On the complexity of polynomial system solving

Transcription

1 Bruno Grenet LIX École Polytechnique partly based on a joint work with Pascal Koiran & Natacha Portier XXVèmes rencontres arithmétiques de Caen Île de Tatihou, June 30. July 4., 2014

2 Is there a (nonzero) solution? X 2 + Y 2 Z 2 = 0 XZ + 3 XY + YZ + Y 2 = 0 XZ Y 2 = 0 2 / 27

3 Is there a (nonzero) solution? X 2 + Y 2 Z 2 = 0 XZ + 3 XY + YZ + Y 2 = 0 XZ Y 2 = 0 PolSys K Input: f 1,..., f s K[X 1,..., X n ] Question: Is there a K n s.t. f(a) = 0? 2 / 27

4 Is there a (nonzero) solution? X 2 + Y 2 Z 2 = 0 XZ + 3 XY + YZ + Y 2 = 0 XZ Y 2 = 0 PolSys K Input: f 1,..., f s K[X 1,..., X n ] Question: Is there a K n s.t. f(a) = 0? Lower and upper bounds in terms of complexity classes K: Either Z or F q for some q = p s Variants: Homogeneity, number of polynomials 2 / 27

5 Some complexity classes EXP PSPACE PH Π 2 Σ 2 AM MA NP conp Definition P Deterministic polynomial time NP, conp Non-deterministic polynomial time MA, AM Merlin-Arthur, Arthur-Merlin Σ 2, Π 2,PH Polynomial hierarchy PSPACE (Non-)deterministic polynomial space EXP Deterministic exponential time P 3 / 27

6 Homogeneous systems HomPolSys K Input: f 1,..., f s K[X 0,..., X n ], homogeneous Question: Is there a nonzero a K n+1 s.t. f(a) = 0? 4 / 27

7 Homogeneous systems HomPolSys K Input: f 1,..., f s K[X 0,..., X n ], homogeneous Question: Is there a nonzero a K n+1 s.t. f(a) = 0? Proposition For K = Z or F q, PolSys K and HomPolSys K are polynomial-time equivalent. 4 / 27

8 Homogeneous systems HomPolSys K Input: f 1,..., f s K[X 0,..., X n ], homogeneous Question: Is there a nonzero a K n+1 s.t. f(a) = 0? Proposition For K = Z or F q, PolSys K and HomPolSys K are polynomial-time equivalent. Proof. PolSys K P m HomPolSys K : Homogenization HomPolSys K P m PolSys K : New polynomial i X i Y i 1, where Y 0,..., Y n are fresh variables 4 / 27

9 Glimpse of Elimination Theory f 1,..., f s K[X 0,..., X n ], f i = For which γ i,α is there a root? α =d i γ i,α X α 5 / 27

10 Glimpse of Elimination Theory f 1,..., f s K[X 0,..., X n ], f i = α =d i γ i,α X α For which γ i,α is there a root? There exist R 1,..., R h K[γ] s.t. R 1 (γ) = 0. = a 0, R h (γ) = 0 f 1 (a) = 0. f s (a) = 0 5 / 27

11 Two Polynomials m n P = p i X i, Q = q j X j : i=0 j=0 6 / 27

12 Two Polynomials P = m p i X i, Q = i=0 n q j X j : j=0 p m p Res(P, Q) = det p m p 0 q n q Sylvester Matrix q n q 0 6 / 27

13 Two Polynomials P = m p i X i Y m i, Q = i=0 n q j X j Y n j : j=0 p m p Res(P, Q) = det p m p 0 q n q Sylvester Matrix q n q 0 6 / 27

14 More generally f 1,..., f n+1 K[X 0,..., X n ] a unique resultant polynomial Sylvester matrix Macaulay matrices (exponential size) 7 / 27

15 More generally f 1,..., f n+1 K[X 0,..., X n ] a unique resultant polynomial Sylvester matrix Macaulay matrices (exponential size) s polynomials > n + 1 variables several polynomials needed 7 / 27

16 More generally f 1,..., f n+1 K[X 0,..., X n ] a unique resultant polynomial Sylvester matrix Macaulay matrices (exponential size) s polynomials > n + 1 variables several polynomials needed s polynomials < n + 1 variables trivial 7 / 27

