Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal

Math Help - Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page 282

  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, page 282

    I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions.

    Example 13 on page 282 (see attachment) reads as follows:

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

    In the third line of the explanation - see page 282 of attachment - we read:

    "But  [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 because  u^2 \notin \mathbb{Q} ... ... "

    Can someone explain why it follows that  u^2 \notin \mathbb{Q} \Longrightarrow  [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1

    Peter
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,789
    Thanks
    693

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

    [\mathbb{Q}(k) : \mathbb{Q}] = 1 implies that the minimal polynomial for k over \mathbb{Q}[x] has degree 1. Hence the minimal polynomial is x-k which implies k \in \mathbb{Q}.
    Thanks from Bernhard
    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, page

    Thanks SlipEternal

    Just one further point of clarification:

    I can see why the minimal polynomial is x - k , but why exactly does that imply that  k \in \mathbb{Q} ?

    Peter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,789
    Thanks
    693

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

    Quote Originally Posted by Bernhard View Post
    Thanks SlipEternal

    Just one further point of clarification:

    I can see why the minimal polynomial is x - k , but why exactly does that imply that  k \in \mathbb{Q} ?

    Peter
    Since (x-k) \in \mathbb{Q}[x], that means that the coefficients of each term are elements of \mathbb{Q}. Hence -k \in \mathbb{Q}, so k \in \mathbb{Q}.
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 14th 2013, 06:44 PM
  2. Replies: 2
    Last Post: September 30th 2013, 08:00 PM
  3. Replies: 2
    Last Post: September 30th 2013, 07:58 PM
  4. Replies: 2
    Last Post: September 28th 2013, 11:02 AM
  5. Replies: 3
    Last Post: September 28th 2013, 09:53 AM

Search Tags


/mathhelpforum @mathhelpforum