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Page 1: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

A gentle introduction to Elimination Theory

March 2018 @ METU

Zafeirakis Zafeirakopoulos

Page 2: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Disclaimer

I Elimination theory is a very wide area of research.

I We will see only parts of it

I through the lens of computation (polynomial system solving)

Z.Zafeirakopoulos 2

Page 3: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Disclaimer

I Elimination theory is a very wide area of research.

I We will see only parts of it

I through the lens of computation (polynomial system solving)

Z.Zafeirakopoulos 2

Page 4: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Disclaimer

I Elimination theory is a very wide area of research.

I We will see only parts of it

I through the lens of computation (polynomial system solving)

Z.Zafeirakopoulos 2

Page 5: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Intro

Membership

Z.Zafeirakopoulos 3

Page 6: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

Definition (Ideal)

Given a ring R, an ideal I ∈ R is a subset of R such that

I ∀a ∈ I, ∀c ∈ R : ca ∈ II ∀a, b ∈ I : a + b ∈ I

Definition (Ideal Membership)

Input a ring R, an ideal I ⊆ R and an element f ∈ ROutput True if f ∈ I, False otherwise

Z.Zafeirakopoulos 4

Page 7: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

Definition (Ideal)

Given a ring R, an ideal I ∈ R is a subset of R such that

I ∀a ∈ I, ∀c ∈ R : ca ∈ II ∀a, b ∈ I : a + b ∈ I

Definition (Ideal Membership)

Input a ring R, an ideal I ⊆ R and an element f ∈ ROutput True if f ∈ I, False otherwise

Z.Zafeirakopoulos 4

Page 8: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership in Euclidean domains is easy

I Division gives unique remainder

I By division we obtain a linear combination f = r +∑gi∈I

qigi

I f ∈ I if and only if r = 0

Note R[x ] is a Euclidean domain

Z.Zafeirakopoulos 4

Page 9: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership in Euclidean domains is easy

I Division gives unique remainder

I By division we obtain a linear combination f = r +∑gi∈I

qigi

I f ∈ I if and only if r = 0

Note R[x ] is a Euclidean domain

Z.Zafeirakopoulos 4

Page 10: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership in Euclidean domains is easy

I Division gives unique remainder

I By division we obtain a linear combination f = r +∑gi∈I

qigi

I f ∈ I if and only if r = 0

Note R[x ] is a Euclidean domain

Z.Zafeirakopoulos 4

Page 11: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership in Euclidean domains is easy

I Division gives unique remainder

I By division we obtain a linear combination f = r +∑gi∈I

qigi

I f ∈ I if and only if r = 0

Note R[x ] is a Euclidean domain

Z.Zafeirakopoulos 4

Page 12: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

A(t) =

1 1 1

t 2t 2

t + 1 0 2t

←−

−t

+

←−−−−−−

−(t+1)

+

1 1 1

0 t 2− t

0 −t − 1 t − 1

←−+ | · (−1)

1 1 1

0 t 2− t

0 1 −1

←−←− ←−

−t

+ | : 2

1 1 10 1 −10 0 1

.

def Gauss(M):

for col in range(len(M[0])):

for row in range(col+1, len(M)):

r = [(rowValue * (-(M[row][col] / M[col][col]))) for rowValue in M[col]]

M[row] = [sum(pair) for pair in zip(M[row], r)]

Z.Zafeirakopoulos 4

Page 13: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

Emmy Nother1920s

NoteIf we want to manipulate ideals, we have tobe able to decide membership.If we can, then we can also

I decide equality of ideals

I Arithmetic of ideals

Z.Zafeirakopoulos 4

Page 14: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

Grete Herman1940s

I Proved that a bound to decide membership would be doublyexponential in the degree.

I The linear combination has huge coefficients.

I Indication that Grobner bases have bad complexity.

Z.Zafeirakopoulos 4

Page 15: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

Grete Herman1940s

I Proved that a bound to decide membership would be doublyexponential in the degree.

I The linear combination has huge coefficients.

I Indication that Grobner bases have bad complexity.

Z.Zafeirakopoulos 4

Page 16: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

Wolfgang Grobner1940s

I Worked with Nother

I Several contributions

I Did not invent the bases bearing his name

Z.Zafeirakopoulos 4

Page 17: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership is hard because remainder is not unique

I For some sets of divisors, remainder is unique

I Every ideal (in a Notherian ring) has such a set of generators.

I Buchberger proved it.

I What is even better, he proved it constructively.

Z.Zafeirakopoulos 4

Page 18: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership is hard because remainder is not unique

I For some sets of divisors, remainder is unique

I Every ideal (in a Notherian ring) has such a set of generators.

I Buchberger proved it.

I What is even better, he proved it constructively.

Z.Zafeirakopoulos 4

Page 19: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership is hard because remainder is not unique

I For some sets of divisors, remainder is unique

I Every ideal (in a Notherian ring) has such a set of generators.

I Buchberger proved it.

I What is even better, he proved it constructively.

Z.Zafeirakopoulos 4

Page 20: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership is hard because remainder is not unique

I For some sets of divisors, remainder is unique

I Every ideal (in a Notherian ring) has such a set of generators.

I Buchberger proved it.

I What is even better, he proved it constructively.

Z.Zafeirakopoulos 4

Page 21: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

I Membership is hard because remainder is not unique

I For some sets of divisors, remainder is unique

I Every ideal (in a Notherian ring) has such a set of generators.

I Buchberger proved it.

I What is even better, he proved it constructively.

Z.Zafeirakopoulos 4

Page 22: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Membership - A tale of computation

def Groebner(ideal):

updated = True

while updated:

updated = False

for f in ideal:

for g in ideal:

r = S_polynomial(f,g).divide(ideal)

if not r.is_zero():

ideal.append(r)

updated = True

if updated: break

if updated: break

return ideal

I Bruno Buchberger was a student of Grobner

Thesis An Algorithm for Finding the Basis Elements of the ResidueClass Ring Modulo a Zerodimensional Polynomial Ideal

Z.Zafeirakopoulos 4

Page 23: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Monomial (Order)

Definition (Term Monoid)

Given a set of variables x1, x2, . . . , xd we consider the multiplicativemonoid T =

{xα11 xα2

2 · · · xαdd : α ∈ Nd

}.

