4 of [1] if C is nonsepcial or Theorem 1.2 of [16] if C is special. In particular, C passes through a set Γ of n 1 general points in H (note that n r + 2 by (7), and that n 1 by (5)). Since d n by (6), and the hyperplane section of a general rational curve of degree d is a general set of d points by Corollary 1.5 of [15], we may thus let C be a general rational curve of degree d passing through Γ. By construction, C C is of degree d + d = d and genus g + n 1 = g. It thus remains to see it is a BN-curve; by Theorem 1.9 of [12], it suffices to see n r + 2 and d + n g + r. The first of these inequalities is just (7), while the second becomes upon rearrangement (8). 10
Definition 2.12. We say that r, d, g, d, g, n, and t satisfy Y (r, d, g, d, g, n, t) if g 0 (9) (r + 1)d rg r 2 r 0 (10) (2r 3)d (r 2) 2 g 2r 2 + 3r 9 0 (11) g g n + 1 0 (12) r(d d ) (r 1)(g g ) + (r 1)n r 2 + 1 0 (13) n 1 0 (14) d n t 0 (15) r(d d ) (r 4)(g g ) 2n 2r + 2 0 (16) 2n + d + g d g r 2 0. (17) Proposition 2.13. Let r 4 and t 0, and d, g, d, and g be integers which satisfy (r + 1)d rg r(r + 1) 0, and suppose there exists an integer n satisfying Y (r,, d, g, d, g, n, t). Then there exists a BNcurve C C P r of degree d and genus g, with C a general BN-curve in a hyperplane, and C a general BN-curve of degree d and genus g transverse to H such that C H is a set of d general points in H, at least t of which do not lie on C. Proof. Let n be the minimal such integer. Note that (11) and (15), together with our assumption that t 0, imply Additionally, (15) implies (2r 3)d (r 2) 2 (g d + n) 2r 2 + 3r 9 0. (11 ) d n 0. (15 ) Since the left-hand sides of (11) and (15) are nonincreasing in n, it follows that n is also the minimal integer satisfying the system of inequalities (11 ), (12), (13), (14), (15 ), (16), (17). Additionally, note that (11) also implies (2r 3)(d + 1) (r 2) 2 g 2r 2 + 3r 9 0. We conclude the desired curve exists by applying Theorem 1.1 and Remar 1.2 of [13]. Lemma 2.14. Let X P r be a subscheme of codimension at least 2, and X be a set of r + 2 points which are general in some nondegenerate component of the smooth locus. Then for every integer m 1, there exists a BN-curve C of degree rm and genus (r + 1)(m 1) whose intersection with X is exactly the reduced scheme. Proof. We argue by induction on m. When m = 1, we first note that Aut P r acts transitively on sets of r + 2 points in linear general position. Applying an automorphism to a rational normal curve, we may thus find a rational normal curve C passing through. It is a classical fact (and also an immediate consequence of Theorem 1.3 of [1]) that N C O P 1(r + 2) (r 1), and so N C ( ) O (r 1) P has 1 11
vanishing cohomology and is generated by global sections. We may thus deform C to a curve passing through which avoids any (excess) intersection with any subvariety of codimension at least 2. For the inductive step, we let C 0 be a BN-curve of degree r(m 1) and genus (r + 1)(m 2) whose intersection with X is exactly the reduced scheme. We then let 0 be a set of r + 2 general points on C 0. Applying our inductive hypothesis again, we may find a rational normal curve C 1 whose intersection with X C 0 is exactly the reduced scheme. Taing C = C 0 C 1 completes the proof, as this is a BN-curve by Theorem 1.6 of [12]. Definition 2.15. We say that r, d, g, d, g, n, m, and t satisfy Z(r, d, g, d, g, n, m, t) if (r + 1)d rg r 2 r 0 (18) g (r + 1)m 0 (19) (2r 3)(d rm) (r 2) 2 (g (r + 1)m) 2r 2 + 3r 9 0 (20) g g n + 1 0 (21) r(d d ) (r 1)(g g ) + (r 1)n r 2 + 1 0 (22) n 1 0 (23) (d rm) n t 0 (24) r(d d ) (r 4)(g g ) 2n 2r + 2 0 (25) 2n + d + g d g r 2 0 (26) 2(d rm) (r 3)(g (r + 1)m 