A gentle introduction to Elimination Theory Zafeirakis...
Transcript of A gentle introduction to Elimination Theory Zafeirakis...
A gentle introduction to Elimination Theory
March 2018 @ METU
Zafeirakis Zafeirakopoulos
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
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
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
Intro
Membership
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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
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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
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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
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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
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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
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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
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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)]
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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
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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.
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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.
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Membership - A tale of computation
Wolfgang Grobner1940s
I Worked with Nother
I Several contributions
I Did not invent the bases bearing his name
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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.
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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.
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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
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
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.
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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
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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.
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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.
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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.
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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.
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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 ].
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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.
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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.
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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.
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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
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 .
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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 .
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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.
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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.
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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.
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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.
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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.
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Elimination Ideal
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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 ]〉 .
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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 ]〉 .
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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⟩
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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⟩
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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 }
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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?
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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?
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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?
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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?
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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?
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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)
).
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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
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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
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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
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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
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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
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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
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
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
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
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
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
Resultants
Z.Zafeirakopoulos 18
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
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
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
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
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
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
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
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
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
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
Implicitization
Z.Zafeirakopoulos 26
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
Implicitization
Example (Folium of Descartes)
x =3t2
t3 + 1, u =
3t
t3 + 1
p(x , y) = x3 − 3xy + y3
Z.Zafeirakopoulos 28
Implicitization
Example (Folium of Descartes)
x =3t2
t3 + 1, u =
3t
t3 + 1
p(x , y) = x3 − 3xy + y3
Z.Zafeirakopoulos 28
Number of roots
I Roots of the resultant are projections of roots.
I Bezout bound:∏
i diI Is it tight?
Z.Zafeirakopoulos 29
Number of roots
I Roots of the resultant are projections of roots.
I Bezout bound:∏
i di
I Is it tight?
Z.Zafeirakopoulos 29
Number of roots
I Roots of the resultant are projections of roots.
I Bezout bound:∏
i diI Is it tight?
Z.Zafeirakopoulos 29
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
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
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
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
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
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
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
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
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
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
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
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
BKK
I Does this imply something for the resultant?
I Can we have a resultant for these (toric) roots?
Z.Zafeirakopoulos 34
Because regularity is boring
Multiplicities
Z.Zafeirakopoulos 35
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
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
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
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
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
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
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
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
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
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
Dual Space
Z.Zafeirakopoulos 38
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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