Let alpha = Squareroot (2 + Squareroot 2)(3 + Squareroot 3) (positive real squareroot s for concreteness) and consider the extension E = Q(alpha). Show that a = (2 + Squareroot 2)(3 + Squareroot 3) is not a square in F = Q(Squareroot 2, Squareroot 3). [If a = c^2, c F, then a phi (a) = (2 + Squareroot 2)^2 (6) = (c phi c)^2 for the automorphism phi Gal(F/Q) fixing Q(Squareroot 2). Since c phi c = N_F/Q (Squareroot 2) (c) Q(Squareroot 2) conclude that this implies Squareroot 6 Q (Squareroot 2), a contradiction.] Conclude from (a) that [E: Q] = 8. Prove that the roots of the minimal polynomial over Q for alpha are the 8 elements plusminus Squareroot 2 plusminus Squareroot 2) (3 plusminus Squareroot 3). Let beta = Squareroot (2 - Squareroot 2)(3 + Squareroot 3). Show that alpha beta = Squareroot 2(3 + Squareroot 3) F so that beta E. Show similarly that the other roots are also elements of E so that E is a Galois extension of Q. Show that the elements of the Galois group are precisely the maps determined by mapping alpha to one of the eight elements in (b). Let sigma Gal(E/Q) be the automorphism which maps alpha to beta. Show that since sigma(alpha^2) = beta^2 that sigma(Squareroot 2) = - Squareroot 2 and sigma(Squareroot 3) = Squareroot 3. From alpha beta = Squareroot 2(3 + Squareroot 3) conclude that sigma(alpha beta) = - alpha beta and hence sigma(beta) = -alpha. Show that sigma is an element of order 4 in Gal(E/Q).
