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