Minimal Polynomials Saravanan Vijayakumaran sarva@ee.iitb.ac.in - PowerPoint PPT Presentation

Minimal Polynomials Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay October 9, 2014 1 / 13

Factoring x q − x over a Field F q and F p Example F = { 0 , 1 , y , y + 1 } ⊂ F 2 [ y ] under + and ∗ modulo y 2 + y + 1 x 4 − x = x ( x − 1 )( x − y )( x − y − 1 ) x ( x + 1 )[ x 2 − x ( y + y + 1 ) + y 2 + y ] = x ( x + 1 )( x 2 + x + 1 ) = The prime subfield of F is F 2 . x , x + 1 , x 2 + x + 1 ∈ F 2 [ x ] are called the minimal polynomials of F Example F 5 = { 0 , 1 , 2 , 3 , 4 } x 5 − x = x ( x − 1 )( x − 2 )( x − 3 )( x − 4 ) The prime subfield of F 5 is F 5 . x , x − 1 , x − 2 , x − 3 , x − 4 ∈ F 5 [ x ] are called the minimal polynomials of F 5 2 / 13

Factoring x q − x over a Field F q and F p • Let F q be a finite field with characteristic p • F q has a subfield isomorphic to F p • Consider the polynomial x q − x ∈ F q [ x ] • Since the prime subfield contains ± 1, x q − x ∈ F p [ x ] • x q − x factors into a product of prime polynomials g i ( x ) ∈ F p [ x ] x q − x = � g i ( x ) i The g i ( x ) ’s are called the minimal polynomials of F q • There are two factorizations of x q − x deg g i ( x ) � � � x q − x = ( x − β ) = g i ( x ) = ⇒ g i ( x ) = ( x − β ij ) β ∈ F q i j = 1 • Each β ∈ F q is a root of exactly one minimal polynomial of F q , called the minimal polynomial of β 3 / 13

Properties of Minimal Polynomials (1) Let F q be a finite field with characteristic p . Let g ( x ) be the minimal polynomial of β ∈ F q . g ( x ) is the monic polynomial of least degree in F p [ x ] such that g ( β ) = 0 Proof. • Let h ( x ) ∈ F p [ x ] be a monic polynomial of least degree such that h ( β ) = 0 • Dividing g ( x ) by h ( x ) , we get g ( x ) = q ( x ) h ( x ) + r ( x ) where deg r ( x ) < deg h ( x ) • Since r ( x ) ∈ F p [ x ] and r ( β ) = 0, by the least degree property of h ( x ) we have r ( x ) = 0 = ⇒ h ( x ) divides g ( x ) • Since g ( x ) is irreducible and deg h ( x ) = deg g ( x ) • Since both h ( x ) and g ( x ) are monic, h ( x ) = g ( x ) 4 / 13

Properties of Minimal Polynomials (2) Let F q be a finite field with characteristic p . Let g ( x ) be the minimal polynomial of β ∈ F q . For any f ( x ) ∈ F p [ x ] , f ( β ) = 0 ⇐ ⇒ g ( x ) divides f ( x ) Proof. • ( ⇐ =) If g ( x ) divides f ( x ) , then f ( x ) = a ( x ) g ( x ) = ⇒ f ( β ) = a ( β ) g ( β ) = 0 • (= ⇒ ) Suppose f ( x ) ∈ F p [ x ] and f ( β ) = 0 • Dividing f ( x ) by g ( x ) , we get f ( x ) = q ( x ) g ( x ) + r ( x ) where deg r ( x ) < deg g ( x ) • Since r ( x ) ∈ F p [ x ] and r ( β ) = 0, by the least degree property of g ( x ) we have r ( x ) = 0 = ⇒ g ( x ) divides f ( x ) 5 / 13

Linearity of Taking p th Power Let F q be a finite field with characteristic p . • For any α ∈ F q , p α = 0 • For any α, β ∈ F q p � p � ( α + β ) p = α j β p − j = α p + β p � j j = 0 • For any integer n ≥ 1, ( α + β ) p n = α p n + β p n i = 0 g i x i ∈ F q [ x ] , • For any g ( x ) = � m g 0 + g 1 x + g 2 x 2 + · · · + g m x m � p n � [ g ( x )] p n = 1 x p n + g p n 2 x 2 p n + · · · + g p n g p n 0 + g p n m x mp n = 6 / 13

