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Galois Groups

  1. Feb 6, 2007 #1
    Hey there,
    firstly I hope that this is the right place to discuss such things. if not, could you direct me somewhere else?
    Ok, I have to construct the Galois Group of f= (x^2-2x-1)^3 (x^2+x+1)^2 (x+1)^4 and then represent it as a permutation group of the roots.

    first I constructed the splitting field extension S:Q (where S= summation symbol and Q = field of Rational numbers)

    The splitting field i Came up with was Q(sqrt (2), sqrt (-3)):Q, and the degree of this splitting field is 4...am I correct here? is this the splitting field?

    The Galois group represented as a permutation group I ended up getting was
    { e (the identity), (sqrt(-3),-sqrt(-3)),(sqrt(2),-sqrt(2)),(sqrt(2),-sqrt(2))(sqrt(-3),-sqrt(-3))}
    isomorphic to the Klein4 group....
    am i doing this right?? it just seems abit simple a result for an initial function that wasn't that simple ! :uhh:
     
  2. jcsd
  3. Feb 6, 2007 #2

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    The splitting field for a polynomial is the same as the splitting field for the product of its irreducible, non-linear factors. Thus the splitting field of f(x) is the same as that of (x^2-2x-1)(x^2+x+1).
     
  4. Feb 7, 2007 #3
    Excellent, thank you very much. the point you have made will help me in the future too ! Is my permutation group correct...not entirely sure as to whether this is all that's required!
     
  5. Feb 7, 2007 #4

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    You know that every permutation must take a root of one polynomial to a root of the same polynomial, so the ones you've written are the only possible ones. However, it may be that not all of those are valid. Have you seen the connection between the degree of a splitting field and the size of its Galois group?
     
  6. Feb 7, 2007 #5
    Yes, the degree of the splitting field is 4, so there are 4 elements in the Galois Group too, so I think I'm done. Thank you for your advice!
     
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