1) (r 1)(r + 2) 0 (27) m 0. (28) Proposition 2.16. Let r 4 and t 0, and d, g, d, and g be integers which satisfy (r + 1)d rg r(r + 1) 0, and suppose there exist integers n and m satisfying Z(r, d, g, d, g, n, m, t). Then there exists an interior BN-curve C 1 C 2 C P r of degree d and genus g, with C a general BN-curve in a hyperplane, and C 1 C 2 a BN-curve of degree d and genus g transverse to H such that C 2 H is a set of general points in H, at least t of which do not lie on C 2, and C 1 is either a BN-curve which is general independent from C 2 H or C 1 =. Proof. By Proposition 2.13, there exists a BN-curve C 2 C P r of degree d rm and genus g (r + 1)m, with C a general BN-curve in a hyperplane H, and C 2 a general BN-curve of degree d rm and genus g (r + 1)m transverse to H such that C 2 H is a set of d rm general points in H, at least t of which do not lie on C. Write Γ = C 2 C. By (20), the bundle N C 2 ( 1) satisfies interpolation. In particular, using (27), we have H 1 (N C 2 ( 1)( )) = 0 where C 2 is a set of r + 2 general points on C 2. We have the exact sequences 0 N C 2 C C 2 ( Γ ) N C 2 C ( ) N C 2 C C 0 0 N C 2 ( 1)( ) N C 2 C C 2 ( Γ ) 0 0 N C /H N C 2 C C N H C (Γ) O C (1)(Γ) 0, 12
where the s denote punctual sheaves, which in particular have vanishing H 1. Since Lemma 3.2 of [12] gives H 1 (N C /H) = 0, we conclude H 1 (N C 2 C ( )) = 0 provided H1 (O C (1)(Γ)) = 0. But since C H is a general BN-curve of degree d d and genus g +1 g n, we have either H 0 (O C (1)) = r or H 1 (O C (1)) = 0. In the second case, H 1 (O C (1)(Γ)) = 0 is immediate. In the first case, this implies via (26) that dim H 1 (O C (1)) = r χ(o C (1)) = n 2 (2n + d + g d g r 2) n. Since twisting up by a general point drops the dimension of H 1 when that dimension is positive (the Serre dual of the familiar statement that twisting down by a general point drops the dimension of H 0 when that dimension is positive), we conclude H 1 (O C (1)(Γ)) = 0 and thus H 1 (N C 2 C ( )) = 0. If m = 0, we tae C 1 = ; as H 1 (N C 2 C ( )) = 0, we have H 1 (N C 1 C 2 C ) = H1 (N C 2 C ) = 0. In particular, [C 1 C 2 C = C 2 C ] is a smooth point of the Hilbert scheme, and thus an interior curve as desired. If m 1, we apply Lemma 2.14 to let C 1 be a BN-curve of degree rm and genus (r+1)(m 1) whose intersection with C 2 is exactly. We then deform C 1 to be general in some component of the space of BN-curves passing through. By Theorem 1.6 of [12], both C 1 C 2 and C 1 C 2 C are BN-curves. Moreover, using our assumption that H 1 (N C 2 C ( )) = 0, Lemmas 3.2, 3.3, and 3.4 of [12] imply C 1 C 2 C is an interior curve, as desired. In Appendix A, we give code in sage to chec or create algebraic expressions for all inequalities that appear in this section. 3 Space Curves In this section, we prove Theorem 1.3 for space curves (r = 3); this will serve as one of the base cases for our larger inductive argument. Our argument here will also be by induction, this time on n. Recall that, as mentioned in the introduction, it suffices to prove maximal ran for T, and we may tae ɛ = 0. Our base case will be n = 0, for which this is a result of Ballico and Ellia if C is a BN-curve [3], and a well-nown classical fact if C is a general rational curve of degree d < r (in this case the restriction map is always surjective). We therefore assume for our inductive argument that n 1. Write d i = deg D i and g i = genus D i, and suppose without loss of generality that d 1 d 2 d n ; write d and g for the degree and genus of C. By our inductive hypothesis, the subscheme T n 1 := C ((D 1 D 2 D n 1 ) P r ) P r satisfies maximal ran for polynomials of degrees and 1. Moreover, if Λ P 3 is a general plane, and 2, then Theorem 1.3 of [17] implies T n 1 Λ = C Λ Λ satisfies maximal ran for polynomials of degree. Since T n 1 is positive-dimensional, an application of Theorem 1.5 of [17] completes the proof unless 3 and (d n, g n ) {(8, 5), (9, 6), (10, 7)}, and dim H 0 (O P 3( 1)) > dim H 0 (O Tn 1 ( 1)) and dim H 0 (O Λ ()) < 8 + dim H 0 (O C Λ ()), 13
