A gentle introduction to Elimination Theory. March METU. Zafeirakis Zafeirakopoulos

Transcription

1 A gentle introduction to Elimination Theory March METU Zafeirakis Zafeirakopoulos

2 Disclaimer Elimination theory is a very wide area of research. Z.Zafeirakopoulos 2

3 Disclaimer Elimination theory is a very wide area of research. We will see only parts of it Z.Zafeirakopoulos 2

4 Disclaimer Elimination theory is a very wide area of research. We will see only parts of it through the lens of computation (polynomial system solving) Z.Zafeirakopoulos 2

5 Intro Membership Z.Zafeirakopoulos 3

6 Membership - A tale of computation Definition (Ideal) Given a ring R, an ideal I R is a subset of R such that a I, c R : ca I a, b I : a + b I Z.Zafeirakopoulos 4

7 Membership - A tale of computation Definition (Ideal) Given a ring R, an ideal I R is a subset of R such that a I, c R : ca I a, b I : a + b I Definition (Ideal Membership) Input a ring R, an ideal I R and an element f R Output True if f I, False otherwise Z.Zafeirakopoulos 4

8 Membership - A tale of computation I Membership in Euclidean domains is easy I Division gives unique remainder Z.Zafeirakopoulos 4

9 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 + X qi gi gi I Z.Zafeirakopoulos 4

10 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 + X qi gi gi I I f I if and only if r = 0 Z.Zafeirakopoulos 4

11 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 + X qi gi gi I I f I if and only if r = 0 Note R[x] is a Euclidean domain Z.Zafeirakopoulos 4

12 Membership - A tale of computation A(t) = t 2t 2 t t t (t+1) t 2 t + 0 t 1 t 1 + ( 1) t 2 t t : 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

13 Membership - A tale of computation Emmy Nöther 1920s Note If we want to manipulate ideals, we have to be able to decide membership. If we can, then we can also decide equality of ideals Arithmetic of ideals Z.Zafeirakopoulos 4

14 Membership - A tale of computation Grete Herman 1940s Proved that a bound to decide membership would be doubly exponential in the degree. The linear combination has huge coefficients. Indication that Gröbner bases have bad complexity. Z.Zafeirakopoulos 4

15 Membership - A tale of computation Grete Herman 1940s Proved that a bound to decide membership would be doubly exponential in the degree. The linear combination has huge coefficients. Indication that Gröbner bases have bad complexity. Z.Zafeirakopoulos 4

16 Membership - A tale of computation Wolfgang Gröbner 1940s Worked with Nöther Several contributions Did not invent the bases bearing his name Z.Zafeirakopoulos 4

17 Membership - A tale of computation Membership is hard because remainder is not unique Z.Zafeirakopoulos 4

18 Membership - A tale of computation Membership is hard because remainder is not unique For some sets of divisors, remainder is unique Z.Zafeirakopoulos 4

19 Membership - A tale of computation Membership is hard because remainder is not unique For some sets of divisors, remainder is unique Every ideal (in a Nötherian ring) has such a set of generators. Z.Zafeirakopoulos 4

20 Membership - A tale of computation Membership is hard because remainder is not unique For some sets of divisors, remainder is unique Every ideal (in a Nötherian ring) has such a set of generators. Buchberger proved it. Z.Zafeirakopoulos 4

21 Membership - A tale of computation Membership is hard because remainder is not unique For some sets of divisors, remainder is unique Every ideal (in a Nötherian ring) has such a set of generators. Buchberger proved it. What is even better, he proved it constructively. Z.Zafeirakopoulos 4

22 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 Bruno Buchberger was a student of Gröbner Thesis An Algorithm for Finding the Basis Elements of the Residue Class Ring Modulo a Zerodimensional Polynomial Ideal Z.Zafeirakopoulos 4

23 Monomial (Order) Definition (Term Monoid) Given a set of variables x 1, x 2,..., x d we consider the multiplicative monoid T = { x α 1 1 x α 2 2 x α d d : α N d}. Z.Zafeirakopoulos 5

24 Monomial (Order) Definition (Term Monoid) Given a set of variables x 1, x 2,..., x d we consider the multiplicative monoid T = { x α 1 1 x α 2 2 x α d d : α N d}. Note that there is a monoid homomorphism between T and N d Z.Zafeirakopoulos 5

25 Monomial (Order) Definition (Term Monoid) Given a set of variables x 1, x 2,..., x d we consider the multiplicative monoid T = { x α 1 1 x α 2 2 x α d d : α N d}. Note that there is a monoid homomorphism between T and N d Definition (Term order) Let be a total order on T. It is called a term order if 0 T for all T T and if a b then ac bc for all a, b, c T. Z.Zafeirakopoulos 5

