eu HOMOTOPY SECTIONS OF ONE DIMENSIONAL IDEALS OVER A LAURENT POLYNOMIAL RING

Let R be a commutative Noetherian ring and I be a one dimensional ideal of the Laurent polynomial ring R[X,X] that contains a doubly monic polynomial. Define I(1) :=< {f(1) : f ∈ I} >. Let I/I be genereted by n ≥ 2 elements over R[X,X]/I . Then any set of n generators of I(1) over R may not be lifted to a set of n generators of I over R[X,X] by describing an example of one dimensional ideal. AMS Subject Classification: 13E05, 13E15, 13C10


Introduction
Let R be a commutative Noetherian ring of finite Krull dimension with identity and I be an ideal of the Laurent polynomial ring R[X, X −1 ].Let M be a finitely generated R-module.We denote µ(M ) to be the least number of elements required to generate M as an R-module.The comparison of the minimal numbers of generators of I and of its module I/I 2 , denoted by µ(I) and µ(I/I 2 ), has been an enliven area of interest (see [1], [5], [6], [11], [14], [15] and [17]).The co-normal bundle I/I 2 of an ideal I in a ring R[X, X −1 ] is an R[X, X −1 ]/Imodule.Many algebraic properties of this module are intertwined with those of the ideal I. Define I(1) =< {f (1) : f ∈ I} >.In general, µ(I(1)) ≤ µ(I) and µ(I/I 2 ) ≤ µ(I) ≤ µ(I/I 2 ) + 1.
Any set of n generators of the maximal ideal (X 1 , X 2 , . . ., X n−1 ) of K[X 1 , X 2 , . . ., X n−1 ] can be lifted (via the map setting X n to 0) to a set of n generators of the maximal ideal It is easy to see that we can perform Euclidean transformations on F 1 , F 2 and using the Euclidean transformations we may assume that F 1 = X 1 and F 2 = 0. Then for these generators, the lift (X 1 , X 2 ) works.
Since lifting of generators is a natural problem which have been studied by many mathematicians in various fashion (see [4], [7], [12], [10], [18]).Here we are giving some examples on lifting problem.Let X be an integral, projective variety of co-dimension two, degree d and dimension r.Suppose Y is its general hyperplane section.The problem of lifting generators of minimal degree σ from the homogeneous ideal of X is studied in ( [12]).Answers in terms of relations between d and σ are known for r = 1, 2. For r = 1, the Laudal's "Generalized trisecant lemma" is given in ( [2], [9]).For r = 2, there is the analogous result in ( [13]).In ( [18]), let X be an integral, projective variety of dimension n and degree d in P N , defined over an algebraically closed field K of characteristic zero.Consider the hyperplane section Y = X K of X, where K ∼ = P N −1 is a general hyperplane in P N .The "Lifting problem" is the problem of finding conditions on d, N, n and s such that any degree s hypersurface in P n−1 containing Y can be lifted to a hypersurface in P N containing X. Motivated by this problem, in [10] Mandal proved that the following: containing a doubly monic polynomial, and I/I 2 is generated by n elements over R[X, X −1 ]/I, where 1) is also generated by n elements, where I( 1) is an ideal of R by putting X = 1.
In ( [4]), lifting problem in various cases has been discussed.If I is an ideal of R[X], then I(0) :=< {f (0) : f ∈ I} > is also an ideal of R so it is very natural to ask whether a set of generators for I(0) can be lifted to a set of generators for I.One interesting consequence of this is in the Laurent case.We consider the problem for one dimensional ideals of Laurent polynomial rings.A natural question is: can we lift any set of n generators of I(1) to a set of n generators of I? The objective of the paper is to come out with a negative answer of the above problem in case of the one dimensional ideals, i.e., a set of n generators of I(1) may not be lifted to a set of n generators of I.

Preliminaries
In this section, we define some terms used in this paper and provide certain standard results without proof.We hope that this will improve the readability and understanding of proof of the paper.
An ideal I of a ring R is called zero dimensional if every prime ideal of R containing I is a maximal ideal, that is, dim(R/I) = 0.For example every maximal ideal is a zero dimensional ideal.Further if I is the intersection of finitely many maximal ideals, then dim(R/I) = 0 and the converse is also true if I is a reduced ideal.An ideal is said to be doubly monic polynomial if coefficient of the highest degree term and the lowest degree term are units.We say that an ideal I is efficiently generated if µ(I) = µ(I/I 2 ).We have a theorem of Mandal [10] about efficient generation of ideals.
Theorem 2.1.[10] Let I be an ideal of R[X, X −1 ] over a commutative Noetherian ring R. Suppose that I contains a doubly monic polynomial and µ(I/I 2 ) ≥ dim(R[X, X −1 ]/I) + 2. Then ideal I is efficiently generated.
The following theorem proved by Kumar and Mandal in [7] shows the positive result in the case of zero dimensional ideal.Theorem 2.2.[7] Let R be a commutative Noetherian ring and I be an ideal of the Laurent polynomial ring R[X, X −1 ] that contains a doubly monic polynomial.Suppose P is a projective R-module of rank Let S : P → I(1) and φ : R[X, X −1 ] → I/I 2 be two surjective homomorphisms such that φ(1) ≡ S mod(X − 1, I(1) 2 ).Then there exists a surjective homomorphism ψ : R[X, X −1 ] → I such that ψ(1) = S and ψ lifts φ.

Main Result
Now we give an example to show that a set of n generators of I(1) may not be lifted to a set of n generators of I in the case of one dimensional ideals.

F 1 + I( 1 ) 2 F 2 + I( 1 ) 2 F 3 +
= α, which is not a member of Q, for φ p (W ) is irreducible over Q of degree greater than 1.Therefore,  