Gr obner basis - What, Why and How? Tushant Mittal Agenda - PowerPoint PPT Presentation

Gr¨ obner basis - What, Why and How? Tushant Mittal

Agenda Motivational Problems 1 2 Monomial Ordering 3 Division Algorithm 4 Gr¨ obner Basis 5 Buchberger’s Algorithm 6 Complexity 7 Applications 2/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Motivational Problems Ideal Membership Problem Given f ∈ k [ x 1 , x 2 , · · · x n ] and an ideal I = < f 1 , f 2 , · · · , f n > , determine if f ∈ I . 3/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Motivational Problems Ideal Membership Problem Given f ∈ k [ x 1 , x 2 , · · · x n ] and an ideal I = < f 1 , f 2 , · · · , f n > , determine if f ∈ I . Solving Polynomial Equations Find all solution in k n of a system of polynomial equations f i ( x 1 , x 2 , · · · , x n ) = 0. In other words, given an ideal I , compute V ( I ). 3/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Motivational Problems Ideal Membership Problem Given f ∈ k [ x 1 , x 2 , · · · x n ] and an ideal I = < f 1 , f 2 , · · · , f n > , determine if f ∈ I . Solving Polynomial Equations Find all solution in k n of a system of polynomial equations f i ( x 1 , x 2 , · · · , x n ) = 0. In other words, given an ideal I , compute V ( I ). Implicitization Problem Given a parametric solution of x i ’s in terms of variables t i i.e. x i = g i ( t 1 , t 2 , · · · , t i ), find a set of polynomials f i such that x i ∈ V ( < f 1 , f 2 , · · · , f n > ). It can be easily observed that this is essentially the inverse of the above question i.e given V ( I ) compute I . 3/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Motivational Problems Ideal Membership Problem Given f ∈ k [ x 1 , x 2 , · · · x n ] and an ideal I = < f 1 , f 2 , · · · , f n > , determine if f ∈ I . Solving Polynomial Equations Find all solution in k n of a system of polynomial equations f i ( x 1 , x 2 , · · · , x n ) = 0. In other words, given an ideal I , compute V ( I ). Implicitization Problem Given a parametric solution of x i ’s in terms of variables t i i.e. x i = g i ( t 1 , t 2 , · · · , t i ), find a set of polynomials f i such that x i ∈ V ( < f 1 , f 2 , · · · , f n > ). It can be easily observed that this is essentially the inverse of the above question i.e given V ( I ) compute I . But an immediate question arises. How do we even store these ideals which are possibly of infinite size ? 3/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Noetherian Ring A Noetherian ring is a ring that satisfies the ascending chain condition on ideals; that is, given any chain of ideals: I 1 ⊆ · · · ⊆ I k − 1 ⊆ I k ⊆ I k +1 ⊆ · · · there exists an n such that: I n = I n +1 = · · · I n + k ∀ k ≥ 0 4/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Noetherian Ring A Noetherian ring is a ring that satisfies the ascending chain condition on ideals; that is, given any chain of ideals: I 1 ⊆ · · · ⊆ I k − 1 ⊆ I k ⊆ I k +1 ⊆ · · · there exists an n such that: I n = I n +1 = · · · I n + k ∀ k ≥ 0 Equivalently, every ideal I in R is finitely generated, i.e. there exist elements a 1 , ..., an in I such that I = < a 1 , a 2 , · · · , a n > 4/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Noetherian Ring A Noetherian ring is a ring that satisfies the ascending chain condition on ideals; that is, given any chain of ideals: I 1 ⊆ · · · ⊆ I k − 1 ⊆ I k ⊆ I k +1 ⊆ · · · there exists an n such that: I n = I n +1 = · · · I n + k ∀ k ≥ 0 Equivalently, every ideal I in R is finitely generated, i.e. there exist elements a 1 , ..., an in I such that I = < a 1 , a 2 , · · · , a n > Theorem (Hilbert Basis Theorem) R is Noetherian ⇒ R [ x ] is Noetherian 4/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Special Cases R = k [ x ] i.e. n = 1. We know that k [ x ] is a PID. Moreover, it is a Euclidean domain and hence, a polynomial g ∈ < f > iff f | g . 