26 Monomial (Order) Definition (Term order) Let be a total order on T. It is called a term order if 0 T for all T T and if a b then ac bc for all a, b, c T. Example (Lexicographic vs DegRevLex) Fix x 1 x 2 x d. 2 x α d d lex x β 1 1 x β 2 2 x β d d if the left-most non-zero entry in β α is positive. x α 1 1 x α 2 x α 1 1 x α 2 2 x α d d drl x β 1 1 x β 2 2 x β d d αi β i or αi = β i and the right-most non-zero entry in β α is positive. Z.Zafeirakopoulos 5 if

27 Monomial (Order) Definition (Term order) Let be a total order on T. It is called a term order if 0 T for all T T and 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[x 1,..., x d ]. Z.Zafeirakopoulos 5

28 Gröbner Bases Fix a term order. Definition Given an ideal I = f 1, f 2,..., f n, a Gröbner basis for I, with respect to the term order, is a set G = {g 1, g 2,..., g m } such that I = G and for every 0 f I we have that lt (g i ) lt (f ) for some i [m]. This is not the only definition. Z.Zafeirakopoulos 6

29 Gröbner Bases Fix a term order. Definition Given an ideal I = f 1, f 2,..., f n, a Gröbner basis for I, with respect to the term order, is a set G = {g 1, g 2,..., g m } such that I = G and for every 0 f I we have that lt (g i ) lt (f ) for some i [m]. This is not the only definition. Other definitions will appear during this series. Z.Zafeirakopoulos 6

30 Gröbner Bases Fix a term order. Definition Given an ideal I = f 1, f 2,..., f n, a Gröbner basis for I, with respect to the term order, is a set G = {g 1, g 2,..., g m } such that I = G and for every 0 f I we have that lt (g i ) lt (f ) for some i [m]. This is not the only definition. Other definitions will appear during this series. A more important property: reduction by G in K[x 1,..., x d ] is unique. Z.Zafeirakopoulos 6

31 Gröbner Bases Fix a term order. Definition Given an ideal I = f 1, f 2,..., f n, a Gröbner basis for I, with respect to the term order, is a set G = {g 1, g 2,..., g m } such that I = G and for every 0 f I we have that lt (g i ) lt (f ) for some i [m]. This is not the only definition. Other definitions will appear during this series. A more important property: reduction by G in K[x 1,..., x d ] is unique. Reduce means to take the remainder after we divide as much as possible with elements of G. Z.Zafeirakopoulos 6

32 A criterion Definition (S-polynomial) Fix a term order and let f, g K[x 1,..., x d ]. The S-polynomial of f and g is S f,g = lcm (lt(f ), lt(g)) f lt(f ) lcm (lt(f ), lt(g)) g lt(g) Z.Zafeirakopoulos 7

33 A criterion Definition (S-polynomial) Fix a term order and let f, g K[x 1,..., x d ]. The S-polynomial of f and g is S f,g = lcm (lt(f ), lt(g)) f lt(f ) lcm (lt(f ), lt(g)) g lt(g) Theorem (Buchberger) A (finite) set G is a Gröbner basis of G if and only if S f,g is reduced to 0 by G for all pairs f, g G. Z.Zafeirakopoulos 7

34 Example Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. Z.Zafeirakopoulos 8

35 Example Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. Then G = { x 2 2y, y 2} Z.Zafeirakopoulos 8

36 Example Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. Then G = { x 2 2y, y 2} S x 2 +(y 1) 2 1,y 2 = y 4 2y 3 = y 2 ( y 2 + 2y) y 2 0 Z.Zafeirakopoulos 8

37 Example Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. Then G = { x 2 2y, y 2} S x 2 +(y 1) 2 1,y 2 = y 4 2y 3 = y 2 ( y 2 + 2y) y 2 0 We interreduce elements of the GB to obtain x 2 2y and y 2. Z.Zafeirakopoulos 8

38 Example Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. Then G = { x 2 2y, y 2} S x 2 +(y 1) 2 1,y 2 = y 4 2y 3 = y 2 ( y 2 + 2y) y 2 0 We interreduce elements of the GB to obtain x 2 2y and y 2. We tend to say that a GB is a nice choice of a generators. Z.Zafeirakopoulos 8

39 Elimination Ideal Z.Zafeirakopoulos 9

40 Elimination ideal Definition Let I K[x 1,..., x d ] be an ideal. Then we define the i-th elimination ideal of I as I i = I K[x i+1,..., x d ]. Z.Zafeirakopoulos 10

