University of Paderborn Department of Mathematics Diploma Thesis ...

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58 CHAPTER 3. OPERATIONS ON STABLE IDEALSin I provides the ideal sat xn−1 ,x n(I). Recall that by Remark 3.18 we know the sequence isuniquely determined, if we choose the monomials with respect to the proof of Lemma 3.17.In particular, we compute saturated stable ideals I (i) , 0 ≤ i ≤ l, where I (0) := I,I (i) := (I (i−1) ) con (the contraction is always performed via the monomial m i−1 ) and I (l) =sat xn−1 ,x n(I):}{{} I (0) (I (0) ) con = I (1) . . . (I (l−2) ) con = I (l−1) (I (l−1) ) con = }{{} I (l) .Isat xn−1 ,xn (I)On the other hand, we find monomials m ′ 1, . . . , m ′ k ∈ M and saturated stable ideals J (j) ,0 ≤ j ≤ k, where J (0) := sat xn−1 ,x n(J), J (j) := (J (j−1) ) exp (expansion performed via themonomial m ′ i−1) and J (k) = J:}{{} J (0) (J (0) ) exp = J (1) . . . (J (k−2) ) exp = J (k−1) (J (k−1) ) exp = }{{} J (k) .sat xn−1 ,xn (J) JSuppose, we additionally assume that I and J have the same Hilbert polynomial. Then,by Corollary 3.26, we conclude that k = l. Hence we obtain a sequence of contractionsand a sequence of expansionswhich we combine toI = I (0) I (1) . . . I (l−1) I (l) = sat xn−1 ,x n(I)sat xn−1 ,x n(J) = J (0) J (1) . . . J (l−1) J (l) = J,I = I (0) I (1) . . . I (l−1) I (l) = J (0) J (1) . . . J (l−1) J (l) = J,since I (l) = sat xn−1 ,x n(I) = sat xn−1 ,x n(J) = J (0) .We are now able to prove an even stronger result under the assumption that the ideals Iand J do not only have the same double saturation, but also the same Hilbert polynomial– we will see that we can perform pairs of contractions and expansions to compute theideal J from the ideal I (and vice versa).Before we can prove this general assertion, we need a more special result stated in theproposition below.Proposition 3.28. Let I, J ⊂ R be saturated stable ideals with p R/I (z) = p R/J (z) andsat xn−1 ,x n(I) = sat xn−1 ,x n(J). If the ideal J contains only one minimal monomial generatorm with the last variable x n−1 , then I and J are linked by a finite sequence of paired expansionsand contractions. In particular: There is an even number of monomials m 1 , . . . , m kand ideals I (0) , I (1) , . . . , I (k) , where(i) I (0) := I and
3.3. STABLE IDEALS WITH THE SAME DOUBLE SATURATION 59(ii) I (l+1) :=l = 0, . . . , k and I (k) = J.{(I (l) ) con if l mod 2 = 0 (contraction via the monomial m l )(I (l) ) exp if l mod 2 = 1 (expansion via the monomial m l )Proof. By Corollary 3.22, there is a sequence of monomials m ′ 1, . . . , m ′ p ∈ M, whose contractionstake I to sat xn−1 ,x n(I), and a sequence of monomials ˜m 1 , . . . , ˜m q ∈ M, whoseexpansions take sat xn−1 ,x n(I) = sat xn−1 ,x n(J) to J. Note that we can also view ˜m 1 , . . . , ˜m qas a sequence of contractions taking J to sat xn−1 ,x n(J). Hence, by Remark 3.18, bothsequences of monomials are uniquely determined (see also: Proof of Lemma 3.17). ByCorollary 3.26, we conclude that p = q, since otherwise the Hilbert polynomial p R/I (z)does not equal p R/J (z).For technical reasons, which will become clear later, we need to have m ′ i ≠ ˜m j for allmonomials considered in the above sequences. If this was not the case, say m ′ i = ˜m j ,we will show that it suffices to consider the contractions m ′ 1, . . . , m ′ i−1 and the expansions˜m j+1 , . . . , ˜m p .Suppose, we have m ′ i = ˜m j for some 1 ≤ i, j ≤ p. Recall that each of the contractions performedsubtracts exactly 1 from the exponent of x n−1 of some monomial generator. SinceJ only contains one monomial m with last variable x n−1 and p R/I (z) = p R/J (z), it followsj = p − i + 1 and m ′ i+1 = ˜m p−i , . . . , m ′ p = ˜m 1 . Because of sat xn−1 ,x n(I) = sat xn−1 ,x n(J),we conclude I (i) = J (p−i+1) . But then we can perform the expansions ˜m p−i+2 , . . . , ˜m pin J (p−i+1) , which provides the ideal J. Hence, without loss of generality, we consider asequence of contractions m ′ 1, . . . , m ′ r taking I to some ideal, in which we perform the expansions˜m 1 , . . . , ˜m r , which provides the ideal J, and m ′ i ≠ ˜m j for all 1 ≤ i, j ≤ r.On a rather informal level: We take a short cut – we do not take I to its double saturationvia contractions and then perform expansions in sat xn−1 ,x n(I) = sat xn−1 ,x n(J), to reach theideal J. Before we contract any monomial m ′ i, which is equal to ˜m p−i+1 , we expand themonomials ˜m p−i+2 , . . . , ˜m p , until we reach the ideal J. Now we have to prove that we canperform pairs of expansions and contractions as mentioned in the proposition above.Since we can now assume m ′ i ≠ ˜m j for all 1 ≤ i, j ≤ r, we will be able to prove the following:We start with the contraction of m ′ 1 in I (0) := I, which provides the ideal I (1) = (I (0) ) con .Then we expand the monomial ˜m 1 in I (1) (we have to show that this is possible). I (2) willbe the ideal (I (1) ) exp , i.e. ˜m 1 expanded in I (1) . We claim that we can continue this processuntil we get the ideal I (2r) = I (k) , which equals J.Since ˜m 1 is expandable in sat xn−1 ,x n(I) = sat xn−1 ,x n(J), we know that ˜m 1 is contained inthe set of minimal generators of sat xn−1 ,x n(J). Because of ˜m 1 ≠ m ′ i for all 1 ≤ i ≤ r,the monomial ˜m 1 cannot be removed from the set of minimal generators of sat xn−1 ,x n(I)by m ′ r, . . . , m ′ 2 viewed as a sequence of expansions taking sat xn−1 ,x n(I) to I (1) . Hence, ˜m 1must be a minimal generator of I (1) as well. The other condition for an expansion (seeDefinition 3.5) is, that no element of R( ˜m 1 ) is contained in the set of minimal generatorsof I (1) . We know that no element of R( ˜m 1 ) is contained in the set of minimal generatorsof sat xn−1 ,x n(J). Again, because of ˜m 1 ≠ m ′ i for all 1 ≤ i ≤ r, the sequence m ′ r, . . . , m ′ 2viewed as a sequence of expansions taking sat xn−1 ,x n(I) to I (1) will not include a monomial
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