2 00 5 Inequalities satisfied by the basic sequence of an integral curve in P 3

We show several new inequalities found recently that the basic sequence of the saturated homogeneous ideal I of an integral curve in P must satisfy. Then we compare our results with Cook’s assertions on the generic initial ideal of I, carrying out numerical computations with the use of a computer. The outcome is that we have obtained new restrictions on the generic initial ideal of I. When char(k) = 0, the basic sequence of a homogeneous ideal I in a polynomial ring over an infinite field k is a sequence of the degrees of the minimal generators of the generic initial ideal of I arranged in a suitable order. Introduction Let X be a curve in P := P3k, namely, a locally Cohen-Macaulay equidimensional closed subscheme of P of dimension one, and let I denote the saturated homogeneous ideal of X in a polynomial ring R := k[x1, x2, x3, x4], where k is an infinite field and the linear forms x1, x2, x3, x4 are chosen sufficiently generally. Then there is what we call a Weierstrass basis { el | 1 ≤ i ≤ 3, 1 ≤ l ≤ mi } of I that is characterized by some properties connected with a representation of the multiplication of el by the linear forms x1, x2, x3, x4 (see [7] and [8]). More precisely, { I [1] = I, I [3] = I, I [i] = I ⊕ I [i+1] (i = 1, 2) as k[xi+1, . . . , x4]-modules, and xiI [i+1] ⊂ (xi+1, . . . , x4)I ′〈i〉 ⊕ I , where I [i] is a finitely generated graded k[xi, . . . , x4]-submodule of I and I 〈i〉 (resp. I ′) is a finitely generated graded free k[xi, . . . , x4]-submodule (resp. k[xi+1, . . . , x4]submodule) of I with free basis { el | 1 ≤ l ≤ mi } for each i (1 ≤ i ≤ 3). We havem1 = 1 and deg(e1) = a = m2. Put b := m3. Let nl := deg(e 2 l ) (1 ≤ l ≤ a) and na+l := deg(e 3 l ) 2000 Mathematics Subject Classification. Primary 14H50; Secondary 13P10, 13D02. Basic Sequence of an Integral Curve 2 (1 ≤ l ≤ b). We may assume that the sequence n1, . . . , na (resp. na+1, . . . , na+b) is nondecreasing after changing the order of e1, . . . , e 2 a (resp. e 3 1, . . . , e 3 b) if necessary. The sequence BR(I) := (a;n1, . . . , na;na+1, . . . , na+b) is determined uniquely by X ⊂ P 3 and is called the basic sequence of I or X (see [2] and [6] – [8]). Note that X is projectively Cohen-Macaulay if and only if b = 0 (i.e. the subsequence (na+1, . . . , na+b) is vacuous). In general, a ≤ n1 ≤ · · · ≤ na and n1 ≤ na+1 ≤ · · · ≤ na+b. There are a number of applications of basic sequence to the study of curves in P (see [1] – [5]). In this paper, expanding the ideas in our earlier work [3, Section 1], we show several new inequalities that the components of BR(I) must satisfy when X is integral and BR(I) takes special forms (see Sections 5 and 6). Our main results are Theorem 5.7, Corollary 5.8, Propositions 5.11, 5.12, and Theorem 6.4. They can be summarized together with our related earlier results in the following manner. Theorem. Assume that X is contained in an irreducible surface of degree a ≥ 2 and that b ≥ 1. Let n1, . . . , n ′ ω (ω ≥ 1) be a strictly increasing sequence of integers such that {n1, . . . , n ′ ω} = { ni | 1 ≤ i ≤ a } and let tl := #{ i | ni = n ′ l, 1 ≤ i ≤ a }. Then, the following inequalities hold, where an additional assumption char(k) = 0 is needed in our proofs of (ii) – (iv). (i) We have ni ≤ ni+1 ≤ ni + 1 for all 1 ≤ i ≤ a− 1. (ii) Suppose that ω > 1. Let m be an integer with 1 ≤ m < ω, t := ∑m l=1 tl, and a ′ be an integer with t + 1 ≤ a ≤ a. If ni = nt + i − t for all t ≤ i ≤ a ′ and na+1 < na′ , then t(t− 1)/2 > p, where p := max{ i | na+i < na′ , 1 ≤ i ≤ b }. (iii) Let a be an integer with 2 ≤ a ≤ a. If ni = n1 + i − 1 for all 1 ≤ i ≤ a , then na+1 ≥ na′. (iv) Let a be an integer with 3 ≤ a ≤ a. If ni = n1 + i − 2 for all 2 ≤ i ≤ a , then na+1 ≥ na′. (v) If ni = n1 + i− 1 for all 1 ≤ i ≤ a, then b > 2 and na+j 6= na+1 + j − 1 for some 1 ≤ j ≤ b. If further there is an integer b0 with 0 < b0 < b such that na+b0 +1 < na+b0+1, then b0 > 2 and na+j 6= na+1 + j − 1 for some 1 ≤ j ≤ b0. (vi) Suppose that b = 1 and that X does not contain any line as an irreducible component. Delete the terms n2, . . . , n ′ ω from the sequence (a, n1, . . . , na, na+1) and then rearrange the remaining terms so that they make a nondecreasing sequence. Let (n 1, . . . , n ′′ a+3−ω) denote the resulting sequence. Then na+1 ≤ a− 2 + a ∑