Definition (Term order)

Let ≤ be a total order on T. It is called a term order if

I 0 ≤ T for all T ∈ T and

I if a ≤ b then ac ≤ bc for all a, b, c ∈ T.

Z.Zafeirakopoulos 5

Page 24: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Monomial (Order)

Definition (Term Monoid)

Given a set of variables x1, x2, . . . , xd we consider the multiplicativemonoid T =

{xα11 xα2

2 · · · xαdd : α ∈ Nd

}.

Note that there is a monoid homomorphism between T and Nd

Definition (Term order)

Let ≤ be a total order on T. It is called a term order if

I 0 ≤ T for all T ∈ T and

I if a ≤ b then ac ≤ bc for all a, b, c ∈ T.

Z.Zafeirakopoulos 5

Page 25: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Monomial (Order)

Definition (Term Monoid)

Given a set of variables x1, x2, . . . , xd we consider the multiplicativemonoid T =

{xα11 xα2

2 · · · xαdd : α ∈ Nd

}.

Note that there is a monoid homomorphism between T and Nd

Definition (Term order)

Let ≤ be a total order on T. It is called a term order if

I 0 ≤ T for all T ∈ T and

I if a ≤ b then ac ≤ bc for all a, b, c ∈ T.

Z.Zafeirakopoulos 5

Page 26: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Monomial (Order)

Definition (Term order)

Let ≤ be a total order on T. It is called a term order if

I 0 ≤ T for all T ∈ T and

I if a ≤ b then ac ≤ bc for all a, b, c ∈ T.

Example (Lexicographic vs DegRevLex)

Fix x1 ≤ x2 ≤ · · · ≤ xd .

I xα11 xα2

2 · · · xαdd ≤lex xβ11 xβ22 · · · x

βdd if the left-most non-zero

entry in β − α is positive.

I xα11 xα2

2 · · · xαdd ≤drl x

β11 xβ22 · · · x

βdd if

I∑αi ≤

∑βi or

I∑αi =

∑βi and the right-most non-zero entry in β − α is

positive.

Z.Zafeirakopoulos 5

Page 27: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Monomial (Order)

Definition (Term order)

Let ≤ be a total order on T. It is called a term order if

I 0 ≤ T for all T ∈ T and

I if a ≤ b then ac ≤ bc for all a, b, c ∈ T.

A term order induces an order on (monomials and thus on)polynomials in K[x1, . . . , xd ].

Z.Zafeirakopoulos 5

Page 28: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Grobner Bases

Fix a term order.

DefinitionGiven an ideal I = 〈f1, f2, . . . , fn〉, a Grobner basis for I , withrespect to the term order, is a set G = {g1, g2, . . . , gm} such thatI = 〈G 〉 and for every 0 6= f ∈ I we have that lt (gi ) | lt (f ) forsome i ∈ [m].

I This is not the only definition.

I Other definitions will appear during this series.

Z.Zafeirakopoulos 6

Page 29: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Grobner Bases

Fix a term order.

DefinitionGiven an ideal I = 〈f1, f2, . . . , fn〉, a Grobner basis for I , withrespect to the term order, is a set G = {g1, g2, . . . , gm} such thatI = 〈G 〉 and for every 0 6= f ∈ I we have that lt (gi ) | lt (f ) forsome i ∈ [m].

I This is not the only definition.

I Other definitions will appear during this series.

Z.Zafeirakopoulos 6

Page 30: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Grobner Bases

Fix a term order.

DefinitionGiven an ideal I = 〈f1, f2, . . . , fn〉, a Grobner basis for I , withrespect to the term order, is a set G = {g1, g2, . . . , gm} such thatI = 〈G 〉 and for every 0 6= f ∈ I we have that lt (gi ) | lt (f ) forsome i ∈ [m].

I This is not the only definition.

I Other definitions will appear during this series.

A more important property:

reduction by G in K[x1, . . . , xd ] is unique.

Z.Zafeirakopoulos 6

Page 31: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Grobner Bases

Fix a term order.

DefinitionGiven an ideal I = 〈f1, f2, . . . , fn〉, a Grobner basis for I , withrespect to the term order, is a set G = {g1, g2, . . . , gm} such thatI = 〈G 〉 and for every 0 6= f ∈ I we have that lt (gi ) | lt (f ) forsome i ∈ [m].

I This is not the only definition.

I Other definitions will appear during this series.

A more important property:

reduction by G in K[x1, . . . , xd ] is unique.

Reduce means to take the remainder after we divide as much aspossible with elements of G .Z.Zafeirakopoulos 6

Page 32: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

A criterion

Definition (S-polynomial)

Fix a term order and let f , g ∈ K[x1, . . . , xd ]. The S-polynomial off and g is

Sf ,g =lcm (lt(f ), lt(g))

lt(f )f − lcm (lt(f ), lt(g))

lt(g)g

Theorem (Buchberger)

A (finite) set G is a Grobner basis of 〈G 〉 if and only if Sf ,g isreduced to 0 by G for all pairs f , g ∈ G .

Z.Zafeirakopoulos 7

Page 33: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

A criterion

Definition (S-polynomial)

Fix a term order and let f , g ∈ K[x1, . . . , xd ]. The S-polynomial off and g is

Sf ,g =lcm (lt(f ), lt(g))

lt(f )f − lcm (lt(f ), lt(g))

lt(g)g

Theorem (Buchberger)

A (finite) set G is a Grobner basis of 〈G 〉 if and only if Sf ,g isreduced to 0 by G for all pairs f , g ∈ G .

Z.Zafeirakopoulos 7

Page 34: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Example

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f1, f2〉.

Then

G ={x2 − 2y , y2

}Sx2+(y−1)2−1,y2 = y4 − 2y3 = y2(−y2 + 2y)→y2 0

We interreduce elements of the GB to obtain x2 − 2y and y2.We tend to say that a GB is a ”nice” choice of a generators.

