Splitting fields of representations of generalized symmetric groups, 4

This post if an aside on cyclotomic fields and Tchebysheff polynomials. Though it seems certain this material is known, I know of no reference.

Let n denote a positive integer divisible by 4, let r=\cos(2\pi/n), s=\sin(2\pi/n), and let d=n/4. If

T_1(x)=x,\ \ T_2(x)=2x^2-1,\ \ T_3(x)=4x^3-3x,\ \  T_4(x)=8x^4-8x^2+1,\ \ ...,

denote the Tchebysheff polynomials (of the 1st kind), defined by


then T_d(r)=0.

Let \zeta_n=exp(2\pi i/n) and let F_n={\mathbb{Q}}(\zeta_n) denote the cyclotomic field of degree \phi(n) over {\mathbb{Q}}. If \sigma_j\in Gal(F_n/{\Bbb{Q}}) is defined by \sigma_j(\zeta_n)=\zeta_n^j then

Gal(F_n/{\Bbb{Q}})\cong ({\Bbb{Z}}/n{\Bbb{Z}})^\times,

where \sigma_j\longmapsto j.

Lemma: Assume n is divisible by 4.

  1. {\mathbb{Q}}(r) is the maximal real subfield of F_n, Galois over {\mathbb{Q}} with


    where $\tau$ denotes complex conjugation. Under the canonical isomorphism

    Gal(F_n/{\Bbb{Q}})\cong ({\Bbb{Z}}/n{\Bbb{Z}})^\times,

    we have

    Gal({\Bbb{Q}}(r)/{\Bbb{Q}})\cong ({\Bbb{Z}}/n{\Bbb{Z}})^\times/\{\pm 1\}.

  2. If n is divisible by 8 then r and s are conjugate roots of T_d. In particular, s\in {\mathbb{Q}}(r) and T_d(s)=0.

  3. We have \sigma_j(r)=T_j(r).
  4. If n\geq 4 is a power of 2 then T_d is the minimal polynomial of {\mathbb{Q}}(r). Furthermore, in this case

    \cos(\pi/4)=\sqrt{2}/2,\ \  \cos(\pi/8)=\sqrt{2+\sqrt{2}}/2,\ \  \cos(\pi/16)=\sqrt{2+\sqrt{2+\sqrt{2}}}/2,\ \ ... .

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