University of Paderborn Department of Mathematics Diploma Thesis ...

3.3. STABLE IDEALS WITH THE SAME DOUBLE SATURATION 55Proof. Let m := x a 00 · . . . · x a n−1n−1 , a n−1 ≥ 1. By the proof of Lemma 3.17, we know thatmis contractible in I. Hence, by the definition of contractions (see Definition 3.9), thex n−1mcontraction of = x a 00 ·. . .·x a n−1−1n−1 removes only the monomial x a 00 ·. . .·x a n−1−1n−1 ·x n−1 = mx n−1from the set of minimal generators of I and includes the monomial x a 00 ·. . .·x a n−1−1n−1 into theset of minimal generators. Since m is a minimal generator of I, the monomial x a 00 ·. . .·x a n−1−1n−1must also be a minimal generator of the ideal I con (otherwise we have a contradiction). Wehave to show that we can expand x a 00 · . . . · x a n−1−1n−1 in I con . Since x a 00 · . . . · x a n−1−1n−1 is aminimal generator of I con , the only other condition for an expansion is that none of theelements of R(x a 00 · . . . · x a n−1−1n−1 ) is contained in the set of minimal generators of I con .Assume that x a 00 · . . . · x at−1t · x a t+1+1t+1 · . . . · x a n−1−1n−1 ∈ R(x a 00 · . . . · x a n−1−1n−1 ) is contained in theset of minimal generators of I con . Then the monomial x a 00 · . . . · x at−1t · x a t+1+1t+1 · . . . · x a n−1−1n−1must have already been contained in the ideal I. Since I is stable, it followsx a 00 · . . . · x at−1t· x a t+1+1t+1 · . . . · x a n−1−1n−1· x t = x a 00 · . . . · x a n−1−1n−1 = m ∈ I,x t+1 x n−1mwhich contradicts the fact that is contractible in I and, hence, cannot be containedx n−1in I. It is clear by the definition of an expansion (see Definition 3.5) that the expansion ofmin I con indeed provides (I con ) exp = I.x n−1By the preceding lemma, now we can also view the sequence x 1 x 2 2, x 1 x 2 , x 1 of contractionstaking I to sat x2 ,x 3(I) of Example 3.20 in reversed order x 1 , x 1 x 2 , x 1 x 2 2 as a sequence ofexpansions taking sat x2 ,x 3(I) to I. Thus, the preceding lemma always provides a finitesequence of expansions, taking the double saturation of some saturated stable ideal to theideal itself.Corollary 3.22. Given two saturated stable ideals I, J ⊂ R with the same double saturation,there is a finite sequence of contractions and expansions taking I to J.Proof. By Lemma 3.17, there is a finite sequence of contractions x A 1, . . . , x A k ∈ M takingI to sat xn−1 ,x n(I). Again by Lemma 3.17, there is a finite sequence of contractionsx B 1, . . . , x B l ∈ M taking J to satxn−1 ,x n(J). Since we can view x B l , . . . , xB 1as a sequenceof expansions taking sat xn−1 ,x n(J) to J and sat xn−1 ,x n(J) = sat xn−1 ,x n(I), the sequence ofcontractions x A 1, . . . , x A k and the sequence of expansions xB l, . . . , xB 1takes I to J.The question, which will be of interest to us in the following, is: What effect do expansionsand contractions have on the Hilbert polynomial of R/I for I ⊂ R a saturated stable ideal?This question is important, since we will later use contractions and expansions to computeall saturated stable ideals to a given Hilbert polynomial. The proposition below characterizesthose expansions and contractions, which do not change the Hilbert polynomial. Asa corollary, we will see that an expansion or a contraction of a monomial can only changethe Hilbert polynomial by adding or subtracting 1.