41 Elimination ideal Definition Let I K[x 1,..., x d ] be an ideal. Then we define the i-th elimination ideal of I as I i = I K[x i+1,..., x d ]. Theorem (Elimination Property of Gröbner Bases) Let k [d] and fix a lexicographic order such that x i x j for all i < k and k < j. If G is a Gröbner basis of I (for the term order we fixed), then I k = G K[x k+1,..., x d ]. Z.Zafeirakopoulos 10

42 Elimination ideal Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. Z.Zafeirakopoulos 11

43 Elimination ideal Let f 1 = x 2 + (y 1) 2 1 f 2 = y 2 and I = f 1, f 2. We saw that a GB for I is { x 2 2y, y 2}. Thus I x = { x 2 2y, y 2} K[y] = y 2 Z.Zafeirakopoulos 11

44 Variety and Vanishing Ideal Let K be an algebraically closed field. Definition (Variety) Let I be an ideal in K[x 1,..., x d ]. Then { } V (I ) = x K d : f (x) = 0 for all f I Definition (Vanishing) Let V K d be a variety. Then I (V ) = {f K[x 1,..., x d ] : f (x) = 0 for all x V } Z.Zafeirakopoulos 12

45 0-dim What does it mean for the GB that the variety is 0-dim? Z.Zafeirakopoulos 13

46 0-dim What does it mean for the GB that the variety is 0-dim? What does it mean for the elimination ideal? Z.Zafeirakopoulos 13

47 0-dim What does it mean for the GB that the variety is 0-dim? What does it mean for the elimination ideal? What does it mean for solving? Z.Zafeirakopoulos 13

48 0-dim What does it mean for the GB that the variety is 0-dim? What does it mean for the elimination ideal? What does it mean for solving? What does it remind us? Z.Zafeirakopoulos 13

49 0-dim What does it mean for the GB that the variety is 0-dim? What does it mean for the elimination ideal? What does it mean for solving? What does it remind us? Z.Zafeirakopoulos 13

50 Variety of the Elimination Ideal For f 1,..., f n K[x 1,..., x d ], we write f i in the form f i = h i (x 2,..., x d )x N i 1 + terms of x 1-degree less than N i, for each 1 i n. Consider the projection π : K n K n 1 : π ( (c 1, c 2,..., c n ) ) = (c 2, c 3,..., c n ). Theorem (Elimination Theorem) Let I 1 be the first elimination ideal of an ideal I K[x 1,..., x n ]. Then V (I 1 ) = π ( V (I ) ) ( V (h 1,..., h m ) V (I 1 ) ). Z.Zafeirakopoulos 14

51 Modeling complementary sequences T = (a, 0, a) (a, 0, a) (0, a, 0) (0, b, 0) 0 T = (a, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (a0 + 0a) + (a0 0a) + ( 0a a0) + (0b b0) AF T (2) = a 2 a Z.Zafeirakopoulos 15

52 Modeling complementary sequences T = (a, 0, a) (a, 0, a) (0, a, 0) (0, b, 0) 0 x1 T = (a, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (ax 1 + x 1 a) + (a0 0a) + ( 0a a0) + (0b b0) AF T (2) = a 2 a Z.Zafeirakopoulos 15

53 Modeling complementary sequences T = (a, 0, a) (a, 0, a) (0, a, 0) (0, b, 0) 0 x1, x2 T = (a, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (ax 1 + x 1 a) + (ax 2 x 2 a) + ( 0a a0) + (0b b0) AF T (2) = a 2 a Z.Zafeirakopoulos 15

54 Modeling complementary sequences T = (a, 0, a) (a, 0, a) (0, a, 0) (0, b, 0) 0 x1, x2, x3 T = (a, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (ax 1 + x 1 a) + (ax 2 x 2 a) + ( x 3 a a0) + (0b b0) AF T (2) = a 2 a 2 + x Z.Zafeirakopoulos 15

55 Modeling complementary sequences T = (a, 0, a) (a, 0, a) (0, a, 0) (0, b, 0) 0 x1, x2, x3, x4 T = (a, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (ax 1 + x 1 a) + (ax 2 x 2 a) + ( x 3 a ax 4 ) + (0b b0) AF T (2) = a 2 a 2 + x 3 x Z.Zafeirakopoulos 15

56 Modeling complementary sequences T = (a, 0, a) (a, 0, a) (0, a, 0) (0, b, 0) 0 x1, x2, x3, x4, x5 T = (a, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (ax 1 + x 1 a) + (ax 2 x 2 a) + ( x 3 a ax 4 ) + (x 5 b b0) AF T (2) = a 2 a 2 + x 3 x 4 x 5 0 Z.Zafeirakopoulos 15

