A bit of trouble with Galois groups

1. Nov 19, 2004

Euclid

Is the Galois group of F=Q(sqrt2,3i) the maps {id, tau , sigma, gamma}, where
(1) id is the identity
(2) tau maps sqrt2 to -sqrt2 and leave 3i alone
(3) sigma leaves sqrt2 alone and maps 3i to -3i
(4) gamma maps sqrt2 to -sqrt2 and 3i to -3i ?
If so, the what are the fixed fields of the subgroups?
If I'm not mistaken the (proper nontrivial) subgroups are H={id, tau}, J={id, sigma}, K={id, gamma}. It appears that the fixed field of H is Q(3i) and the fixed field of J is Q(sqrt2). But it also appears that the fixed field of K is just Q, which is also the fixed field of Gal(F/Q). But F/Q is Galois since F is the splitting field of a seperable polynomial, so we can't have two distinct groups associated to the same intermediate field.
What am I doing wrong?

2. Nov 20, 2004

Hurkyl

Staff Emeritus
Try writing down an arbitrary element of F: it would be a Q-linear combination of 4 elements:

1, 3i, sqrt2, 3isqrt2

So, apply K to this arbitrary element and see if that gets you anywhere...