University of Paderborn Department of Mathematics Diploma Thesis ...

More documents

Recommendations

Info

54 CHAPTER 3. OPERATIONS ON STABLE IDEALSand we obtain I (2) = (x 0 , 1) = K[x 0 , x 1 , x 2 ] = sat x1 ,x 2(I).Example 3.20. Consider the ideal I := (x 0 , x 2 1, x 1 x 3 2) ⊂ K[x 0 , x 1 , x 2 , x 3 ]. Since the generatorsin I g := {x 0 , x 2 1, x 1 x 3 2} fulfill Theorem 2.7 and do not contain x 3 , I is a saturatedstable ideal. The double saturation sat xn−1 ,x n(I) is the ideal generated by the set of monomials{x 0 , x 1 }, since we have to set x 2 := 1. This provides the monomials x 0 , x 2 1, x 1 , wherewe may forget about x 2 1, since it is divisible by x 1 . Now we proceed with respect to theproof of the lemma:• We have to choose the first monomial of I g containing the variable x n−1 = x 2 , i.e. x 1 x 3 2.In the terminology of the proof, we get x A = x 1 x 2 2. Since x 1 x 2 2 is not contained in I gand L(x 1 x 2 2) = {x 0 x 2 2, x 2 1x 2 } ⊂ I, we can contract x 1 x 2 2 (Note that L(x 1 x 2 2) is not asubset of the set of minimal generators of I – if we had the condition L(x 1 x 2 2) ⊂ I g ,the monomial x 1 x 2 2 would not be contractible at all. Because of this, the definitionof a contraction and the contractibility of a monomial differs from the definition ofthe contraction in [16]). We will denote the ideal, which we obtain after performingthe contraction, by I (1) . It is generated minimally byI (1)g = I g ∪ {x 1 x 2 2}\{x 1 x 3 2} = {x 0 , x 2 1, x 1 x 2 2}.• Next, we have to choose x 1 x 2 2. Again, the monomial x 1 x 2 is not contained in I g(1) andL(x 1 x 2 ) = {x 0 x 2 , x 2 1} ⊂ I (1) . Hence, x 1 x 2 is contractible in I g(1) and it provides theideal I (2) generated byI (2)g = I (1)g ∪ {x 1 x 2 }\{x 1 x 2 2} = {x 0 , x 2 1, x 1 x 2 }.• In the last step, we have to choose x 1 x 2 ∈ I g(2) with L(x 1 ) = {x 0 } ⊂ I (2) . Hence, theideal I (3) , which we obtain after performing the contraction of x 1 in I g(2) is generatedbyI g (3) = I g (2) ∪ {x 1 }\{x 2 1, x 1 x 2 } = {x 0 , x 1 },which shows I (3) = sat x2 ,x 3(I).In Example 3.16, we saw that a contraction followed by an expansion of the same monomialin some stable ideal I ⊂ R will not always give the same ideal I again. If we restrictourselves to the type of contractions, which have been used above to compute the doublesaturations, the lemma below shows that the contraction of such a monomial followed bythe expansion of the same monomial will always provide the same ideal.Lemma 3.21. Let I ⊂ R be a saturated stable ideal with I sat xn−1 ,x n(I) and m ∈ I thefirst minimal monomial generator of I with the last variable x n−1 . Then the contractionmfollowed by the expansion of provides (I con ) exp = I.x n−1
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.
Page 1:
University of PaderbornDepartment o
Page 5:
”Im großen Garten der Geometriek
Page 9:
AbstractThe main result of this the
Page 12 and 13:
Glossary of MuPAD procedures 129Bib
Page 14 and 15: This thesis is organized in four ch
Page 16 and 17: 6described in Chapter 4.Additionall
Page 18 and 19: 8 CHAPTER 1. NOTATIONS AND PREREQUI
Page 20 and 21: 10 CHAPTER 1. NOTATIONS AND PREREQU
Page 30 and 31: 20 CHAPTER 2. STABLE IDEALSExample
Page 32 and 33: 22 CHAPTER 2. STABLE IDEALSset, whe
Page 34 and 35: 24 CHAPTER 2. STABLE IDEALSm ∈ M
Page 36 and 37: 26 CHAPTER 2. STABLE IDEALSRemark a
Page 38 and 39: 28 CHAPTER 2. STABLE IDEALSby Lemma
Page 40 and 41: 30 CHAPTER 2. STABLE IDEALSwith Hil
Page 42 and 43: 32 CHAPTER 2. STABLE IDEALSfor all
Page 44 and 45: 34 CHAPTER 2. STABLE IDEALS• We h
Page 46 and 47: 36 CHAPTER 2. STABLE IDEALSis stabl
Page 48 and 49: 38 CHAPTER 2. STABLE IDEALSWe will
Page 50 and 51: 40 CHAPTER 2. STABLE IDEALSAs a cor
Page 52 and 53: 42 CHAPTER 2. STABLE IDEALSTheorem
Page 54 and 55: 44 CHAPTER 2. STABLE IDEALSBinomial
Page 56 and 57: 46 CHAPTER 3. OPERATIONS ON STABLE
Page 86 and 87: 76 CHAPTER 4. ALGORITHMS FOR STABLE
Page 110 and 111: 100 CHAPTER 4. ALGORITHMS FOR STABL
Page 112 and 113: 102 CHAPTER 4. ALGORITHMS FOR STABL
Page 114 and 115:
104 CHAPTER 4. ALGORITHMS FOR STABL
Page 116 and 117:
Page 118 and 119:
Page 120 and 121:
Page 122 and 123:
Page 124 and 125:
Page 126 and 127:
Page 128 and 129:
Page 130 and 131:
Page 132 and 133:
Page 134 and 135:
Page 137 and 138:
Glossary of NotationsK a field of c
Page 139:
Glossary of MuPAD procedurescompute
Page 142:
[13] F.S. Macaulay, Some properties
show all

University of Paderborn Department of Mathematics Diploma Thesis ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?