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HW11_solns

# HW11_solns - d β Z we have h a,b i β h d i 3 I = 2 r sx...

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MATH 3360 Sketch of solutions for Assignment 11 3.1 1. ( m 0 + m 1 x + m 2 x 2 )( g 0 + g 1 x ). 2. n = 4, k = 3. 3. G = g 0 g 1 0 0 0 g 0 g 1 0 0 0 g 0 g 1 4. ( m 0 , m 1 , m 2 ) · G . 5. 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 ). 3.2 [graded out of 7 points] 1. x 6 + x 3 + 1. 2. Check that ( x + x 3 + x 6 + x 7 )( x 6 + x 3 + 1) 0 mod( x 9 - 1). 3. 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 4. Check that H · (0 , 1 , 0 , 1 , 0 , 0 , 1 , 1 , 0) (0 , 0 , 0)(mod 2). 5. 010110, using the fact that x 7 + x 6 + x 3 + x = ( x 3 + 1)( x 4 + x 3 + x ). 6. n = 9, k = 6. 3.3 [graded out of 5 points] 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 . Additional Problem 1 It is clearly cyclic. g ( x ) = x + 1. 1

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Additional Problem 3 [graded out of 8 points] 1. For any ra + sb, r 0 a + s 0 b ∈ h a, b i , ( r + r 0 ) a + ( s + s 0 ) b ∈ h a, b i and ( kr ) a + ( ks ) b ∈ h a, b i ∀ k R . 2. d = gcd( a, b ). By Bezout, we know that r, s Z such that ra + sb = d , so h d i ⊆ h a, b i . Also, for any element r 0 a + s 0 b ∈ h a, b i , r 0 a + s 0 b = ( r 0 a d + s 0 b d ) d . Since r 0 a d + s 0 b d Z , we have h a, b i ⊆ h d i . 3. I = { 2 r + sx | r, s Z [ x ] } . If f ( x ) I , f (0) = 2 r + s ·
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Unformatted text preview: d β Z , we have h a,b i β h d i . 3. I = { 2 r + sx | r,s β Z [ x ] } . If f ( x ) β I , f (0) = 2 r + s Β· 0 = 2 r β‘ 0 mod 2. So, I β { f ( x ) β Z [ x ] | f (0) β‘ 0 mod 2 } . For the other direction, note that for f ( x ) to be in { f ( x ) β Z [ x ] | f (0) β‘ 0 mod 2 } , implies that the constant term is even, so f ( x ) = 2 r + x Β· h ( x ), where h ( x ) is any polynomial.Thus, we get { f ( x ) β Z [ x ] | f (0) β‘ 0 mod 2 } β I . 4. Since 2 β I , the only way for I to be principal is if I = h 2 i (impossible since x / β h 2 i ) or I = h 1 i (impossible since 1 / β I ). 2...
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