5/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Special Cases R = k [ x ] i.e. n = 1. We know that k [ x ] is a PID. Moreover, it is a Euclidean domain and hence, a polynomial g ∈ < f > iff f | g . Linear Algebra techniques can be used efficiently when the degree of the polynomials is restricted to 1 irrespective of n. 5/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Special Cases R = k [ x ] i.e. n = 1. We know that k [ x ] is a PID. Moreover, it is a Euclidean domain and hence, a polynomial g ∈ < f > iff f | g . Linear Algebra techniques can be used efficiently when the degree of the polynomials is restricted to 1 irrespective of n. We will generalize both the idea of division and a basis to solve the problem for the general case. 5/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Monomial Ordering We will use the notation x α to represent � n i x α i where α = ( α 1 , α 2 , · · · , α n ). i Definition (admissible ordering of monomials) A total ordering on all monomials is an ordering for which holds: x α < x β ⇒ ∀ δ : x α x δ < x β x δ . ∀ α : 1 < x α . 6/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Monomial Ordering We will use the notation x α to represent � n i x α i where α = ( α 1 , α 2 , · · · , α n ). i Definition (admissible ordering of monomials) A total ordering on all monomials is an ordering for which holds: x α < x β ⇒ ∀ δ : x α x δ < x β x δ . ∀ α : 1 < x α . A few popular orderings are: 1. Lexicographical ordering: In which we compare x α and x β thus: if the first k − 1 indices agree, α i = β i , i ≤ k − 1 and the k th differ, we decide based on that index α k ≤ β k ⇒ α ≤ β , and the reverse. 6/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Monomial Ordering We will use the notation x α to represent � n i x α i where α = ( α 1 , α 2 , · · · , α n ). i Definition (admissible ordering of monomials) A total ordering on all monomials is an ordering for which holds: x α < x β ⇒ ∀ δ : x α x δ < x β x δ . ∀ α : 1 < x α . A few popular orderings are: 1. Lexicographical ordering: In which we compare x α and x β thus: if the first k − 1 indices agree, α i = β i , i ≤ k − 1 and the k th differ, we decide based on that index α k ≤ β k ⇒ α ≤ β , and the reverse. 2. Graded lexicographical order: in which the order is by the degree of the monomials and ties are broken using lexicographical ordering. 6/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Preliminary Definitions i a i x α i be a polynomial. Associated with it are the following Let f = � definitions 7/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Preliminary Definitions i a i x α i be a polynomial. Associated with it are the following Let f = � definitions Definition (Multidegree) multideg ( f ) = max i α i 7/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Preliminary Definitions i a i x α i be a polynomial. Associated with it are the following Let f = � definitions Definition (Multidegree) multideg ( f ) = max i α i Definition (Leading Coefficient) LC ( f ) = a multideg ( f ) 7/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Preliminary Definitions i a i x α i be a polynomial. Associated with it are the following Let f = � definitions Definition (Multidegree) multideg ( f ) = max i α i Definition (Leading Coefficient) LC ( f ) = a multideg ( f ) Definition (Leading Monomial) LM ( f ) = x multideg ( f ) 7/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Preliminary Definitions i a i x α i be a polynomial. Associated with it are the following Let f = � definitions Definition (Multidegree) multideg ( f ) = max i α i Definition (Leading Coefficient) LC ( f ) = a multideg ( f ) Definition (Leading Monomial) LM ( f ) = x multideg ( f ) Definition (Leading Term) LT ( f ) = LC ( f ) LT ( f ) 7/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Example Let f = 7 x 3 y 2 z + 2 x 2 yz 4 + 9 xy 4 + 3 yz 7 + 2. Using the lex ordering, multideg ( f ) = (3 , 2 , 1) LC ( f ) = 7 LM ( f ) = x 3 y 2 z LT ( f ) = 7 x 3 y 2 z 8/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Example Let f = 7 x 3 y 2 z + 2 x 2 yz 4 + 9 xy 4 + 3 yz 7 + 2. Using the lex ordering, multideg ( f ) = (3 , 2 , 1) LC ( f ) = 7 LM ( f ) = x 3 y 2 z LT ( f ) = 7 x 3 y 2 z Whereas using the grlex ordering we would get, multideg ( f ) = (0 , 0 , 7) LC ( f ) = 3 LM ( f ) = yz 7 LT ( f ) = 3 yz 7 8/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Division Algorithm The division algorithm is essentially the same as the one in the univariate case but there is a small change which has to be made. To see this, let us look at an example, 9/18 08/04/2017 Tushant Mittal Indian Institute of Technology

Division Algorithm The division algorithm is essentially the same as the one in the univariate case but there is a small change which has to be made. To see this, let us look at an example, a 1 : x + y a 2 : 1 r xy + 1 ) x 2 y + xy 2 + y 2 y 2 + 1 9/18 08/04/2017 Tushant Mittal Indian Institute of Technology