57 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, x 1, a) (a, x 2, a) (x 3, a, x 4 ) (x 5, b, x 6 ) AF T (1) = (ax 1 + x 1 a) + (ax 2 x 2 a) + ( x 3 a ax 4 ) + (x 5 b bx 6 ) AF T (2) = a 2 a 2 + x 3 x 4 x 5 x 6 Z.Zafeirakopoulos 15

58 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() abx 5 abx b2 x 2 1 x 3x b2 x 2 1 x 3x b2 x 2 1 x 4x b2 x 2 1 x 4x 5 x b2 x 2 1 x 4x b2 x 1 x 2 3 x b2 x 1 x 2 3 x b 2 x 1 x 2 4 x b2 x 1 x 2 4 x b2 x 1 x 2 5 x b2 x 1 x 2 5 b2 x 1 x 5 x b2 x 1 x 2 6 b2 x 3 x 2 5 b2 x 3 x b2 x 4 x 2 5 x2 6 b 2 x 4 x b2 x 4 x 5 x 6 b 2 x 4 x6 2, ax ax 4x 5 x ax bx2 1 x2 4 x bx2 1 x2 4 x bx2 1 x bx2 1 x bx 1x 3 x bx 1 x 3 x bx 1x 4 x5 2 x bx 1x 4 x 5 x bx2 3 x bx2 3 x bx2 4 x bx2 4 x bx2 5 x bx 5x bx bx 6, ax 3 ax 4 x 5 x bx2 1 x2 4 x bx2 1 x2 4 x bx2 1 x bx2 1 x bx 1x 3 x bx 1x 3 x bx 1x 4 x5 2 x bx 1 x 4 x 5 x bx2 3 x bx2 3 x bx2 4 x bx2 4 x bx2 5 x bx 5x bx bx 6, ax4 2 ax2 5 x bx2 1 x 4x bx1 2 x 4x bx 1x4 2 x bx 1x4 2 x bx 1x5 2 x bx 1x 5 x bx 4x5 2 x bx 4x 5 x6 2 bx 4x 5 + bx 4 x 6, ax 4 x5 2 ax bx2 1 x2 4 x bx2 1 x 5x bx 1x 4 x 2 5 x bx 1x 4 x 6 bx 2 4 x bx2 5 x bx 5x 2 6, ax 4x 2 6 ax bx2 1 x2 4 x bx 2 1 x 5x bx 1x 4 x 5 x bx 1x 4 x 5 + bx 2 4 x 5 bx 5 x 2 6, bx2 1 x2 3 x 5 + bx 2 1 x2 4 x 5 bx 2 1 x 5x 2 6 bx2 1 x 5 bx 2 3 x 5 bx 2 4 x 5 + bx 5 x bx 5, bx 2 1 x2 3 x 6 + bx 2 1 x2 4 x 6 bx 2 1 x 5x 2 6 bx2 1 x 6 bx 2 3 x 6 bx 2 4 x 6 + bx 5 x bx 6, bx 2 1 x2 5 x 6 bx 2 1 x 5x 2 6 Z.Zafeirakopoulos 16

59 Modeling complementary sequences The ideal I is 2-dim. The elimination ideal is 0-dim Z.Zafeirakopoulos 17

60 Modeling complementary sequences The ideal I is 2-dim. The elimination ideal is 0-dim We eliminated the parameters That s good because we want the equations to hold for all values of the parameters Z.Zafeirakopoulos 17

61 Modeling complementary sequences The ideal I is 2-dim. The elimination ideal is 0-dim We eliminated the parameters That s good because we want the equations to hold for all values of the parameters Live demo? Z.Zafeirakopoulos 17

62 Resultants Z.Zafeirakopoulos 18

63 Common factors Let f 1, f 2 K[x]. Then f 1 and f 2 have a common factor if and only if there are polynomials A, B K[x] such that: A and B are not both zero. deg(a) deg(f 2 ) 1 and deg(b) deg(f 2 ) 1 Af 1 + Bf 2 = 0 Z.Zafeirakopoulos 19

64 Common factors Let f 1, f 2 K[x]. Then f 1 and f 2 have a common factor if and only if there are polynomials A, B K[x] such that: A and B are not both zero. deg(a) deg(f 2 ) 1 and deg(b) deg(f 2 ) 1 Af 1 + Bf 2 = 0 Now, if we expand Af 1 + Bf 2 and force all coefficients to be 0, we get a linear system. Z.Zafeirakopoulos 19

65 Sylvester Resultant Syl(f 1, f 2 ) = f 1,d1 f 1, f 1,d1 f 1,0 f 2,d2 f 2, f 2,d2 f 2,0 d 2 d 1 Definition The resultant res x (f 1, f 2 ) is the determinant of Syl (f 1, f 2 ). Z.Zafeirakopoulos 20

