University of Paderborn Department of Mathematics Diploma Thesis ...

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.