University of Paderborn Department of Mathematics Diploma Thesis ...

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68 CHAPTER 3. OPERATIONS ON STABLE IDEALSare minimal generators of J. We already know that m is a minimal generator of J. It is clearthat x 0 , x 1 , . . . , x n−r−2 are minimal generators of J. Consider the monomial x ar+1n−r−1 ∈ J.If we assume that it is not a minimal generator of J, it follows that there is a monomialgenerator x k n−r−1 ∈ J, k ≤ a r . But then, this generator also divides m, which is a contradiction.If x arn−r−1 · x a r−1+1n−r were not a minimal generator of J, there would be a generatorx k n−r−1 · x l n−r ∈ J, k ≤ a r , l ≤ a r−1 . Again, m would also be divisible by this monomial,which is a contradiction.Iteration of this argument provides that all monomials in the above set are minimal generators.Now assume, there is a monomial generator ˜m of J, which is not contained in theabove set. Since the above set generates a lexicographic and stable ideal and ˜m cannotcontain the variable x n−1 , it follows either, that ˜m divides any of the above monomials,which contradicts the fact that they are all minimal generators, or ˜m is divisible by anyof the above monomials, which is a contradiction to the fact that ˜m is assumed to be aminimal generator. Hence, J is generated by the monomials in the above set, which provesclaim 2.It follows by Theorem 2.25 that J is the unique lexicographic ideal L p associated to theHilbert polynomial p(z) (which is again associated to the Hilbert series H(t) of R/L H ).Since the expansions and contractions described in Remark 3.31, which we used to computeJ from L H , do not change the double saturation, it follows sat xn−1 ,x n(L H ) = sat xn−1 ,x n(J) =sat xn−1 ,x n(L p ).Example 3.36. Let R := K[x 0 , x 1 , x 2 , x 3 ] and H(t) = 1 − t − t2 + t 4 + t 5 − t 6. Then the(1 − t) 4unique lexicographic ideal L H , such that H R/LH (t) = H(t), is L H = (x 0 , x 2 1, x 1 x 2 2, x 4 2) andp R/LH (z) = 6. To compute L p from L H , we first we contract x 1 x 2 in L H , which provides theideal (x 0 , x 2 1, x 1 x 2 , x 4 2). Next, we contract x 1 , and obtain (x 0 , x 1 , x 4 2). Then we expand x 4 2,which provides (x 0 , x 2 1, x 5 2), and finally we obtain the ideal (x 0 , x 2 1, x 6 2) from the expansionof x 5 2. Indeed, the ideal (x 0 , x 2 1, x 6 2) equals the unique lexicographic ideal L pR/LH associatedto the Hilbert polynomial p R/LH (z).The proof of the preceding theorem suggests a method to compute the lexicographic idealL p from the lexicographic ideal L H . Since we know by the theorem that these ideals havethe same double saturation, it follows that we can also compute L H from L p :Corollary 3.37. Let p R/I (z) be the Hilbert polynomial of R/I for some saturated stableideal I ⊂ R. Then the lexicographic ideal L H associated to any Hilbert series H(t), suchthat p R/LH (z) = p R/I (z), can be obtained from L p by a sequence of paired expansions andcontractions.Proof. By Theorem 3.35, we know sat xn−1 ,x n(L H ) = sat xn−1 ,x n(L p ). Since p R/LH (z) =p R/Lp (z), the assertion follows from Theorem 3.32.Remark 3.38. Corollary 3.37 implies that to a given Hilbert polynomial p(z) we can findall Hilbert series associated to the Hilbert polynomial p(z) (in other words: We computeall Hilbert functions to a given Hilbert polynomial). We give the algorithm in pseudo code.
3.3. STABLE IDEALS WITH THE SAME DOUBLE SATURATION 69Algorithm 3.39. (Computation of all Hilbert series to a given Hilbert polynomial) Letp(z) be the Hilbert polynomial of R/I for I ⊂ R a saturated homogeneous ideal.Input. A Hilbert polynomial p(z).1. Compute the unique lexicographic ideal L p of Theorem 2.25 from p(z).2. Use pairs of contractions and expansions to compute all saturated lexicographic idealsL H of Theorem 2.21 with p R/LH (z) = p(z).3. Use the formula of Theorem 2.17 to compute the Hilbert series of any such lexicographicideal L H .Output. All Hilbert series H R/J (t) for some saturated homogeneous ideal J ⊂ R withp(z) = p R/J (z).Proof. (Correctness) The algorithm works correct by Theorem 2.25, Corollary 3.37 andTheorem 2.17.Step 2 of the algorithm does not specify any details of how to use the pairs of contractionsand expansion to find all lexicographic ideals L H in a systematic way – this will becomeclear, when we present the source code to this step within Chapter 4.Another consequence of Corollary 3.37 and Theorem 3.35 gives us a characterization of theunique lexicographic ideal L p associated to a Hilbert polynomial p(z), which we introducedin Theorem 2.25:Corollary 3.40. Let p(z) ≠ 0 be the Hilbert polynomial of R/I for some saturated stableideal I ⊂ R.(i) Among all saturated lexicographic ideals L H associated to Hilbert series H(t), suchthat p(z) = p R/LH (z), there is exactly one saturated lexicographic ideal with onlyone minimal generator containing the last variable x n−1 . This ideal is the uniquelexicographic ideal L p associated to the Hilbert polynomial p(z) by Theorem 2.25.(ii) The Hilbert function of R/L p is minimal among all Hilbert functions h R/J associatedto the Hilbert polynomial p(z), where J ⊂ R is a saturated homogeneous ideal.In particular: For any saturated lexicographic ideal L H associated to a Hilbert seriesH(t), such that p(z) = p R/LH (z), we have h R/Lp (i) ≤ h R/LH (i) for all i ∈ N 0 .Proof. The first assertion follows from Corollary 3.37 and the proof of Theorem 3.35. Thesecond one follows from the fact that we can compute any such lexicographic ideal L H fromL p by a sequence of paired contractions and expansions: When we proved Proposition 3.23and Corollary 3.26, we characterized the effect of such expansions and contractions onthe values of the Hilbert function. Let m denote the last monomial generator of L p andmd := deg m. The contraction of in L p provides h L con(i) = hp Lp (i) + 1 for all i ≥ d − 1.x n−1
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