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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. 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....
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This note was uploaded on 02/16/2010 for the course MATH 113 taught by Professor Ogus during the Spring '08 term at Berkeley.
 Spring '08
 OGUS
 Math

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