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

Math Help - Field Theory - Element u transcendental of F

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    559
    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,  F \subseteq E a subfield of E, and  \alpha \in E an element of E.

    In E let

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

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

    Then

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

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

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

    (4)  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  F[ \alpha ]  \subseteq E  and E is a field,  F[ \alpha ]  is an integral domain, and  F( \alpha )  is simply the field of quotients of  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

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

    Hence  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  F( \alpha )   but Nicholson implies that the relation   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; October 8th 2013 at 11:42 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,932
    Thanks
    782

    Re: Field Theory - Element u transcendental of F

    I don't see that implication. I see that Nicholson uses that definition specifically for when u is transcendental, but I do not see any indication by him that it would not be true when u is algebraic. The proof that it is true for when u is algebraic should be identical to the one where it is transcendental, possibly even easier. Given any f \in F[x], can you show that 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: December 15th 2012, 08:08 PM
  2. Replies: 3
    Last Post: June 5th 2012, 09:07 AM
  3. Replies: 1
    Last Post: April 2nd 2012, 05:32 AM
  4. Replies: 1
    Last Post: April 1st 2012, 09:14 AM
  5. Field Extensions: Algebraic and Transcendental Elements
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 1st 2010, 04:04 PM

Search Tags


/mathhelpforum @mathhelpforum