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Unformatted text preview: Math 235 â€“ Algebra I Solutions to Assignment 6 October 30, 2007 Solution of 1. (1) By Fermatâ€™s Little Theorem, we know that 2 10 â‰¡ 1 (mod 11), so (2 19808 + 6) 1 + 1 â‰¡ ((2 10 ) 1981 Â· 2 2 + 6) 1 + 1 (mod 11) â‰¡ (1 Â· 4 1 + 6) 1 + 1 (mod 11) â‰¡ (3 + 6) 1 + 1 (mod 11) â‰¡ ( 2) 1 + 1 (mod 11) â‰¡  6 + 1 (mod 11) â‰¡  5 â‰¡ 6 (mod 11) (2) We remark that 5 Â· 29 = 145, so that 12 2 = 144 â‰¡  1 (mod 29). It follows that 12 4 â‰¡ 12 8 â‰¡ 12 16 â‰¡ 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 Â· 12 8 Â· 12 16 â‰¡ 12 Â· 1 Â· 1 â‰¡ 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 Â· f ( x ) + 1 Â· 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 ) ....
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 Fall '07
 Goren
 Algebra, Ring, Prime number, Euclidean algorithm, BB BB BB

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