17 More generally f 1,..., f n+1 K[X 0,..., X n ] a unique resultant polynomial Sylvester matrix Macaulay matrices (exponential size) s polynomials > n + 1 variables several polynomials needed s polynomials < n + 1 variables trivial Resultant K Input: f 1,..., f n+1 K[X 0,..., X n ], homogeneous Question: Is there a nonzero a K n+1 s.t. f(a) = 0? 7 / 27

18 Upper bounds

19 Hilbert s Nullstellensatz Theorem Let f 1,..., f s K[X 1,..., X n ]. Then a K, f(a) 0 q 1,..., q s K[X], 1 = q 1 f 1 + +q s f s. 9 / 27

20 Hilbert s Nullstellensatz Theorem Let f 1,..., f s K[X 1,..., X n ]. Then a K, f(a) 0 q 1,..., q s K[X], 1 = q 1 f 1 + +q s f s. Sketch of an algorithm. Write q i = q i,α X α where the q i,α s are indeterminates. α D q i f i = 1 is a linear system of D n equations on sd n variables. i Linear systems can be solved in logarithmic space. Do not store the linear system, but compute entries on demand. = PolSys K can be solved in space poly(n log D, log s). 9 / 27

21 Polynomial System Solving in PSPACE a K, f(a) 0 q 1,..., q s s.t. 1 = q 1 f q s f s. Theorem [Kollár 88, Fitchas-Galligo 90] The q i s can be chosen such that deg(q i ) max(3, d) n. 10 / 27

22 Polynomial System Solving in PSPACE a K, f(a) 0 q 1,..., q s s.t. 1 = q 1 f q s f s. Theorem [Kollár 88, Fitchas-Galligo 90] The q i s can be chosen such that deg(q i ) max(3, d) n. Corollary For K = Z or F q, PolSys K belongs to PSPACE. 10 / 27

23 Polynomial System Solving in PSPACE a K, f(a) 0 q 1,..., q s s.t. 1 = q 1 f q s f s. Theorem [Kollár 88, Fitchas-Galligo 90] The q i s can be chosen such that deg(q i ) max(3, d) n. Corollary For K = Z or F q, PolSys K belongs to PSPACE. More specifically, PolSys K DSPACE((n log d log s) O(1) ). 10 / 27

24 Computing the resultant Theorem The resultant is computable in polynomial space. [Canny 87] 11 / 27

25 Computing the resultant Theorem The resultant is computable in polynomial space. Proof idea. The resultant can be expressed as a gcd of n determinants of Macaulay matrices. [Canny 87] Macaulay matrices can be represented by polynomial-size boolean circuits. The determinant can be computed in logarithmic space. 11 / 27

26 Computing the resultant Theorem The resultant is computable in polynomial space. Proof idea. The resultant can be expressed as a gcd of n determinants of Macaulay matrices. [Canny 87] Macaulay matrices can be represented by polynomial-size boolean circuits. The determinant can be computed in logarithmic space. Theorem [Koiran-Perifel 07] The same holds true in Valiant s algebraic model of computation. 11 / 27

27 Macaulay matrices f 1,...,f n+1 K[X 0,..., X n ], homogeneous, of degrees d 1,..., d n D = i (d i 1), M n D = {X α 0 0 Xα n n : α α n = D} 12 / 27

28 Macaulay matrices f 1,...,f n+1 K[X 0,..., X n ], homogeneous, of degrees d 1,..., d n D = i (d i 1), M n D = {X α 0 0 Xα n n : α α n = D} Definition The first Macaulay matrix is defined as follows: Its rows and columns are indexed by M n D; The row indexed by X α represents X α X d i i f i, where i = min{j : d j α j }. Other Macaulay matrices are defined by reordering the f i s. 12 / 27

29 Macaulay matrices f 1,...,f n+1 K[X 0,..., X n ], homogeneous, of degrees d 1,..., d n D = i (d i 1), M n D = {X α 0 0 Xα n n : α α n = D} Definition The first Macaulay matrix is defined as follows: Its rows and columns are indexed by M n D; The row indexed by X α represents X α X d i i f i, where i = min{j : d j α j }. Other Macaulay matrices are defined by reordering the f i s. Resultant : GCD of the determinants of n Macaulay matrices 12 / 27