66 Sylvester Resultant Theorem If f, g K[x] then the resultant Res(f, g, x) K[x] is an integer polynomial in the coefficients of f and g. Z.Zafeirakopoulos 21

67 Sylvester Resultant Theorem If f, g K[x] then the resultant Res(f, g, x) K[x] is an integer polynomial in the coefficients of f and g. Theorem 1 gcd(f, g) K[x] Res(f, g, x) = 0 Z.Zafeirakopoulos 21

68 Resultants and Elimination ideals Theorem Let f, g K[x 1,..., x d ] and c = (c 2,..., c d ) C d 1 satisfy the following: f (x 1, c) C[x 1 ] has degree deg(f ) g(x 1, c) C[x 1 ] has degree p deg(g) Then Res(f, g, x 1 )(c) = lt(f )(c) deg(g) p Res (f (x 1, c), g(x 1, c), x 1 ) Z.Zafeirakopoulos 22

69 Resultants and Elimination ideals Theorem (ExtensionTheorem) Let I = f 1, f 2,..., f n C[x 1, x 2,..., x d ] and let I 1 be the first elimination ideal of I. We write f i in the form f i = h i (x 2,..., x d )x N i 1 + terms of x 1 degree less than N i, and g i C[x 2,..., x d ] is not zero. If (c 2,..., c d ) V (I 1 ) and (c 2,..., c d ) V (h 1, h 2,..., h n ) then there exist c 1 C such that (c 1, c 2,..., c d ) V (I ) Z.Zafeirakopoulos 23

70 Elimination ideal vs Resultant Theorem Let I = f 1, f 2 K[x 1,..., x n ] and R = res x1 (f 1, f 2 ). Then V (R) = V (h 1, h 2 ) π ( V (I ) ). Z.Zafeirakopoulos 24

71 Elimination ideal vs Resultant Theorem Let I = f 1, f 2 K[x 1,..., x n ] and R = res x1 (f 1, f 2 ). Then V (R) = V (h 1, h 2 ) π ( V (I ) ). Theorem If f 1, f 2 K[x, y] and R = res x (f 1, f 2 ) is not identically zero, then V (I 1 ) = π ( V (I ) ). Z.Zafeirakopoulos 24

72 Resultant System Definition Let f 1,..., f n K[x 1,..., x d ] and introduce n new variables u i. Consider the resultant R i = Res x1 (f i, i j u j f j ). The resultant system RS x1 (f 1,..., f n ) is the set of coefficients of R i seen as a polynomial in variables u 1,..., u n. Z.Zafeirakopoulos 25

73 Implicitization Z.Zafeirakopoulos 26

74 Implicitization Given parameterization x 0 = α 0 (t),..., x n = α n (t), t := (t 1,..., t n ), compute the smallest algebraic variety containing the closure of the image of α : R n R n+1 : t α(t), α := (α 0,..., α n ). This is contained in the variety defined by the ideal p(x 0,..., x n ) p(α 0 (t),..., α n (t)) = 0, t. When this is a principal ideal we wish to compute its defining polynomial p(x). Z.Zafeirakopoulos 27

75 Implicitization Example (Folium of Descartes) x = 3t2 t 3 + 1, u = 3t t Z.Zafeirakopoulos 28

76 Implicitization Example (Folium of Descartes) x = 3t2 t 3 + 1, u = 3t t p(x, y) = x 3 3xy + y 3 Z.Zafeirakopoulos 28

77 Number of roots Roots of the resultant are projections of roots. Z.Zafeirakopoulos 29

78 Number of roots Roots of the resultant are projections of roots. Bezout bound: i d i Z.Zafeirakopoulos 29

79 Number of roots Roots of the resultant are projections of roots. Bezout bound: i d i Is it tight? Z.Zafeirakopoulos 29

80 Newton Polytope Definition Given a polynomial f = α N d c α x α 1 1 x α 2 2 x α d d K[x 1, x 2,..., x d ], the support of f is Sup(f ) = { α N d : c α 0 } and its Newton polytope is the convex hull of its support NP (f ) = CH {Sup(f )}. Z.Zafeirakopoulos 30

81 Newton Polytope Definition Given a polynomial f = α N d c α x α 1 1 x α 2 2 x α d d K[x 1, x 2,..., x d ], the support of f is Sup(f ) = { α N d : c α 0 } and its Newton polytope is the convex hull of its support NP (f ) = CH {Sup(f )}. Example f = x 3 y 3x 2 + 2xy xy y Sup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)} y Z.Zafeirakopoulos 30 x

82 Newton Polytope Definition Given a polynomial f = α N d c α x α 1 1 x α 2 2 x α d d K[x 1, x 2,..., x d ], the support of f is Sup(f ) = { α N d : c α 0 } and its Newton polytope is the convex hull of its support NP (f ) = CH {Sup(f )}. Example f = x 3 y 3x 2 + 2xy xy y Sup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)} y Z.Zafeirakopoulos 30 x

