By Perrine C. D.

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Thus it remains to show that deg rha (X) = dcg ma(X), In fact we show that rhaiX) = ma(X). 12. Consider a/. , there is an automorphism a/ of E with at {a) = at and at | F = id. But B = Fix(Gal(E/F)) so a, | B = id, and hence a, e Gal(E/B). , X — at is a factor of rhaiX) for each /, and so rhaiX) = niaiX), as required. 8 The Fundamental Theorem of Galois Theory In this section we reach our first main goal. Before we get there, we have some work to do. We let F* denote the set of nonzero elements of F, and recall that F* is a group under multiplication.

The construction in part (4) of this example always gave 53. Here is a case where the Galois group is Z/3Z. Let f be a primitive ninth root of 1 and consider the polynomial f(X) = (X - (? + r ' ) ) ( ^ - ( ? ' + r ' ) ) ( ^ - ( ? ' + r ' ) ) = X^ — 3X + 1 € Q[X]. , a root in Q, and it does not). Let AQ, AI, and X2 be the roots of this polynomial, in the order written, and let E = F(Ao, Ai, A2) be its splitting field. Then (E/Q) = deg f(X) = 3, and since | Gal(E/Q)| = (E/Q), the Galois group must be isomorphic to Z/3Z.

Suppose J F C / I ) > 0. Then fi(X) has an irreducible factor gi(X) of degree greater than 1. Let ai be a root of gi(X) in Ei and let 0^2 be a root of aoigiiX)) in E2. Then Ei D F(ai) and E2 2 F(a2). 1, there is an isomorphism ai : F(ai) -^ F(cr2) with ai | F = CTO and cri(ai) = 012. 7 Normal, Separable, and Galois Extensions 25 Set B = F(ai), and consider / i ( Z ) e B[X]. Now gi(X) e B[X] has the factor X — Qfi, so fi(X) has at least k + I irreducible factors in B[X] and hence d^{fi) < d^ifi).