or equivalently, ( ) + 2 n 1 ( 1)d g + 2 + d i (29) 3 i=1 ( ) + 2 7 + d. (30) 2 Note also that in this case, D n P 3 is the general complete intersection of 11 d n quadrics (c.f. Theorem 1.6 of [14]); in particular, we may specialize it to the union of the general complete intersection of 3 quadrics plus d n 8 additional general points. We may thus reduce to the case d n = 8. Since 3, the inequality (30) implies d ( ) ( +2 2 7 3+2 ) 2 7 = 3; in particular, C cannot be a degenerate rational curve. We thus have ρ(d, g, 3) 0, or upon rearrangement: g 4d 12. (31) 3 Combining (31) with our assumption that d i d n 8 for all i, condition (29) implies ( ) + 2 3 7 d + 8n 2. 3 3 Rearranging and combining this with (30), we obtain In particular, if n 2, then 2 + 3 12 2 d 3 + 3 2 + 2 + 12 48n. (32) 6 14 2 + 3 12 2 3 + 3 2 + 2 84, 6 14 which does not hold for any 3; consequently, n = 1. In this case, 2 + 3 12 2 3 + 3 2 + 2 36, 6 14 which does not hold for any 5; consequently, {3, 4}. For each, equation (32) gives upper and lower bounds on d; for each such d, equations (31) and (29) then give upper and lower bounds respectively on g. Using these bounds, it thus remains only to consider the case where n = 1, and D 1 = D n is a canonical curve; and either = 3 and or = 4 and (d, g) {(3, 0), (4, 0), (4, 1), (5, 2), (6, 4)}, (d, g) = (8, 6). If = 3 and (d, g) {(4, 0), (5, 2), (6, 4)}, then using our inductive hypothesis that C satisfies maximal ran for cubics, we see that C lies on at most a 7 dimensional family of 14
cubics. But since D 1 P 3 is a general complete intersection of 3 cubics, it contains 7 general points. Consequently, C (D 1 P 3 ) does not lie on any cubics, and so satisfies maximal ran for cubics. If = 3 and (d, g) {(3, 0), (4, 1)}, then partition D 1 P 3 = A B into two sets of 4 points, and let Q be a general quadric containing A. Since any subset of 7 points of D 1 P 3 are general, Q is smooth and does not contain any point of B; moreover, A is a set of 4 general points on Q, while B is a set of 4 general points in P 3 (although not independently general from A!). We now specialize C to a curve of type (a, 2) on Q, where { 1 if (d, g) = (3, 0); a = 2 if (d, g) = (4, 1). The exact sequence of sheaves 0 I B P 3(1) I A B C P 3(3) I A Q (3 a, 1) 0 gives rise to the long exact sequence in cohomology H 1 (I B P 3(1)) H 1 (I A B C P 3(3)) H 1 (I A Q (3 a, 1)). Since A and B are general sets of 4 points in Q and P 3 respectively, dim H 0 (O P 3(1)) = 4 and dim H 0 (O Q (3 a, 1)) = 8 2a 4, while H 1 (O P 3(1)) = H 1 (O Q (3 a, 1)) = 0, we conclude that H 1 (I B P 3(1)) = H 1 (I A Q (3 a, 1)) = 0, which implies H 1 (I A B C P 3(3)) = 0 as desired. If = 4 and (d, g) = (8, 6), then partition D 1 P 3 = A B into two sets of 4 points, and let S be a general cubic containing A. As in the previous cases, S is smooth and does not contain any point of B; moreover, A is a set of 4 general points on S, while B is a set of 4 general points in P 3. Write S as the blowup of P 2 at six points, L for the pullbac of the class of a line in P 2 to S, and E 1, E 2,..., E 6 for the six exceptional divisors. By Lemma 9.3 of [14] plus results of [10], we may specialize C to a curve on S of class The exact sequence of sheaves 6L E 1 E 2 2E 3 2E 4 2E 5 2E 6. 