83 Newton Polytope Definition Given a polynomial f = α N d c α x α 1 1 x α 2 2 x α d d K[x 1, x 2,..., x d ], the support of f is Sup(f ) = { α N d : c α 0 } and its Newton polytope is the convex hull of its support NP (f ) = CH {Sup(f )}. Example f = x 3 y 3x 2 + 2xy xy y Sup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)} y f = x 2 3y 2 + 2xy + 2x y + 1 Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)} Z.Zafeirakopoulos 30 x

84 Newton Polytope Definition Given a polynomial f = α N d c α x α 1 1 x α 2 2 x α d d K[x 1, x 2,..., x d ], the support of f is Sup(f ) = { α N d : c α 0 } and its Newton polytope is the convex hull of its support NP (f ) = CH {Sup(f )}. Example f = x 3 y 3x 2 + 2xy xy y Sup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)} y f = x 2 3y 2 + 2xy + 2x y + 1 Sup(f ) = {(2, 0), (0, 2), (1, 1), (1, 0), (0, 1), (0, 0)} Z.Zafeirakopoulos 30 x

85 Mixed Volume Let P 1, P 2,..., P k R d be polytopes and λ 1, λ 2,..., λ k R 0. Theorem (Minkowski) Then there exist V α1,α 2,...,α k = α 1 +α 2 + +α k =d 0, such that Vol (λ 1 P 1 λ 2 P 2 λ k P k ) ( ) d V α1,α α 1, α 2,..., α 2,...,α k λ α 1 1 λα 2 2 λα k k k Definition The mixed volume MV (P 1, P 2,..., P d ) is the coefficient of λ 1 λ 2... λ d in Vol (λ 1 P 1 λ 2 P 2 λ d P d ). Z.Zafeirakopoulos 31

86 The BKK bound Theorem (Bernstein, Khovanskii, Kushnirenko) Let f 1, f 2,..., f d C[x 1, x 2,..., x d ]. Z.Zafeirakopoulos 32

87 The BKK bound Theorem (Bernstein, Khovanskii, Kushnirenko) Let f 1, f 2,..., f d C[x 1, x 2,..., x d ]. Then the number of isolated solutions to the polynomial system f 1 (x) = = f d (x) = 0 with (x 1, x 2,..., x d ) (C {0}) d Z.Zafeirakopoulos 32

88 The BKK bound Theorem (Bernstein, Khovanskii, Kushnirenko) Let f 1, f 2,..., f d C[x 1, x 2,..., x d ]. Then the number of isolated solutions to the polynomial system f 1 (x) = = f d (x) = 0 with (x 1, x 2,..., x d ) (C {0}) d is (counting multiplicities) Z.Zafeirakopoulos 32

89 The BKK bound Theorem (Bernstein, Khovanskii, Kushnirenko) Let f 1, f 2,..., f d C[x 1, x 2,..., x d ]. Then the number of isolated solutions to the polynomial system f 1 (x) = = f d (x) = 0 with (x 1, x 2,..., x d ) (C {0}) d is (counting multiplicities) bounded by the mixed volume of the Newton polytopes of f 1, f 2,..., f d. Z.Zafeirakopoulos 32

90 The BKK bound f 1 = 1 + αx + βy 2 f 2 = x + γy 4 Bezout bound: deg(f 1 ) deg(f 2 ) = 8 2s+4t 4t 2s t s+t V (snp(f 1 ) tnp(f 2 )) = s 2 + ( 2 1) 2st Z.Zafeirakopoulos 33

91 The BKK bound f 1 = 1 + αx + βy 2 f 2 = x + γy 4 Bezout bound: deg(f 1 ) deg(f 2 ) = 8 2s+4t 4t 2s t s+t V (snp(f 1 ) tnp(f 2 )) = s 2 + ( 2 1) 2st MV (NP(f 1 ), NP(f 2 )) = 2! 2 = 4 Z.Zafeirakopoulos 33

92 BKK Does this imply something for the resultant? Can we have a resultant for these (toric) roots? Z.Zafeirakopoulos 34

93 Because regularity is boring Multiplicities Z.Zafeirakopoulos 35

94 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Z.Zafeirakopoulos 36

95 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Z.Zafeirakopoulos 36

96 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Z.Zafeirakopoulos 36

97 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 Dual Space: f 1, f 2 = 1, x, 2 2 x + y, 2 3 x + x y Z.Zafeirakopoulos 36

