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38 CHAPTER 2. STABLE IDEALSWe will compute the ideal L p in a few specific examples:Example 2.26. Let R := K[x 0 , x 1 , x 2 , x 3 ], i.e. n = 3.(i) Let p(z) = 2z + 1 be the given Hilbert polynomial. Now we proceed as in 1.11to compute m 0 and m( 1 : In ) the( terminology ) of the theorem ( we( get r = ) 1 and thusz + 1 z + 1 − 2z z − m0m 1 = 1!·2 = 2. Since −= 2z−1 and − = m 0 , we2 21)1obtain m 0 = 2 and a 1 = 2 as well as a 0 = 0. Since the generators of L p are x n−r−2 =x 0 , x ar+1n−r−1 = x 3 1 and x arn−r−1 ·x a r−1n−r = x 2 1 ·x 0 2 = x 2 1, we write L p = (x 0 , x 3 1, x 2 1) = (x 0 , x 2 1).Indeed, L p is a saturated lexicographic ideal and p R/Lp (z) = p(z).( ) ( ) ( ( )z + 1 z + 1 − 3 z z − 4(ii) Let p(z) = 3z + 1. We write p(z) = −+ − ,22 1)1i.e. r = 1, m 1 = 3, m 0 = 4, and therefore a 1 = 3, a 0 = 4 − 3 = 1. Thus, the ideal L pis generated by the monomials x 0 , x 4 1 and x 3 1x 2 .(iii) As the last example, we consider the non-linear Hilbert polynomial p(z) = 2z 2 +2z+1of K[x 0 , x 1 , x 2 , x 3 , x 4 ]/(x 2 0, x 2 1). From 1.11 we know that m 2 = 4, m 1 = 6 and m 0 = 12.Thus, we have r = 2, a 2 = 4, a 1 = 6 − 4 = 2, a 0 = 6 which provides L p =(x 0 , x 5 1, x 4 1x 3 2, x 4 1x 2 2x 6 3).That the ideal L p of the last example indeed provides p R/Lp (z) = 2z 2 +2z +1 is not clear atthis time. It will not be proved now. In the next section we state an algorithm to computethe Hilbert polynomial of any stable ideal. Then, we use this algorithm to compute theHilbert polynomial of R/L p and show that it equals 2z 2 + 2z + 1.Furthermore, in Chapter 3, we will present a link between the ideals L H and L p defined inthis and the preceding section. In detail, we will be able to prove that if H(t) is a Hilbertseries and p(z) the associated Hilbert polynomial, the unique lexicographic ideals L H andL p have the same double saturation.2.7 Hilbert polynomials of stable idealsIn Chapter 2, Section 2.4, Theorem 2.17 provided a formula for the computation of theHilbert series of a stable ideal. As a corollary of the following theorem and this formula,we obtain an algorithm to determine the Hilbert polynomial of a given stable ideal.Theorem 2.27. (Hilbert polynomials of stable ideals) Let d ≥ 1 and H R/I (t) =g(t)(1 − t) dk∑be the Hilbert series of R/I for some homogeneous ideal I ⊂ R, where g(t) = C i · t i ,C i ∈ Z, 0 ≤ i ≤ k, with g(1) ≠ 0. Then the Hilbert polynomial p R/I (z) is given byk∑( )z + d − 1 − jp R/I (z) = C j ·.d − 1j=0i=0
2.7. HILBERT POLYNOMIALS OF STABLE IDEALS 39∞∑( ) d − 1 + iProof. We know that· t i =d − 1i=0∞∑( ) d − 1 + iMultiplying the equation· t i =d − 1i=0H R/I (t) of the theorem:g(t)∞∑( ) d − 1 + j(1 − t) = (C d 0 + C 1 · t + . . . + C k · t k ) ·· t jd − 1j=0∞∑( ) d − 1 + j=· (C 0 · t j + C 1 · t j+1 . . . + C k · t j+k ).d − 1j=0For i ≥ k, the coefficient of t i in H R/I (t) is( ) ( ) ( )d − 1 + i d − 1 + i − 1d − 1 + i − k·C 0 +·C 1 +. . .+·C k =d − 1d − 1d − 11(1 − t) d and g(t) = C 0 + C 1 · t + . . . + C k · t k .1by g(t) gives the Hilbert series(1 − t)dk∑( )d − 1 + i − jC j·.d − 1Since the coefficient of t i in H R/I (t) is the value of the Hilbert function h R/I (i) by definitionof the Hilbert series, we obtaink∑( )d − 1 + i − jh R/I (i) = C j ·d − 1for all i ≥ k. Since h R/I (i) = p R/I (i) for all i ≫ 0, we get p R/I (z) =which proves the assertion.j=0j=0k∑( )z + d − 1 − jC j·,d − 1Note that the number d of the theorem is the Krull dimension of R/I. The theorem doesnot state anything about the case d = 0. If the Krull dimension of R/I equals zero, R/Idescribes the empty set (as projective scheme when viewed from the standpoint of algebraicgeometry). Thus, d = 0 represents the trivial case, which will not be of any furtherinterest to us in the following.Furthermore, note that the result of the preceding theorem is valid for any homogeneousideal I ⊂ R. We will not make use of it in this generality. However, to see that it reallyworks for any homogeneous ideal, we examine one small example where the ideal I is notstable.Example 2.28. Let I := (x 2 0, x 3 1) ⊂ K[x 0 , x 1 , x 2 ]. The reduced Hilbert series of R/I isH R/I (t) = 1 + 2t + 2t2 + t 3, i.e. d = 1, C 0 = 1, C 1 = 2, C 2 = 2 and C 3 = 1 in the1 − tterminology of the theorem. Thus, we obtain the Hilbert polynomial3∑( ) z − j3∑p R/I (z) = C j · = C j = 6.0j=0j=0j=0
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