0 I B P 3(1) I A B C P 3(3) I A S (3L 2E 1 2E 2 E 3 E 4 E 5 E 6 ) 0 gives rise to the long exact sequence in cohomology H 1 (I B P 3(1)) H 1 (I A B C P 3(4)) H 1 (I A S (6L 3E 1 3E 2 2E 3 2E 4 2E 5 2E 6 )). Since the cone of effective curves on S is spanned by the 27 lines, the Naai-Moishezon criterion implies 9L 4E 1 4E 2 3E 3 3E 4 3E 5 3E 6 is ample. Consequently, by Kodaira vanishing, 6L 3E 1 3E 2 2E 3 2E 4 2E 5 2E 6 has no higher cohomology. In particular, by the Riemann Roch theorem for surfaces, dim H 0 (O S (6L 3E 1 3E 2 2E 3 2E 4 2E 5 2E 6 )) = 4. Since A and B are general sets of 4 points in S and P 3 respectively, we conclude that H 1 (I B P 3(1)) = H 1 (I A Q (6L 3E 1 3E 2 2E 3 2E 4 2E 5 2E 6 )) = 0, which implies H 1 (I A B C P 3(4)) = 0 as desired, thus completing the inductive step. 15
4 Quadrics In this section, we prove Theorem 1.3 for quadrics ( = 2); this will serve as another base case for our larger inductive argument. As in the previous section, we suppose ɛ = 0 and see to show T satisfies maximal ran. Our argument here will also be by induction, this time on r, using a construction due to Ballico [2]. The base case of r = 3 was done in Section 3, and the case n = 0 was done by Ballico in [2], so we suppose for our inductive argument that r 4 and n 1. Write d and g for the degree and genus of C. If g d + 2, then C cannot be a degenerate rational curve, and so C is a BN-curve and ( ) r + 2 2d + 1 g 2d + 1 g (r 1)(g d 2) [(r + 1)d rg r(r + 1)] = r 2 + 3r 1. 2 In particular, by results of Ballico [2], C does not lie on any quadrics. In particular, we conclude that C (D 1 D 2 D n ) P r also does not lie on any quadrics, as desired. We thus suppose g d + 1 for the remainder of this section. For our inductive argument, we pic a hyperplane H P r+1, transverse to P r P r+1, and write Λ = P r H for the corresponding hyperplane in P r. If C is nonspecial, then we invoe Corollary 4.2 of [17] if C is a BN-curve (respectively application of an automorphism of projective space if C is a degenerate rational curve) to degenerate C to a general BN-curve (respectively general rational curve) C Λ. We also invoe Corollary 4.2 of [17] to degenerate each D i to D i D i where D i H and D i is transverse to H, such that deg D i = r + 1. Note that (D 1 D n) P r is a collection of r + 1 general points, since Aut P r acts transitively on collections of r + 1 points in linear general position. Proposition 2.3 then implies Theorem 1.3 for T, as desired. If C is special (so g + r d 1 0), then we invoe Corollary 4.2 of [17] to degenerate each D i to a general BN-curve Di H. Writing d = d r 1 and g = g r 1, we have g = [(r + 1)d rg r(r + 1)] + (r + 1)(g + r d 1) 0 ρ(d, g, r 1) = [(r + 1)d rg r(r + 1)] + (g + r d 1) 0. We may therefore let C Λ be a general BN-curve of degree d and genus g. Since Aut Λ acts transitively on collections of r + 1 points in linear general position, we may therefore construct a reducible curve C C P r, where C Λ P r is as above, C P r is a general BN-curve of degree r + 1 and genus 1 transverse to Λ, and C Λ = C C is a set of r + 1 general points in Λ. This curve has degree d + r + 1 = d and genus g + r + 1 = g, and is a BN-curve by Theorem 1.9 of [12] since our assumption that g d + 1 implies d + (r + 1) g + r; this curve is thus a specialization of C. With (, d, g, h, ɛ 0, ɛ 0) = (2, r + 1, 1, 0, 0, 0), we note that (1) is an equality. In particular, r, d, g, and h satisfy either I(r, 2, d, g, r + 1, 1, h, 0, 0, 0) or S(r, 2, d, g, r + 1, 1, h, 0, 0, 0). Applying Proposition 2.2 thus yields the desired result. 5 Curves of Extreme Degree: n = 0 In this section, we deal with the easy cases of Theorem 1.3 when n = 0; i.e. with those cases where the Brill Noether number is far from zero or the restriction map is far from being an isomorphism. As mentioned in the introduction, we may assume ɛ = 0. 16
