Bourbaki

Theorem 2 (Gauss)

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

Algebra, chapter V, § 11. Abelian extensions

— Let QQ be an algebraic closure of QQ and let n≥1n \geq 1 be an integer.

a) The cyclotomic polynomial Φn(X)\Phi_n(X) is irreducible in Q[X]Q[X].

b) The degree of Rn(Q)R_n(Q) over QQ is φ(n)\varphi(n).

c) The homomorphism χn\chi_n of Gal(Qˉ/Q)\mathrm{Gal}(\bar{Q}/Q) into (Z/nZ)∗(\mathbf{Z}/n\mathbf{Z})^* is surjective and defines by passage to quotients an isomorphism of Gal(Rn(Q)/Q)\mathrm{Gal}(R_n(Q)/Q) onto (Z/nZ)∗(\mathbf{Z}/n\mathbf{Z})^*.

Taking account of Prop. 6, we need only prove c). Every integer rr prime to nn is a product of prime numbers p1,…,psp_1, \ldots, p_s not dividing nn; so it is enough to show that for every prime number pp not dividing nn, the mapping x↦xpx \mapsto x^p of μn(Q)\mu_n(Q) into itself extends to an automorphism of Rn(Q)R_n(Q). It suffices to prove that if ζ\zeta is a primitive nn-th root of unity, the minimal polynomial ff of ζ\zeta over QQ is equal to the minimal polynomial gg of ζp\zeta^p over QQ.

We shall assume that f≠gf \neq g and argue by contradiction. The polynomials ff and gg are monic irreducible in Q[X]Q[X] and divide Xn−1X^n - 1, so there exists u∈Q[X]u \in Q[X] such that Xn−1=fguX^n - 1 = fg u (IV, p. 13, Prop. 13). Lemma 2 (V, p. 82) shows that f,gf, g and uu have integer coefficients. Let us denote by vˉ\bar{v} the polynomial with coefficients in Fp\mathbf{F}_p obtained from a polynomial v∈Z[X]v \in \mathbf{Z}[X] by reduction mod pp. We thus have Xn−1=fˉgˉuˉX^n - 1 = \bar{f} \bar{g} \bar{u} in Fp[X]\mathbf{F}_p[X].

Further we have g(ζp)=0g(\zeta^p) = 0 and so g(Xp)g(X^p) is a multiple of f(X)f(X) in Q[X]\mathbf{Q}[X]. By Lemma 2 there exists h∈Z[X]h \in \mathbf{Z}[X] such that g(Xp)=f(X)⋅h(X)g(X^p) = f(X) \cdot h(X). Now we have v(Xp)=v(X)pv(X^p) = v(X)^p for every polynomial v∈Fp[X]v \in \mathbf{F}_p[X]. By reduction mod pp we thus obtain gˉp=fˉhˉ\bar{g}^p = \bar{f} \bar{h}. If vv is an irreducible polynomial in Fp[X]\mathbf{F}_p[X] dividing fˉ\bar{f}, it must then divide gˉ\bar{g}. Since fˉgˉ\bar{f} \bar{g} divides Xn−1X^n - 1, we conclude that v2v^2 divides Xn−1X^n - 1 in Fp[X]\mathbf{F}_p[X]. This is absurd because the polynomial Xn−1X^n - 1 is separable in Fp[X]\mathbf{F}_p[X].

It may be shown that for every abelian extension EE of finite degree over QQ there exists an integer n≥1n \geq 1 such that EE is isomorphic to a subextension of Rn(Q)R_n(Q). \* In other words, the field Q(μ∞(C))Q(\mu_{\infty}(C)) is an abelian closure of QQ. *(« Kronecker-Weber theorem »).

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

Cited by

Nothing.

Cites

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