• Views 163
• Citations 0
• ePub 3
• PDF 28
`Journal of MathematicsVolume 2018, Article ID 9316901, 11 pageshttps://doi.org/10.1155/2018/9316901`
Research Article

## Intersection of Ideals in a Polynomial Ring over a Dual Valuation Domain

Université Marien Ngouabi, Faculté des Sciences et Technique Département de Mathématiques, BP: 69, Brazzaville, Congo

Correspondence should be addressed to Andre S. E. Mialebama Bouesso; moc.liamg@seduetnias

Received 17 September 2018; Revised 23 October 2018; Accepted 29 October 2018; Published 15 November 2018

Copyright © 2018 Regis F. Babindamana and Andre S. E. Mialebama Bouesso. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let be a valuation domain and let be a dual valuation domain. We propose a method for computing a strong Gröbner basis in ; given polynomials , a method for computing a generating set for is given; and, finally, given two ideals and of , we propose an algorithm for computing a generating set for .

#### 1. Introduction

The theory of Gröbner bases initially introduced over the fields (see [1]) is well developed over a great family of rings (see [28]). However, many rings are still not part of this family. In this paper, we are interested in the ring of dual valuation domain where is a valuation domain. This ring is well known in the case of as a typical example of Weil bundle in differential geometry (see the example on page 2 in [9]). Athough the ring has no zero divisors, the ring is noetherian and has zero divisors. In this paper, we study this ring in detail and we propose an algorithm for computing strong Gröbner basis for ideals of . We generalize Schreyer’s theorem which enables us to compute a strong Gröbner basis for when is a strong Gröbner basis for some ideal . This important result is used later to compute a generating set for for any arbitrary subset . The last part of this paper is devoted in the computation of a generating set of the intersection of two ideals; that is, given two ideals and of , it is well known that a polynomial if there exist such thatObserve that if one can find such that then the polynomial can be easily found. The goal of this paper is to compute the set of all the syzygies which will directly lead to computing a generating set for .

#### 2. Basic Notions

Definition 1 (dual valuation domain). (i)A commutative ring is called valuation domain if it is a domain and whenever , we have or .(ii)If is a valuation domain, the ring , also denoted as , is called dual valuation domain.(iii)Elements of are of the form where and . We call the real part of and call the imaginary part of .

Throughout this work, we write for the dual valuation domain and for the polynomial ring .

Lemma 2. An element divides if and only if the following hold in : (i) divides .(ii) divides .

Proof. Assume that divides in ; then there exists such that . That is, ; this leads toThe first equation shows that divides and ; replacing this in the second equation, we have .
Conversely, if divides , then we can set . Since by hypothesis divides also , there exists such that ; that is, . We can then write Observe that ; this leads to That is, .

Example 3. Let us consider the valuation domain which is nothing but the localization of at the multiplicative subset . In , let us check the divisibility between and . Observe that but and ; this means that , where .

Definition 4 (monomial ordering). (1)A total ordering on the set of monomials is said to be a monomial ordering if we have .(2)A monomial ordering is called(i)global if ;(ii)local if ;(iii)mixte if it is neither global nor local. For details on monomial ordering, see [10, 11].

Definition 5 (monomial ordering for modules). Let be a free -module of rank , and let be the canonical basis of . (1)A monomial in is the product of a monomial in with a basis element , that is, an element of the form where . We denote by the set of all monomials in .(2)Accordingly, a term in is the product of a monomial in with an element in .(3)For monomials and coefficients , the least common multiple of two terms is given by(4)A monomial ordering on is a total ordering such that if and are monomials in , then, , we have .(5)A monomial ordering in can be extended in as a module ordering. Denoting a module ordering by , we say that if one of the following conditions holds:(a) or and . In this case, we say that the priority is given to monomials.(b) or and . In this case, we say that the priority is given to components.

Definition 6 (notations). Let be a nonzero polynomial in and let be a finitely generated ideal of . With respect to a monomial ordering , we have the following(i)The multidegree of denoted as is defined as .(ii)The leading coefficient of denoted as is defined as .(iii)The leading monomial of denoted as is defined as .(iv)The leading term of denoted as is defined as .(v)The tail of is defined as .(vi)The leading ideal of is defined as .

Lemma 7. Let and be monomials in . An element belongs to if and only if there exists such that in and in .

Proof. The proof follows from the proof of Lemma 2.2 in [6].

Lemma 8. Let . Then the quotient ideal where And .