98 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 Dual Space: f 1, f 2 = 1, x, 2 2 x + y, 2 3 x + x y # = 4 Z.Zafeirakopoulos 36

99 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 Dual Space: f 1, f 2 = 1, x, 2 2 x + y, 2 3 x + x y # = 4 #{1, y } = 2 Z.Zafeirakopoulos 36

100 and an Algebraic Problem Given an ideal I, find a basis for R /I f 1 = x 2 + (y 1) 2 1 f 2 = y 2 I = f 1, f 2 Z.Zafeirakopoulos 37

101 and an Algebraic Problem Given an ideal I, find a basis for R /I f 1 = x 2 + (y 1) 2 1 f 2 = y 2 I = f 1, f 2 V (I ) = ζ = (0, 0) Z.Zafeirakopoulos 37

102 and an Algebraic Problem Given an ideal I, find a basis for R /I f 1 = x 2 + (y 1) 2 1 f 2 = y 2 I = f 1, f 2 V (I ) = ζ = (0, 0) µ(ζ) := dim K R /I Z.Zafeirakopoulos 37

103 and an Algebraic Problem Given an ideal I, find a basis for R /I f 1 = x 2 + (y 1) 2 1 f 2 = y 2 I = f 1, f 2 V (I ) = ζ = (0, 0) µ(ζ) := dim K R /I GB gives us a basis for R /I Z.Zafeirakopoulos 37

104 Dual Space Z.Zafeirakopoulos 38

105 Dual Space of a Polynomial Ring Definition (Dual Space of a Polynomial Ring) Let R = K[x 1,..., x d ]. Then ˆR := {λ : R K λ is linear}. Z.Zafeirakopoulos 39

106 Dual Space of a Polynomial Ring Definition (Dual Space of a Polynomial Ring) Let R = K[x 1,..., x d ]. Then ˆR := {λ : R K λ is linear}. ˆR is infinite dimensional Z.Zafeirakopoulos 39

107 Dual Space of a Polynomial Ring Definition (Dual Space of a Polynomial Ring) Let R = K[x 1,..., x d ]. Then ˆR := {λ : R K λ is linear}. ˆR is infinite dimensional Example Let ζ = (ζ 1,..., ζ d ) K d and a = (a 1,..., a d ) N d. Define Then a ζ ˆR. ζ a : R K p (dx 1 ) a 1... (dx d ) a d (p)(ζ). Z.Zafeirakopoulos 39

108 Dual Space of a Polynomial Ring Definition (Dual Space of a Polynomial Ring) Let R = K[x 1,..., x d ]. Then ˆR := {λ : R K λ is linear}. ˆR is infinite dimensional Example Let ζ = (ζ 1,..., ζ d ) K d and a = (a 1,..., a d ) N d. Define Then a ζ ˆR. ζ a : R K p (dx 1 ) a 1... (dx d ) a d (p)(ζ). ˆR and K[[ ζ ]] are isomorphic as K-vector spaces Z.Zafeirakopoulos 39

109 Dual Space of a Polynomial Ring Definition I R, I := { λ ˆR λ(f ) = 0 } f I. Z.Zafeirakopoulos 40

110 Dual Space of a Polynomial Ring Definition I R, I := { λ ˆR λ(f ) = 0 } f I. I is a (not necessarily finite dimensional) subspace of ˆR Z.Zafeirakopoulos 40

111 Dual Space of a Polynomial Ring Definition I R, I := { λ ˆR λ(f ) = 0 } f I. I is a (not necessarily finite dimensional) subspace of ˆR Theorem (Marinari, Mora and Möller, 95; Mourrain, 97) Let ζ V (I ) be an isolated point and Q ζ be its associated primary component. Then Q ζ = I K[ ζ ] Z.Zafeirakopoulos 40

112 Dual Space of a Polynomial Ring Definition I R, I := { λ ˆR λ(f ) = 0 } f I. I is a (not necessarily finite dimensional) subspace of ˆR Theorem (Marinari, Mora and Möller, 95; Mourrain, 97) Let ζ V (I ) be an isolated point and Q ζ be its associated primary component. Then Q ζ = I K[ ζ ] Q ζ is a finite dimensional subspace of I Z.Zafeirakopoulos 40

113 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Z.Zafeirakopoulos 41

114 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 Dual Space: f 1, f 2 = 1, x, 2 2 x + y, 2 3 x + x y Z.Zafeirakopoulos 41

115 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 Dual Space: f 1, f 2 = 1, x, 2 2 x + y, 2 3 x + x y # = 4 Z.Zafeirakopoulos 41