30 Large determinants Theorem [G.-Koiran-Portier 10-13] Deciding the nullity of the determinant of a matrix represented by a boolean circuit is PSPACE-complete (over any field). 13 / 27

31 Large determinants Theorem [G.-Koiran-Portier 10-13] Deciding the nullity of the determinant of a matrix represented by a boolean circuit is PSPACE-complete (over any field). Proof idea. Let M be a PSPACE Turing Machine and G x M its graph of configurations, with initial configuration c i and accepting configuration c a ; G x M can be represented by a boolean circuit; There exists a path c i c a in G x M iff M accepts x; Let A adjacency matrix of G x M: det(a) 0 c i c a. 13 / 27

32 Large determinants Theorem [G.-Koiran-Portier 10-13] Deciding the nullity of the determinant of a matrix represented by a boolean circuit is PSPACE-complete (over any field). Proof idea. Let M be a PSPACE Turing Machine and G x M its graph of configurations, with initial configuration c i and accepting configuration c a ; G x M can be represented by a boolean circuit; There exists a path c i c a in G x M iff M accepts x; Let A adjacency matrix of G x M: det(a) 0 c i c a. Theorem [Malod 11] The same holds true in Valiant s algebraic model of computation. 13 / 27

33 Arthur, Merlin and PolSys Z Theorem [Koiran 96] Under the Extended Riemann Hypothesis, PolSys Z is in AM. 14 / 27

34 Arthur, Merlin and PolSys Z Theorem [Koiran 96] Under the Extended Riemann Hypothesis, PolSys Z is in AM. L NP iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), (x, y) V. 14 / 27

35 Arthur, Merlin and PolSys Z Theorem [Koiran 96] Under the Extended Riemann Hypothesis, PolSys Z is in AM. L NP iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), (x, y) V. L MA iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), Pr r Σ p( x )((x, y, r) V) 2/3. 14 / 27

36 Arthur, Merlin and PolSys Z Theorem [Koiran 96] Under the Extended Riemann Hypothesis, PolSys Z is in AM. L NP iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), (x, y) V. L MA iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), Pr r Σ p( x )((x, y, r) V) 2/3. L AM iff there exists V P and a polynomial p s.t. for all x, x L Pr r Σ p( x )( y Σ p( x ) (x, r, y) V) 2/3. 14 / 27

37 Arthur, Merlin and PolSys Z Theorem [Koiran 96] Under the Extended Riemann Hypothesis, PolSys Z is in AM. L NP iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), (x, y) V. L MA iff there exists V P and a polynomial p s.t. for all x, x L y Σ p( x ), Pr r Σ p( x )((x, y, r) V) 2/3. L AM iff there exists V P and a polynomial p s.t. for all x, x L Pr r Σ p( x )( y Σ p( x ) (x, r, y) V) 2/3. NP MA AM 14 / 27

38 Polynomial system mod primes Let f = (f 1,..., f s ), with f i Z[X 1,..., X n ]; Let π f (x) be the set of prime numbers x, s.t. f has a root mod p. 15 / 27

39 Polynomial system mod primes Let f = (f 1,..., f s ), with f i Z[X 1,..., X n ]; Let π f (x) be the set of prime numbers x, s.t. f has a root mod p. Theorem There exist polynomial-time computable A and x 0 s.t. If f has no root in C, then π f (x 0 ) A; [Koiran 96] If f has a root in C, then π f (x 0 ) 8A(log A + 3). 15 / 27

40 Polynomial system mod primes Let f = (f 1,..., f s ), with f i Z[X 1,..., X n ]; Let π f (x) be the set of prime numbers x, s.t. f has a root mod p. Theorem There exist polynomial-time computable A and x 0 s.t. If f has no root in C, then π f (x 0 ) A; [Koiran 96] If f has a root in C, then π f (x 0 ) 8A(log A + 3). Proof idea. Using Hilbert s Nullstellensatz, there exists b Z and q i Z[X] such that q 1 f q s f s = b, with log b = exp(s, d). Using effective quantifier elimination, consider a root a of f such that Q(a) = Q/ R where R is small. Roots of R in F p yield roots of f in F p. Use an Effective Chebotarev Density Theorem (ERH) to prove that R has many roots. 15 / 27

