Lindsey, Hyesu, Liz

Term Order Via Matrices

1) Explain how vectors/ matrices can be used to define term orders 2) Prove the resulting order is a term order

Definitions

In R[X], let 𝑋𝛼 = 𝑥1𝑎1𝑥2

𝑎2 ⋯𝑥𝑛𝑎𝑛 and 𝑋𝛽 = 𝑥1

𝑏1𝑥2𝑏2 ⋯𝑥𝑛

𝑏𝑛.

Then we define 𝛼 =

𝑎1𝑎2⋮𝑎𝑛

and 𝛽 =

𝑏1𝑏2⋮𝑏𝑛

. Also, define 𝑚 × 𝑛 matrix

A=

𝑢1𝑢2⋮𝑢𝑚

where 𝑢𝑗 = (𝑢𝑗1, 𝑢𝑗2,…, 𝑢𝑗𝑛) and 𝑚 ≥ 𝑛.

Now, we can define an order <𝑢 in R[𝑥1, 𝑥2, … , 𝑥𝑛- such that

𝑋𝛼 <𝑢 𝑋𝛽 ⟺ A𝛼 <𝑢 A𝛽

Example

In R x, y, z , let 𝑋𝛼=𝑥2𝑦3 and 𝑋𝛽 = 𝑥𝑦𝑧. Then,

𝛼 =230 and 𝛽 =

111 .

Let 𝑢1 = 2 3 4 , 𝑢2 = 8 1 5 , and 𝑢3 = −5 11 7 . Then,

A =2 3 48 1 5−5 11 7

.

Consider

A𝛼 =131923

and A𝛽 =91413

.

We see that 𝑨𝜷 < 𝑨𝜶. Thus, 𝑿𝜷 < 𝑿𝜶.

Review:

A Term Order on R[X] is a total order on the set R of power

products in R[X] such that

1) 1 < 𝑋𝛼 ∀ 𝑋𝛼∈ 𝑅 𝑋 , 𝑋𝛼 ≠ 1;

2) If 𝑋𝛼 < 𝑋𝛽, 𝑡ℎ𝑒𝑛 𝑋𝛼 𝑋𝛾 < 𝑋𝛽 𝑋𝛾 , ∀ 𝑋𝛾 ∈ 𝑅 𝑋 .

Key term orders

1) In R[X], lexicographic (lex) term order

2) In R[X], degree lexicographic (deglex) term order

3) In R[X], degree reverse lexicographic (degrevlex) term order

Key Term Orders

Lex Deglex Degrevlex

Given a linear order on the

variables,

a) The power product with

the most of the largest

variables is the largest

b) If tied, consider second

largest variable

c) etc…

Given a linear order on the

individual variables,

a) The power product of the

highest degree is the

largest.

b) If tied, the larger power

product is the one with

more of the largest

variable

c) If still tied, consider the

next largest variable and

so on.

d) etc…

Given a linear order on the

variables

a) The power product of the

highest degree is the

largest

b) If tied, more of the

smallest variable gives a

smaller term

c) If still tied, more of the

next smallest variable

gives smaller terms

d) etc…

Lexicographical Term Order Via Matrices

1 0 ⋯ 0 00 1 0 ⋮ ⋮ ⋮ 0⋮0

⋮0

⋱ 00⋯

10

001

𝑎1𝑎2⋮𝑎𝑛

=

𝑎1𝑎2⋮𝑎𝑛

1 0 ⋯ 0 00 1 0 ⋮ ⋮ ⋮ 0⋮0

⋮0

⋱ 00⋯

10

001

𝑏1𝑏2⋮𝑏𝑛

=

𝑏1𝑏2⋮𝑏𝑛

𝑥1𝑎1𝑥2

𝑎2⋯𝑥𝑛𝑎𝑛

𝑥1𝑏1𝑥2

𝑏2⋯𝑥𝑛𝑏𝑛

𝑥1 > 𝑥2> ⋯ > 𝑥𝑛

In R[𝑥1,𝑥2,…, 𝑥𝑛-, lex is given by the 𝑛 × 𝑛 identity matrix

applied to the vectors 𝛼 and 𝛽 :