Definition 5.1. We say that integers r,, d, and g satisfy U(r,, d, g) if ( ) r r + r + d 1 0 ( 1)d g ( ) r + r + 0 g + r d 1 0 (r + 1)d rg r(r + 1) 0. In this section, we will show that Theorem 1.3 holds unless r,, d, and g satisfy U(r,, d, g). Since ρ(d, g, r) = (r + 1)d rg r(r + 1) 0 by assumption, Ballico and Ellia have proven Theorem 1.3 when n = 0 in the case d g + r, and ( ) ( ) ( ) ( ) r + r + 1 r + r + 1 r r + = and r + are integers, it remains to prove Theorem 1.3 when either d r ( ) r + r + or ( 1)d + 1 g r + = ( r + Since we have already proven Theorem 1.3 for r = 3 and = 2, we suppose r 4 and 3. For this, it suffices by Proposition 2.2 to show that r,, d, and g satisfy I(r,, d, g, d, g, 0, 0, 0, 0) or S(r,, d, g, d, g, 0, 0, 0, 0). By assumption either (1) holds with, or (2) holds with. It thus remains to show that if (2) holds with >, then (1) holds with, or in other words that d > r ( ) r + r + ( 1)d + 1 g ( ) r + r +. So assume d r (r+ ) r+ + 1. Since 3 and r 4 by assumption, we obtain d r ( ) ( ) ( ) r + + r 1 3 + r 1 r + + 1 = + 1 + 1 = 1 r 1 r 1 6 r3 + 1 2 r2 + 1 3 r + 1 r; in particular, C is a BN-curve, and so ρ(d, g, r) = (r + 1)d rg r(r + 1) 0. Thus, ( 1)d + 1 g ( 1)d + 1 g as desired. (r + 1)d rg r(r + 1) r r 2r 1 = d + r + 2 r ( ( ) r 2r 1 r r + r r + = ( ) r + r + + ( ) r + r +, ) + 1 (r 1)( 3) + (r 4) r + 17 + r + 2 ( ) r + r ). + r + r2 1 r (33)
6 Curves of Extreme Degree: n > 0 In this section, we deal with the easy cases of Theorem 1.3 when n > 0; i.e. with those cases where the restriction map is far from being an isomorphism. Definition 6.1. We say that integers r,, d, g, h, and ɛ 0 satisfy V (r,, d, g, h, ɛ 0 ) if ( ) r + d + 1 g + h + ɛ 0 0 ( ) r r + r + d r r + 1 h + r 1 0 g 0 (r + 1)d rg r(r + 1) 0 h r 1 0. In this section, we show it suffices to verify Theorem 1.3 when r,, d, g, h, and ɛ 0 satisfy V (r,, d, g, h, ɛ 0 ). First, we note that if d + 1 g + h + ɛ 0 ( ) r+ < 0, then T satisfies maximal ran if and only if T {p} does where p is a general point. We may thus increase ɛ (and therefore h) until ( ) r + d + 1 g + h + ɛ 0 0. (34) By assumption we have g 0, and since n 1 we have h d 1 r +1. It thus suffices to verify Theorem 1.3 when V (r,, d, g, h, ɛ 0 ) is satisfied, or (34) holds and C is a degenerate rational curve, or (34) holds and r r + ( ) r + d r h + r 1 < 0. (35) r + 1 We first verify Theorem 1.3 when (34) and (35) hold. Fix a hyperplane H P r. Applying Proposition 2.9, we may degenerate each D i to curves D i, and the points {p ɛ0 +1, p ɛ0 +2,..., p ɛ } to points {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ}, with so that are sets of general points, and #({p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H) + #(D i H) = h h, {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} H H and {p ɛ 0 +1, p ɛ 0 +2,..., p ɛ} P r H P r D 1 H, D 2 H,..., D n H H and D 1 P r H, D 2 P r H,..., D n P r H P r are sets of subsets of hyperplane sections of independantly general BN-curves, provided that r, h, and h satisfy J(r, h, h ). Moreover, we can assume the second of these sets is a set of subsets 18
