University of Paderborn Department of Mathematics Diploma Thesis ...

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66 CHAPTER 3. OPERATIONS ON STABLE IDEALSProof of Claim 1. By the algorithm described in Remark 3.31, we successively contract thelargest monomial generators of L H (with respect to the lexicographic order) with the lastvariable x n−1 , until there is only one monomial generator left, which contains x n−1 . Let mbe the first monomial in L H with respect to the lexicographic order, such that m containsmlast variable x n−1 . Then we contract in L H , which provides some ideal L conHx .n−1Assume L conH is not a lexicographic ideal. Since the only effect of the contraction of min L H is removing m from the set of minimal generators and includingx n−1mx n−1into the set ofminimal generators, L conH is not a lexicographic ideal, if [Lconsubset of [R] deg m−1 . Hence, there is at least one monomial ˜m ∈ [R] deg m−1 with ˜m >H ] deg m−1 is not a lexicographicmand ˜m /∈ [L conH ] deg m−1. It follows ˜m · x n−1 > m. Since m ∈ [L conH ] deg m and [L conH ] deg mis a lexicographic subset of [R] deg m , we obtain ˜m · x n−1 ∈ [L conH ] deg m, and in particular˜m·x n−1 ∈ L H . Therefore, there is a minimal monomial generator ¯m·x n−1 in L H , such that˜m · x n−1 is divisible by ¯m · x n−1 . Note that the monomial generator contains x n−1 , sinceotherwise it would divide ˜m, which is a contradiction to the fact that ˜m /∈ [L conH ] deg m−1.Furthermore, we have deg( ¯m · x n−1 ) ≤ deg( ˜m · x n−1 ) = deg m. But this contradicts thefact that we have chosen m to be the first monomial generator in L H with the last variablex n−1 .It follows that L conHx n−1is a lexicographic ideal. Iteration of the process proves that any ofthe contractions performed successively due to Remark 3.31 provides a lexicographic ideal.Since any of the expansions described in Remark 3.31 to take L H to J only increases theexponent of the last and only monomial generator containing the last variable x n−1 , theseexpansions will always give a lexicographic ideal. It follows that J is a lexicographic ideal,which proves Claim 1.Since we know that J is a lexicographic ideal with only one minimal monomial generatorcontaining x n−1 , set m := x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 1n−2 · x a 0n−1 to be this monomialwith suitable r, a 0 , a 1 , . . . , a r−2 , a r−1 , a r .Claim 2. The set of minimal generators of J is given by{x 0 , x 1 , . . . , x n−r−2 ,x ar+1n−r−1,x arn−r−1 · x a r−1+1n−r ,x arn−r−1 · x a r−1n−r · x a r−2+1n−r+1 , . . . ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2n−3 · x a 1+1n−2 ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 1n−2 · x a 0n−1 = m}.Proof of Claim 2. First we show that each of the monomials in the above set is containedin J. Then, since J is stable and lexicographic, we will be able to conclude that all
3.3. STABLE IDEALS WITH THE SAME DOUBLE SATURATION 67monomials are minimal generators of J and that J cannot contain any other minimalmonomial generators.We already know that m is a minimal generator of J. Since J is stable, we know byTheorem 2.7 thatm· x n−2 = x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 1+1n−2 · x a 0−1n−1x n−1is also contained in J. Since m is the only minimal generator of J containing the variablex n−1 , it even follows thatis contained in J. Because ofx arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2n−3 · x a 1+1n−2x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2+1n−3 · x a 1n−1 > x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2n−3 · x a 1+1n−2 ∈ Jwith respect to the lexicographic order and since J is lexicographic, we concludex arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2+1n−3 · x a 1n−1 ∈ J.Since m is the only minimal generator of J containing last variable x n−1 , it even followsx arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2+1n−3 ∈ J.By the same arguments, it follows that all monomials in{x ar+1n−r−1,x arn−r−1 · x a r−1+1n−r ,x arn−r−1 · x a r−1n−r · x a r−2+1n−r+1 , . . . ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2n−3 · x a 1+1n−2 ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 1n−2 · x a 0n−1 = m},are contained in J. It remains to show that x 0 , x 1 , . . . , x n−r−2 ∈ J. Because of x ar+1n−r−1 ∈ Jand x 0 · x arn−1 > xn−r−1, ar+1 the monomial x 0 · x arn−1 must also be contained in J. Hence, thereis a monomial generator of J, such that x 0 · x arn−1 is divisible by this generator. Since m isthe only monomial generator in J containing the last variable x n−1 , it follows x 0 ∈ J. Bythe same arguments we obtain x 1 , . . . , x n−r−2 ∈ J.Next, we want to show that all monomials in{x 0 , x 1 , . . . , x n−r−2 ,x ar+1n−r−1,x arn−r−1 · x a r−1+1n−r ,x arn−r−1 · x a r−1n−r · x a r−2+1n−r+1 , . . . ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 2n−3 · x a 1+1n−2 ,x arn−r−1 · x a r−1n−r · x a r−2n−r+1 · . . . · x a 1n−2 · x a 0n−1 = m},
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University of PaderbornDepartment o
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