Degree Lexicographical Term Order

1 1 ⋯ 1 11 0 0 ⋯ 00 1⋮00

0

⋮0

0 0⋱0⋯

010

⋮001

𝑎1𝑎2⋮𝑎𝑛

=

𝑎1 + 𝑎2 +⋯+ 𝑎𝑛𝑎1𝑎2⋮𝑎𝑛

1 1 ⋯ 1 11 0 0 ⋯ 00 1⋮00

0⋮0

0 0⋱0⋯

010

⋮001

𝑏1𝑏2⋮𝑏𝑛

=

𝑏1 + 𝑏2 +⋯+ 𝑏𝑛𝑏1𝑏2⋮𝑏𝑛

𝑥1𝑎1𝑥2

𝑎2⋯𝑥𝑛𝑎𝑛

𝑥1𝑏1𝑥2

𝑏2⋯𝑥𝑛𝑏𝑛

𝑥1 > 𝑥2> ⋯ > 𝑥𝑛

In R[𝑥1,𝑥2,…, 𝑥𝑛-, deglex is given by the 𝑚 × 𝑛 matrix applied to the

vectors 𝛼 and 𝛽 :

Degree Lexicographical Term Order Continued…

1 1 ⋯ 1 1 1 0 0 ⋯ 0 0 1

⋮ 0

0⋯

0 0⋱0

0 1

⋮00

𝑎1𝑎2⋮𝑎𝑛

=

𝑎1 + 𝑎2 +⋯+ 𝑎𝑛𝑎1𝑎2⋮

𝑎𝑛−1

1 1 ⋯ 1 1 1 0 0 ⋯ 0 0 1

⋮ 0

0⋯

0 0⋱0

0 1

⋮00

𝑏1𝑏2⋮𝑏𝑛

=

𝑏1 + 𝑏2 +⋯+ 𝑏𝑛𝑏1𝑏2⋮

𝑏𝑛−1

𝑥1𝑎1𝑥2

𝑎2⋯𝑥𝑛𝑎𝑛

𝑥1𝑏1𝑥2

𝑏2 ⋯𝑥𝑛𝑏𝑛

𝑥1 > 𝑥2> ⋯ > 𝑥𝑛

We can simplify this to an 𝑛 × 𝑛 matrix applied to the vectors 𝛼

and 𝛽 :

Degree Reverse Lexicographical Term Order

In R[𝑥1,𝑥2,…, 𝑥𝑛-, degrevlex is given by the 𝑚 × 𝑛 matrix applied to

the vectors 𝛼 and 𝛽 :

1 1 ⋯ 1 10 0 ⋯ 0 −1 0 ⋯ ⋮ 0−1

0−10

0 −1⋰0⋯

0⋮0

0⋮00

𝑎1𝑎2⋮𝑎𝑛

=

𝑎1 + 𝑎2 +⋯+ 𝑎𝑛−𝑎𝑛−𝑎𝑛−1

⋮−𝑎1

1 1 ⋯ 1 10 0 ⋯ 0 −1 0 ⋯ ⋮ 0−1

0−10

0 −1⋰0⋯

0⋮0

0⋮00

𝑏1𝑏2⋮𝑏𝑛

=

𝑏1 + 𝑏2 +⋯+ 𝑏𝑛−𝑏𝑛−𝑏𝑛−1

⋮−𝑏1

𝑥1𝑎1𝑥2

𝑎2⋯𝑥𝑛𝑎𝑛

𝑥1𝑏1𝑥2

𝑏2⋯𝑥𝑛𝑏𝑛

𝑥1 > 𝑥2> ⋯ > 𝑥𝑛

Degree Reverse Lexicographical Term Order Continued…

We can simplify this to an 𝑛 × 𝑛 matrix applied to the vectors 𝛼 and 𝛽 :

