Recall the Construction 2.3.17 and letT denote the set of generators of a given border basis. In this section we shall investigate box and segment border basis schemes with order ideals which have special shapes. Our main interest is to give an exact number of non-trivial polynomials of the set T for each border basis scheme. We will need the number of non-trivial defining equations of those schemes as we look into the ideal- theoretic complete intersection (see Definition 2.1.10) property of them. Throughout this section we let P denote the polynomial ring K[x1, ..., xn].
Let us start with the formal definition of a box border basis scheme. Definition 2.5.1. Let d1, ..., dn be integers > 2.
(a) The order ideal B(d1, ..., dn) = {xα11· · · xnαn ∈ Tn | αi < di f or i = 1, ..., n} is
called the box of size (d1, ..., dn). If it is clear which size we are dealing with, we
simply write B.
(b) Let I be a zero-dimensional ideal of P. AB-border basis of I is called a box border basis of I.
(c) TheB-border basis scheme is called the box border basis scheme and is denoted by BB(d1,...,dn) or simply by BB.
Lemma 2.5.2. Let B(d1, ..., dn) be the box of size (d1, ..., dn). The number of terms in
the order ideal is
The number of terms in the border is |∂B| = ν = n X i=1 ( n−1 Y j=1 dij),
where the set {di1, ..., din−1} is a subset of {d1, ..., dn}.
Proof. The first part that|B| = d1· · · dn is clear. Now we prove the second part of the
claim. Let ti ∈ B with
log(ti) = (α1, ..., αk, ..., αn).
We have xkti ∈ ∂B if and only if αk+1 = dkwith αj ∈ {0, 1, ..., dj−1} and j ∈ {1, ..., n}.
By fixing the kth-component of log(t
i) as dk− 1 and letting the other components of
this vector be one of the possible values from the set{0, 1, ..., dj − 1}, we can find the
number of border elements that are calculated by multiplying the order ideal elements with the indeterminate xk. Therefore for one indeterminate xk the number of border
elements is d1· · · dk−1dk+1· · · dn. Clearly, the set {d1, ..., dk−1, dk+1, ..., dn} has (n − 1)
elements and is a subset of{d1, ..., dn}. By summing up these multiplications for every
indeterminate, we find the desired result.
Lemma 2.5.3. LetB(d1, ..., dn) be the box of size (d1, ..., dn). The number of non-trivial
polynomials in the generating set T of I(BB) is
µ((n− 1)ν − ν′) with ν′ = m X i=1 ( n−2 Y j=1 dij),
where m = n(n− 1)/2 and {di1, ..., din−2} is a subset of {d1, ..., dn}.
Proof. Let log(tq) = (α1, ..., αk, ..., αn) ∈ Kn. As a result of the definition of generic
multiplication matrices, the entries of the generic multiplication matrices are non- trivial if and only if xktq is in ∂B or xltq ∈ ∂B. Our aim is to find the number of order
ideal elements with this property. In order to find the non-trivial entries of the matrix AkAl− AlAk, we fix the kth-component of log(tq) as dk− 1 and for j 6= k, l let αj be
from{1, ..., dj− 1}. Thus from this product of generic multiplication matrices we have
(d1· · · dk−1dk+1· · · dl· · · dn) + (dk− 1)d1· · · dk−1dk+1· · · dl−1dl+1· · · dn.
multiplication with xk or xl. We compute this for every pair xk, xl ∈ {x1, ..., xn} with
l 6= k. The number of such pairs is n(n − 1)/2. Therefore we have
(d2d3· · · dn+ (d1− 1)d3d4· · · dn) +· · · + (d1· · · dn−1+ (dn− 1)d1· · · dn−2)).
This is equal to the following.
((n− 1)(d2d3· · · dn+· · · + dk−1dk+1· · · dn))− · · · − (d3d4· · · dn+· · · + d1· · · dn−2)
= (n− 1)ν − ν′
Since τkl
pq is computed for every tp ∈ B, the result is µ((n − 1)ν − ν′).
Example 2.5.4. Consider the box order ideal B(2, 2, 2, 2). By Lemma 2.5.2, we have µ = 24 = 16 and by Lemma 2.5.2, we have ν = n(2· 2 · 2) = 32. By Lemma 2.5.3, the
number of non-trivial indeterminates is
µ((n− 1)ν − ν′) = 16((4− 1)32 − ν′)
where ν′ = m· (2 · 2) with m = 4 2
= 6. Therefore the number of non-trivial elements in the generating set of I(BB) is
µ((n− 1)ν − ν′) = 16(3· 32 − 12) = 1152.
Definition 2.5.5. Let I be a zero-dimensional ideal of P.
(a) The order ideal O = {1, x1, ..., xµ−11 } is called the segment order ideal.
(b) An O-border basis of I is called a segment border basis of I.
(c) The O-border basis scheme is called the segment border basis scheme.
Lemma 2.5.6. Let O denote the segment order ideal {1, x1, ..., xµ−11 } and let BO be a
segment border basis scheme. Then the number of terms in ∂O is (n − 1)µ + 1 and the number of non-trivial polynomials in T is µ2(n(n−1)
2 ).
Proof. The first part is a result of Definition 2.2.1. For the second claim, recall that there are n-different generic multiplication matrices with µ2 entries and one uses pairs
of them to constructT . Since the shape of the order ideal is a segment, n − 1 generic multiplication matrices have neither 0 nor 1 as entry. Therefore the matrix [Aj,Ai]
(see Equation (2.5)) for every pair Ai,Aj ∈ {A1, ...,An} does not have trivial entries.
