Bourbaki

Exercise 8, § 10

Tag 06NW, permanent. Cite it as [Bourbaki, Tag 06NW].

Algebra, V, p. 156

Let K be a field and N a Galois extension of K of finite degree nn. a) Let pp be a prime number dividing nn and let p′p' be the highest power of pp dividing nn. Show that there exist subextensions LiL_i of N for 0≤i≤r0 \leq i \leq r such that for 0≤i≤r−10 \leq i \leq r - 1, LiL_i contains Li−1L_{i-1}, L0=NL_0 = N, LiL_i is a Galois extension of degree pp of Li−1L_{i-1}, and [Li:K][L_i : K] is not divisible by pp (cf. I, p. 77, Th. 1 and I, p. 78, Th. 2). b) Let x∈Nx \in N be such that N is generated by x and its conjugates. Show that there exists a subextension E of N such that x∉Ex \notin E and [N:E]=p′[N : E] = p'. (Consider the Sylow p-groups HjH_j of Gal(N/K)\mathrm{Gal}(N/K) and the (normal) subgroup H generated by their union ; if FjF_j (resp. F) is the subfield of invariants of H, (resp. H) show that x∉Fx \notin F and deduce that there is an index j such that x∉Fjx \notin F_j; then we can take E=FjE = F_j.) Conclude that there exists a subextension L of N such that L(x)L(x) is a Galois extension of degree pp of L (use I, p. 77, Prop. 12). c) Conversely, let Ω\Omega be an algebraic closure of N and suppose that there exists a subextension L of Ω\Omega such that L(x)L(x) is Galois of degree pp over L. Show that then pp divides nn. (Observe that N∩L(x)=(N∩L)(x)N \cap L(x) = (N \cap L)(x) and that (N∩L)(x)(N \cap L)(x) is a Galois extension of degree pp of N∩LN \cap L.)

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