University of Paderborn Department of Mathematics Diploma Thesis ...

60 CHAPTER 3. OPERATIONS ON STABLE IDEALSof R( ˜m 1 ) into the set of minimal generators of sat xn−1 ,x n(I). Hence, we can expand themonomial ˜m 1 in I (1) , which provides the ideal I (2) = (I (1) ) exp .Now, we proceed by induction on r. In the case r = 1, we have just proved that we can contractm ′ 1 in I, which provides I (1) , and then expand ˜m 1 in I (1) . Since I (1) = sat xn−1 ,x n(I) =sat xn−1 ,x n(J), the expansion of ˜m 1 in I (1) indeed provides the ideal I (2) = J.Let r > 1. Again by induction, we may assume that we have performed the contractionexpansionpairs(m ′ 1, ˜m 1 ), (m ′ 2, ˜m 2 ), . . . (m ′ i−1, ˜m i−1 ),which provide some ideal I (2i−2) (since we have proved above that we can perform thecontraction-expansion pair (m ′ 1, ˜m 1 )). We have to show that we can contract m ′ i in I (2i−2)and expand ˜m i in I (2i−1) = (I 2i−2 ) con . Since m ′ i is contractible in the ideal obtained byperforming the contractions m ′ 1, . . . , m ′ i−1 in I, it suffices to show that the expansions˜m 1 , . . . , ˜m i−1 did not hurt the condition for a contraction. Since we have m ′ i ≠ ˜m j for all1 ≤ j ≤ r, the monomial m ′ i · x n−1 cannot have been removed from the set of generatorsby any of the expansions performed, for an expansions removes only the monomial itself,which is expanded (Note: We only contract monomials with the last variable x n−1 , sincewe have chosen them to take I to sat xn−1 ,x n(I)).The second condition for a contraction is that m ′ i is not contained in I (2i−2) . If m ′ i wouldhave been included into the ideal by an expansion of some monomial ˜m k , k ∈ {1, . . . , i−1},then m ′ i is a multiple of ˜m k , which contradicts the fact that m ′ i·x n−1 is a minimal generator,which has not been removed yet.The last condition for a contraction is L(m ′ i) ⊂ I. We know, L(m ′ i) is contained inthe ideal, which we obtain after performing the contractions m ′ 1, . . . , m ′ i−1 in I. Again,it suffices to show that none of the expansions removes an element of L(m ′ i) from theideal. But if an expansion, say the expansion of ˜m k , k ∈ {1, . . . , i − 1}, removed suchan element, the monomial ˜m k must itself have been contained in L(m ′ i). Since ˜m k musthave been a minimal generator (otherwise it could not have been expanded), the monomialm ′ i · x n−1 cannot fulfill the condition of stable ideals (see Theorem 2.7), because we candivide m ′ i · x n−1 by x n−1 and multiply it by a suitable variable x l , such that the monomialobtained is divisible by ˜m k and no other generator. Hence, if the minimal generator ˜m khad been removed, either the ideal obtained is no longer stable, which is a contradiction,or the monomial ˜m k cannot have been a minimal generator, which is also a contradiction.Hence, we perform the contraction via the monomial m ′ i in I 2i−2 , which provides I (2i−1) .The fact that ˜m i is expandable in I (2i−1) , follows from the same arguments as stated above,when we proved that ˜m 1 is expandable in I (1) . Thus, all contraction-expansion pairs canbe performed.It remains to show that I (2r) = J, i.e. that the last contraction-expansion pair provides theideal J. We know that I (2r) is a saturated stable ideal with the same double saturationand the same Hilbert polynomial as I and J. Hence, by the choice of the monomials beingcontracted and expanded during the process taking I to I (2r) , the set of minimal generatorsof I (2k) must be a subset of the set of minimal generators of J. Since I (2r) and J do havethe same Hilbert polynomial, there is no monomial generator in J, which is not containedin I (2r) . Thus, it follows I (2r) = J.