HW12sol - elements { a 1 , a 2 , . . . , a n } . If any...

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MATH 3360 – Applicable Algebra. HW 12 Solutions (1) Problem 3.1. Solution: (a) ( m 0 + m 1 x + m 2 x 2 ) · ( g 0 + g 1 x ). (b) n = 4 , k = n - r = 4 - 1 = 3. (c) G = g 0 g 1 0 0 0 g 0 g 1 0 0 0 g 0 g 1 (d) ( m 0 , m 1 , m 2 ) · G . (e) Both give ( g 0 m 0 , g 1 m 0 + g 0 m 1 , g 0 m 2 + g 1 m 1 , g 1 m 2 ). (2) Problem 3.2. [graded] Solution: (a) x 9 - 1 divided by 1 + x 3 gives h ( x ) = x 6 + x 3 + 1. (b) Check that ( x + x 3 + x 6 + x 7 )( x 6 + x 3 + 1) 0 mod x 9 - 1 . (c) H = 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 (d) Check that H · (0 , 1 , 0 , 1 , 0 , 0 , 1 , 1 , 0) T (0 , 0 , 0) mod 2. (e) 010110, using the fact that x 7 + x 6 + x 3 + x = ( x 3 + 1)( x 4 + x 3 + x ). (f) n = 9 , k = 9 - 3 = 6. (3) Problem 3.3. [graded] Solution: Note that the generating polynomial must divide gcd(1 + x 7 , 1 + x + x 3 ) = 1 + x + x 3 . Thus G = 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 So, the basis are the rows of G. (4) Problem 3.6. [graded] Solution: By the definition of an ideal I R . Also, since 1 I , we know that r · 1 = r I for all r R . Thus, I = R . (5) Problem 3 page 390. Solution: Let a be any element of G , where G is a finite group of order n . Consider the set of
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Unformatted text preview: elements { a 1 , a 2 , . . . , a n } . If any element in this set is the identity, we are done. Otherwise, there are two elements in the set above, say a s and a t , that are the same. Then a t-s is the identity element and is part of the set. Hence, we are done. (6) Problem 4 page 390. [graded] Solution: Let d be the order of a and let m be such that a m = 1. Say, m = dq + r , where 0 r < d . Then, 1 = a m = a dq + r = ( a d ) q a r = a r , which implies that r = 0 since d was the smallest positive power of a that gives 1. Thus, d | m . 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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