This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 235 Algebra I Solutions to Assignment 6 October 30, 2007 Solution of 1. (1) By Fermats Little Theorem, we know that 2 10 1 (mod 11), so (2 19808 + 6) 1 + 1 ((2 10 ) 1981 2 2 + 6) 1 + 1 (mod 11) (1 4 1 + 6) 1 + 1 (mod 11) (3 + 6) 1 + 1 (mod 11) ( 2) 1 + 1 (mod 11)  6 + 1 (mod 11)  5 6 (mod 11) (2) We remark that 5 29 = 145, so that 12 2 = 144  1 (mod 29). It follows that 12 4 12 8 12 16 1 (mod 29). One way to compute 12 25 quickly is to express 25 in base 2, 25 = 1 + 8 + 16, giving 12 25 = 12 1 12 8 12 16 12 1 1 12 (mod 29) . Solution of 2. (1) Write f ( x ) = x 4 x 3 x 2 + 1 , g ( x ) = x 3 1 as polynomials in Q [ x ] . By applying the Euclidian Algorithm, one finds ( f ( x ) , g ( x )) = x 1 = ( x + 1) f ( x ) + (2 x 2 ) g ( x ) Here are the steps: f ( x ) = ( x 1) g ( x ) ( x 2 x ); g ( x ) = ( x + 1)( x 2 x ) + ( x 1); x 2 x = x ( x 1) + 0 . (2) Write f ( x ) = x 5 + x 4 + 2 x 3 x 2 x 2 , g ( x ) = x 4 + 2 x 3 + 5 x 2 + 4 x + 4 as polynomials in Q [ x ] . By applying the Euclidian Algorithm, one finds ( f ( x ) , g ( x )) = x 2 + x + 2 = 1 4 ( x + 2) f ( x ) + 1 4 (3 x x 2 ) g ( x ) . Here are the steps: f ( x ) = ( x 1) g ( x ) ( x 3 + x 2); g ( x ) = ( x + 2)( x 3 + x 2) + 4( x 2 + x + 2); x 3 + x 2 = ( x 1)( x 2 + x + 2) + 0 . 1 (3) We have this time f ( x ) = x 4 + 3 x 3 + 2 x + 4 , g ( x ) = x 2 1 , as polynomials in Z / 5 Z [ x ] . One easily verifies that f ( x ) = ( x 2 + 3 x + 1) g ( x ) in Z / 5 Z [ x ], and since g ( x ) is monic, ( f ( x ) , g ( x )) = x 2 1 = 0 f ( x ) + 1 g ( x ) . (d) In this case the two polynomials are f ( x ) = 4 x 4 + 2 x 3 + 6 x 2 + 4 x + 5 , g ( x ) = 3 x 3 + 5 x 2 + 6 x with coefficients in Z / 7 Z . Their GCD is found to be ( f ( x ) , g ( x )) = x 1 = 3( x 1) f ( x ) + 3( x 2 x + 3) g ( x ) , and the steps are: f ( x ) = xg ( x ) + 5( x 2 2 x + 1); g ( x ) = (3 x + 4)( x 2 2 x + 1) + 4( x 1); x 2 2 x + 1 = ( x 1)( x 1) + 0 . (Remember that the GCD is monic!) (5) In this case we have f ( x ) = x 3 ix 2 + 4 x 4 i, g ( x ) = x 2 + 1 in C [ x ]. This time the GCD turns out to be ( f ( x ) , g ( x )) = x i = 1 3 f ( x ) 1 3 ( x i ) g ( x ) And the steps are: f ( x ) = ( x i ) g ( x ) + 3( x i ); g ( x ) = ( x i )( x + i ) + 0 . (6) Finally, put f ( x ) = x 4 + x + 1 , g ( x ) = x 2 + x + 1 in Z / 2 Z [ x ]. Once again, by performing the Division Algorithm, one finds that f ( x ) = ( x 2 + x ) g ( x ) + 1 , so that ( f ( x ) , g ( x )) = 1 = f ( x ) ( x 2 + x ) g ( x ) ....
View Full
Document
 Fall '07
 Goren
 Algebra

Click to edit the document details