University of Paderborn Department of Mathematics Diploma Thesis ...

3.4. STABLE IDEALS WITH THE SAME HILBERT POLYNOMIAL 71Remark 3.42. Let I, J ⊂ R be saturated stable ideals with p R/I (z) = c = p R/J (z),c ∈ N. We know that the double saturation of I and J equals R. Hence, if any of thecontractions within the sequence of paired contractions and expansions used to take I toJ provided the whole ring R, we would not be able to perform any expansion any moreand could never reach the ideal J. We claim that this case cannot appear, if we followthe procedure in the proof of Theorem 3.32. By Lemma 3.17, there is either a sequence ofmonomials whose contractions take I to its double saturation, and a sequence of monomialswhose contractions take J to its double saturation. These contractions are exactly those,which we used in Proposition 3.28 and Theorem 3.32 to perform pairs of contractions andexpansions to compute one ideal from the other. Since the double saturation of I and Jequals R, in both cases the last contraction performed before reaching R is the contractionof 1 (i.e. x n−1 is removed from the set of generators and replaced by 1, which provides thewhole ring R). Hence, the last contraction taking I to its double saturation equals thelast contraction taking J to its double saturation. But in the proof of Proposition 3.28,we showed that we can omit these contractions, when we compute J from I by pairs ofcontractions and expansions. Thus, when using the pairs of expansions and contractionsof Proposition 3.28 and Theorem 3.32, it cannot happen that any of the ideals computedwithin the process equals the whole ring R.Next, we will prove by induction that all saturated stable ideals with the same Hilbertpolynomial are linked by finite sequences of contractions and expansions.Theorem 3.43. All saturated stable ideals with the same Hilbert polynomial are linked byfinite sequences of contractions and expansions.Proof. Let I, J ⊂ R be saturated stable ideals with p R/I (z) = p R/J (z). We may assumep R/I (z) ≠ 0, because otherwise there is nothing to prove. As mentioned above, we doinduction on the degree d := deg p R/I (z).The assertion is true for d = 0 by Lemma 3.41. Now assume d > 0. Since I and J aresaturated stable ideals in R, we know that the variable x n does not appear in any of theirmonomial generators. Hence, we can view I and J as possibly non-saturated stable idealsĪ and ¯J in ¯R := K[x 0 , . . . , x n−1 ].Claim. p ¯R/ Ī(z) = p ¯R/ ¯J(z).Proof of the Claim. By the exact sequence0 −−−→ R/I(−1)x n−−−→ R/I −−−→ R/(I + xn R) −−−→ 0,it follows p R/(I+xnR)(z) = p R/I (z) − p R/I (z − 1). Similarly we obtain from0 −−−→ R/J(−1)p R/(J+xnR)(z) = p R/J (z) − p R/J (z − 1). Sincex n−−−→ R/J −−−→ R/(J + xn R) −−−→ 0R/(I + x n R) ∼ = (R/x n R)/((I + x n R)/x n R) ∼ = ¯R/Ī