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When is a Galois group not faithful

  1. May 27, 2012 #1
    Hi, I'm looking at proposition 1.14(c) of Artin's Algebra.

    It says

    if we have K a splitting field for polynomial f from F[x], with roots a_1,...,a_n,
    then the Galois group G(K/F) acts faithfully on the set of roots.

    I look at faithful as the symmetries in the roots completely represent the group.
    That is, no root is fixed by any group element (besides the identity (edited)).

    When should I worry about this, are there any ways to construct a relevant counterexample if we drop a condition?

    (So for actions, transitive and faithful is like surjectivity and injectivity respectively?)
     
    Last edited: May 28, 2012
  2. jcsd
  3. May 28, 2012 #2

    This cannot be correct as any element in K, and also the roots of f, are fixed by the identity automorphism.

    The action is faithful if [itex]\,\forall 1\neq g\in Gal\left(K/F\right)\,\,\exists \,1\leq i\leq n\,\,s.t.\,\,g(a_i)\neq a_i\,[/itex] , which of course is true.

    DonAntonio




     
  4. May 28, 2012 #3

    Hurkyl

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    A transitive action is like a one-element set. More precisely, there is only one orbit.

    A faithful action means you have "enough" elements to distinguish group elements -- two group elements are equal if and only if they have the same values.
     
  5. May 28, 2012 #4
    Yes thank you, I edited it. So I was reading some more and looks like part (c) is less interesting than parts (a) and (b) of prop 1.14, and it appears part (c) is true for nearly all finitely generated extensions, not just splitting fields. But I may not get to this soon, I'm more interested right now in just learning how to apply Galois theory.
     
  6. May 28, 2012 #5
    Also, I guess I meant I was thinking out loud about mnemonics for remembering transitive and faithful. Once I remember they're like surjectivity and injectivity, but for group actions, I can remember the definitions.
     
  7. May 28, 2012 #6

    mathwonk

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    I have not taught this lately but I seem to recall that faithfulness fails only in characteristic p > 0. and you are right in your definition of faithfulness if you omit the identity element as pointed out above. I.e. to me a faithful action is one where the subgroup fixing all elements consists of just the identity. Indeed if you look in the index of the book you find the definition of a faithful action on lines 12-13 page 183 of Artin.
     
  8. May 28, 2012 #7

    If by "faithful" the OP meant "transitive" then it is easy: The Galois group on a Galois extension is always transitive on the set of roots

    of any irreducible polynomial over the base field which has at least one (and thus all) root in the upper field.
     
  9. May 29, 2012 #8

    mathwonk

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    I have never heard of faithful meaning transitive.
     
  10. May 29, 2012 #9

    Neither have I.

    DonAntonio
     
  11. May 29, 2012 #10

    mathwonk

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    ok, nowicit.
     
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