41 Arthur-Merlin protocol Theorem Let U be a universe and {S x U : x Σ } a collection of sets s.t. for all x, either S x α U or S x 4α U, and S x NP. Then the following problem is in AM: Given x, does S x 4α U? 16 / 27

42 Arthur-Merlin protocol Theorem Let U be a universe and {S x U : x Σ } a collection of sets s.t. for all x, either S x α U or S x 4α U, and S x NP. Then the following problem is in AM: Given x, does S x 4α U? Proof idea. 4α 1: Arthur chooses y U at random, and asks Merlin a certificate that y S x. If S x U, Pr(y S x ) 1. α 1: Consider a set T of size 4α U and a family of universal hash functions h : U T. 1. Arthur chooses h and t T at random. 2. Merlin must return y S x s.t. h(y) = t, with a certificate 16 / 27

43 Arthur-Merlin protocol Theorem Let U be a universe and {S x U : x Σ } a collection of sets s.t. for all x, either S x α U or S x 4α U, and S x NP. Then the following problem is in AM: Given x, does S x 4α U? Proof idea. 4α 1: Arthur chooses y U at random, and asks Merlin a certificate that y S x. If S x U, Pr(y S x ) 1. α 1: Consider a set T of size 4α U and a family of universal hash functions h : U T. 1. Arthur chooses h and t T at random. 2. Merlin must return y S x s.t. h(y) = t, with a certificate Proof (PolSys Z AM). U = {p x 0 : p is prime}, S f = π f (x 0 ). 16 / 27

44 Lower bounds

45 Lower bounds for non-square systems Proposition [Folklore] For K = Z or F p, PolSys K & HomPolSys K are NP-hard. 18 / 27

46 Lower bounds for non-square systems Proposition [Folklore] For K = Z or F p, PolSys K & HomPolSys K are NP-hard. Proof. Case HomPolSys Fp, with p 2: BoolSys Boolean variables u 1,..., u n Equations u i = True u i = u j u i = u j u k 18 / 27

47 Lower bounds for non-square systems Proposition [Folklore] For K = Z or F p, PolSys K & HomPolSys K are NP-hard. Proof. Case HomPolSys Fp, with p 2: BoolSys Boolean variables u 1,..., u n Equations u i = True u i = u j u i = u j u k HomPolSys K Variables (over F p ) X 0 and X 1,..., X n Polynomials X 2 0 X 2 i for every i > 0 and X 0 (X i + X 0 ) X 0 (X i + X j ) (X i + X 0 ) 2 (X j + X 0 ) (X k + X 0 ) 18 / 27

48 Lower bound for the resultant in char. 0 Proposition [Heintz-Morgenstern 93] Resultant Z is NP-hard. 19 / 27

49 Lower bound for the resultant in char. 0 Proposition [Heintz-Morgenstern 93] Resultant Z is NP-hard. Proof. Partition Z : Input: S = {u 1,..., u n } Z Question: Does there exist S S, u i = u j? i S j/ S 19 / 27

50 Lower bound for the resultant in char. 0 Proposition Resultant Z is NP-hard. Proof. Partition Z : [Heintz-Morgenstern 93] Input: S = {u 1,..., u n } Z Question: Does there exist S S, u i = u j? i S j/ S X 2 1 X2 0 = 0. X 2 n X 2 0 = 0 u 1 X u n X n = 0 19 / 27

51 Lower bound for the resultant in char. 0 Proposition Resultant Z is NP-hard. Proof. Partition Z : [Heintz-Morgenstern 93] Input: S = {u 1,..., u n } Z Question: Does there exist S S, u i = u j? i S j/ S X 2 1 X2 0 = 0. X 2 n X 2 0 = 0 u 1 X u n X n = 0 Note. Partition Fp P 19 / 27

52 Hardness in positive characteristics HomPolSys Fp is NP-hard: # homogeneous polynomials # variables HomPolSys K Variables X 0 and X 1,..., X n over F p Polynomials X 2 0 X 2 i for every i > 0 and X 0 (X i + X 0 ) X 0 (X i + X j ) (X i + X 0 ) 2 (X j + X 0 ) (X k + X 0 ) 20 / 27

53 Hardness in positive characteristics HomPolSys Fp is NP-hard: # homogeneous polynomials # variables Two strategies: Reduce the number of polynomials Increase the number of variables HomPolSys K Variables X 0 and X 1,..., X n over F p Polynomials X 2 0 X 2 i for every i > 0 and X 0 (X i + X 0 ) X 0 (X i + X j ) (X i + X 0 ) 2 (X j + X 0 ) (X k + X 0 ) 20 / 27

