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Math Help - The Galois Group

  1. #1
    Ant
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    The Galois Group

    Hi,

    I've been reading some history and some background to Galois theory and these usually talk about the "group of an equation". I'm assuming this is the same as the "Galois group" which I've been taught. However, the galois group, as i know it, is a group of embeddings (injective ring homomorphism) under composition, whereas the "group of an equation" is always talked about as being a group of permutations of roots. I'm suspect these two things to be the same and I'm trying to see the connection. Could anyone say if the follow is correct or if I'm on the right lines?

    Let K be a field. Let f(X) be an irreducible polynomial in  K[X]. Let \alpha be a root of f(x) in some field extension L/K.

    So the definition of the Galois group is \{K-embeddings, L \to L\}.

    But we know each K embedding is uniquely defined by its behaviour on the root \alpha. Futhermore, Artin's Extension theorem, implies that any K embedding will send \alpha to another root \beta in the splitting field of f over K. (note that beta may equal alpha). If, infact it turns out that \beta is actually in L. Then this particular K-embedding (sending alpha to beta) is in the Galois group.

    So in this way, each K embedding corresponds to a permutation of a root of f(x). i.e. alpha being sent to beta.

    So the Galois group of K embeddings from L to L is equivalent to a, particular, subgroup of the group of permutation of f's roots.

    Is this above correct?

    Thanks for any clarification!
    Last edited by Ant; March 20th 2013 at 04:03 PM.
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    Re: The Galois Group

    Yes. Given any polynomial, say, over Q, we know it splits into linear factors "somewhere". So, we can enumerate the roots, in whatever fashion we like, so that embeddings correspond to permutations of {1,...,n}. The problem with this using this perspective in computing the Galois group G is that we might only find G up to conjugacy.

    From what books did you learn Galois theory? I've read a few books on the subject and all but one (Lang's Algebra) take the permutation approach.
    Thanks from Ant
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  3. #3
    Ant
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    Re: The Galois Group

    Quote Originally Posted by spudwish View Post
    Yes. Given any polynomial, say, over Q, we know it splits into linear factors "somewhere". So, we can enumerate the roots, in whatever fashion we like, so that embeddings correspond to permutations of {1,...,n}. The problem with this using this perspective in computing the Galois group G is that we might only find G up to conjugacy.

    From what books did you learn Galois theory? I've read a few books on the subject and all but one (Lang's Algebra) take the permutation approach.
    Thanks for your reply.

    What do you mean by "we might only find the galois group up to conjugacy"?

    I've mostly been learning from the lecture course given at my university. The reading I've been doing has mostly not been too technical in nature but historical. Thanks again!
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    Re: The Galois Group

    Do you know what conjugacy is? With the permutation approach, the end result depends on how we number the roots. For f(x)=x^3+1, which root is (-1)^(1/3), is it the first, or what? It's arbitrary. Consider the polynomial (x^2+x+1)(x^2+x+2); its Galois group is S2 x S2, where S2 is the group of permutations of the set {1,2}. Well, the polynomial has four distinct roots, so in reality the Galois group is the product of the groups of permutations of {i,j} resp. {k,l}, for any choice of i,j,k,l from {1,2,3,4}.
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