Bourbaki

Exercise 4, § 11

Algebra, V, p. 161

Let K be a field of characteristic p, q a prime number ≠p\neq p such that K contains the q-th roots of unity and let ζ\zeta be a primitive q-th root of unity. a) Let E be a cyclic extension of K of degree qeq^e and σ\sigma a K-automorphism of E generating Gal(E/K)\mathrm{Gal}(E/K). Let F be the field of degree qe−1q^{e-1} over K such that K⊂F⊂EK \subset F \subset E; there exists 0∈E0 \in E, a root of an irreducible polynomial Xq−aX^q - a in F[X]F[X] such that E=F(0)E = F(0) and θσm=ζθ\theta^{\sigma^m} = \zeta \theta (where m=qe−1m = q^{e-1}). Show that also E=K(0)E = K(0) and 0′′=βθ0'' = \beta \theta where β∈F\beta \in F is such that ασ−1=βq\alpha^{\sigma^{-1}} = \beta^q and NF/K(β)=ζN_{F/K}(\beta) = \zeta (observe that 0′′=βθk0'' = \beta \theta^k with 0<k<q0 < k < q and β∈F\beta \in F, by Ex. 1 ; by calculating θσmq\theta^{\sigma^{mq}}, show that kmq−1≡0(modq)k^{mq} - 1 \equiv 0 \pmod{q} and deduce that k=1k = 1). b) Conversely, let F be a cyclic extension of degree qe−1q^{e-1} of K with e≥2e \geq 2 and let aa be a K-automorphism of F generating Gal(F/K)\mathrm{Gal}(F/K). If there exists an element β∈F\beta \in F such that NF/K(β)=ζN_{F/K}(\beta) = \zeta and if α∈F\alpha \in F is such that ασ−1=βq\alpha^{\sigma^{-1}} = \beta^q, show that for every c∈K∗c \in K^* the polynomial Xq−cαX^q - c \alpha is irreducible in F[X]F[X]. If 8 is one of the roots of this polynomial in an algebraic closure Ω\Omega of F, show that there exists a K-homomorphism σ‾\overline{\sigma} of E=F(0)E = F(0) into Ω\Omega, extending a and such that 8′′=βθ8'' = \beta \theta; deduce that E is a cyclic extension of K of degree qeq^e, that σ‾\overline{\sigma} generates Gal(E/K)\mathrm{Gal}(E/K) and that E=K(0)E = K(0). Finally show that every cyclic extension of degree qeq^e of K, containing F, is the root field of a polynomial Xq−cαX^q - c \alpha in F[X]F[X], for an appropriate c∈K∗c \in K^* (use Th. 3 of V, p. 85). c) Take K to be the field Q of rational numbers. The polynomial X2+1X^2 + 1 is irreducible in Q[X]Q[X]; if i is one of its roots, then F=Q(i)F = Q(i) is a cyclic extension of Q of degree 2 but there exists no cyclic extension of degree 4 of Q containing F.

Solution

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Transcribed from the printed volume. Not translated.