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Tuesday, March 8, 2016

UTM Ideals Varieties and Algorithm - Chapter 2 Section 2 Exercise 1

Problem:


Solution:

It is a really simple sorting exercise, so let's do some warmup.

Part (a)
lex order: x3+x2+2x+3yz2+z
LM(f)=x3
LT(f)=x3
multideg(f)=(3,0,0)
(3, 0, 0)
(2, 0, 0)
(1, 0, 0)
(0, 1, 0)
(0, 0, 2)
(0, 0, 1)

grlex order: x3+x2z2+2x+3y+z
LM(f)=x3
LT(f)=x3
multideg(f)=(3,0,0)
(3, 0, 0)
(2, 0, 0)
(0, 0, 2)
(1, 0, 0)
(0, 1, 0)
(0, 0, 1)

grevlex order: x3+x2z2+2x+3y+z
LM(f)=x3
LT(f)=x3
multideg(f)=(3,0,0)
(3, 0, 0)
(2, 0, 0)
(0, 0, 2)
(1, 0, 0)
(0, 1, 0)
(0, 0, 1)

Part (b)
lex order: 3x5yz4+2x2y8xy4+xyz3
LM(f)=x5yz4
LT(f)=3x5yz4
multideg(f)=(5,1,4)
(5, 1, 4)
(2, 8, 0)
(1, 4, 0)
(1, 1, 3)

grlex order: 3x5yz4+2x2y8xy4+xyz3
LM(f)=x5yz4
LT(f)=3x5yz4
multideg(f)=(5,1,4)
(5, 1, 4)
(2, 8, 0)
(1, 4, 0)
(1, 1, 3)

grevlex order: 2x2y83x5yz4xy4+xyz3
LM(f)=x2y8
LT(f)=2x2y8
multideg(f)=(2,8,0)
(2, 8, 0)
(5, 1 ,4)
(1, 4, 0)
(1, 1, 3)

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