University of Paderborn Department of Mathematics Diploma Thesis ...

4.7. COMPUTING ALL HILBERT FUNCTIONS TO A HILBERT POLYNOMIAL 119p R/I (i). This algorithm is more efficient than that sketched above.The following lemma suggests, how we find such an upper bound, such that the value ofthe Hilbert function equals the value of the Hilbert polynomial.Lemma 4.26. Let I ⊂ R be a saturated homogeneous ideal, h R/I the Hilbert functionand p R/I (z) the Hilbert polynomial of R/I. Let k > 0 denote the degree of the last monomialgenerator of the lexicographic ideal L p associated to p R/I (z) by Theorem 2.25. Thenh R/I (i) = p R/I (i) for all i ≥ k.Proof. By Theorem 2.21, there is a unique lexicographic ideal L H associated to the Hilbertseries H R/I (t) of R/I, such that h R/I (i) = h R/LH (i) for all i ∈ Z. By Corollary 3.40,we know that the Hilbert function h R/Lp of R/L p is minimal among all Hilbert functionsassociated to the Hilbert polynomial p R/I (z). If k denotes the degree of the last monomialgenerator of L p , then by Remark 1.20 and by Gotzmann’s Persistence Theorem 2.31, weobtain h R/Lp (i) = p R/Lp (i) for all i ≥ k. In the proof of Corollary 3.40 we saw thath R/LH (i) ≥ h R/Lp (i) for i < k and h R/LH (i) = h R/Lp (i) for all i ≥ k. Hence, we obtainh R/I (i) = h R/LH (i) = h R/Lp (i) = p R/Lp (i) = p R/I (i)for all i ≥ k, which proves the assertion.Now we combine the above lemma with Remark 4.25 to give an efficient algorithm (firstin pseudo code and later in source code) to compute the values of a Hilbert function froma given Hilbert series.Algorithm 4.27. Let H R/I (t) =g(t) be the reduced Hilbert series of R/I for I ⊂ R(1 − t)da saturated homogeneous ideal, g(t) and d as in Remark 4.25.Input. A reduced Hilbert series H R/I (t) and i > 0.1. Compute the Hilbert polynomial p R/I (z) of R/I from H R/I (t).2. Compute k, the degree of the last monomial generator of the unique lexicographicideal L p associated to p R/I (z).3. For j < k use the formulas stated in Remark 4.25 to compute h R/I (j).4. For k ≤ j ≤ i compute h R/I (j) = p R/I (j).Output. h R/I (j) for 0 ≤ j ≤ i.Proof. (Correctness) The algorithm works correct by Remark 4.25, where we showed, howwe can obtain the values h R/I (j) for 0 ≤ j ≤ k − 1, and by Lemma 4.26, which guaranteesthat h R/I (j) = p R/I (j) for all j ≥ k.