of hyperplane sections of independantly general nonspecial BN-curves provided that r, h, and h satisfy N(r, h, h ). Applying Proposition 2.2, it this remains to show there exists an integer h so that r,, d, g, h, and h satisfy { I(r,, d, g, d, g, h, h N(r, h, h ) if = 3;, ɛ 0, ɛ 0 ) and J(r, h, h ) otherwise. Since these upper and lower bounds on h are all integers, we just have to chec: ( ) r + r + ( 1)d 1 + g ɛ 0 d + h r ( ) r + r + ; { h h r h if = 3, 2r+2 r + r + 1 h otherwise; ( ) { r + r + h r h if = 3, 2r+2 ( 1)d 1 + g ɛ 0 h otherwise; h r + d + h r ( ) r + r + 1 r +. The first of these is just (34) upon rearrangement. The second is immediate if h 2r + 2, which implies r + h h r h h. And when h < 2r + 2, the second follows from our r+1 2r+2 assumption that h r 1 0, which implies r + h h = h r h r+1 2r+2. The final of these inequalities follows from (35), which implies r + h < d + h r (r+ ) r+1 r+ + 1. It thus remains to show the third of these inequalities. By separately considering the cases when C is a degenerate rational curve and when C is a BN-curve, we always have (r + 1)d rg (r + 1) 0. Combining this with (34), we conclude r 2r 1 r r 1 ( d + 1 g + h + ɛ 0 ( )) r + + (r + 1)d rg (r + 1) r r 1 0; or upon rearrangement, r 2r 1 r r 1 h ( ) r + r + + ( 1)d + 1 g + ɛ 0 rɛ r 0 + 2r + 1 + (r 2r 1) r r 1 0. (r+ r+ ) It thus remains to show { r 2r 1 r r 1 h h r h 2r+2 h if = 3; otherwise. 19
Since h r h 2r+2 r+2 2r+2 h, this reduces in turn to r 2r 1 r r 1 { r+2 2r+2 if = 3; 1 otherwise. This is clear, this completing the verification of Theorem 1.3 when (34) and (35) hold. It thus remains to verify Theorem 1.3 when (34) holds and C is a degenerate rational curve, but (35) does not hold. First, we claim this implies = 3. Indeed, if C is a degenerate rational curve, then in particular g = 0; since ɛ 0 2, the condition (34) gives ( ) r + h d + 1. Since d r 1, this yields d + We conclude ( ) r r + r + d r r + 1 h d + = r r + 1 r (( ) r + r + 1 ( ) r + r ( ) r + r + 1 = r ( ) r + r + 1 > r r + 1 r r + 1 h + r 1 < ( ) r + ) d + 1 r r 1 r + 1 r r 1 r + 1 r2 r 2 r + 1 r + 1 d r r r + 1 r2 r 2 r + 2. r + 1 (r 1) r r r + 1 r ( ) r + r + r ( ) r + r + 1 + r2 r + 1 r + 1 ( ( )) 1 r + = (r + 1)(r + ) (r + )(r 2 r + 1) (r r) ( ( )) 1 r + (r + 1)(r + ) (r + )(r 2 r + 1) (r r) 4 1 ( = 24(r + 1) ( 4) 4 (r 4) + 3( 4) 3 (r 4) 2 + 3( 4) 2 (r 4) 3 + ( 4)(r 4) 4 + 4( 4) 4 + 33( 4) 3 (r 4) + 57( 4) 2 (r 4) 2 + 31( 4)(r 4) 3 + 3(r 4) 4 + 84( 4) 3 + 341( 4) 2 (r 4) + 299( 4)(r 4) 2 + 66(r 4) 3 + 644( 4) 2 + 1199( 4)(r 4) + 441(r 4) 2 + 1740( 4) ) + 1170(r 4) + 1056 0, 20
and so (35) holds if 4. Next we suppose = 3. In this case, ɛ 0 = 0, and so (34) yields We conclude Consequently, r r + 3 h r3 + 6r 2 + 11r 6 3d. d + r r + 1 h d + r ( ) r 3 r + 1 + 6r 2 + 11r 3d 6 = r4 + 6r 3 + 11r 2 2r 1 6r + 6 r + 1 d r4 + 6r 3 + 11r 2 2r 1 (r 1) 6r + 6 r + 1 = r4 + 6r 3 r 2 + 18r 6. 6r + 6 ( ) r + 3 d 3 r 2)(r 4) h + r 1 r(r r + 1 3r + 3 0. Thus, (35) holds unless all of the above inequalities are equalities; as r 4 this forces r = 4 and d = 3 and h = 25. It thus remains to consider the case (r,, d, g, h) = (4, 3, 3, 0, 25). In this case, we let H P 3 P 4 be the hyperplane containing C. Applying Proposition 2.9, we may degenerate each D i to curves D i, and the points {p 1, p 2,..., p ɛ } to points {p 1, p 2,..., p ɛ}, with so that are sets of general points, and #({p 1, p 2,..., p ɛ} H) + #(D i H) = 15, {p 1, p 2,..., p ɛ} H H and {p 1, p 2,..., p ɛ} P 4 H P 4 D 1 H, D 2 H,..., D n H H and D 1 P 4 H, D 2 P 4 H,..., D n P 4 H P 4 are sets of subsets of hyperplane sections of independantly general nonspecial BN-curves. Applying Proposition 2.3 then yields the desired result. 7 The Inductive Argument In this section, we give our inductive argument to prove Theorem 1.3. We begin with the case of cubic polynomials ( = 3) with n = 0. In this case, combining Propositions 2.2, 2.11, and 2.13 (with t = 0), we see that it suffices to chec: 21
