Results 1 to 4 of 4

Math Help - Splitting Fields and Galois Groups

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    29

    Splitting Fields and Galois Groups

    Hi,

    I've been set a question and think I've managed about half of it but stuck on some other bits..

    Show that f := x^3 - 3x + 1 is irreducible over \mathbb{Q}. Let a\in\mathbb{C} be given as a root of f. Show that (1-a)^{-1} is also a root of f. Hence, find a splitting field L for f over \mathbb{Q}. Identity the Galois group \tau(L : \mathbb{Q}). (You must justify any claim that a particular map is an isomorphism. Be sure to state carefully
    any facts that you assume in doing this.)

    first part is easy just reducing by \mathbb{Z}_{2} to get \bar{f} := x^3 +x + 1\in\mathbb{Z}\ \mathbb{Z}_{2}(x) which has no zeros in \mathbb{Z}_{2} and thus irreducible over \mathbb{Z} and hence over \mathbb{Q} also.

    I think I can identify the Galois group using Prop 22.4 which gives \Delta= -27(-1)^2 - 4(-3)^3 = 81 , a perfect square thus the Galois Group is \mathbb{A}_{3}\cong C_{3}

    I can't think how to show that (1-a)^{-1} is also a root of f. For the splitting field all I can think to do is say that as a\in\mathbb{C} is a root of fthen f splits in \mathbb{C}(a) and so \mathbb{C}(a) is the splitting field for f over \mathbb{Q}.

    Any help for this last bit or if you can point out any errors I've already made would be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2010
    Posts
    29

    Re: Splitting Fields and Galois Groups

    As I posted just thought that to show (1-a)^{-1} is also a root of f you would say something along the lines of f factors into f:= (x-a)(x^2+bx-\frac{1}{a}) for some b\in\mathbb{C} and somehow this ends up with (1-a)^{-1} as a root... I know this is obviously not correct but am I along the right lines??
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,367
    Thanks
    736

    Re: Splitting Fields and Galois Groups

    one can verify \frac{1}{1-a} is a root directly:

    f(\frac{1}{1-a}) = \left(\frac{1}{1-a}\right)^3 - \frac{3}{1-a} + 1

    = \frac{1}{(1-a)^3} - \frac{3(1-a)^2}{(1-a)^3} + \frac{(1-a)^3}{(1-a^3)}

    = \frac{1 - 3 + 6a - 3a^2 + 1 - 3a + 3a^2 - a^3}{(1-a)^3}

    = \frac{-(1 - 3a + a^3)}{(1-a)^3} = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    29

    Re: Splitting Fields and Galois Groups

    Haha, cheers; I'm a fool. Other than that does the rest seem OK?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Splitting Fields
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 14th 2011, 11:40 AM
  2. splitting field and galois groups
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 13th 2010, 08:24 PM
  3. Splitting fields
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 24th 2010, 09:11 AM
  4. Extension fields / splitting fields proof...
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 19th 2007, 07:29 AM
  5. Fixed fields for Galois groups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 8th 2006, 08:44 AM

Search Tags


/mathhelpforum @mathhelpforum