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.