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52 CHAPTER 3. OPERATIONS ON STABLE IDEALS3.3 Stable ideals with the same double saturationThis section is dedicated to the proof that all saturated stable ideals with the same doublesaturation and the same Hilbert polynomial are linked by finite sequences of paired expansionsand contractions. As in the preceding section, where we presented a modified versionof the definition of a contraction, some of the results presented here had to be modified andhad to be proved in a different way from that, discussed in [16]. Wherever modificationshad to be made, these have been documented within the framework of remarks.Recall from 2.14 that the double saturation sat xn−1 ,x n(I) of a stable ideal I ⊂ R canbe computed as follows: First compute the saturation sat xn (I) of I in R by Theorem2.13. The ideal obtained is still stable and can be viewed as an ideal in K[x 0 , . . . , x n−1 ].Hence, by 2.14, we obtain the ideal sat xn−1 ,x n(I) by computing the saturation of sat xn (I)in K[x 0 , . . . , x n−1 ].Lemma 3.17. Let I ⊂ R be a saturated stable ideal, I g the set of its minimal monomialgenerators and sat xn−1 ,x n(I) its double saturation. There is a finite sequence of contractionstaking I to sat xn−1 ,x n(I).Proof. Let the monomials of I g be listed in descending lexicographic order. Furthermore,let x A · x a n−1n−1 = x a 00 · . . . · x arr · x a n−1n−1 with 0 ≤ r < n − 1 be the first monomial of I g(with respect to the lexicographic order) containing the variable x n−1 . Note that we mayassume that x n does not appear in the monomial, since I is saturated. We claim thatx A is contractible in I g . Since x A · x a n−1n−1 is an element of I g , the monomial x A cannot becontained in I. The only other condition for a contraction is L(x A ) ⊂ I. Since I is stableand x A · x a n−1n−1 ∈ I, we know in particular that all monomials of the formx a 00 · . . . · x a i−1i−1 · xa i+1i · x a i+1i+1 · . . . · xa j−1j−1 · xa j−1j · x a j+1j+1 · . . . · xar r · x a n−1n−1are contained in I for 0 ≤ i < j ≤ n − 1. In the lexicographic order, we havex a 00 · . . . · x a i−1i−1 · xa i+1i · x a i+1i+1 · . . . · xa j−1j−1 · xa j−1j · x a j+1j+1 · . . . · xar r · x a n−1n−1 > x A · x a n−1n−1for all such monomials. Since we have chosen x A · x a n−1n−1where x n−1 appears, it followsto be the first monomial generator,x a 00 · . . . · x a i−1i−1 · xa i+1i · x a i+1i+1 · . . . · xa j−1j−1 · xa j−1j · x a j+1j+1 · . . . · xar rfor all 0 ≤ i < j ≤ r. In particular, we obtain L(x A ) ⊂ I, since the left-shift L(x A ) iscontained in the ideal generated by such monomials. This proves that x A is contractiblein I g .Note that the contraction of the monomial x A replaces x A · x a n−1n−1 by x A in the ideal considered.Hence, in the next step, we search for the next monomial generator (with respectto the lexicographic order) containing x n−1 and iterate the above procedure. Since thereis only a finite number of monomials in I g containing the variable x n−1 , we can repeat∈ I
3.3. STABLE IDEALS WITH THE SAME DOUBLE SATURATION 53such contractions, until there is no monomial left in I g containing x n−1 . Additionally, itfollows from the definition of a contraction (see 3.9) that the ideal obtained has the samedouble saturation as the ideal I we started with. Hence, we obtain a finite sequence ofcontractions taking I to sat xn−1 ,x n(I).In the preceding part of the proof, we omitted one special case, i.e. the case that x A·x a n−1n−1 =x a n−1n−1 , a n−1 > 1. In this situation, we proceed as described above until the variable x n−1appears in the set of monomial generators of an ideal. In the following step, we contract1. By the definition of the contraction (see 3.9), the contraction of 1 removes the variablex n−1 from the set of monomial generators and inserts 1. Hence, the ideal obtained is thewhole ring R, which obviously is the correct result, since the double saturation of a stableideal containing the monomial x a n−1n−1 equals R (we will state a simple example below, wherethis case appears).Remark 3.18. Note that the proof of Lemma 3.17 is constructive. The first importantobservation is that the monomials, which have to be contracted to take the ideal I to itsdouble saturation sat xn−1 ,x n(I) (i.e. the sequences of contractions) are uniquely determined,if we follow the algorithmic scheme of the proof. The second important aspect is that eachof the contractions can be performed successively, such that it subtracts exactly one fromthe exponent of x n−1 in some monomial of I g . The last fact will be used in the proofof the theorem that all saturated stable ideals with the same double saturation and thesame Hilbert polynomial are linked by finite sequences of expansions and contractions. Wewill compute the double saturation of a saturated stable ideal by this new method in twoconcrete examples. First, we treat the case, where we deal with an ideal, whose doublesaturation equals the whole ring R.Example 3.19. Let I := (x 0 , x 2 1) ⊂ K[x 0 , x 1 , x 2 ]. I is saturated and stable. The doublesaturation of I is sat x1 ,x 2(I) = (x 0 , 1) = K[x 0 , x 1 , x 2 ]. With respect to the preceding proof,we have to proceed as follows:• The first monomial to be contracted is x 1 . The left-shift of x 1 is the set L(x 1 ) = {x 0 },which is contained in the ideal I. Since x 1 /∈ I, we may contract x 1 . This providesthe set of generatorsI g (1) := I g ∪ {x 1 }\{x 2 1} = {x 0 , x 1 }.Let I (1) be the ideal generated minimally by I g(1) . Note, that we have indeed decreasedthe exponent of x 1 by 1 as stated in the proof of the lemma. The ideal I is stillsaturated and stable and it has the same double saturation as the ideal I, we startedwith.• The next monomial to be contracted is 1, since in the terminology of the proof above,we have x A · x n−1 = 1 · x 1 . By Definition 3.2 we have L(1) = ∅, which is a subsetof I. Here, the special case of Definition 3.2 becomes apparent (in order to contract1, we have to guarantee that the left-shift of 1 is always a subset of the given ideal).Contraction providesI g (2) := I g (1) ∪ {1}\{x 1 } = {x 0 , 1},
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