Polynomial Resultants Henry Woody May 2, 2016 The Resultant - PowerPoint PPT Presentation

Polynomial Resultants Henry Woody Polynomial Resultants Henry Woody May 2, 2016

The Resultant Polynomial Resultants Henry Woody Definition: For f ( x ) = a n x n + ... + a 1 x + a 0 , g ( x ) = b m x m + ... + b 1 x + b 0 ∈ F [ x ], � Res( f , g , x ) = a m n b n ( α i − β j ) , m i , j where f ( α i ) = 0 for 1 ≤ i ≤ n , and g ( β j ) = 0 for 1 ≤ j ≤ m .

Common Factor Lemma Polynomial Resultants Henry Woody Let f ( x ) , g ( x ) ∈ F [ x ] have degrees n and m , both greater than zero, respectively. Then f ( x ) and g ( x ) have a non-constant common factor if and only if there exist nonzero polynomials A ( x ) , B ( x ) ∈ F [ x ] such that deg( A ( x )) ≤ m − 1, deg( B ( x )) ≤ n − 1 and A ( x ) f ( x ) + B ( x ) g ( x ) = 0.

Proof Polynomial Resultants Henry Woody ( = ⇒ ) Assume h ( x ) ∈ F [ x ] is a common factor of f ( x ) and g ( x ), then f ( x ) = h ( x ) f 1 ( x ) and g ( x ) = h ( x ) g 1 ( x ). Consider, g 1 ( x ) f ( x ) + ( − f 1 ( x )) g ( x ) = g 1 ( x )( h ( x ) f 1 ( x )) − f 1 ( x )( h ( x ) g 1 ( x )) = 0

Proof Polynomial Resultants Henry Woody ( ⇐ = ) Assume A ( x ) and B ( x ) exist. Assume, contrary to the lemma, that f ( x ) and g ( x ) share no non-constant factors. Then gcd( f ( x ) , g ( x )) = r ( x ) f ( x ) + s ( x ) g ( x ) = 1

Polynomial Resultants Henry Woody Let f ( x ) = a n x n + ... + a 1 x + a 0 , a n � = 0 g ( x ) = b m x m + ... + b 1 x + b 0 , b m � = 0 A ( x ) = c m − 1 x m − 1 + ... + c 1 x + c 0 , B ( x ) = d n − 1 x n − 1 + ... + d 1 x + d 0 . And A ( x ) f ( x ) + B ( x ) g ( x ) = 0

The Sylvester Matrix Polynomial Resultants Syl( f , g , x ) Henry Woody a n b m   a n − 1 a n b m − 1 b m   ... ...   a n − 2 a n − 1 b m − 2 b m − 1     . . . . ... ...   . . . . . . a n . . b m     . . . .  . . . .  . . a n − 1 . . b m − 1     a 0 a 1 b 0 b 1     . . ... ...  . .  . . a 0 b 0     ... ...   a 1 b 1   a 0 b 0

Properties of the Sylvester Matrix Polynomial Resultants Henry Woody The determinant of the Sylvester matrix Syl( f , g , x ) is a polynomial in the coefficients a i , b j of the polynomials f ( x ) and g ( x ). Further, det(Syl( f , g , x )) = Res( f , g , x ) . For f ( x ) , g ( x ) ∈ F [ x ], there exist polynomials A ( x ) , B ( x ) ∈ F [ x ] so that A ( x ) f ( x ) + B ( x ) g ( x ) = Res( f , g , x ).

Applications: The Discriminant Polynomial Resultants Henry Woody For a polynomial f ( x ) ∈ F [ x ], where f ( x ) = a n x n + ... + a 1 x + a 0 , the discriminant is given by D = ( − 1) n ( n − 1) / 2 Res( f , f ′ , x ) , a n where f ′ ( x ) is the derivative of f ( x ).

Discriminant Example Polynomial Resultants Henry Woody Let f ( x ) = ax 2 + bx + c , then f ′ ( x ) = 2 ax + b . � � a 2 a 0 D = ( − 1) 2(2 − 1) / 2 � � = − 1 a ( a ( b 2 ) − b (2 ab ) + c (4 a 2 )) � � b b 2 a � � a � � 0 c b � � = − 1 a ( ab 2 − 2 ab 2 + 4 a 2 c ) = − 1 a ( − ab 2 + 4 a 2 c ) = b 2 − 4 ac ,

Applications: Elimination Polynomial Resultants Henry Woody Res i : F [ x 1 , ..., x n ] × F [ x 1 , ..., x n ] → F [ x 1 , ..., x i − 1 , x i +1 , ..., x n ], where Res i is the resultant relative to the variable x i .

Elimination Example Polynomial Resultants Henry Woody Let f ( x , y ) = x 2 y 2 − 25 x 2 + 9 and g ( x , y ) = 4 x + y be two polynomials in F [ x , y ].

Partial Solutions Polynomial Resultants Henry Woody Theorem: If ( α 1 , ..., α i − 1 , α i +1 , ..., α n ) is a solution to a homogeneous system of polynomials in F [ x 1 , ..., x i − 1 , x i +1 , ..., x n ] obtained by taking resultants of polynomials in F [ x 1 , ..., x n ] with respect to x i , then there exists α i ∈ E , where E is the field in which all polynomials in the system split, such that ( α 1 , ..., α i , ..., α n ) is a solution to the system in F [ x 1 , ..., x n ].

The End Polynomial Resultants Henry Woody Thank You. This is the end. Questions? Comments?