1. Factor x6 - 1 over Z7; factor x5 - 1 over Z11.
Solution: Since the multiplicative group Z7× has order 6, each nonzero element of Z7 is a root of x6 - 1. Thus Z7 itself is the splitting field of x6 - 1. (Of course, this can also be proved directly from Theorem 6.5.2.) Thus we have the factorization x6 - 1 = x(x-1)(x+1)(x-2)(x+2)(x-3)(x+3).
In solving the second half of the problem, looking for roots of x5 - 1 in Z11 is the same as looking for elements of order 5 in the multiplicative group Z11×. Theorem 6.5.10 shows that Z11× is cyclic, so it contains 4 elements of order 5, which means the x5 - 1 must split over Z11. To look for a generator for Z11×, we might as well start with 2. The powers of 2 are 22 = 4, 23 = 8, 24 = 5, 25 = -1, so 2 is in fact a generator of Z11×. The even powers of 2 have order 5, and these are 4, 5, 26 = 9, and 28 = 3. Therefore x5 - 1 = (x-1)(x-3)(x-4)(x-5)(x-9).
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