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University of Paderborn Department of Mathematics Diploma Thesis ...

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40 CHAPTER 2. STABLE IDEALSAs a corollary <strong>of</strong> Theorem 2.17 and Theorem 2.27 we obtain the following algorithm tocompute the Hilbert polynomial <strong>of</strong> a stable ideal I.Algorithm 2.29. (Computation <strong>of</strong> the Hilbert polynomial <strong>of</strong> a stable ideal) Let I ⊂ R bea stable ideal and I g = {m 1 , . . . , m r } a set <strong>of</strong> minimal generators <strong>of</strong> I, d i := deg m i thedegree and l i the index <strong>of</strong> the last variable <strong>of</strong> the monomial m i , 1 ≤ i ≤ r.The algorithm proceeds in three steps:Input. I g = {m 1 , . . . , m r }, i.e. the set <strong>of</strong> minimal generators <strong>of</strong> a stable ideal.1. Compute the Hilbert series H R/I (t) <strong>of</strong> R/I, i.e.H R/I (t) = 1 − ∑ ri=1 (1 − t)li · t d i(1 − t) n+1 .2. Compute the reduced Hilbert series ˜H R/I (t), i.e.˜H R/I (t) =∑ ki=0 C i · t i(1 − t) d ,where the numerator does not vanish for t = 1.3. Compute the Hilbert polynomial <strong>of</strong> R/I by using the preceding theorem, i.e.p R/I (z) =k∑( )z + d − 1 − jC j ·.d − 1j=0Output. The Hilbert polynomial p R/I (z) <strong>of</strong> R/I.Pro<strong>of</strong>. (Correctness) The correctness <strong>of</strong> the algorithm follows from Theorem 2.17 andTheorem 2.27.We will now use the algorithm in a concrete example. Recall that in Example 2.26 (iii)we did not prove that the computed ideal has the stated Hilbert polynomial. We now givethe pro<strong>of</strong> by using the new algorithm to compute its Hilbert polynomial.Example 2.30. Let R := K[x 0 , x 1 , x 2 , x 3 , x 4 ], and let I be the unique lexicographic ideal,which we obtained in Example 2.26 (iii), i.e. I = (x 0 , x 5 1, x 4 1x 3 2, x 4 1x 2 2x 6 3). We want to computethe Hilbert polynomial according to the preceding algorithm. Thus, we proceed in the threesteps mentioned above:1. The non-reduced Hilbert series H R/I (t) <strong>of</strong> R/I is1 − (1 − t) 0 · t − (1 − t) · t 5 − (1 − t) 2 · t 7 − (1 − t) 3 · t 12(1 − t) 5 .

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