Results 1 to 4 of 4

Math Help - Galois extension

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    7

    Galois extension

    Explain why \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2) is not a Galois extension. Find the smallest extension field E/\mathbb{Q}(^3\sqrt2) so that E/\mathbb{Q} is Galois. Determine the isomorphism class of \text{Gal}(E/\mathbb{Q}).

    I know how to prove that \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2) is not a Galois extension. But I don't know how to do the next two parts of this question. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by vaevictis59 View Post
    Explain why \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2) is not a Galois extension. Find the smallest extension field E/\mathbb{Q}(^3\sqrt2) so that E/\mathbb{Q} is Galois. Determine the isomorphism class of \text{Gal}(E/\mathbb{Q}).

    I know how to prove that \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2) is not a Galois extension. But I don't know how to do the next two parts of this question. Thanks in advance.
    Notice that the minimal polynomial of \sqrt[3]{2} over \mathbb{Q} is x^3 - 2. The roots of this polynomial is \sqrt[3]{2},\zeta\sqrt[3]{2},\zeta^2\sqrt[3]{2}. Thus, the splitting field of this polynomial is E=\mathbb{Q}(\zeta,\sqrt[3]{2}). Notice that E/\mathbb{Q} is Galois with \mathbb{Q}(\sqrt[3]{2})\subset E and it is the smallest extension because it is splitting field of the minimal polynomial. Now we see that \text{Gal}(E/\mathbb{Q}) = S_3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    7
    Quote Originally Posted by ThePerfectHacker View Post
    Notice that the minimal polynomial of \sqrt[3]{2} over \mathbb{Q} is x^3 - 2. The roots of this polynomial is \sqrt[3]{2},\zeta\sqrt[3]{2},\zeta^2\sqrt[3]{2}. Thus, the splitting field of this polynomial is E=\mathbb{Q}(\zeta,\sqrt[3]{2}). Notice that E/\mathbb{Q} is Galois with \mathbb{Q}(\sqrt[3]{2})\subset E and it is the smallest extension because it is splitting field of the minimal polynomial. Now we see that \text{Gal}(E/\mathbb{Q}) = S_3.
    How do we know this is isomorphic to S_3 and not \mathbb{Z}_6? I don't see how we know which group of order 6 it is isomorphic to.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    There are only two groups of order 6.

    A cyclic (abelian) one.

    And D_6 \cong S_3 which is non abelian.

    Write out the automorphisms it is pretty clear it is not cyclic nor abelian.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Galois Extension
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 13th 2010, 03:04 PM
  2. finite Extension of Fp Galois
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 19th 2010, 10:33 PM
  3. Galois extension
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 27th 2009, 05:41 PM
  4. Galois extension over Q
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 21st 2008, 04:08 PM
  5. Galois Extension
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 29th 2008, 04:21 PM

Search Tags


/mathhelpforum @mathhelpforum