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# Links in Gr¨ obner fans

In document Preface (Sider 59-63)

Recall that NP(F) means the (outer) normal cone of a polyhedron P at the faceF. In the following definition we implicitly use Proposition 4.5.2.

Definition 5.7.1 Let P ⊆ Rn be a polyhedron and ω ∈ P. We define the tangent cone atω to be dual cone linkω(P) :=NP(F) where F is the face of P containing ω in its relative interior. Since this does not depend on ω but only onP.

Why we chose the weird notation linkω(P) should become clear soon.

Example 5.7.2 Let P = conv((0,0),(0,1),(1,0),(1,1)) ⊆ R2. The tangent cone at (1,1) is the negative orthant R2≤0.

Lemma 5.7.3 Let P ⊆Rn be a polyhedron and ω ∈P. Then u ∈linkω(P) if and only ifω+εu∈P for all ε >0 sufficiently small. Furthermore, for ε >0 sufficiently small

Proof. Left to the reader. 2

Example 5.7.4 LetP = conv((0,0),(0,1),(1,0),(1,1))⊆R2. Then link(9

10,1)(P) =R×R≤0 and

This is also what the lemma states forv= (1,1),u= (−1,0) and ε= 101 .

Lexicographic Lexicographic

Lexicographic

b a

c

Figure 11: The intersection of the Gr¨obner fan of the ideal of Example 4.0.22 with the triangle conv{(1,0,0),(0,1,0),(0,0,1)}. The intersection of the link at the point (3,4,2).

Definition 5.7.5 LetF be a polyhedral complex andω∈supp(F). We define the link ofF atω:

Again, the link does not depend on ω but only the face containing ω in its relative interior.

Lemma 5.7.6 The link of a polyhedral complex at a point is a polyhedral fan.

We will not prove this lemma, but rather see that it is always true for Gr¨obner fans.

Example 5.7.7 The link of a fan in a point is shown in Figure 9.

Proposition 5.7.8 [13, Proposition 1.13] Let I ⊆ k[x1, . . . , xn] be an ideal and u, v ∈ Rn. Suppose that I is homogeneous or u ∈ Rn>0. Then for ε > 0 sufficiently small

inu+εv(I) = inv(inu(I)).

Proof. Let≺be a term ordering. We claim thatu+εv∈C(≺v)u(I). Notice that (≺v)umight not be a term ordering, but by our discussion in Remark 5.5.4 this is not a problem because (≺v)uagrees with some term ordering on a homogeneous generating set for I. We will use Corollary 4.2.4 to show u+εv ∈ C(≺v)u(I).

Letg∈ G(≺v)u(I). It suffices to prove that in(≺v)

u(g) = in(≺v)

u(inu+εv(g)). But this follows from in(≺v)

u(inu+εv(g)) = in(≺v)

u(inv(inu(g))) for ε >0 sufficiently small and in(≺v)

u(inv(inu(g))) = inv(inv(inu(g))) = inv(inu(g)) = in(≺v)

u(g).

We apply Corollary 4.4.4 which says

inu+εv(I) =hinu+εv(f) :f ∈ G(≺v)u(I)i=hinv(inu(f)) :f ∈ G(≺v)u(I)i

=hinv(g) :g∈ Gv(inu(I))i= inv(inu(I)).

Here the second equality is true when ε is sufficiently small and the third is obtained by applying Corollary 4.4.4 a second time usingu ∈ C(≺v)u(I). The last equality again follows from Corollary 4.4.4 usingv∈Cv(inu(I)). 2 Example 5.7.9 The following is a reduced Gr¨obner basis for the initial ideal in(3,4,2)(I) ofI of Example 4.0.22

{c7, bc5, b2, ac6, abc3 1850

19281ac5, a2c4, a2bc2, a3c2 980

19281ac5, a3b 916

19281ac5, a4c, a5}

This was computed with Algorithm 4.4.5. The Gr¨obner fan of this ideal equals the link at the point (3,4,2) of the Gr¨obner fan of I. It is shown in Figure 11.

Corollary 5.7.10 Let I ⊆k[x1, . . . , xn] be an ideal and letu∈Rn>0. Then linku(Gfan(I)) = Gfan(inu(I)).

5.8 “Very homogeneous” ideals

Clearly, the initial ideal inω(I) is ω-homogeneous, but if ω comes from a cone in the Gr¨obner fan which is not just a ray, the ideal would me homogeneous with respect to many more vectors.

Definition 5.8.1 Let I ⊆k[x1, . . . , xn] be an ideal. We call the set{ω ∈Rn : inω(I) =I}the homogeneity space ofI.

Lemma 5.8.2 The homogeneity space of an ideal I ⊆k[x1, . . . , xn]is a linear subspace ofRn.

Proof. We wish to apply Proposition 4.4.9. We choose v = 0 ∈ Rn and ≺ to be the lexicographic term ordering. Now the proposition tells us that u ∈ Rn is in the homogeneity space of I if and only if ∀g ∈ G(I) : inu(g) = g.

