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A bit of trouble with Galois groups

  1. Nov 19, 2004 #1
    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. jcsd
  3. Nov 20, 2004 #2

    Hurkyl

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    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...
     
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