Results 1 to 7 of 7

Math Help - Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Example 13 from Nicholson: Introduction to Abstract Algebra, Section 6.2, page 282 reads as follows: (see attachment)

    ------------------------------------------------------------------------------------------------------------------

    Example 13: If  u = \sqrt[3]{2} show that  \mathbb{Q}(u) = \mathbb{Q}(u)^2

    ------------------------------------------------------------------------------------------------------------------

    The solution comes down to the following:

    Given   \mathbb{Q}(u) \supseteq \mathbb{Q}(u)^2 \supseteq \mathbb{Q}

    so  [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = [ \mathbb{Q}(u) \ : \  \mathbb{Q}(u)^2] \ [ \mathbb{Q}(u)^2 \ : \   \mathbb{Q} ]

    Now Nicholson shows that  [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = 3  and   [ \mathbb{Q}(u)^2 \ : \   \mathbb{Q} ] = 3

    so  [ \mathbb{Q}(u) \ : \  \mathbb{Q}(u)^2] = 1

    Then Nicholson (I think) concludes that   \mathbb{Q}(u) =  \mathbb{Q}(u)^2

    ----------------------------------------------------------------------------------------------------------

    My problem is as follows:

    How (exactly) does it follow that:

     [ \mathbb{Q}(u) \ : \  \mathbb{Q}(u)^2] = 1   \Longrightarrow  \mathbb{Q}(u) =  \mathbb{Q}(u)^2

    Can someone help?

    Peter
    Attached Files Attached Files
    Last edited by Bernhard; October 15th 2013 at 05:46 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,794
    Thanks
    694

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Think minimal polynomials.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Quote Originally Posted by SlipEternal View Post
    Think minimal polynomials.
    Thanks SlipEternal ... well, given that  [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] =  1 the degree of minimal polynomial of the root u over  \mathbb{Q}(u)^2 is 1.

    Thus I imagine the minimal polynomial is of the form x - u ... ... (the coefficients of the polynomial should be in  \mathbb{Q}(u)^2 ???)

    How does this lead us to  \mathbb{Q}(u) = \mathbb{Q}(u)^2  ?

    I think I need a bit more help?

    Peter
    Last edited by Bernhard; October 15th 2013 at 06:01 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,794
    Thanks
    694

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Quote Originally Posted by Bernhard View Post
    Thanks SlipEternal ... well, given that  [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] =  1 the degree of minimal polynomial of the root u over  \mathbb{Q}(u)^2 is 1.

    Thus I imagine the minimal polynomial is of the form x - u ... ... (the coefficients of the polynomial should be in  \mathbb{Q}(u)^2 ???)

    How does this lead us to  \mathbb{Q}(u) = \mathbb{Q}(u)^2  ?

    I think I need a bit more help?

    Peter
    That implies that u \in \mathbb{Q}(u^2) since u is the coefficient of x^0 in that minimal polynomial.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Quote Originally Posted by SlipEternal View Post
    That implies that u \in \mathbb{Q}(u^2) since u is the coefficient of x^0 in that minimal polynomial.
    Sorry SlipEternal ... still not completely following.

    Let me restate where I am (thanks to your help)

    We have that the degree of the minimal polynomial m(u) of  \mathbb{Q}(u)   over  \mathbb{Q}(u^2)   is 1.

    Thus, the minimal polynomial m(u) is a monic polynomial of degree 1 with coefficients in  \mathbb{Q}(u^2)   such that m(u) = 0.

    So then  m(u) = a_0  +  a_1 u  =  0 where   a_0, a_1 \in \mathbb{Q}(u^2)

    How do we (formally) show that  u \in \mathbb{Q}(u^2)

    Peter
    Last edited by Bernhard; October 15th 2013 at 06:50 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,794
    Thanks
    694

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    A monic polynomial means a_1 = 1.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    oh ... of course!

    So  m(u) = a_0 + u = 0 where  a_0 \in \mathbb{Q}(u^2)

    So  u = -a_0

    and  a_0 and of course  - a_0 \in \mathbb{Q}(u^2)

    Thanks!

    Peter
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: October 15th 2013, 05:18 PM
  2. Replies: 3
    Last Post: October 14th 2013, 07:45 PM
  3. Replies: 3
    Last Post: October 14th 2013, 06:44 PM
  4. Replies: 2
    Last Post: September 30th 2013, 07:58 PM
  5. Replies: 2
    Last Post: September 28th 2013, 11:02 AM

Search Tags


/mathhelpforum @mathhelpforum