1. Factor x^{6} - 1 over Z_{7}; factor x^{5} - 1 over Z_{11}.
Solution: Since the multiplicative group Z_{7}^{×} has order 6, each nonzero element of Z_{7} is a root of x^{6} - 1. Thus Z_{7} itself is the splitting field of x^{6} - 1. (Of course, this can also be proved directly from Theorem 6.5.2.) Thus we have the factorization x^{6} - 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 x^{5} - 1 in Z_{11} is the same as looking for elements of order 5 in the multiplicative group Z_{11}^{×}. Theorem 6.5.10 shows that Z_{11}^{×} is cyclic, so it contains 4 elements of order 5, which means the x^{5} - 1 must split over Z_{11}. To look for a generator for Z_{11}^{×}, we might as well start with 2. The powers of 2 are 2^{2} = 4, 2^{3} = 8, 2^{4} = 5, 2^{5} = -1, so 2 is in fact a generator of Z_{11}^{×}. The even powers of 2 have order 5, and these are 4, 5, 2^{6} = 9, and 2^{8} = 3. Therefore x^{5} - 1 = (x-1)(x-3)(x-4)(x-5)(x-9).
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