Chapter
3
The Arrow Grading
This chapter introduces a grading (see Definition 3.1.1) on the polynomial ring where the vanishing ideal of a border basis scheme is defined. We shall call this grading the arrow grading (see Definition 3.2.1). Before looking into the properties of the arrow grading, in Section 3.1 we give a short discussion on Zm-graded polynomial rings where
m is a positive integer. Our focus point in this section is gradings defined by matrices, whose properties we recall in Definition 3.1.6.
The second section (see Section 3.2) is devoted to investigating the arrow grading in detail. First we illustrate that the arrow grading is not like any of the special gradings that we define in the first section (see Example 3.2.12). This shows that it is not a trivial work to deal with the problems we introduce in the rest of the thesis. Then we prove that the vanishing ideal of a border basis scheme is homogenous with respect to the arrow grading. Although it is a known fact (see for example Lemma 4.1 [Huib09]), we give an alternative proof to the known ones. Moreover, we remark that the torus action (see page 208 of [Hai98] and page 363 of [MilSt05]) on BO results in the arrow
grading (see Remark 3.2.8). Since the arrow grading is not of positive type, we easily show that there might exist more than one maximal homogenous ideal in BO. This
contradicts to the claims on page 363 of [MilSt05] (see Counterexamples 3.2.12 and 3.2.16).
Consequently, the main idea delivered in this chapter is to show that giving an algebraic proof of any claim on a border basis scheme is not straightforward by showing the peculiarity of the grading which is defined on the coordinate ring of a border basis scheme.
3.1
Gradings
In this section we discuss gradings and gradings defined by a matrix. Unless otherwise stated, throughout this section let K be a field. Let P denote the polynomial ring K[x1, ..., xn].
Definition 3.1.1. Let R be a ring. Let (M, +) be a monoid. The ring R is called an M-graded ring if there exists a family of additive subgroups{Rm}m∈M such that the
following properties are satisfied.
a) R =Lm∈MRm
b) RmRm′ ⊆ Rm+m′ for all m, m′ ∈ M
The elements of Rm are called the homogenous elements of degree m of R and for
every r ∈ Rmit is written deg(r) = m. If R is an M -graded ring, then for every element
r ∈ R we have r =Pm∈Mrm such that for each m∈ M we have rm ∈ Mm. For every
m ∈ M the element rm is called the homogeneous component of degree m.
Remark 3.1.2. Since 0 is included in every additive subgroup, we say that 0 ∈ R is homogenous of every degree.
Example 3.1.3. Consider the polynomial ring P = K[x1, ..., xn]. We equip P with an
Nn-grading by letting P
α denote an additive subgroup of P, so that for each polynomial
f ∈ Pα, all monomials in the support of f are of degree α∈ Nn. This grading is called
the standard grading of P.
Definition 3.1.4. Let R be a graded ring. An ideal of R is called a homogeneous ideal if it can be generated by homogenous elements.
Lemma 3.1.5. Let I be an ideal of a graded polynomial ring. The ideal I is homoge- nous if and only if the homogenous components of each polynomial f ∈ I are included in I, as well.
Proof. This follows from [KrRo00], Proposition 1.7.10.
Definition 3.1.6. Let m be a positive integer. Let P be Zm-graded polynomial ring
where K ⊆ P0 and x1, ...., xn are homogeneous elements. Let W denote a matrix from
the set Matm,n(Z). Let w1, ..., wm denote the rows of W (see Definitions 4.1.6 and 4.2.4
(a) If for every i∈ {1, ..., n} the ith-column of W has the degree of the indeterminate
xi, then W is called the degree matrix of the grading.
(b) We say P is graded by W if P is Zm-graded and the degrees of the indeterminates
are the columns of W with K⊆ P0.
(c) Let P be graded by W. The grading on P is said to be of non-negative type if there exists a tuple (α1, ..., αm) ∈ Zm such that the entries of α1w1 + ... + αmwm
which correspond to non-zero columns of W are positive.
(d) Let P be graded by W. The grading on P is called of positive type if there exists a tuple (α1, ..., αm)∈ Zm such that all entries of α1w1 + ... + αmwm are positive.
(e) Let P be graded by W. The grading on P is called non-negative if the first non-zero element in each non-zero column of W is positive.
(f) Let P be graded by W. The grading on P defined by W is called positive if no column of W is zero and the first non-zero element in each column is positive. Remark 3.1.7. If the grading given is positive (resp. non-negative) then it is of positive type (resp. of non-negative type).
Now we give some examples for gradings defined by matrices.
Example 3.1.8. Let P be graded by W = (1, ..., 1). The grading on P is a positive grading and it is also of positive type. Actually, it is equal to the standard grading on P (see Example 3.1.3 ).
Example 3.1.9. Let P denote K[x1, x2] and be graded by W = −1
0
0 −1
!
. Then the grading defined on P is of non-negative type, since for α = (α1, α2) = (−1, −1) ∈ Z2,
we have α1w1+ α2w2 = (1, 1)
Example 3.1.10. Let P denote K[x1, x2] and be graded by W = −1
1 1 −1 ! . Then we have a(−1, 1) + b(1, −1) = (−a + b, −b + a).
Since for any a, b∈ Z \ {0} the first and the second entry of this tuple (−a + b, −b + a) can not be simultaneously larger than 0, this implies that the grading on P given by W is neither of positive type nor of non-negative type.
We want to mention one more grading before ending this section. Let R be a ring and I be an ideal of it. For all n∈ N we let grn
I(R) := In/In+1 and grI0 := R/I.
Lemma 3.1.11. grI(R) =Ln∈NgrIn(R) is an N-graded ring.
Proof. Let x = a + In+1 ∈ grn
I(R) y = b + Im+1 ∈ grIm(R). Then we have
xy = ab + In+m+1 ∈ grm+nI .
Definition 3.1.12. The ring grI(R) is called graded ring of I.