Z.Zafeirakopoulos 8

Page 35: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Example

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f1, f2〉. Then

G ={x2 − 2y , y2

}

Sx2+(y−1)2−1,y2 = y4 − 2y3 = y2(−y2 + 2y)→y2 0

We interreduce elements of the GB to obtain x2 − 2y and y2.We tend to say that a GB is a ”nice” choice of a generators.

Z.Zafeirakopoulos 8

Page 36: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Example

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f1, f2〉. Then

G ={x2 − 2y , y2

}Sx2+(y−1)2−1,y2 = y4 − 2y3 = y2(−y2 + 2y)→y2 0

We interreduce elements of the GB to obtain x2 − 2y and y2.We tend to say that a GB is a ”nice” choice of a generators.

Z.Zafeirakopoulos 8

Page 37: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Example

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f1, f2〉. Then

G ={x2 − 2y , y2

}Sx2+(y−1)2−1,y2 = y4 − 2y3 = y2(−y2 + 2y)→y2 0

We interreduce elements of the GB to obtain x2 − 2y and y2.

We tend to say that a GB is a ”nice” choice of a generators.

Z.Zafeirakopoulos 8

Page 38: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Example

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f1, f2〉. Then

G ={x2 − 2y , y2

}Sx2+(y−1)2−1,y2 = y4 − 2y3 = y2(−y2 + 2y)→y2 0

We interreduce elements of the GB to obtain x2 − 2y and y2.We tend to say that a GB is a ”nice” choice of a generators.

Z.Zafeirakopoulos 8

Page 39: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Elimination Ideal

Z.Zafeirakopoulos 9

Page 40: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Elimination ideal

DefinitionLet I ⊆ K[x1, . . . , xd ] be an ideal. Then we define the i-thelimination ideal of I as

Ii = I ∩K[xi+1, . . . , xd ].

Theorem (Elimination Property of Grobner Bases)

Let k ∈ [d ] and fix a lexicographic order such that xi ≤ xj for alli < k and k < j . If G is a Grobner basis of I (for the term orderwe fixed), then

Ik = 〈G ∩K[xk+1, . . . , xd ]〉 .

Z.Zafeirakopoulos 10

Page 41: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Elimination ideal

DefinitionLet I ⊆ K[x1, . . . , xd ] be an ideal. Then we define the i-thelimination ideal of I as

Ii = I ∩K[xi+1, . . . , xd ].

Theorem (Elimination Property of Grobner Bases)

Let k ∈ [d ] and fix a lexicographic order such that xi ≤ xj for alli < k and k < j . If G is a Grobner basis of I (for the term orderwe fixed), then

Ik = 〈G ∩K[xk+1, . . . , xd ]〉 .

Z.Zafeirakopoulos 10

Page 42: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Elimination ideal

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f 1, f2〉.

We saw that a GB for I is{x2 − 2y , y2

}. Thus

Ix ={x2 − 2y , y2

}∩K[y ] =

⟨y2⟩

Z.Zafeirakopoulos 11

Page 43: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Elimination ideal

Let

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

and I = 〈f 1, f2〉.

We saw that a GB for I is{x2 − 2y , y2

}. Thus

Ix ={x2 − 2y , y2

}∩K[y ] =

⟨y2⟩

Z.Zafeirakopoulos 11

Page 44: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Variety and Vanishing Ideal

Let K be an algebraically closed field.

Definition (Variety)

Let I be an ideal in K[x1, . . . , xd ]. Then

V (I ) ={x ∈ Kd : f (x) = 0 for all f ∈ I

}

Definition (Vanishing)

Let V ⊆ Kd be a variety. Then

I (V ) = {f ∈ K[x1, . . . , xd ] : f (x) = 0 for all x ∈ V }

Z.Zafeirakopoulos 12

Page 45: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

0-dim

I What does it mean for the GB that the variety is 0-dim?

I What does it mean for the elimination ideal?

I What does it mean for solving?

I What does it remind us?

Z.Zafeirakopoulos 13

Page 46: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

0-dim

I What does it mean for the GB that the variety is 0-dim?

I What does it mean for the elimination ideal?

I What does it mean for solving?

I What does it remind us?

Z.Zafeirakopoulos 13

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0-dim

I What does it mean for the GB that the variety is 0-dim?

I What does it mean for the elimination ideal?

I What does it mean for solving?

I What does it remind us?

Z.Zafeirakopoulos 13

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0-dim

I What does it mean for the GB that the variety is 0-dim?

I What does it mean for the elimination ideal?

I What does it mean for solving?

I What does it remind us?

Z.Zafeirakopoulos 13

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0-dim

I What does it mean for the GB that the variety is 0-dim?

I What does it mean for the elimination ideal?

I What does it mean for solving?

I What does it remind us?

Z.Zafeirakopoulos 13

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Variety of the Elimination Ideal

For f1, . . . , fn ∈ K[x1, . . . , xd ], we write fi in the form

fi = hi (x2, . . . , xd)xNi1 + terms of x1-degree less than Ni ,

for each 1 ≤ i ≤ n. Consider the projection π : Kn → Kn−1:

π((c1, c2, . . . , cn)

)= (c2, c3, . . . , cn) .

Theorem (Elimination Theorem)

Let I1 be the first elimination ideal of an ideal I E K[x1, . . . , xn].Then

V (I1) = π(V (I )

)∪(V (h1, . . . , hm) ∩ V (I1)

).

Z.Zafeirakopoulos 14

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (a0 + 0a) + (a0− 0a) + (−0a− a0) + (0b − b0)

AFT (2) = a2 − a2 + 02 − 02

Z.Zafeirakopoulos 15

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→ x1

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (ax1 + x1a) + (a0− 0a) + (−0a− a0) + (0b − b0)

AFT (2) = a2 − a2 + 02 − 02

Z.Zafeirakopoulos 15

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→ x1, x2

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (ax1 + x1a) + (ax2 − x2a) + (−0a− a0) + (0b − b0)

AFT (2) = a2 − a2 + 02 − 02

Z.Zafeirakopoulos 15

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→ x1, x2, x3

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (ax1 + x1a) + (ax2 − x2a) + (−x3a− a0) + (0b − b0)

AFT (2) = a2 − a2 + x30− 02

Z.Zafeirakopoulos 15

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→ x1, x2, x3, x4

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (ax1 + x1a) + (ax2 − x2a) + (−x3a− ax4) + (0b − b0)