1 1 ⋯ 1 10 0 ⋯ 0 −10 ⋯ ⋮ 0

0−1

0 −1⋰0

0 ⋯

0⋮0

𝑎1𝑎2⋮𝑎𝑛

=

𝑎1 + 𝑎2 +⋯+ 𝑎𝑛−𝑎𝑛−𝑎𝑛−1

⋮−𝑎2

1 1 ⋯ 1 10 0 ⋯ 0 −10 ⋯ ⋮ 0

0−1

0 −1⋰0

0 ⋯

0⋮0

𝑏1𝑏2⋮𝑏𝑛

=

𝑏1 + 𝑏2 +⋯+ 𝑏𝑛−𝑏𝑛−𝑏𝑛−1

⋮−𝑏2

𝑥1𝑎1𝑥2

𝑎2⋯𝑥𝑛𝑎𝑛

𝑥1𝑏1𝑥2

𝑏2⋯𝑥𝑛𝑏𝑛

𝑥1 > 𝑥2> ⋯ > 𝑥𝑛

Properties of order <𝑢

1. Transitive relations: If 𝑋𝛼 <𝑢 𝑋𝛽 and 𝑋𝛽 <𝑢 𝑋𝛾 , then 𝑋𝛼 <𝑢

𝑋𝛾 ∀ 𝑋𝛼 , 𝑋𝛽 , and 𝑋𝛾.

2. If 𝑋𝛼 <𝑢 𝑋𝛽 then 𝑋𝛾𝑋𝛼 <𝑢 𝑋𝛾𝑋𝛽 ∀ 𝑋𝛼 , 𝑋𝛽 , and 𝑋𝛾 .

3. If the vectors 𝑢1, 𝑢2, … , 𝑢𝑚 span ℚ𝑛, then the order, <𝑢, is a total

order.

4. If the vectors 𝑢1, 𝑢2, … , 𝑢𝑚 span ℚ𝑛, then the order, <𝑢, is a term

order if and only if for all i, the first 𝑢𝑗 such that 𝑢𝑗𝑖 ≠ 0 satisfies 𝑢𝑗𝑖 > 0.

Proof of Property 2): If 𝑋𝛼 <𝑢 𝑋𝛽 then 𝑋𝛾𝑋𝛼 <𝑢 𝑋𝛾𝑋𝛽 ∀ 𝑋𝛼 , 𝑋𝛽 , and 𝑋𝛾.

Proof: Suppose 𝑋𝛼 <𝑢 𝑋𝛽. Then 𝐴𝛼 <𝑢 𝐴𝛽 . Multiply 𝑋𝛼 by 𝑋𝛾, and get

𝑋𝛾𝑋𝛼 = 𝑋𝛾+𝛼 .

Then, A(𝛾 + 𝛼 ) = 𝐴𝛾 + 𝐴𝛼 .

Since 𝐴𝛼 <𝑢 𝐴𝛽 , 𝐴𝛾 + 𝐴𝛼 <𝑢 𝐴𝛾 + 𝐴𝛽 = A 𝛾 + 𝛽 .

Hence, by the definition of the order <𝑢,

𝑋𝛾𝑋𝛼 = 𝑋𝛾+𝛼 <𝑢 𝑋𝛾+𝛽= 𝑋𝛾𝑋𝛽 .

∴ 𝑋𝛾𝑋𝛼 <𝑢 𝑋𝛾𝑋𝛽 .

∎

Proof of Property 3)

If the vectors 𝑢1, 𝑢2, … , 𝑢𝑚 span ℚ𝑛, then the order <𝑢, is a total order.

Proof: Suppose 𝑢1, 𝑢2, … , 𝑢𝑚 span ℚ𝑛. Then m ≥ 𝑛.

⟸ Assume 𝑋𝛼 = 𝑋𝛽. This implies 𝛼 = 𝛽 . Then for any 𝑚 × 𝑛 matrix

𝐴, 𝐴𝛼 = 𝐴𝛽 .

∴the order of 𝑋𝛼 = the order of 𝑋𝛽.

Definition of Total Order: The order of 𝑋𝛼 = the order 𝑋𝛽 ⟺ 𝑋𝛼 = 𝑋𝛽.

Proof of Property 3) continued…

(⟹) Suppose the order of 𝑋𝛼 = the order of 𝑋𝛽. Then 𝐴𝛼 = 𝐴𝛽 .

Case 1: If 𝑚 = 𝑛, then there exist 𝐴−1 by the Invertible Matrix Theorem. So

𝐴𝛼 = 𝐴𝛽 ⟹ 𝐴−1𝐴𝛼 = 𝐴−1𝐴𝛽 ⟹ 𝛼 = 𝛽 . Therefore, 𝑋𝛼 = 𝑋𝛽.