116 A Geometric Problem Given two curves, find (projections of) intersections (with multiplicity). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Resultant: res x (f 1, f 2 ) = y 4 deg(res x (f 1, f 2 )) = 4 Elimination Ideal: GB of f 1, f 2 K[y] = y 2 deg(g) = 2 Dual Space: f 1, f 2 = 1, x, 2 2 x + y, 2 3 x + x y # = 4 #{1, y } = 2 Z.Zafeirakopoulos 41

117 Deflation Definition Starting from a system f and an approximation ζ of ζ, construct a new system, in which the singularity ζ is obviated. Example Let f = { x1 2 + x2 + x31, x1 + x2 2 + x31, x1 + x2 + x3 2 1 } Approximate root ζ = (0.95, 0.08, 0.05). Z.Zafeirakopoulos 42

118 Deflation Definition Starting from a system f and an approximation ζ of ζ, construct a new system, in which the singularity ζ is obviated. Example Let f = { x1 2 + x2 + x31, x1 + x2 2 + x31, x1 + x2 + x3 2 1 } Approximate root ζ = (0.95, 0.08, 0.05). Compute a dual basis (1, d d d 3 ) Z.Zafeirakopoulos 42

119 Deflation Definition Starting from a system f and an approximation ζ of ζ, construct a new system, in which the singularity ζ is obviated. Example Let f = { x1 2 + x2 + x31, x1 + x2 2 + x31, x1 + x2 + x3 2 1 } Approximate root ζ = (0.95, 0.08, 0.05). Compute a dual basis (1, d d d 3 ) Let D(d, λ 2, λ 3 ) = (1, d 1 λ 2 d 2, d2 λ 3 d3) and initial point (0.95, 0.08, 0.05, 0.955,.894) Z.Zafeirakopoulos 42

120 Deflation Definition Starting from a system f and an approximation ζ of ζ, construct a new system, in which the singularity ζ is obviated. Example Let f = { x1 2 + x2 + x31, x1 + x2 2 + x31, x1 + x2 + x3 2 1 } Approximate root ζ = (0.95, 0.08, 0.05). Compute a dual basis (1, d d d 3 ) Let D(d, λ 2, λ 3 ) = (1, d 1 λ 2 d 2, d2 λ 3 d3) and initial point (0.95, 0.08, 0.05, 0.955,.894) After 15 iterations of ζ = ζ JDf (ζ, λ 2, λ 3 ) we obtain (1.0, , , 1.0, 1.0) Z.Zafeirakopoulos 42

121 Deflation Definition Starting from a system f and an approximation ζ of ζ, construct a new system, in which the singularity ζ is obviated. Example Let f = { x1 2 + x2 + x31, x1 + x2 2 + x31, x1 + x2 + x3 2 1 } Approximate root ζ = (0.95, 0.08, 0.05). Compute a dual basis (1, d d d 3 ) Let D(d, λ 2, λ 3 ) = (1, d 1 λ 2 d 2, d2 λ 3 d3) and initial point (0.95, 0.08, 0.05, 0.955,.894) After 15 iterations of ζ = ζ JDf (ζ, λ 2, λ 3 ) we obtain (1.0, , , 1.0, 1.0) The same accuracy (15 digits) would be achieved after 27 iterations of the original system. Z.Zafeirakopoulos 42

122 What s Next? Macaulay resultant Macaulay Matrix Extraneous factor Computing the elimination ideal using resultants. compare V (I1 ) and V (Res) compare V (I 1 ) and V (Res) compare I 1 and Res Dual bases Directional multiplicity Deflation Sparse Elimination Theory. Z.Zafeirakopoulos 43

123 Heterogeneous Algorithms for Combinatorics, Geometry and Number Theory HALCYON project TÜBITAK 3501 Position for a master s or PhD student Team: Busra Sert (MSGSU) Basak Karakas (Ege U) Duration: until October 2020 MathData project TÜBITAK 3001 Position for a master s or undergrad student Duration: until March 2019 NMK School on Integer Partitions: May Z.Zafeirakopoulos 44

124 Heterogeneous Algorithms for Combinatorics, Geometry and Number Theory HALCYON project TÜBITAK 3501 Position for a master s or PhD student Team: Busra Sert (MSGSU) Basak Karakas (Ege U) Duration: until October 2020 MathData project TÜBITAK 3001 Position for a master s or undergrad student Duration: until March 2019 NMK School on Integer Partitions: May Z.Zafeirakopoulos 44

125 Z.Zafeirakopoulos Thank You 44 Heterogeneous Algorithms for Combinatorics, Geometry and Number Theory HALCYON project TÜBITAK 3501 Position for a master s or PhD student Team: Busra Sert (MSGSU) Basak Karakas (Ege U) Duration: until October 2020 MathData project TÜBITAK 3001 Position for a master s or undergrad student Duration: until March 2019 NMK School on Integer Partitions: May