AFT (2) = a2 − a2 + x3x4 − 02

Z.Zafeirakopoulos 15

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→ x1, x2, x3, x4, x5

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (ax1 + x1a) + (ax2 − x2a) + (−x3a− ax4) + (x5b − b0)

AFT (2) = a2 − a2 + x3x4 − x50

Z.Zafeirakopoulos 15

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Modeling complementary sequences

T = (a, 0, a)|(a, 0,−a)|(0,−a, 0)|(0, b, 0) 0→ x1, x2, x3, x4, x5, x6

T = (a, x1, a)|(a, x2,−a)|(x3,−a, x4)|(x5, b, − x6)

AFT (1) = (ax1 + x1a) + (ax2 − x2a) + (−x3a− ax4) + (x5b − bx6)

AFT (2) = a2 − a2 + x3x4 − x5x6

Z.Zafeirakopoulos 15

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Modeling complementary sequences

R.<a,b,x1,x2,x3,x4,x5,x6> = PolynomialRing(QQ,order="lex")

f = 2*a*x1 - a * x3 - a*x4+b*x5-b*x6

g= x3*x4-x5*x6

S=[f,g, x1^3-x1,x2^3-x2,x3^3-x3,x4^3-x4,x5^3-x5,x6^3-x6]

I = R*S

I.groebner_basis()abx5−abx6+

43b2x21 x3x

25 +

43b2x21 x3x

26 +

43b2x21 x4x

25−

163b2x21 x4x5x6+

43b2x21 x4x

26 +

16b2x1x

23 x

25 +

16b2x1x

23 x

26 +

16b2x1x

24 x

25 + 1

6b2x1x

24 x

26 −

23b2x1x

25 x

26 + 1

2b2x1x

25 − b2x1x5x6 + 1

2b2x1x

26 − b2x3x

25 − b2x3x

26 −

23b2x4x

25 x

26 −

b2x4x25+

143b2x4x5x6−b2x4x

26 , ax1−

12ax4x5x6− 1

2ax4− 2

3bx21 x

24 x5+

23bx21 x

24 x6+

23bx21 x5−

23bx21 x6+

13bx1x3x5−

13bx1x3x6− 1

3bx1x4x

25 x6 +

13bx1x4x5x

26 + 1

6bx23 x5−

16bx23 x6 +

23bx24 x5−

23bx24 x6 +

16bx25 x6−

16bx5x

26 −

16bx5 +

16bx6, ax3 − ax4x5x6 − 4

3bx21 x

24 x5 + 4

3bx21 x

24 x6 + 4

3bx21 x5 −

43bx21 x6 + 2

3bx1x3x5 − 2

3bx1x3x6 − 2

3bx1x4x

25 x6 +

23bx1x4x5x

26+

13bx23 x5−

13bx23 x6+

43bx24 x5−

43bx24 x6+

13bx25 x6−

13bx5x

26−

43bx5+

43bx6, ax

24−ax25 x

26+

43bx21 x4x5−

43bx21 x4x6 +

23bx1x

24 x5−

23bx1x

24 x6 +

23bx1x

25 x6−

23bx1x5x

26 + 1

3bx4x

25 x6−

13bx4x5x

26 − bx4x5 + bx4x6, ax4x

25 −

ax4+43bx21 x

24 x6−

43bx21 x5x

26 −

23bx1x4x

25 x6+

23bx1x4x6−bx24 x6−

13bx25 x6+

43bx5x

26 , ax4x

26 −ax4− 4

3bx21 x

24 x5+

43bx21 x5x

26 + 2

3bx1x4x5x

26 −

23bx1x4x5+bx24 x5−bx5x

26 , bx

21 x

23 x5+bx21 x

24 x5−bx21 x5x

26 −bx21 x5−bx23 x5−bx24 x5+

bx5x26 + bx5, bx

21 x

23 x6 + bx21 x

24 x6 − bx21 x5x

26 − bx21 x6 − bx23 x6 − bx24 x6 + bx5x

26 + bx6, bx

21 x

25 x6 − bx21 x5x

26 −

bx25 x6+bx5x26 , x

31 −x1, x

32 −x2, x

33 −x3, x3x4−x5x6, x3x5x6−x4x

25 x

26 , x

34 −x4, x

24 x5x6−x5x6, x

35 −x5, x

36 −x6

Z.Zafeirakopoulos 16

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Modeling complementary sequences

I The ideal I is 2-dim.

I The elimination ideal is 0-dim

I We eliminated the parameters

I That’s good because we want the equations to hold for allvalues of the parameters

Live demo?

Z.Zafeirakopoulos 17

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Modeling complementary sequences

I The ideal I is 2-dim.

I The elimination ideal is 0-dim

I We eliminated the parameters

I That’s good because we want the equations to hold for allvalues of the parameters

Live demo?

Z.Zafeirakopoulos 17

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Modeling complementary sequences

I The ideal I is 2-dim.

I The elimination ideal is 0-dim

I We eliminated the parameters

I That’s good because we want the equations to hold for allvalues of the parameters

Live demo?

Z.Zafeirakopoulos 17

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Resultants

Z.Zafeirakopoulos 18

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Common factors

Let f1, f2 ∈ K[x ]. Then f1 and f2 have a common factor if and onlyif there are polynomials A, B ∈ K[x ] such that:

I A and B are not both zero.

I deg(A) ≤ deg(f2)− 1 and deg(B) ≤ deg(f2)− 1

I Af1 + Bf2 = 0

Now, if we expand Af1 + Bf2 and force all coefficients to be 0, weget a linear system.

Z.Zafeirakopoulos 19

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Common factors

Let f1, f2 ∈ K[x ]. Then f1 and f2 have a common factor if and onlyif there are polynomials A, B ∈ K[x ] such that:

I A and B are not both zero.

I deg(A) ≤ deg(f2)− 1 and deg(B) ≤ deg(f2)− 1

I Af1 + Bf2 = 0

Now, if we expand Af1 + Bf2 and force all coefficients to be 0, weget a linear system.

Z.Zafeirakopoulos 19

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Sylvester Resultant

Syl(f1, f2) =

f1,d1 · · · · · · · · · · · · f1,0. . .

