Results 1 to 4 of 4

Math Help - Determining a Galois Group

  1. #1
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485

    Determining a Galois Group

    I am trying to compute the Galois group for the polynomial x^4-6x^2+7, but I'm having a bit of trouble. I know that x^4-6x^2+7=(x^2-(3-\sqrt{2}))(x^2-(3+\sqrt{2})) over \mathbb{Q}(\sqrt{2}). The polynomial will then split over \mathbb{Q}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}}), which is an extension of degree 8.

    Now I have the automorphisms \tau_1: \sqrt{2}\mapsto -\sqrt{2}, \tau_2: \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}} and \tau_3: \sqrt{3-\sqrt{2}}\mapsto -\sqrt{3-\sqrt{2}}. I know that \tau_1^2=\tau_2^2=\tau_3^2=e. But I can't seem to get relations between them (and in particular, I would like to find an element of order 4 to show that this is the dihedral group). For example, I am interested in powers of \tau_1 \tau_2 but I'm not sure how to run the computation.

    Any suggestions would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by roninpro View Post
    I am trying to compute the Galois group for the polynomial x^4-6x^2+7, but I'm having a bit of trouble. I know that x^4-6x^2+7=(x^2-(3-\sqrt{2}))(x^2-(3+\sqrt{2})) over \mathbb{Q}(\sqrt{2}). The polynomial will then split over \mathbb{Q}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}}), which is an extension of degree 8.

    Now I have the automorphisms \tau_1: \sqrt{2}\mapsto -\sqrt{2}, \tau_2: \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}} and \tau_3: \sqrt{3-\sqrt{2}}\mapsto -\sqrt{3-\sqrt{2}}. I know that \tau_1^2=\tau_2^2=\tau_3^2=e. But I can't seem to get relations between them (and in particular, I would like to find an element of order 4 to show that this is the dihedral group). For example, I am interested in powers of \tau_1 \tau_2 but I'm not sure how to run the computation.

    Any suggestions would be much appreciated.
    let a = \sqrt{3+\sqrt{2}} and b = \sqrt{3-\sqrt{2}}. note that \sqrt{2} \in \mathbb{Q}(a,b). so the splitting field is just \mathbb{Q}(a,b).
    as you said, the Galois group is D_8, the dihedra group of order 8. just define \sigma(a)=b, \ \sigma(b)=-a and \tau(a)=a, \ \tau(b)=-b. then o(\sigma)=4, \ o(\tau)=2, \ o(\tau \sigma)=2. thus the Galois group is D_8.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by roninpro View Post
    I am trying to compute the Galois group for the polynomial x^4-6x^2+7, but I'm having a bit of trouble. I know that x^4-6x^2+7=(x^2-(3-\sqrt{2}))(x^2-(3+\sqrt{2})) over \mathbb{Q}(\sqrt{2}). The polynomial will then split over \mathbb{Q}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}}), which is an extension of degree 8.

    Now I have the automorphisms \tau_1: \sqrt{2}\mapsto -\sqrt{2}, \tau_2: \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}} and \tau_3: \sqrt{3-\sqrt{2}}\mapsto -\sqrt{3-\sqrt{2}}. I know that \tau_1^2=\tau_2^2=\tau_3^2=e. But I can't seem to get relations between them (and in particular, I would like to find an element of order 4 to show that this is the dihedral group). For example, I am interested in powers of \tau_1 \tau_2 but I'm not sure how to run the computation.

    Any suggestions would be much appreciated.

    I'm interested in knowing how did you come up with those three automorphisms, as you didn't define them

    over a basis of the splitting field over the rationals...

    I mean: (1) How can you actually know the three automorphisms (i.e., how can you tell they're even different, to

    begin with)?, and (2) How can you possibly find out any relation between them if you don't know the maps?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    Quote Originally Posted by tonio View Post
    I'm interested in knowing how did you come up with those three automorphisms, as you didn't define them

    over a basis of the splitting field over the rationals...

    I mean: (1) How can you actually know the three automorphisms (i.e., how can you tell they're even different, to

    begin with)?, and (2) How can you possibly find out any relation between them if you don't know the maps?

    Tonio
    I guess I didn't write down my maps completely. For \tau_1, I meant that \sqrt{2}\mapsto \sqrt{2}, and everything else is fixed (i.e. \sqrt{3+\sqrt{2}}\mapsto \sqrt{3+\sqrt{2}}, and so on). For \tau_2, I meant that \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}} and fixes everything else, etc... So the fixed field for \tau_2, \tau_3 would be \mathbb{Q}(\sqrt{2}). It then seemed reasonable to me that something like \tau_1\tau_2 would be an element of order 4, but I didn't know how to compute the powers.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Galois group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 25th 2010, 10:39 PM
  2. galois group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 19th 2010, 04:10 AM
  3. galois group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 14th 2010, 12:26 AM
  4. Galois group
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 28th 2009, 01:40 AM
  5. Galois group
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 10th 2009, 12:48 AM

Search Tags


/mathhelpforum @mathhelpforum