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 subfield of Q(X)
invariant under the field automorphism X -> 1/X is Q(X+1/X). In practice
you probably know how to do this, e.g. (z^3+1/z^3) = (z+1/z)^3 - 3(z+1/z)
etc. The point is that all the conjugates of x already lie in
Q(x), so that Q(x) is already a Galois extension of Q. In particular,
all the Galois conjugates of x are _real_.
On the other hand, your typical K_(n+1)/K_n is not Galois, and its
Galois closure not real, since the Galois closure necessarily includes
zeta*a_n^(1/b_n) too, where zeta is a primitive b_n - th root of
unity. This is not real unless b_n=2.
Thus the only radical extensions of Q which are both Galois and real
are chains of quadratic extensions. In particular, such fields have
degree 2^n over Q for some n. Thus the possible q are limited to
those with \EulerPhi(q) = a power of 2. Of course if
q = Prod(p^(e_p)) is the prime factorization of q, then
\EulerPhi(q) = Prod(p^(e_p-1)*(p-1)), so for all p>2 we have e_p=1
and p-1=power of 2. Thus
The only values of x = cos(p Pi/q) expressible using real radicals only
are those with q = 2^a * (a product of distinct Fermat primes).
Presumably, of course, these are limited to 3, 5, 17, 257, and 65537.
dave