. . .

f1,d1 · · · · · · · · · · · · f1,0f2,d2 · · · · · · f2,0

. . .. . .

. . .. . .

f2,d2 · · · · · · · · · f2,0

d2 d1

DefinitionThe resultant resx (f1, f2) is the determinant of Syl (f1, f2).

Z.Zafeirakopoulos 20

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Sylvester Resultant

TheoremIf f , g ∈ K[x ] then the resultant Res(f , g , x) ∈ K[x ] is an integerpolynomial in the coefficients of f and g .

Theorem

1 6= gcd(f , g) ∈ K[x ]⇔ Res(f , g , x) = 0

Z.Zafeirakopoulos 21

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Sylvester Resultant

TheoremIf f , g ∈ K[x ] then the resultant Res(f , g , x) ∈ K[x ] is an integerpolynomial in the coefficients of f and g .

Theorem

1 6= gcd(f , g) ∈ K[x ]⇔ Res(f , g , x) = 0

Z.Zafeirakopoulos 21

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Resultants and Elimination ideals

TheoremLet f , g ∈ K[x1, . . . , xd ] and c = (c2, . . . , cd) ∈ Cd−1 satisfy thefollowing:

I f (x1, c) ∈ C[x1] has degree deg(f )

I g(x1, c) ∈ C[x1] has degree p ≤ deg(g)

Then

Res(f , g , x1)(c) = lt(f )(c)deg(g)−pRes (f (x1, c), g(x1, c), x1)

Z.Zafeirakopoulos 22

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Resultants and Elimination ideals

Theorem (ExtensionTheorem)

Let I = 〈f1, f2, . . . , fn〉 ⊆ C[x1, x2, . . . , xd ] and let I1 be the firstelimination ideal of I . We write fi in the form

fi = hi (x2, . . . , xd)xNi1 + terms of x1 − degree less than Ni ,

and gi ∈ C[x2, . . . , xd ] is not zero.If (c2, . . . , cd) ∈ V (I1) and (c2, . . . , cd) 6∈ V (h1, h2, . . . , hn) thenthere exist c1 ∈ C such that (c1, c2, . . . , cd) ∈ V (I )

Z.Zafeirakopoulos 23

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Elimination ideal vs Resultant

TheoremLet I = 〈f1, f2〉 ∈ K[x1, . . . , xn] and R = resx1 (f1, f2). Then

V (R) = V (h1, h2) ∪ π(V (I )

).

TheoremIf f1, f2 ∈ K[x , y ] and R = resx (f1, f2) is not identically zero, then

V (I1) = π(V (I )

).

Z.Zafeirakopoulos 24

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Elimination ideal vs Resultant

TheoremLet I = 〈f1, f2〉 ∈ K[x1, . . . , xn] and R = resx1 (f1, f2). Then

V (R) = V (h1, h2) ∪ π(V (I )

).

TheoremIf f1, f2 ∈ K[x , y ] and R = resx (f1, f2) is not identically zero, then

V (I1) = π(V (I )

).

Z.Zafeirakopoulos 24

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Resultant System

DefinitionLet f1, ..., fn ∈ K[x1, . . . , xd ] and introduce n new variables ui .Consider the resultant

Ri = Resx1(fi ,∑i 6=j

uj · fj).

The resultant system RSx1(f1, ..., fn) is the set of coefficients of Ri

seen as a polynomial in variables u1, ..., un.

Z.Zafeirakopoulos 25

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Implicitization

Z.Zafeirakopoulos 26

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Implicitization

Given parameterization

x0 = α0(t), . . . , xn = αn(t), t := (t1, . . . , tn),

compute the smallest algebraic variety containing the closure ofthe image of

α : Rn → Rn+1 : t 7→ α(t), α := (α0, . . . , αn).

This is contained in the variety defined by the ideal

〈p(x0, . . . , xn) | p(α0(t), . . . , αn(t)) = 0, ∀t〉.

When this is a principal ideal we wish to compute its definingpolynomial p(x).Z.Zafeirakopoulos 27

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Implicitization

Example (Folium of Descartes)

x =3t2

t3 + 1, u =

3t

t3 + 1

p(x , y) = x3 − 3xy + y3

Z.Zafeirakopoulos 28

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Implicitization

Example (Folium of Descartes)

x =3t2

t3 + 1, u =

3t

t3 + 1

p(x , y) = x3 − 3xy + y3

Z.Zafeirakopoulos 28

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Number of roots

I Roots of the resultant are projections of roots.

I Bezout bound:∏

i diI Is it tight?

Z.Zafeirakopoulos 29

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Number of roots

I Roots of the resultant are projections of roots.

I Bezout bound:∏

i di

I Is it tight?

Z.Zafeirakopoulos 29

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Number of roots

I Roots of the resultant are projections of roots.

I Bezout bound:∏

i diI Is it tight?

Z.Zafeirakopoulos 29

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Newton Polytope

DefinitionGiven a polynomial

f =∑α∈Nd

cαxα11 xα2

2 · · · xαdd ∈ K[x1, x2, . . . , xd ],

the support of f is Sup(f ) ={α ∈ Nd : cα 6= 0

}and its Newton

polytope is the convex hull of its support NP (f ) = CH {Sup(f )}.

Example

f = x3y − 3x2 + 2xy2 + 21xy − ySup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)}

f = x2 − 3y2 + 2xy + 2x − y + 1Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)}

x

y

Z.Zafeirakopoulos 30

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Newton Polytope

DefinitionGiven a polynomial

f =∑α∈Nd

cαxα11 xα2

2 · · · xαdd ∈ K[x1, x2, . . . , xd ],

the support of f is Sup(f ) ={α ∈ Nd : cα 6= 0

}and its Newton

polytope is the convex hull of its support NP (f ) = CH {Sup(f )}.

Example

f = x3y − 3x2 + 2xy2 + 21xy − ySup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)}

f = x2 − 3y2 + 2xy + 2x − y + 1Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)}

x

y

Z.Zafeirakopoulos 30

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Newton Polytope

DefinitionGiven a polynomial

f =∑α∈Nd

cαxα11 xα2

2 · · · xαdd ∈ K[x1, x2, . . . , xd ],

the support of f is Sup(f ) ={α ∈ Nd : cα 6= 0

}and its Newton

polytope is the convex hull of its support NP (f ) = CH {Sup(f )}.