Proof. Let ; then ; that is, there exists such that . (i)If , then .(ii)If , then ; in this case, for some ; since , then for some . This means that . Conversely let where is described in the hypothesis. (i)If , then ; in this case, there exists such that ; that is, ; therefore .(ii)If , then and ; that is, ; in this case, we have ; therefore .

Throughout this work, we denote with basis , where .

Theorem 9 (division’s algorithm). Let be a monomial ordering on and . There exist and such that the following conditions hold: (i).(ii).(iii) or no term occurring in is divisible by any of .

Example 10. Let , where . Let . The goal is to divide by in using the lexicographic ordering. Set . Observe that and using Lemma 2, we have ; this leads to and . Again and using Lemma 2, we have ; this leads to and . Since , then we move to the next polynomial. Observe that and using Lemma 2, we have ; this leads to and . Since neither nor is divisible by any term occurring in , we will send each term of in the remainder and get and . Thus, .

#### 3. Gröbner Bases

Definition 11. A set of polynomials is called strong Gröbner basis for with respect to a monomial ordering if there exists such that .

Proposition 12 (ideal membership problem). Let be a strong Gröbner basis for and let be a monomial ordering. Let ; then if and only if the remainder of the division of by is zero.

Proof. The proof follows from [12].

Definition 13 (S-vectors). Let be nonzero polynomial vectors and let be a monomial ordering. The S-vector of and is defined as follows: (1)If and , then where ,.(2)ElseIf , we call the S-polynomial of and .

Example 14. Let where and let us compute the S-polynomial for and using the lexicographic ordering in , where (1) and .Observe that but ; in this case, the S-polynomial is given by the formula and we have .(2) and .Observe that and ; in this case, we compute the S-polynomial by the formulawhere , since 2 and 5 are units. In this case, we have

Definition 15 (extended S-vector). Let be a polynomial vector such that ; then we define the extended S-vector of as .

Example 16. If , then, using the lexicographic ordering in , we have .

Definition 17 (syzygy). Let be nonzero polynomials of and let be the canonical basis for . By a syzygy on , we mean an element of the kernel of the homomorphism of modules In the other words, a syzygy on is an element such that The set of all syzygies on is denoted by .

Definition 18 (Schreyer’s ordering). Given a monomial ordering on and nonzero polynomials , we define Schreyer’s ordering (or induced ordering) on induced by and as follows: if and only if (1) or(2) and .

Theorem 19 (Buchberger’s criterion). Let be nonzero polynomials of and let be a monomial ordering. Then form a strong Gröbner basis for if and only if the following conditions hold: (1)For each pair of integers , the remainder of the division’s algorithm of by is zero.(2)The remainder of the divisoin’s algorithm of by is zero for each .

Proof. Assume that is a Gröbner basis for . Since and , by Proposition 12, we see straightforwardly that each remainder by division’s algorithm (see Algorithm 1) will be zero.
Assume that the remainder of each S-polynomial and each extended S-polynomial is zero; using Algorithm 1, we have Observe by definition that, for some , we have and then Therefore is a syzygy for . Set Observe by Algorithm 1 that This means that and by Schreyer’s ordering ; therefore Observe also that , where , and are coefficients as in the definition of S-polynomials. For each , we have ; that is, and then is a syzygy for . Set Observe by Algorithm 1 that ; this means that and since , by Schreyer’s ordering, we conclude that ; therefore Let  us  prove  that   form  a  strong  Gröbner  basis  for   Let ; then for some . Let us prove that there exists such that . Let ; with respect to Schreyer’s ordering induced by and , let us divide by (listed in some order). We have where and the remainder . Let ; multiplying by , we get We transform as (1)If , then it is clear that ; that is, and by Lemma 7 there exists such that .(2)If , then there exists such that and .Case 1. Assume that there exists such that ; then it is clear that . Observe that for some polynomial with . This means that , which contradict the fact that no term occurring in the remainder is divisible by any of and .
Case 2. , we have . In this case, we choose the biggest such that . Since , then it is clear that ; by Lemma 7, there exists such that ; that is, and . Of course which means that ; therefore ; that is,(i)Assume that ; in this case, we have ; that is, there exists such that Observe that and ; using and , we see that which contradict the fact that no term occurring in the remainder is divisible by any of .(ii)Assume that and ; in this case, we have and ; using , we see that which contradict the fact that no term occurring in the remainder is divisible by any of .(iii)Assume that and ; in this case, we haveand , where . Since by hypothesis (i.e., for some ), we have ; using , we see that , which contradict the fact that no term occurring in the remainder is divisible by any of .