Case 2: If 𝑚 > 𝑛, then there exist 𝑛 × 𝑛 matrix 𝐵 = ,𝑣 1, 𝑣 2, … , 𝑣 𝑛] where

{𝑣 1, 𝑣 2, … , 𝑣 𝑛+ ⊆ *𝑢1, 𝑢2, … , 𝑢𝑚+ and whose rows are from the rows of 𝐴. Then,

𝐴𝛼 = 𝐴𝛽 ⟹ 𝐵𝛼 = 𝐵𝛽 , and since 𝐵 is an 𝑛 × 𝑛 matrix, by IMT, 𝐵−1 exists. Hence,

B𝛼 = 𝐵𝛽 ⟹ 𝐵−1 𝐵𝛼 = 𝐵−1𝐵𝛽 ⟹ 𝛼 = 𝛽 . Therefore, 𝑋𝛼 = 𝑋𝛽.

∴ the order <𝑢, is a total order.

∎

Proof of Property 4) If the vectors 𝑢1, 𝑢2, … , 𝑢𝑚 span ℚ𝑛, then the order, <𝑢, is a term order if and

only if for all i, the first 𝑢𝑗 such that 𝑢𝑗𝑖 ≠ 0 satisfies 𝑢𝑗𝑖 > 0.

Proof: Suppose 𝑢1, 𝑢2, … , 𝑢𝑚 span ℚ𝑛. Then, for each column, there exist at least one

nonzero element. Let 𝑢𝑗𝑖 be the first non-zero entry of the ith column. So 𝑢𝑗𝑖 ≠ 0.

Note that

1 = 𝑋0 ⟹ 0 =0⋮0 and 𝑋𝑒 𝑖 ⟹ 𝑒 𝑖 =

0⋮010⋮0

.

A Term Order on R[X]

0) A total order on the set R of power products in R[X] such that

1) 1 < 𝑋𝛼 ∀ 𝑋𝛼∈ 𝑅 𝑋 , 𝑋𝛼 ≠ 1;

2) If 𝑋𝛼 < 𝑋𝛽, 𝑡ℎ𝑒𝑛 𝑋𝛼 𝑋𝛾 < 𝑋𝛽 𝑋𝛾, ∀ 𝑋𝛾 ∈ 𝑅 𝑋

(Property 3 Done!)

(Property 2 Done!)

⇦ Only need to show

Proof of Property 4) continued…

Recall, A=

𝑢1𝑢2⋮𝑢𝑚

.

Note that ∀ 𝑗 = 1,2,… ,𝑚 𝑢𝑗0 = 0, and 𝑢𝑗𝑒𝑖 = 𝑢𝑗𝑖 ≠ 0.

That is, A0 =

0⋮00⋮0

and 𝐴𝑒 𝑖 =

0⋮0𝑢𝑗𝑖⋮

𝑢𝑚𝑖

Proof of Property 4) continued… (⟹) Now, suppose the order <𝑢 is a term order.

Then, 1 <𝑢 𝑋𝑒 𝑖 and so

A0 =

0⋮00⋮0

<𝑢 𝐴𝑒 𝑖 =

0⋮0𝑢𝑗𝑖⋮

𝑢𝑚𝑖

∴ 0 < 𝑢𝑗𝑖.

Proof of Property 4) continued… (⟸) Suppose 𝑢𝑗𝑖>0. Then

A0 =

0⋮00⋮0

<𝑢 𝐴𝑒 𝑖 =

0⋮0𝑢𝑗𝑖⋮

𝑢𝑚𝑖

,

which implies 1 <𝑢 𝑋𝑒 𝑖 .

By Transitivity, we know that all power products have greater order than 1.

∴ <𝑢 is a term order.

∎

Objectives Accomplished! 1) Explain how vectors/ matrices can be used to define term orders

2) Prove the resulting order is a term order

References An Introduction to Gro bner Bases, William W. Adams, Phillippe

Loustaunau, Graduate Studies in Mathematics, vol.3 (p18-p24)

Term Ordering on the Polynomial Ring, Lorenzo Robbiano,

Genova, Italy, 1985, (p513-p517)

Algorithms in Singular, Hans Schonemann, Computer Science

Journal of Moldova, vol4, no3.(12), 1996 (p315-341)

Thank you!