Example

f = x3y − 3x2 + 2xy2 + 21xy − ySup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)}

f = x2 − 3y2 + 2xy + 2x − y + 1Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)}

x

y

Z.Zafeirakopoulos 30

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Newton Polytope

DefinitionGiven a polynomial

f =∑α∈Nd

cαxα11 xα2

2 · · · xαdd ∈ K[x1, x2, . . . , xd ],

the support of f is Sup(f ) ={α ∈ Nd : cα 6= 0

}and its Newton

polytope is the convex hull of its support NP (f ) = CH {Sup(f )}.

Example

f = x3y − 3x2 + 2xy2 + 21xy − ySup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)}

f = x2 − 3y2 + 2xy + 2x − y + 1Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)}

x

y

Z.Zafeirakopoulos 30

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Newton Polytope

DefinitionGiven a polynomial

f =∑α∈Nd

cαxα11 xα2

2 · · · xαdd ∈ K[x1, x2, . . . , xd ],

the support of f is Sup(f ) ={α ∈ Nd : cα 6= 0

}and its Newton

polytope is the convex hull of its support NP (f ) = CH {Sup(f )}.

Example

f = x3y − 3x2 + 2xy2 + 21xy − ySup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)}

f = x2 − 3y2 + 2xy + 2x − y + 1Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)}

x

y

Z.Zafeirakopoulos 30

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Mixed Volume

Let P1,P2, . . . ,Pk ⊆ Rd be polytopes and λ1, λ2, . . . , λkR≥0.

Theorem (Minkowski)

Then there exist Vα1,α2,...,αk≥ 0, such that

Vol (λ1P1 ⊕ λ2P2 ⊕ · · · ⊕ λkPk)

=∑

α1+α2+···+αk=d

(d

α1, α2, . . . , αk

)Vα1,α2,...,αk

λα11 λ

α22 · · ·λ

αkk

DefinitionThe mixed volume MV (P1,P2, . . . ,Pd) is the coefficient ofλ1λ2 . . . λd in Vol (λ1P1 ⊕ λ2P2 ⊕ · · · ⊕ λdPd).

Z.Zafeirakopoulos 31

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The BKK bound

Theorem (Bernstein, Khovanskii, Kushnirenko)

Let f1, f2, . . . , fd ∈ C[x1, x2, . . . , xd ].

Then the number of isolatedsolutions to the polynomial system f1(x) = · · · = fd(x) = 0 with(x1, x2, . . . , xd) ∈ (C− {0})d is (counting multiplicities) boundedby the mixed volume of the Newton polytopes of f1, f2, . . . , fd .

Z.Zafeirakopoulos 32

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The BKK bound

Theorem (Bernstein, Khovanskii, Kushnirenko)

Let f1, f2, . . . , fd ∈ C[x1, x2, . . . , xd ]. Then the number of isolatedsolutions to the polynomial system f1(x) = · · · = fd(x) = 0 with(x1, x2, . . . , xd) ∈ (C− {0})d

is (counting multiplicities) boundedby the mixed volume of the Newton polytopes of f1, f2, . . . , fd .

Z.Zafeirakopoulos 32

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The BKK bound

Theorem (Bernstein, Khovanskii, Kushnirenko)

Let f1, f2, . . . , fd ∈ C[x1, x2, . . . , xd ]. Then the number of isolatedsolutions to the polynomial system f1(x) = · · · = fd(x) = 0 with(x1, x2, . . . , xd) ∈ (C− {0})d is (counting multiplicities)

boundedby the mixed volume of the Newton polytopes of f1, f2, . . . , fd .

Z.Zafeirakopoulos 32

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The BKK bound

Theorem (Bernstein, Khovanskii, Kushnirenko)

Let f1, f2, . . . , fd ∈ C[x1, x2, . . . , xd ]. Then the number of isolatedsolutions to the polynomial system f1(x) = · · · = fd(x) = 0 with(x1, x2, . . . , xd) ∈ (C− {0})d is (counting multiplicities) boundedby the mixed volume of the Newton polytopes of f1, f2, . . . , fd .

Z.Zafeirakopoulos 32

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The BKK bound

f1 = 1 + αx + βy2

f2 = x + γy4

Bezout bound:

deg(f1) deg(f2) = 8

t s+t

2s

4t

2s+4t

V (sNP(f1)⊕ tNP(f2)) = s2 +(21

)2st

MV (NP(f1),NP(f2)) = 2! 2 = 4

Z.Zafeirakopoulos 33

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The BKK bound

f1 = 1 + αx + βy2

f2 = x + γy4

Bezout bound:

deg(f1) deg(f2) = 8

t s+t

2s

4t

2s+4t

V (sNP(f1)⊕ tNP(f2)) = s2 +(21

)2st

MV (NP(f1),NP(f2)) = 2! 2 = 4

Z.Zafeirakopoulos 33

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BKK

I Does this imply something for the resultant?

I Can we have a resultant for these (toric) roots?

Z.Zafeirakopoulos 34

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Because regularity is boring

Multiplicities

Z.Zafeirakopoulos 35

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 36

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 36

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 36

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 36

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

⟩I # = 4

I #{1, ∂y} = 2

Z.Zafeirakopoulos 36

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

⟩I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 36

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and an Algebraic Problem

Given an ideal I , find a basis for R /I

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

I I = 〈f1, f2〉

I V (I ) = ζ = (0, 0)

I µ(ζ) := dimK R /I

I GB gives us a basis for R /I

Z.Zafeirakopoulos 37

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and an Algebraic Problem

Given an ideal I , find a basis for R /I

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

I I = 〈f1, f2〉I V (I ) = ζ = (0, 0)

I µ(ζ) := dimK R /I

I GB gives us a basis for R /I

Z.Zafeirakopoulos 37

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and an Algebraic Problem

Given an ideal I , find a basis for R /I

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

I I = 〈f1, f2〉I V (I ) = ζ = (0, 0)

I µ(ζ) := dimK R /I

I GB gives us a basis for R /I

Z.Zafeirakopoulos 37

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and an Algebraic Problem

Given an ideal I , find a basis for R /I

I f1 = x2 + (y − 1)2 − 1

I f2 = y2

I I = 〈f1, f2〉I V (I ) = ζ = (0, 0)

I µ(ζ) := dimK R /I

I GB gives us a basis for R /I

Z.Zafeirakopoulos 37

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Dual Space

Z.Zafeirakopoulos 38

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Dual Space of a Polynomial Ring

Definition (Dual Space of a Polynomial Ring)

Let R = K[x1, . . . , xd ]. Then R := {λ : R → K | λ is linear}.