Lemma 7.1. Let r 4, and d and g be integers satisfying U(r, 3, d, g). Then either: There exist integers d and n satisfying X(r, d, g, d, n) and I(r, 3, d, g, d, 0, 0, 0, 0, 0); There exist integers d and n satisfying X(r, d, g, d, n) and S(r, 3, d, g, d, 0, 0, 0, 0, 0); There exist integers d, g, and n sastisfying Y (r, d, g, d, g, n, 0) and I(r, 3, d, g, d, g, 0, 0, 0, 0); or There exist integers d, g, and n sastisfying Y (r, d, g, d, g, n, 0) and S(r, 3, d, g, d, g, 0, 0, 0, 0). Proof. This will be deferred to Appendix B.1. Next, we consider the case of cubic polynomials with n > 0. In this case, combining Propositions 2.2, 2.9, 2.11, and 2.13 (with t = 0), we see that it suffices to chec: Lemma 7.2. Let r 4, and d, g, and h be integers satisfying V (r, 3, d, g, h, 0). Then either: There exist integers d, h, and n satisfying X(r, d, g, d, n), I(r, 3, d, g, d, 0, h, h, 0, 0), and K(r, h, h ); There exist integers d, h, and n satisfying X(r, d, g, d, n), I(r, 3, d, g, d, 0, h, h, 0, 0), and M(r, h, h ); There exist integers d, h, and n satisfying X(r, d, g, d, n), I(r, 3, d, g, d, 0, h, h, 0, 0), and N(r, h, h ); There exist integers d, g, h, and n satisfying Y (r, d, g, d, g, n, 0), I(r, 3, d, g, d, g, h, h, 0, 0), and K(r, h, h ); There exist integers d, g, h, and n satisfying Y (r, d, g, d, g, n, 0), I(r, 3, d, g, d, g, h, h, 0, 0), and L(r, h, h ); or 22
There exist integers d, g, h, and n satisfying Y (r, d, g, d, g, n, 0), I(r, 3, d, g, d, g, h, h, 0, 0), and N(r, h, h ). Proof. This will be deferred to Appendix B.2. Next, we consider the case of quartic polynomials ( = 4) with n = 0. In this case, combining Propositions 2.2, 2.13 (with t = 0), and 2.16 (again with t = 0), we see that it suffices to chec: Lemma 7.3. Let r 4, and d and g be integers satisfying U(r, 4, d, g). Then either: There exist integers d, g, and n sastisfying Y (r, d, g, d, g, n, 0) and I(r, 4, d, g, d, g, 0, 0, 0, 0); There exist integers d, g, and n sastisfying Y (r, d, g, d, g, n, 0) and S(r, 4, d, g, d, g, 0, 0, 0, 0); There exist integers d, g, m, and n sastisfying or Z(r, d, g, d, g, m, n, 0) and I(r, 4, d, g, d, g, 0, 0, 0, 0); There exist integers d, g, m, and n sastisfying Z(r, d, g, d, g, m, n, 0) and S(r, 4, d, g, d, g, 0, 0, 0, 0). Proof. This will be deferred to Appendix B.3. Next, we consider the case of quartic polynomials with n > 0. In this case, combining Propositions 2.2, 2.9, 2.11, 2.13 (with t = 0), and 2.16 (again with t = 0), we see that it suffices to chec: Lemma 7.4. Let r 4, and d, g, and h be integers satisfying V (r, 4, d, g, h, 0). Then either: There exist integers d, h, and n satisfying X(r, d, g, d, n), I(r, 4, d, g, d, 0, h, h, 0, 0), and J(r, h, h ); There exist integers d, g, h, and n satisfying Y (r, d, g, d, g, n, 0), I(r, 4, d, g, d, g, h, h, 0, 0), and J(r, h, h ); There exist integers d, g, h, and n satisfying Y (r, d, g, d, g, n, 0), I(r, 4, d, g, d, g, h, h, 0, 0), and K(r, h, h ); 23
There exist integers d, g, h, m, and n satisfying or Z(r, d, g, d, g, m, n, 0), I(r, 4, d, g, d, g, h, h, 0, 0), and J(r, h, h ); There exist integers d, g, h, m, and n satisfying Z(r, d, g, d, g, m, n, 0), I(r, 4, d, g, d, g, h, h, 0, 0), and K(r, h, h ). Proof. This will be deferred to Appendix B.4. Next, we consider the case of polynomials of higher degree ( 5) with n = 0. In this case, combining Propositions 2.2, 2.13 (with t = 2), and 2.16 (again with t = 2), we see that it suffices to chec: Lemma 7.5. Let r 4 and 5, and d and g be integers satisfying U(r,, d, g). Then either: There exist integers d, g, and n sastisfying Y (r, d, g, d, g, n, 2) and I(r,, d, g, d, g, 0, 0, 0, 0); There exist integers d, g, and n sastisfying Y (r, d, g, d, g, n, 2) and S(r,, d, g, d, g, 0, 0, 0, 0); There