Bourbaki

Proposition 2

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

Algebra, chapter V, § 11. Abelian extensions

— Let K be algebraically closed and let p be its characteristic exponent. There exists an isomorphism of μ_∞(K) onto the group Q/Z[1/p].

We have denoted by Z[1/p] the subring of Q generated by 1/p, that is, the set of rational numbers of the form a/p^n with a ∈ Z and n ≥ 1; thus we have Z[1/p] = Z if K is of characteristic 0.

Let (v_n), be the strictly increasing sequence consisting of all the integers which are not divisible by p, if p ≠ 1; put X = v_1 v_2 ... v_n and denote by H_n the group of λ_n-th roots of unity; we have H_{n+1} ⊃ H_n and μ_∞(K) = U H_n. Since H_n is cyclic of order X, (Th. 1), there exists a sequence (a_n), of roots of unity such that a_n generates H_n and a_n = α_n^{v_n + 1}.

Further let β_n be the residue class mod Z[1/p] of 1/λ_n and let H'_n be the cyclic subgroup of Q/Z[1/p] generated by β_n. It is clear that β_n = v_{n+1} β_{n+1} and H'_n is of order λ_n because X is not divisible by p if p ≠ 1. Thus there exists for all n ≥ 1 an isomorphism φ_n : H_n → H'_n such that φ_n(α_n) = β_n and the relations α_n = α_{n+1}^{v_n + 1}, β_n = v_{n+1} β_{n+1}, show that φ_n+, extends φ_n. Finally there exists a unique isomorphism φ of μ_∞(K) onto Q/Z[1/p] extending the isomorphisms φ_n, that is, φ(α_n) = β_n for all n ≥ 1.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Transcribed from the printed edition, 2003, Springer. Not translated.