University of Paderborn Department of Mathematics Diploma Thesis ...

32 CHAPTER 2. STABLE IDEALSfor all j ≥ ˜d, which showsh R/I (j + 1) = h R/I (j) 〈j〉 = h R/J (j) 〈j〉 = h R/J (j + 1)for all j ≥ ˜d. Hence, we obtain h R/I (j) = h R/J (j) for all j ≥ 0, which proves (i).(ii) We have to show that we can always find a monomial m as described in step (2)(ii)of the algorithm. Assume, at some stage of computation we cannot find a monomial mof degree d in ¯R. Let I := (M) denote the ideal generated by the monomials, which arecontained in M at this time. We know that the numerator of H(t) still contains t d . Sincewe know from (i) that h I (d) ≤ h J (d), we must have h I (d) < h J (d). Hence, there is atleast one more monomial of degree d in J than in I. Since we cannot find a monomialm ∈ ¯R = K[x 0 , . . . , x n−1 ], there is at least one monomial generator of J containing thevariable x n , which is a contradiction to the assumption that J is saturated and stable.(iii) At each stage of computation, the monomial m chosen in step (2)(ii) is the largestmonomial of ¯R, which is not yet contained in the ideal I. Hence, I is a lexicographic idealof ¯R at each stage of computation, which shows that L H is lexicographic in ¯R. Becauseany division and multiplication by a monomial generator m of L H in the sense of Definition2.8 produces a monomial m ′ with m ′ > m in lexicographic order, we conclude that L His stable. In step (2)(ii), we only choose monomials in ¯R not containing the variable x n .Hence, by Theorem 2.13, L H is saturated.(iv) The algorithm terminates, if we have g(t) = 1. Thus, one minus the sum of allpolynomials (1 − t) l · t d added to g(t) in step (2)(iii) equals the numerator of the Hilbertseries H R/J (t), which shows H R/J (t) = H R/LH (t).Note that it is necessary, to assume that the Hilbert series H(t) of Theorem 2.21 belongsto R/J for J ⊂ R a saturated (stable) ideal. Otherwise, we could choose the idealJ := (x 2 0, x 0 x 1 , x 0 x 2 , x 2 1, x 1 x 2 , x 2 2) ⊂ R := K[x 0 , x 1 , x 2 ], which is generated by all monomialsin x 0 , x 1 , x 2 of degree 2. Hence, the numerator of the Hilbert series H R/J (t) will contain−6t 2 and we would have to choose 6 monomials in ¯R = K[x 0 , x 1 ] to eliminate the term,but there are only three monomials of degree 2 in K[x 0 , x 1 ], which can be included intothe set M in the algorithm. In this case the algorithm cannot give a correct result.We will look at two examples, to see how the algorithm works in detail.Example 2.23. Let R := K[x 0 , x 1 , x 2 ] and I := (x 2 0, x 0 x 1 , x 2 1). The (non reduced) Hilbertseries of R/I is given by H(t) = 1 − 3t2 + 2t 3(1 − t) 3 . If the algorithm is correct, it should returnthe ideal L H = I, for I is a saturated stable ideal fulfilling the conditions of Theorem 2.21.We will now follow the steps of the algorithm:• At the start, we have M := ∅ and d = 2. The first monomial in lexicographic ordersuch that the ideal generated by M is a stable ideal is m = x 2 0. Thus,M := M ∪ {x 2 0} = {x 2 0},