I R is infinite dimensional

Example

Let ζ = (ζ1, . . . , ζd) ∈ Kd and a = (a1, . . . , ad) ∈ Nd . Define

∂aζ : R −→ Kp 7→ (dx1)a1 . . . (dxd)ad (p)(ζ).

Then ∂aζ ∈ R.

I R and K[[∂ζ ]] are isomorphic as K-vector spaces

Z.Zafeirakopoulos 39

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Dual Space of a Polynomial Ring

Definition (Dual Space of a Polynomial Ring)

Let R = K[x1, . . . , xd ]. Then R := {λ : R → K | λ is linear}.

I R is infinite dimensional

Example

Let ζ = (ζ1, . . . , ζd) ∈ Kd and a = (a1, . . . , ad) ∈ Nd . Define

∂aζ : R −→ Kp 7→ (dx1)a1 . . . (dxd)ad (p)(ζ).

Then ∂aζ ∈ R.

I R and K[[∂ζ ]] are isomorphic as K-vector spaces

Z.Zafeirakopoulos 39

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Dual Space of a Polynomial Ring

Definition (Dual Space of a Polynomial Ring)

Let R = K[x1, . . . , xd ]. Then R := {λ : R → K | λ is linear}.

I R is infinite dimensional

Example

Let ζ = (ζ1, . . . , ζd) ∈ Kd and a = (a1, . . . , ad) ∈ Nd . Define

∂aζ : R −→ Kp 7→ (dx1)a1 . . . (dxd)ad (p)(ζ).

Then ∂aζ ∈ R.

I R and K[[∂ζ ]] are isomorphic as K-vector spaces

Z.Zafeirakopoulos 39

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Dual Space of a Polynomial Ring

Definition (Dual Space of a Polynomial Ring)

Let R = K[x1, . . . , xd ]. Then R := {λ : R → K | λ is linear}.

I R is infinite dimensional

Example

Let ζ = (ζ1, . . . , ζd) ∈ Kd and a = (a1, . . . , ad) ∈ Nd . Define

∂aζ : R −→ Kp 7→ (dx1)a1 . . . (dxd)ad (p)(ζ).

Then ∂aζ ∈ R.

I R and K[[∂ζ ]] are isomorphic as K-vector spaces

Z.Zafeirakopoulos 39

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Dual Space of a Polynomial Ring

Definition

∀I E R, I⊥ :={λ ∈ R | λ(f ) = 0 ∀f ∈ I

}.

I I⊥ is a (not necessarily finite dimensional) subspace of R

Theorem (Marinari, Mora and Moller, 95; Mourrain, 97)

Let ζ ∈ V (I ) be an isolated point and Qζ be its associated primarycomponent. Then

Q⊥ζ = I⊥ ∩K[∂ζ ]

I Q⊥ζ is a finite dimensional subspace of I⊥

Z.Zafeirakopoulos 40

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Dual Space of a Polynomial Ring

Definition

∀I E R, I⊥ :={λ ∈ R | λ(f ) = 0 ∀f ∈ I

}.

I I⊥ is a (not necessarily finite dimensional) subspace of R

Theorem (Marinari, Mora and Moller, 95; Mourrain, 97)

Let ζ ∈ V (I ) be an isolated point and Qζ be its associated primarycomponent. Then

Q⊥ζ = I⊥ ∩K[∂ζ ]

I Q⊥ζ is a finite dimensional subspace of I⊥

Z.Zafeirakopoulos 40

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Dual Space of a Polynomial Ring

Definition

∀I E R, I⊥ :={λ ∈ R | λ(f ) = 0 ∀f ∈ I

}.

I I⊥ is a (not necessarily finite dimensional) subspace of R

Theorem (Marinari, Mora and Moller, 95; Mourrain, 97)

Let ζ ∈ V (I ) be an isolated point and Qζ be its associated primarycomponent. Then

Q⊥ζ = I⊥ ∩K[∂ζ ]

I Q⊥ζ is a finite dimensional subspace of I⊥

Z.Zafeirakopoulos 40

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Dual Space of a Polynomial Ring

Definition

∀I E R, I⊥ :={λ ∈ R | λ(f ) = 0 ∀f ∈ I

}.

I I⊥ is a (not necessarily finite dimensional) subspace of R

Theorem (Marinari, Mora and Moller, 95; Mourrain, 97)

Let ζ ∈ V (I ) be an isolated point and Qζ be its associated primarycomponent. Then

Q⊥ζ = I⊥ ∩K[∂ζ ]

I Q⊥ζ is a finite dimensional subspace of I⊥

Z.Zafeirakopoulos 40

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 41

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 41

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

⟩I # = 4

I #{1, ∂y} = 2

Z.Zafeirakopoulos 41

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A Geometric Problem

Given two curves, find (projections of) intersections (with multiplicity).

f1 = x2 + (y − 1)2 − 1f2 = y2

I Resultant:resx (f1, f2) = y4

deg(resx (f1, f2)) = 4

I Elimination Ideal:GB of 〈f1, f2〉 ∩K[y ] =

⟨y2⟩

deg(g) = 2

I Dual Space:〈f1, f2〉⊥ =

⟨1, ∂x , 2∂

2x + ∂y , 2∂

3x + ∂x∂y

⟩I # = 4I #{1, ∂y} = 2

Z.Zafeirakopoulos 41

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Deflation

DefinitionStarting from a system f and an approximation ζ∗ of ζ, constructa new system, in which the singularity ζ is obviated.

