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Prove an extension is not normal

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    let b be a square root of 1+i, show that Q(b):Q is not a normal extension. Also, what is the Galois group of the extension?

    2. Relevant equations


    3. The attempt at a solution
    so b = +/- (1+i)^(1/2), and it's minimal polynomial is x^4+4 which has roots -(2)^1/2 and 2^(1/2) that are not in Q(b) and therefore the extension is not normal. In a proper splitting field, x^4+4 splits into (x-(1+i))(x+(1+i))(x-2^(1/2))(x+2^(1/2)), the Galois group would have a map that permutes the roots of the first two factors and a map that would permute the roots of the second two factors, therefore it would be isomorphic to Z_2 x Z_2.

    Is this correct?
     
  2. jcsd
  3. Nov 5, 2016 #2

    andrewkirk

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    That doesn't look right to me.

    We have ##b^2=1+i## so that ##b^2-1=i## so that ##(b^2-1)^2=-1##. Hence a monic 4th degree polynomial with root ##b## is ##(x^2-1)^2+1=x^4-2x^2+2##. Since the minimal polynomial is unique, I don't think your polynomial can also be minimal.

    The structure of your argument is otherwise OK. There will be four roots of the minimal polynomial, two of which are in ##\mathbb Q(b)## and two of which are not.
     
  4. Nov 5, 2016 #3

    andrewkirk

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    On the second part of the question, you appear to have found the Galois group of the extension that includes the other two roots of that minimal polynomial, which is not what was asked. Nor will the Galois group just be ##\mathbb Z_2##. This part of this question has many similarities to that earlier one you asked about an extension based on ##c##, a primitive fourteenth root of 1. Here the extension is based on ##b##, a primitive 16th root of ##2^8##.

    Remember how each fixing automorphism in that other example mapped ##c## to ##c^k## where ##k\geq 1## and ##k\leq## (did you ever end up working out what the upper limit was?).

    Well here the same thing will apply. Each fixing automorphism will map ##b## to a positive integer power of ##b##, up to a certain upper limit.

    If you can solve that other problem, including finding that upper limit, you can solve this one.
     
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