54 Hardness in positive characteristics HomPolSys Fp is NP-hard: # homogeneous polynomials # variables Two strategies: Reduce the number of polynomials Increase the number of variables HomPolSys K Variables X 0 and X 1,..., X n over F p Polynomials X 2 0 X 2 i for every i > 0 and X 0 (X i + X 0 ) X 0 (X i + X j ) (X i + X 0 ) 2 (X j + X 0 ) (X k + X 0 ) 20 / 27

55 A randomized reduction s Define g i = α ij f j, 0 i n: f(a) = 0 = g(a) = 0. j=1 21 / 27

56 A randomized reduction Define g i = s α ij f j, 0 i n: f(a) = 0 = g(a) = 0. j=1 Effective Bertini Theorem: There exists F of degree 3 n+1 s.t. the reciprocal holds as soon as F(α) 0. [Krick-Pardo-Sombra 01] 21 / 27

57 A randomized reduction Define g i = s α ij f j, 0 i n: f(a) = 0 = g(a) = 0. j=1 Effective Bertini Theorem: There exists F of degree 3 n+1 s.t. the reciprocal holds as soon as F(α) 0. [Krick-Pardo-Sombra 01] Schwartz-Zippel Lemma: [DeMillo-Lipton, Zippel, Schwartz, 78-80] Pr s(n+1) α F (F(α) = 0) deg(f) q F q 21 / 27

58 A randomized reduction Define g i = s α ij f j, 0 i n: f(a) = 0 = g(a) = 0. j=1 Effective Bertini Theorem: There exists F of degree 3 n+1 s.t. the reciprocal holds as soon as F(α) 0. [Krick-Pardo-Sombra 01] Schwartz-Zippel Lemma: [DeMillo-Lipton, Zippel, Schwartz, 78-80] Pr s(n+1) α F (F(α) = 0) deg(f) q F q Build an extension L/F p with at least 3 n+2 elements [Shoup 90] 21 / 27

59 A randomized reduction Define g i = s α ij f j, 0 i n: f(a) = 0 = g(a) = 0. j=1 Effective Bertini Theorem: There exists F of degree 3 n+1 s.t. the reciprocal holds as soon as F(α) 0. [Krick-Pardo-Sombra 01] Schwartz-Zippel Lemma: [DeMillo-Lipton, Zippel, Schwartz, 78-80] Pr s(n+1) α F (F(α) = 0) deg(f) q F q Build an extension L/F p with at least 3 n+2 elements Choose the α ij s independently at random in L; [Shoup 90] 21 / 27

60 Define g i = A randomized reduction s α ij f j, 0 i n: f(a) = 0 = g(a) = 0. j=1 Effective Bertini Theorem: There exists F of degree 3 n+1 s.t. the reciprocal holds as soon as F(α) 0. [Krick-Pardo-Sombra 01] Schwartz-Zippel Lemma: [DeMillo-Lipton, Zippel, Schwartz, 78-80] Pr s(n+1) α F (F(α) = 0) deg(f) q F q Build an extension L/F p with at least 3 n+2 elements Choose the α ij s independently at random in L; [Shoup 90] Theorem [G.-Koiran-Portier 10-13] Let p be a prime number. Resultant Fq is NP-hard for degree-2 polynomials for some q = p s, under randomized reductions. 21 / 27

61 Hardness in positive characteristics HomPolSys Fp is NP-hard: # homogeneous polynomials # variables Two strategies: Reduce the number of polynomials Increase the number of variables HomPolSys K Variables X 0 and X 1,..., X n over F p Polynomials X 2 0 X 2 i for every i > 0 and X 0 (X i + X 0 ) X 0 (X i + X j ) (X i + X 0 ) 2 (X j + X 0 ) (X k + X 0 ) 22 / 27

62 Hardness in positive characteristics HomPolSys Fp Two strategies: is NP-hard: # homogeneous polynomials # variables Reduce the number of polynomials Increase the number of variables HomPolSys K Variables X 0 and X 1,..., X n over F p Polynomials X 2 0 X 2 i for every i > 0 and X 0 (X i + X 0 ) X 0 (X i + X j ) (X i + X 0 ) 2 (X j + X 0 ) (X k + X 0 ) f 1,..., f n f n+1,..., f s 22 / 27

