Date: Mon, 25 Nov 96 15:46:28 CST From: rusin (Dave Rusin) To:
grubb Subject: yes, it's trivial So suppose x = cos(p Pi/q) is
expressible using real radicals only, that is, it is contained
in a sequence of field extensions Q \subseteq K1 \subseteq K2
... with each K_(n+1) = K_n( a_n^(1/b_n) ) \subseteq R. What is
the minimal polynomial of x over Q? Well, x is x=(z + 1/z)/2 for
z = exp(Pi i p/q), so evidently x lies in the maximal real
subfield of Q_q = Q(zeta_q). Since the inclusion Q_q intersect R
--> Q_q is known to induce a surjection of Galois groups
Gal(Q_q/Q) --> Gal(Q_q \cap R / Q), the algebraic conjugates of
x are precisely the elements of the form x_m=(z^m + 1/z^m)/2,
where gcd(m,q)=1. In particular, the number of these conjugates,
and hence the degree of the Galois closure of Q(x), is
|Z_q^\times| = \EulerPhi(q). Now, it is well known (meaning I
don't feel like working out the algebra) that x_m (for _any_ m)
can be expressed as a polynomial in x. I guess that's the fact
that the fixed field of the