The Leading Ideal of a Complete Intersection of Height Two,Part III

Let (S, 𝔫) be an s-dimensional regular local ring with s > 2, and let I = (f, g) be an ideal in S generated by a regular sequence f, g of length two. As in 2 Goto , S. , Heinzer , W. , Kim , M.-K. ( 2006 ). The leading ideal of a complete intersection of height two . J. Algebra 298 : 238 – 247 . Google Scholar], 3 Goto , S. , Heinzer , W. , Kim , M.-K. ( 2007 ). The leading ideal of a complete intersection of height two, II . J. Algebra 312 : 709 – 732 . Google Scholar]], we examine the leading form ideal I* of I in the associated graded ring G: = gr_𝔫(S). Let μ _G (I*) = n ≥ 3, and let {ξ₁, ξ₂,…, ξ _n } be a minimal homogeneous system of generators of I* such that ξ₁ = f* and ξ₂ = g*, and c _i : = deg ξ _i  ≤ deg ξ _i+1: = c _i+1 for each i ≤ n ? 1. For m ≤ n, we say that K _m : = (ξ₁,…, ξ _m )G is an ideal generated by part of a minimal homogeneous generating set of I*. Let D _i : = GCD(ξ₁,…, ξ _i ) and d _i  = deg D _i for i with 1 ≤ i ≤ m. Let K _m be perfect with ht _G K _m  = 2. We prove that the following are equivalent: 1. deg ξ _i+1 = deg ξ _i  + (d _i?1 ? d _i ) +1, for all i with 3 ≤ i ≤ m ? 1;

2. deg ξ _i+1 ≤ deg ξ _i  + (d _i?1 ? d _i ) +1, for all i with 3 ≤ i ≤ m ? 1.

Furthermore, if these equivalent conditions hold, then K _m  = I*. Moreover, if e(G/K _m ) = e(G/I*), we prove that K _m  = I*. We illustrate with several examples in the cases where K _m is or is not perfect.