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