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

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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. Let’s 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 α in Z 7 by trying all 7 possibilities, then divide by x - α , 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. (i) True. A polynomial of degree 1 in F [ x ] has the form ax + b with a 6 = 0, and then - b/a is a zero. (j) True; this follows from (g).
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