hw9soln - Math 113 HW #9 Solutions Fraleigh 23.8. We first...

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Unformatted text preview: Math 113 HW #9 Solutions Fraleigh 23.8. We first try to find a single generator. We have 2 11 = 1, so 2 is not a generator. The same problem occurs with 3 and 4. Lets try- 2 = 21 instead. The successive powers of- 2 are- 2, 4,- 8, 16,- 9, 18,- 13, 3,- 6, 12,- 1, 2,- 4, 8,- 16, 9,- 18, 13,- 3, 6,- 12, 1. So- 2 has order 22 and is a generator. By Corollary 6.16, the set of all generators consists of the elements (- 2) n where n is relatively prime to 22, i.e. n = 1, 3, 5, 7, 9, 13, 15, 17, 19, 21. Reading these off from the above list, the generators are- 2,- 8,- 9,- 13,- 6,- 4,- 16,- 18,- 3,- 12. Adding 23 to these to make them positive and sorting, we get 5, 7, 10, 11, 14, 15, 17, 19, 20, 21. Fraleigh 23.10. To do this we look for a root in Z 7 by trying all 7 possibilities, then divide by x- , then repeat this process with the quotient. 0 and 1 are not roots, but 2 is, so we divide to get x 3 + 2 x 2 + 2 x + 1 = ( x- 2)( x 2 + 4 x + 3). Now we see that the quotient x 2 + 4 x + 3 factors over Z , and hence over Z 7 , as ( x + 1)( x + 3). Thus the final answer is ( x- 2)( x + 1)( x + 3). Fraleigh 23.18. Yes, use p = 3. Fraleigh 23.19. Yes, use p = 3 again. Fraleigh 23.25. (a) True because it has degree 1 and we are over a field. (b) True as in (a). (c) True by Eisenstein with p = 3. (d) False, because 5 is a root, so in Z 7 [ x ] we have x 2 +3 = ( x +2)( x- 2). (e) True by a previous problem since the units in a field are the nonzero elements, by the definition of a field. (f) True; this seems to be exactly the same as (e). (?!?) (g) True; this follows from the factor theorem as we showed in class. (h) True. This follows from (g) since a polynomial in F [ x ] can be regarded as a polynomial in E [ x ] also....
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hw9soln - Math 113 HW #9 Solutions Fraleigh 23.8. We first...

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