63 Reduction New variables: Y 1,..., Y s n 1 New system g(x, Y) = 23 / 27

64 Reduction New variables: Y 1,..., Y s n 1 New system f 1 (X). (untouched) f n (X) g(x, Y) = 23 / 27

65 Reduction New variables: Y 1,..., Y s n 1 New system f 1 (X) g(x, Y) =.. (untouched) f n (X) f n+1 (X) +λy1 2 f n+2 (X) Y1 2 +λy2 2 f s 1 (X) Y 2 s n 2 +λy2 s n 1 f s (X) Y 2 s n 1 23 / 27

66 Reduction New variables: Y 1,..., Y s n 1 New system f 1 (X) g(x, Y) =.. (untouched) f n (X) f n+1 (X) +λy1 2 f n+2 (X) Y1 2 +λy2 2 f s 1 (X) Y 2 s n 2 +λy2 s n 1 f s (X) Y 2 s n 1 a root of f = (a, 0) root of g 23 / 27

67 Equivalence? f 1 (a). (a, b) non trivial root of g. f n (a) f n+1 (a) +λb 2 1 f n+2 (a) b 2 1 +λb 2 2 f s 1 (a) b 2 s n 2 +λb2 s n 1 f s (a) b 2 s n 1? = a non trivial root of f 24 / 27

68 Equivalence? f 1 (a). (a, b) non trivial root of g. f n (a) f n+1 (a) +λb 2 1 f n+2 (a) b 2 1 +λb 2 2 f s 1 (a) b 2 s n 2 +λb2 s n 1 f s (a) b 2 s n 1? = a non trivial root of f a = 0 = b = 0 24 / 27

69 Equivalence? f 1 (a). (a, b) non trivial root of g. f n (a) f n+1 (a) +λb 2 1 f n+2 (a) b 2 1 +λb 2 2 f s 1 (a) b 2 s n 2 +λb2 s n 1 f s (a) b 2 s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 24 / 27

70 Equivalence? f 1 (a). (a, b) non trivial root of g. f n (a) f n+1 (a) +λb 2 1 f n+2 (a) b 2 1 +λb 2 2 f s 1 (a) b 2 s n 2 +λb2 s n 1 f s (a) b 2 s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 ɛ i = f n+i (a) 24 / 27

71 Equivalence? (a, b) non trivial root of g ɛ 1 +λb 2 1 ɛ 2 b 2 1 +λb 2 2. ɛ s n 2 b 2 s n 2 +λb2 s n 1 ɛ s n 1 b 2 s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 ɛ i = f n+i (a) 24 / 27

72 Equivalence? (a, b) non trivial root of g ɛ 1 +λb 2 1 ɛ 2 b 2 1 +λb 2 2. ɛ s n 2 b 2 s n 2 +λb2 s n 1 ɛ s n 1 b 2 s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 ɛ i = f n+i (a) B i = b 2 i 24 / 27

73 Equivalence? (a, b) non trivial root of g ɛ 1 +λb 1 ɛ 2 B 1 +λb 2. ɛ s n 2 B s n 2 +λb s n 1 ɛ s n 1 B s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 ɛ i = f n+i (a) B i = b 2 i 24 / 27

74 Equivalence? (a, b) non trivial root of g ɛ 1 +λb 1 ɛ 2 B 1 +λb 2. ɛ s n 2 B s n 2 +λb s n 1 ɛ s n 1 B s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 ɛ i = f n+i (a) B i = b 2 i det = ± ( ɛ 1 + ɛ 2 λ + + ɛ s n λ s n 1) 24 / 27

75 Equivalence? (a, b) non trivial root of g ɛ 1 +λb 1 ɛ 2 B 1 +λb 2. ɛ s n 2 B s n 2 +λb s n 1 ɛ s n 1 B s n 1? = a non trivial root of f a = 0 = b = 0 a 0 = 1 and a i = ±1 ɛ i = f n+i (a) B i = b 2 i det = ± ( ɛ 1 + ɛ 2 λ + + ɛ s n λ s n 1) det = 0? = i, ɛ i = 0 = f 1 (a) = = f s (a) = 0 24 / 27