Test for Membership in F p [ x ] Let F q be a finite field with characteristic p . F q has a subfield isomorphic to F p . For any g ( x ) ∈ F q [ x ] g p ( x ) = g ( x p ) ⇐ ⇒ g ( x ) ∈ F p [ x ] Note that g ( x ) ∈ F p [ x ] ⇐ ⇒ all its coefficients g i belong to F p Proof. g 0 + g 1 x + g 2 x 2 + · · · + g m x m � p � g p ( x ) = 1 x p + g p 2 x 2 p + · · · + g p g p 0 + g p m x mp = g 0 + g 1 x p + g 2 x 2 p + · · · + g m x mp g ( x p ) = ⇒ g p g p ( x ) = g ( x p ) ⇐ i = g i ⇐ ⇒ g i ∈ F p 7 / 13

Roots of Minimal Polynomials Theorem Let F q be a finite field with characteristic p. Let g ( x ) be the minimal polynomial of β ∈ F q . If q = p m , then the roots of g ( x ) are of the form � β, β p , β p 2 , . . . , β p n − 1 � where n is a divisor of m Proof. We need to show that • There is an integer n such that β p i is a root of g ( x ) for 1 ≤ i < n • n divides m • All the roots of g ( x ) are of this form 8 / 13

Roots of Minimal Polynomials Proof continued. • Since g ( x ) ∈ F p [ x ] , g p ( x ) = g ( x p ) • If β is a root of g ( x ) , then β p is also a root • β p 2 , β p 3 , β p 4 , . . . , are all roots of g ( x ) • Let n be the smallest integer such that β p n = β • All elements in the set β, β p , β p 2 , β p 3 , . . . , β p n − 1 are distinct • If β p a = β p b for some 0 ≤ a < b ≤ n − 1, then β p a � p n − b β p b � p n − b ⇒ β p n + a − b = β p n = β � � = = • If n does not divide m , then m = an + r where 0 < r < n β p m = β = ⇒ β p r = β which is a contradiction 9 / 13

Roots of Minimal Polynomials Proof continued. � β, β p , β p 2 , . . . , β p n − 1 � • It remains to be shown that are the only roots of g ( x ) i = 0 ( x − β p i ) • Let h ( x ) = � n − 1 • h ( x ) ∈ F p [ x ] since n − 1 n − 1 n − 1 ( x − β p i ) p = ( x p − β p i + 1 ) = ( x p − β p i ) = h ( x p ) h p ( x ) = � � � i = 0 i = 0 i = 0 • Since g ( x ) is the least degree monic polynomial in F p [ x ] with β as a root, g ( x ) = h ( x ) Note: The roots of a minimal polynomial are said to form a cyclotomic coset 10 / 13

Minimal Polynomials of F 16 The prime subfield of F 16 is F 2 . x 16 + x = x ( x + 1 )( x 2 + x + 1 )( x 4 + x + 1 )( x 4 + x 3 + 1 )( x 4 + x 3 + x 2 + x + 1 ) • The number of primitive elements of F 16 is φ ( 15 ) = 8 • All the roots of x 4 + x + 1 and x 4 + x 3 + 1 are primitive elements • Let α be a root of x 4 + x + 1. F 16 = { 0 , 1 , α, α 2 , . . . , α 14 } • x has root 0 and x + 1 has root 1 • The roots of x 4 + x + 1 are { α, α 2 , α 4 , α 8 } • The roots of x 2 + x + 1 are { α 5 , α 10 } • The roots of x 4 + x 3 + x 2 + x + 1 are { α 3 , α 6 , α 9 , α 12 } • The roots of x 4 + x 3 + 1 are { α 7 , α 14 , α 13 , α 11 } 11 / 13

Minimal Polynomials of F 16 x 16 + x = x ( x + 1 )( x 2 + x + 1 )( x 4 + x + 1 )( x 4 + x 3 + 1 )( x 4 + x 3 + x 2 + x + 1 ) Power Polynomial Tuple � � 0 0 0 0 0 0 � � 1 1 1 0 0 0 � � 0 1 0 0 α α α 2 α 2 � � 0 0 1 0 α 3 α 3 � � 0 0 0 1 α 4 � � 1 + α 1 1 0 0 α 5 α + α 2 � 0 1 1 0 � α 2 + α 3 α 6 � � 0 0 1 1 α 7 1 + α + α 3 � � 1 1 0 1 α 8 1 + α 2 � 1 0 1 0 � α 9 α + α 3 � � 0 1 0 1 α 10 1 + α + α 2 � � 1 1 1 0 α + α 2 + α 3 α 11 � � 0 1 1 1 1 + α + α 2 + α 3 α 12 � � 1 1 1 1 1 + α 2 + α 3 α 13 � � 1 0 1 1 α 14 1 + α 3 � � 1 0 0 1 12 / 13

Questions? Takeaways? 13 / 13