Example

Let f ={x12 + x2 + x31, x1 + x22 + x31, x1 + x2 + x321

}I Approximate root ζ = (0.95, 0.08, 0.05).

I Compute a dual basis (1, d1 − 0.955d2 − 0.894d3)

I Let D(d , λ2, λ3) = (1, d1 − λ2d2, d2− λ3d3) and initial point(0.95, 0.08, 0.05, 0.955, .894)

I After 15 iterations of ζ = ζ∗ − JDf (ζ∗, λ∗2, λ∗3) we obtain

(1.0, 6.938 · 1018, 5.204 · 1017, 1.0, 1.0)

I The same accuracy (15 digits) would be achieved after 27iterations of the original system.

Z.Zafeirakopoulos 42

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Deflation

DefinitionStarting from a system f and an approximation ζ∗ of ζ, constructa new system, in which the singularity ζ is obviated.

Example

Let f ={x12 + x2 + x31, x1 + x22 + x31, x1 + x2 + x321

}I Approximate root ζ = (0.95, 0.08, 0.05).

I Compute a dual basis (1, d1 − 0.955d2 − 0.894d3)

I Let D(d , λ2, λ3) = (1, d1 − λ2d2, d2− λ3d3) and initial point(0.95, 0.08, 0.05, 0.955, .894)

I After 15 iterations of ζ = ζ∗ − JDf (ζ∗, λ∗2, λ∗3) we obtain

(1.0, 6.938 · 1018, 5.204 · 1017, 1.0, 1.0)

I The same accuracy (15 digits) would be achieved after 27iterations of the original system.

Z.Zafeirakopoulos 42

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Deflation

DefinitionStarting from a system f and an approximation ζ∗ of ζ, constructa new system, in which the singularity ζ is obviated.

Example

Let f ={x12 + x2 + x31, x1 + x22 + x31, x1 + x2 + x321

}I Approximate root ζ = (0.95, 0.08, 0.05).

I Compute a dual basis (1, d1 − 0.955d2 − 0.894d3)

I Let D(d , λ2, λ3) = (1, d1 − λ2d2, d2− λ3d3) and initial point(0.95, 0.08, 0.05, 0.955, .894)

I After 15 iterations of ζ = ζ∗ − JDf (ζ∗, λ∗2, λ∗3) we obtain

(1.0, 6.938 · 1018, 5.204 · 1017, 1.0, 1.0)

I The same accuracy (15 digits) would be achieved after 27iterations of the original system.

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Deflation

DefinitionStarting from a system f and an approximation ζ∗ of ζ, constructa new system, in which the singularity ζ is obviated.

Example

Let f ={x12 + x2 + x31, x1 + x22 + x31, x1 + x2 + x321

}I Approximate root ζ = (0.95, 0.08, 0.05).

I Compute a dual basis (1, d1 − 0.955d2 − 0.894d3)

I Let D(d , λ2, λ3) = (1, d1 − λ2d2, d2− λ3d3) and initial point(0.95, 0.08, 0.05, 0.955, .894)

I After 15 iterations of ζ = ζ∗ − JDf (ζ∗, λ∗2, λ∗3) we obtain

(1.0, 6.938 · 1018, 5.204 · 1017, 1.0, 1.0)

I The same accuracy (15 digits) would be achieved after 27iterations of the original system.

Z.Zafeirakopoulos 42

Page 121: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Deflation

DefinitionStarting from a system f and an approximation ζ∗ of ζ, constructa new system, in which the singularity ζ is obviated.

Example

Let f ={x12 + x2 + x31, x1 + x22 + x31, x1 + x2 + x321

}I Approximate root ζ = (0.95, 0.08, 0.05).

I Compute a dual basis (1, d1 − 0.955d2 − 0.894d3)

I Let D(d , λ2, λ3) = (1, d1 − λ2d2, d2− λ3d3) and initial point(0.95, 0.08, 0.05, 0.955, .894)

I After 15 iterations of ζ = ζ∗ − JDf (ζ∗, λ∗2, λ∗3) we obtain

(1.0, 6.938 · 1018, 5.204 · 1017, 1.0, 1.0)

I The same accuracy (15 digits) would be achieved after 27iterations of the original system.

Z.Zafeirakopoulos 42

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What’s Next?

I Macaulay resultantI Macaulay MatrixI Extraneous factor

I Computing the elimination ideal using resultants.I compare V (I1) and V (Res)I compare 〈V (I1)〉 and 〈V (Res)〉I compare 〈I1〉 and 〈Res〉

I Dual basesI Directional multiplicityI Deflation

I Sparse Elimination Theory.

Z.Zafeirakopoulos 43

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Heterogeneous Algorithms for Combinatorics, Geometry andNumber Theory

HALCYON project TUBITAK 3501Position for a master’s or PhD student

Team:Busra Sert (MSGSU)

Basak Karakas (Ege U)Duration: until October 2020

MathData project TUBITAK 3001Position for a master’s or undergrad student

Duration: until March 2019

NMK School on Integer Partitions: May 21-25 2018.

Thank You

Z.Zafeirakopoulos 44

Page 124: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Heterogeneous Algorithms for Combinatorics, Geometry andNumber Theory

HALCYON project TUBITAK 3501Position for a master’s or PhD student

Team:Busra Sert (MSGSU)

Basak Karakas (Ege U)Duration: until October 2020

MathData project TUBITAK 3001Position for a master’s or undergrad student

Duration: until March 2019

NMK School on Integer Partitions: May 21-25 2018.

Thank You

Z.Zafeirakopoulos 44

Page 125: A gentle introduction to Elimination Theory Zafeirakis ...web0.boun.edu.tr/alp.bassa/ankaraistanbul/Zaf1.pdf · Membership - A tale of computation Grete Herman 1940s IProved that

Heterogeneous Algorithms for Combinatorics, Geometry andNumber Theory

HALCYON project TUBITAK 3501Position for a master’s or PhD student

Team:Busra Sert (MSGSU)

Basak Karakas (Ege U)Duration: until October 2020

MathData project TUBITAK 3001Position for a master’s or undergrad student

Duration: until March 2019

NMK School on Integer Partitions: May 21-25 2018.

Thank YouZ.Zafeirakopoulos 44