76 Last step det = ± ( ɛ 1 + ɛ 2 λ + + ɛ N λ N 1) Compute an irreducible polynomial P F p [ξ] of degree N; [Shoup 90] Let L = F p [ξ]/(p) and λ = ξ L. 25 / 27

77 Last step det = ± ( ɛ 1 + ɛ 2 λ + + ɛ N λ N 1) Compute an irreducible polynomial P F p [ξ] of degree N; [Shoup 90] Let L = F p [ξ]/(p) and λ = ξ L. In the extension L, det = 0 ɛ i = 0 for all i. For coefficients in F p instead of L: put P inside the system 25 / 27

78 Last step det = ± ( ɛ 1 + ɛ 2 λ + + ɛ N λ N 1) Compute an irreducible polynomial P F p [ξ] of degree N; [Shoup 90] Let L = F p [ξ]/(p) and λ = ξ L. In the extension L, det = 0 ɛ i = 0 for all i. For coefficients in F p instead of L: put P inside the system Theorem Let p be a prime number. [G.-Koiran-Portier 10-13] Resultant Fp is NP-hard for linear-degree polynomials. 25 / 27

79 Last step det = ± ( ɛ 1 + ɛ 2 λ + + ɛ N λ N 1) Compute an irreducible polynomial P F p [ξ] of degree N; [Shoup 90] Let L = F p [ξ]/(p) and λ = ξ L. In the extension L, det = 0 ɛ i = 0 for all i. For coefficients in F p instead of L: put P inside the system Theorem [G.-Koiran-Portier 10-13] Let p be a prime number. Resultant Fp is NP-hard for linear-degree polynomials. Resultant Fq is NP-hard for degree-2 polynomials for some q = p s. 25 / 27

80 Main results Lower bound Upper bound Z NP-hard AM F p NP-hard PSPACE 26 / 27

81 Main results Lower bound Upper bound Z NP-hard AM F p NP-hard PSPACE Evaluation of the resultant in PSPACE 26 / 27

82 Main results Lower bound Upper bound Z NP-hard AM F p NP-hard PSPACE Evaluation of the resultant in PSPACE Ideal membership problem is EXPSPACE-complete Input: g, f 1,..., f s K[X] Question: Does g belong to f 1,..., f s? [Mayr-Meyer 82] 26 / 27

83 Main results Lower bound Upper bound Z NP-hard AM F p NP-hard PSPACE Evaluation of the resultant in PSPACE Ideal membership problem is EXPSPACE-complete Input: g, f 1,..., f s K[X] Question: Does g belong to f 1,..., f s? PolSys K is NP K complete (BSS model) [Mayr-Meyer 82] 26 / 27

84 Some open questions Reduce the gap between NP and PSPACE in positive characteristics 27 / 27

85 Some open questions Reduce the gap between NP and PSPACE in positive characteristics Derandomize Koiran s theorem: PolSys Z NP? 27 / 27

86 Some open questions Reduce the gap between NP and PSPACE in positive characteristics Derandomize Koiran s theorem: PolSys Z NP? NP-hardness for degree-2 polynomial systems in F p? 27 / 27

87 Some open questions Reduce the gap between NP and PSPACE in positive characteristics Derandomize Koiran s theorem: PolSys Z NP? NP-hardness for degree-2 polynomial systems in F p? Complexity of solving sparse or lacunary polynomial systems? 27 / 27

88 Some open questions Reduce the gap between NP and PSPACE in positive characteristics Derandomize Koiran s theorem: PolSys Z NP? NP-hardness for degree-2 polynomial systems in F p? Complexity of solving sparse or lacunary polynomial systems? Complexity of root finding, especially: Input: f 1,..., f n K[X 0,..., X n ], homogeneous Output: A root a K of f always a solution: PPAD, TFNP,...? 27 / 27

89 Some open questions Reduce the gap between NP and PSPACE in positive characteristics Derandomize Koiran s theorem: PolSys Z NP? NP-hardness for degree-2 polynomial systems in F p? Complexity of solving sparse or lacunary polynomial systems? Complexity of root finding, especially: Input: f 1,..., f n K[X 0,..., X n ], homogeneous Output: A root a K of f always a solution: PPAD, TFNP,...? Thank you! 27 / 27