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

A gentle introduction to Elimination Theory March 2018 @ METU Zafeirakis Zafeirakopoulos

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

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

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

Intro Membership Z.Zafeirakopoulos 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Elimination Ideal Z.Zafeirakopoulos 9

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

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

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

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

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

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

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

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

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

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

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 2 + 0 2 0 2 Z.Zafeirakopoulos 15

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 2 + 0 2 0 2 Z.Zafeirakopoulos 15

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 2 + 0 2 0 2 Z.Zafeirakopoulos 15

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 3 0 0 2 Z.Zafeirakopoulos 15

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 4 0 2 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 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

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

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 6 + 4 3 b2 x 2 1 x 3x 2 5 + 4 3 b2 x 2 1 x 3x 2 6 + 4 3 b2 x 2 1 x 4x 2 5 16 3 b2 x 2 1 x 4x 5 x 6 + 4 3 b2 x 2 1 x 4x 2 6 + 1 6 b2 x 1 x 2 3 x2 5 + 1 6 b2 x 1 x 2 3 x2 6 + 1 6 b 2 x 1 x 2 4 x2 5 + 1 6 b2 x 1 x 2 4 x2 6 2 3 b2 x 1 x 2 5 x2 6 + 1 2 b2 x 1 x 2 5 b2 x 1 x 5 x 6 + 1 2 b2 x 1 x 2 6 b2 x 3 x 2 5 b2 x 3 x 2 6 2 3 b2 x 4 x 2 5 x2 6 b 2 x 4 x5 2 + 14 3 b2 x 4 x 5 x 6 b 2 x 4 x6 2, ax 1 2 1 ax 4x 5 x 6 1 2 ax 4 3 2 bx2 1 x2 4 x 5+ 2 3 bx2 1 x2 4 x 6+ 3 2 bx2 1 x 5 2 3 bx2 1 x 6+ 1 3 bx 1x 3 x 5 1 3 bx 1 x 3 x 6 3 1 bx 1x 4 x5 2 x 6 + 1 3 bx 1x 4 x 5 x6 2 + 6 1 bx2 3 x 5 6 1 bx2 3 x 6 + 3 2 bx2 4 x 5 2 3 bx2 4 x 6 + 6 1 bx2 5 x 6 1 6 bx 5x6 2 1 6 bx 5 + 1 6 bx 6, ax 3 ax 4 x 5 x 6 4 3 bx2 1 x2 4 x 5 + 4 3 bx2 1 x2 4 x 6 + 3 4 bx2 1 x 5 4 3 bx2 1 x 6 + 3 2 bx 1x 3 x 5 2 3 bx 1x 3 x 6 2 3 bx 1x 4 x5 2 x 6 + 2 3 bx 1 x 4 x 5 x6 2 + 1 3 bx2 3 x 5 3 1 bx2 3 x 6+ 4 3 bx2 4 x 5 3 4 bx2 4 x 6+ 1 3 bx2 5 x 6 3 1 bx 5x6 2 3 4 bx 5+ 3 4 bx 6, ax4 2 ax2 5 x2 6 + 4 3 bx2 1 x 4x 5 4 3 bx1 2 x 4x 6 + 2 3 bx 1x4 2 x 5 3 2 bx 1x4 2 x 6 + 3 2 bx 1x5 2 x 6 2 3 bx 1x 5 x6 2 + 1 3 bx 4x5 2 x 6 3 1 bx 4x 5 x6 2 bx 4x 5 + bx 4 x 6, ax 4 x5 2 ax 4 + 4 3 bx2 1 x2 4 x 6 4 3 bx2 1 x 5x 2 6 2 3 bx 1x 4 x 2 5 x 6 + 2 3 bx 1x 4 x 6 bx 2 4 x 6 1 3 bx2 5 x 6 + 4 3 bx 5x 2 6, ax 4x 2 6 ax 4 4 3 bx2 1 x2 4 x 5 + 4 3 bx 2 1 x 5x 2 6 + 2 3 bx 1x 4 x 5 x 2 6 2 3 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 2 6 + 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 2 6 + bx 6, bx 2 1 x2 5 x 6 bx 2 1 x 5x 2 6 Z.Zafeirakopoulos 16

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

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

Resultants Z.Zafeirakopoulos 18

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

Sylvester Resultant Syl(f 1, f 2 ) = f 1,d1 f 1,0...... f 1,d1 f 1,0 f 2,d2 f 2,0............ 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

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

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

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

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

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

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

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

Implicitization Z.Zafeirakopoulos 26

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

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

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

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

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

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

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 2 + 21xy y Sup(f ) = {(3, 1), (2, 0), (1, 2), (1, 1), (0, 1)} y Z.Zafeirakopoulos 30 x

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

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

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

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

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

BKK Does this imply something for the resultant? 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). f 1 = x 2 + (y 1) 2 1 f 2 = y 2 Z.Zafeirakopoulos 36

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

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

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

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

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

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

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

Dual Space Z.Zafeirakopoulos 38

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

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

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

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

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

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

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 1 0.955d 2 0.894d 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, 6.938 10 18, 5.204 10 17, 1.0, 1.0) Z.Zafeirakopoulos 42

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

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 21-25 2018. Z.Zafeirakopoulos 44

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 21-25 2018.