Bourbaki

Exercise 12, § 11

Algebra, V, p. 163

Let K be a field of characteristic q, Ω\Omega an algebraically closed extension of K, x an element of Ω\Omega such that x∉Kx \notin K and M a maximal element of the set of subextensions of Ω\Omega not containing x (V, p. 156, Ex. 9); suppose further that M is perfect so that M(x)M(x) is a Galois extension of prime degree p of M (loc. cit.). a) Show that either p=2p = 2 and Ω=M(i)\Omega = M(i) or for each integer r ≥1\geq 1, there exists a single extension of degree prp^r of M; this extension M, is cyclic over M, Ω\Omega is the union of the M, and the M, are the only extensions of finite degree of M. (Note that if a p-group Γ\Gamma of order prp^r has only one subgroup of order pr−1p^{r-1}, it is necessarily cyclic, arguing by induction on r and using I, p. 133, Ex. 10, and I, p. 77, Cor. of Prop. 11). b) Suppose that Ω≠M(i)\Omega \neq M(i) and that p≠qp \neq q. Show that M contains the p-th roots of unity; we thus have M(x)=M1=M(α)M(x) = M_1 = M(\alpha) with αp∈M\alpha^p \in M. Show that if a is the p-th power of an element of M1M_1, then M1=M(ζ)M_1 = M(\zeta), where ζ\zeta is p2p^2-th root of unity (if a=βpa = \beta^p for β∈M1\beta \in M_1, consider the minimal polynomial of β\beta in M[X]M[X]). c) Under the hypotheses of b) show that if a is not a p-th power of an element of M1M_1 then MrM_r is the splitting field of the polynomial Xpr−1−aX^{p^{r-1}} - a of M1[X]M_1[X] (use b)). d) Under the hypotheses of b), show that if a is the p-th power of an element of M1M_1, there exists y∈M1y \in M_1 such that γ∉M\gamma \notin M and that M2M_2 is the splitting field of Xp−γX^p - \gamma; show that y is not the p-th power of an element of M2M_2 and deduce that MrM_r is the splitting field of the polynomial Xpr−1−γX^{p^{r-1}} - \gamma of M1[X]M_1[X].

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