This is equivalent to saying that all terms of g have the same u-degree. This translates into a set of linear condition of u that must be satisfied. Therefore the homogeneity space is a subspace ofRn. 2

Since the homogeneity space is a linear subspace, it equals its closure. There- fore C0(I) = {ω∈Rn: inω(I) = in0(I)} = {ω∈Rn: inω(I) =I} ={ω ∈ Rn : inω(I) = I}, which is exactly the homogeneity space. Therefore C0(I) is our notation for the homogeneity space ofI.

Example 5.8.3 We wish to compute the homogeneity space ofI = in(2,18,36)(J), where J is the ideal in Example 4.0.22. We compute the following reduced Gr¨obner basis for the initial ideal{c2, bc, b2+c, a3c, a9b, a18}. By the argument of the lemma, the homogeneity space is all vectors which pick the same polyno- mials as initial forms. This translates just into the condition inω(b2+c) =b2+c.

Which meansω·(0,2,0)t=ω·(0,0,1)t. Consequently the homogeneity space is the hyperplane passing through the origin with normal vector (0,2,−1).

In Definition 3.3.1 we defined the lineality space of a cone C. This is a face ofC because (Proposition 3.3.12) it is the intersection of faces ofC (every inequalityAgives rise to a face faceA(C)). By the lineality space of a fan we mean the intersection of all cones in the fan. This is the smallest cone in the fan. We notice that the lineality space of the Gr¨obner fan of an ideal I equals the homogeneity space ofI. (Because the homogeneity space is a cone in the Gr¨obner fan and has no faces by Lemma 5.8.2.)

In the following we will be interested in ideals in k[x1, . . . , xn] with n−1- dimensional homogeneity space. Fix such an idealI and call the homogeneity space L. Let g be an element of a reduced Gr¨obner basis of I. By Proposi- tion 5.3.9 we know thegmust be homogeneous in any grading given by a vector ω ∈L. Let cxα and cxβ be two terms of g we conclude that ω·α=ω·β for all ω∈L. That isα−β is in the orthogonal complementL. In other words the exponent vectors ofglie on a line, or equivalently, the Newton polytope of gis a line segment. We have proved the following lemma.

Lemma 5.8.4 Let I ⊆k[x1, . . . , xn] be an ideal with a n−1-dimensional ho- mogeneity space and≺a term ordering. The Newton polytope of anyg∈ G(I) is either a single point or a line segment. Furthermore, the line segments, asg runs through G(I), are parallel.

Fix a generatorv∈RnforL. We wish to argue that our “very homogeneous”

ideal has at most two reduced Gr¨bner bases. Let G be one reduced Gr¨obner basis of I and suppose we want to compute G(I) with respect to some term ordering ≺. Only one of two things can happen: ≺ will pick the terms with exponent in direction v or in direction −v. As we observed earlier (proof of Proposition 5.3.9) all intermediate polynomials in a run of Buchberger’s algo- rithm on G will also be homogeneous and therefore line segments (or points).

We have proved the following Proposition.

Proposition 5.8.5 Let I ⊆k[x1, . . . , xn]be an ideal with an−1-dimensional homogeneity space. ThenI has only one or two reduced Gr¨obner bases.

Example 5.8.6 The ideal hx−yi ⊆ Q[x, y] is homogeneous in the standard grading and has the reduced Gr¨obner bases {x−y}and {y−x}.

Example 5.8.7 The idealI :=hxy−1i ⊆Q[x, y] is homogeneous in the grad- ing induced by the vector (1,−1). The homogeneity space ofI is span{(1,−1)}. The ideal has only on reduced Gr¨obner basis becausexy has to be larger than

−1 in every term ordering.

LetA∈Rd×nbe a matrix whose rows form a basis of the lineality space of I. Let’s assume that the rowspace contains a positive vector. This matrix gives rise to an A-grading as in Section 5.4. Returning to our ideal I, we notice by Lemma 5.4.3 that itsA-graded Hilbert function equals that of in(I) for any

≺. Therefore, the two initial ideals ofI have the same Hilbert functions.

Our final observation in this subsection is that the Hilbert function

Figure 12: The two staircase diagrams of the initial ideals of Example 5.8.8.

Example 5.8.8 The “very homogeneous” ideal I in Example 5.8.3 has two reduced Gr¨obner bases:

{c2, bc, b2+c, a3c, a9b, a18} and

{c+b2, b3, a3b2, a9b, a18}

The corresponding staircase diagrams are shown in Figure 12. Let’s pickA= 0 1 2

1 0 0

whose rowspace is the homogeneity space of I. The monomials of A-degree

2 2

are {a2c, a2b2}. Looking at the first Gr¨obner basis w.r.t. ≺ (and the corresponding initial ideal in(I) = hc2, bc, b2, a3c, a9b, a18i) we get HAI(

2 2

) = HAin(I)( 2

2

) = 2−1 = 1 because there are two monomials of this A-degree but one is in the initial ideal. For theA-degree

4 2

we get the monomials {a2c2, a2b2a2b2c, a2b4}. But here all monomials are in the initial ideal(s) so the Hilbert function value is 3−3 = 0.

In document Preface (Sider 59-63)

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