exist integers d, g, m, and n sastisfying or Z(r, d, g, d, g, m, n, 2) and I(r,, d, g, d, g, 0, 0, 0, 0); There exist integers d, g, m, and n sastisfying Z(r, d, g, d, g, m, n, 2) and S(r,, d, g, d, g, 0, 0, 0, 0). Proof. This will be deferred to Appendix B.5. Finally, we consider the case of polynomials of higher degree with n > 0. Combining Propositions 2.2, 2.9, 2.13 (with t = 2), and 2.16 (with t = 2), we see that it suffices to chec: Lemma 7.6. Let r 4 and 5, and d, g, and h be integers satisfying V (r,, d, g, h, 2). Then either: There exist integers d, g, h, n, and ɛ 0 satisfying Y (r, d, g, d, g, n, 2), I(r,, d, g, d, g, h, h, 2, ɛ 0), E( 2, ɛ 0, 3), and J(r, h, h ); 24
There exist integers d, g, h, m, n, and ɛ 0 satisfying Z(r, d, g, d, g, m, n, 2), I(r,, d, g, d, g, h, h, 2), E( 2, ɛ 0, 3), and J(r, h, h ); For every integer h satisfying A(r, h, h ), there exist integers d, g, n, and ɛ 0 satisfying or Y (r, d, g, d, g, n, 2), I(r,, d, g, d, g, h, h, 2), and E( 2, ɛ 0, 0); For every integer h satisfying A(r, h, h ), there exist integers d, g, m, n, and ɛ 0, satisfying Z(r, d, g, d, g, m, n, 2), I(r,, d, g, d, g, h, h, 2), and E( 2, ɛ 0, 0). Proof. This will be deferred to Appendix B.6. A Inequalities from Section 2 In this appendix, we give code in sage to chec or create algebraic expressions for all inequalities that appear in Section 2. # File name : inequalities. py def I(r,, d, g, dp, gp, h, hp, e0, e0p, B): # We pass B = binomial ( r +, ) as an extra argument. return [ ( - 1) * dp + 1 - gp + hp + e0p - * B / (r + ), * d - g - ( - 1) * dp + gp + h - hp + e0 - e0p - r * B / ( r + ) ] def S(r,, d, g, dp, gp, h, hp, e0, e0p, B): # We pass B = binomial ( r +, ) as an extra argument. return [ * B / (r + ) - (( - 1) * dp + 1 - gp + hp + e0p ), r * B / (r + ) - ( * d - g - ( - 1) * dp + gp + h - hp + e0 - e0p ) ] def E(e0, e0p, t): return [ e0p - t, e0 - e0p ] def A(r, h, hp): return [ hp, 25
] h - (r + 1) * hp def J(r, h, hp): return [ hp - r - h / (r + 1), h - hp ] def K(r, h, hp): return [ 2 * r + 1 - h, hp, h - hp ] def L(r, h, hp): return [ 3 * r + 2 - h, hp - 2, 2 - hp ] def M(r, h, hp): return [ 3 * r + 2 - h, hp - r - 2, h - hp - r/2 ] def N(r, h, hp): return [ hp - r - h / (r + 1), h - hp - r * h / (2 * r + 2) ] def Np(r, h, hp): # When 3 r + 3 <= h <= 4 r + 3: # then floor (h / (r + 1)) = 3 and floor (h / (2r + 2)) = 1, # so N(r, h, hp) is satisfied if: return [ h - 3 * r - 3, 4 * r + 3 - h, hp - r - 3, h - hp - r ] 26
def X(r, d, g, dp, n): return [ g - n + 1, r * (d - dp) - (r - 1) * g + (r - 1) * n - r **2 + 1, n - 1, dp - n, r + 2 - n, 2 * n + d - dp - g - r - 1 ] def Y(r, d, g, dp, gp, n, t): return [ gp, (r + 1) * dp - r * gp - r **2 - r, (2 * r - 3) * dp - (r - 2) **2 * gp - 2 * r **2 + 3 * r - 9, g - gp - n + 1, r * (d - dp) - (r - 1) * (g - gp) + (r - 1) * n - r **2 + 1, n - 1, dp - n - t, r * (d - dp) - (r - 4) * (g - gp) - 2 * n - 2 * r + 2, 2 * n + d + gp - dp - g - r - 2 ] def Z(r, d, g, dp, gp, m, n, t): return [ gp - (r + 1) * m, (r + 1) * dp - r * gp - r **2 - r, (2 * r - 3) * (dp - r * m) - (r - 2) **2 * (gp - (r + 1) * m) - 2 * r **2 + 3 * r - 9, g - gp - n + 1, r * (d - dp) - (r - 1) * (g - gp) + (r - 1) * n - r **2 + 1, n - 1, (dp - r * m) - n - t, r * (d - dp) - (r - 4) * (g - gp) - 2 * n - 2 * r + 2, 2 * n + d + gp - dp - g - r - 2, 2 * (dp - r * m) - (r - 3) * (gp - (r + 1) * m - 1) - (r - 1) * (r + 2), m ] def U(r,, d, g, B): # We pass B = binomial ( r +, ) as an extra argument. return [ r * B / (r + ) - d - 1, ( - 1) * d - g - * B / (r + ), g + r - d - 1, (r + 1) * d - r * g - r * (r + 1) ] def V(r,, d, g, h, e0, B): # We pass B = binomial ( r +, ) as an extra argument. 27