Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By SlipEternal

Thread: Field Theory - Element u transcendental of F

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    594
    Thanks
    2

    Field Theory - Element u transcendental of F

    In Section 10.2 Algebraic Extensions in Papantonopoulou: Algebra - Pure and Applied, Proposition 10.2.2 on page 309 (see attachment) reads as follows:

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

    10.2.2 Proposition

    Let E be a field, $\displaystyle F \subseteq E $ a subfield of E, and $\displaystyle \alpha \in E $ an element of E.

    In E let

    $\displaystyle F[ \alpha ] = \{ f( \alpha ) \ | \ f(x) \in F[x] \} $

    $\displaystyle F ( \alpha ) = \{ f ( \alpha ) / g ( \alpha ) \ | \ f(x), g(x) \in F[x] \ , \ g( \alpha ) \ne 0 \} $

    Then

    (1) $\displaystyle F[ \alpha ] $ is a subring of E containing F and $\displaystyle \alpha $

    (2) $\displaystyle F[ \alpha ] $ is the smallest such subring of E

    (3) $\displaystyle F( \alpha ) $ is a subfield of E containing F and $\displaystyle \alpha $

    (4) $\displaystyle F( \alpha ) $ is the smallest such subfield of E

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

    Papantonopoulou proves (1) and (2) (see attachment) and then writes:

    " ... ... (3) and (4) are immediate from (1) and (2) since $\displaystyle F[ \alpha ] \subseteq E $ and E is a field, $\displaystyle F[ \alpha ] $ is an integral domain, and $\displaystyle F( \alpha ) $ is simply the field of quotients of $\displaystyle F[ \alpha ] $. "

    [Note: I do not actually follow this statement - can someone help clarify this "immediate" proof]


    ================================================== ==============================================

    However ...

    ... in Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions, page 279 (see attachment) we read:

    " ... ... If u is transcendental over , it is routine to verify that

    $\displaystyle F(u) = \{ f(u){g(u)}^{-1} \ | \ f(x), g(x) in F[x] \ ; \ g(x) \ne 0 $

    Hence $\displaystyle F(u) \cong F(x) $ where F(x) is the field of quotients of the integral domain F[x]. ... ... "

    ================================================== ===============================================

    ***My problem with the above is that Papantonopoulou and Nicholson both give the same expression for $\displaystyle F( \alpha ) $ but Nicholson implies that the relation $\displaystyle F(u) = \{ f(u){g(u)}^{-1} \ | \ f(x), g(x) in F[x] \ ; \ g(x) \ne 0 \} $ is only the case if u is transcendental???

    Can someone please clarify this issue for me.

    Peter
    Last edited by Bernhard; Oct 8th 2013 at 10:42 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,317
    Thanks
    1285

    Re: Field Theory - Element u transcendental of F

    I don't see that implication. I see that Nicholson uses that definition specifically for when $\displaystyle u$ is transcendental, but I do not see any indication by him that it would not be true when $\displaystyle u$ is algebraic. The proof that it is true for when $\displaystyle u$ is algebraic should be identical to the one where it is transcendental, possibly even easier. Given any $\displaystyle f \in F[x]$, can you show that $\displaystyle f(u) \in F[u]$? If so, then you are done.
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Dec 15th 2012, 07:08 PM
  2. Replies: 3
    Last Post: Jun 5th 2012, 08:07 AM
  3. Replies: 1
    Last Post: Apr 2nd 2012, 04:32 AM
  4. Replies: 1
    Last Post: Apr 1st 2012, 08:14 AM
  5. Field Extensions: Algebraic and Transcendental Elements
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 1st 2010, 03:04 PM

Search Tags


/mathhelpforum @mathhelpforum