8. Let f(x) be a polynomial in Q[x] be irreducible over Q, and let F be the splitting field for f(x) over Q. If [F:Q] is odd, prove that all of the roots of f(x) are real.
Solution: Theorem 8.2.6 implies that f(x) has no repeated roots, so Gal(F/Q) has odd order. If u is a nonreal root of f(x), then since f(x) has rational coefficients, its conjugate must also be a root of f(x). It follows that F is closed under taking complex conjugates. Since complex conjugation defines an automorphism of the complex numbers, it follows that restricting the automorphism to F defines a homomorphism from F into F. Because F has finite degree over Q, the homomorphism must be onto as well as one-to-one. Thus complex conjugation defines an element of the Galois group of order 2, and this contradicts the fact that the Galois group has odd order. We conclude that every root of f(x) must be real.
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9. Find an element a with Q( , i) = Q(a).
Solution: It follows from the solution of Problem 8.1.2 that we could take a = / 2 + / 2 i.
Second solution: If we follow the proof of Theorem 8.2.6, we have u = u_{1} = , u_{2} = -, v = v_{1} = i, and v_{2} = -i. The proof shows the existence of an element a with u+av u_{i}+av_{j} for all i and all j 1. To find such an element we need + ai + a(-i) and + ai - + a(-i). The easiest solution is to take a = 1, and so we consider the element a = + i. We have Q Q(a) Q(, i), and since a^{ -1} is in Q(a), we must have (+i)^{ -1} = ( - i) / 3 in Q(a). But then - i belongs, and it follows immediately that and i both belong to Q(a), which gives us the desired equality Q(a) = Q(, i).
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10. Find the Galois group of x^{6}-1 over Z_{7}.
Solution: The Galois group is trivial because x^{6}-1 already splits over Z_{7}.
Comment: Recall that Z_{7} is the splitting field of x^{7}-x = x(x^{6}-1).
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