HW9sol - x x 1 x 2 x 1 x 3 x 2 1 x 3 x 1 x 4 x 3 1 x 4 x 1...

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MATH 3360 – Applicable Algebra. HW 9 Solutions (1) Problem 11 on page 298. Solution: One example can be R = Z / m Z , f ( x ) = x 2 . For a given n , we let m be a product of squares of n distinct prime numbers. (What are the roots?) (2) Problem 18 on page 301. Solution: This follows by induction on Euclid’s Algorithm for polynomials and goes along the same lines as the proof for Bezout’s identity for numbers that can be found on pages 40-41. (3) Problem 23 on page 302. Solution: A solution is r ( x ) = x 6 + x 5 + x 2 and s ( x ) = x 4 + x 3 + x 2 + x + 1. (4) Problem 25(i) on page 302. (graded) Solution: Euclidean algorithm gives: x 5 + 2 x 3 + x 2 + x + 1 = ( x 4 + 2 x 3 + x + 1)( x + 1) - x x 4 + 2 x 3 + x + 1 = ( - x )( - x 3 + x 2 - 1) + 1 So, the gcd( x 5 + 2 x 3 + x 2 + x + 1 , x 4 + 2 x 3 + x + 1) = 1. [Note: the back of the book has a wrong answer.] (5) Problem 44 on page 305. Solution: Use Fermat’s Theorem and the Root Theorem to show that ( x - a ) ± ± ± x p - x for all 0 a p - 1. Then use a degree argument to show that these are the only factors. (6) Problem 46 on page 305. (graded) Solution: This is easily done using the Root Theorem to get that
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Unformatted text preview: x , x + 1 , x 2 + x + 1 , x 3 + x 2 + 1 , x 3 + x + 1 , x 4 + x 3 + 1 , x 4 + x + 1 , x 4 + x 3 + x 2 + x + 1 are irreducibles. Note that x 4 + x 2 + 1 = ( x 2 + x + 1) 2 is not irreducible. (7) Problem 47(iii) on page 305. (graded) Solution: x 7 + x 6 + x 4 + 1 = ( x + 1)( x 2 + x + 1)( x 4 + x 3 + 1). Note that these factors are irreducibles by problem 46 above. (8) Problem 2 on page 363. Solution: Units: 1 , 2 , x + 2 , 2 x + 1. Zero divisors: x , 2 x , x + 1 , 2 x + 2. (9) Problem 14(iii) on page 363. (graded) Solution: This can be done by noting that x 2 ≡ -2 mod x 2 + 2, and so x 9 ≡ ( x 2 ) 4 · x ≡ 16 x mod x 2 + 2 . So the question becomes to find f ( x ) such that 16 x f ( x ) ≡ 1(mod x 2 + 2). Solving this congruence gives f ( x ) =-1 32 x . 1...
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