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# hw11 - we see that bd = 1 It is then not diﬃcult to...

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MATH 3360 HOMEWORK SOLUTION 11 (Some computational problems are omitted.) p426, E12 For this problem, it is not ok to simply check the solution from the back of the book is correct. (There won’t be solutions to check in any exam.) First of all, it is easy to see that t 4 + t + 1 does not have roots in F 4 , so it does not have liner factors. If it is reducible, then the only possibility is that it is the product of two irreducible quadratic polynomials. Now, one way is to check through the list of irreducible quadratic polynomials to see whether any of them is a factor. The other way to do this is to assume that: t 4 + t + 1 = ( t 2 + at + b )( t 2 + ct + d ) and solve for a, b, c, d . For example, by looking at the cubic term, we see that a + c = 0 and by looking at the constant term
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Unformatted text preview: we see that bd = 1. It is then not diﬃcult to discuss all possibilities. p426, E21 Since x 2 = 1 has at most 2 solutions, 2 is the only element of order 2. That means, any element β which is not 0 , 1 , 2 is of order either 13 or 26. If β has order 13, then β 13 = 1, then (2 β ) 13 = 2 13 β 13 = 2 6 = 1, so 2 β does not have order 13. Therefore it must have order 26. Suppose that β has order 26, then β 13 6 = 1. In addition, ( β 13 ) 2 = 1, so β 13 = 2. Then (2 β ) 13 = 2 13 β 13 = 2 14 = 1. So 2 β has order 13. Note that the converse direct is direct from the previous argument, using the fact that 2 × 2 β = β . 1...
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