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

Thread: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

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

    Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

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

    On page 282 the Corollary to Theorem 5 states the following: (see attachment for Theorem 5 and the Corollary)

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

    Corollary. Let $\displaystyle E \supseteq F $ be fields and let $\displaystyle u \in E $ be algebraic over F .

    If $\displaystyle v \in F(u) $, then v is also algebraic over F and $\displaystyle {deg}_F(v) $ divides $\displaystyle {deg}_F(u) $.

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

    The proof begins as follows:

    " Proof. Here $\displaystyle F(u) \supseteq F(v) \supseteq F $ ... ... etc etc


    My problem is as follows:

    How do you show formally and explicitly that $\displaystyle F(u) \supseteq F(v) \supseteq F $

    Would appreciate some help.

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

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,337
    Thanks
    1299

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

    Since $\displaystyle v \in F(u)$, if we consider $\displaystyle F(u)$ as a vector space over $\displaystyle F$ with basis $\displaystyle \{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace (since $\displaystyle v$ can be represented as a linear combination of elements of the basis for $\displaystyle F(u)$). And $\displaystyle \{1\}$ spans a linear subspace of both spaces. So, $\displaystyle \span(\{1\}) \subseteq \span(\{1,v^1, \ldots, v^{\deg(u)-1}\}) \subseteq F(u)$. Since $\displaystyle F(v)$ can be viewed as the linear space spanned by the basis of powers of $\displaystyle v$, we have $\displaystyle F \subseteq F(v) \subseteq F(u)$.
    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
    594
    Thanks
    2

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

    Quote Originally Posted by SlipEternal View Post
    Since $\displaystyle v \in F(u)$, if we consider $\displaystyle F(u)$ as a vector space over $\displaystyle F$ with basis $\displaystyle \{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace (since $\displaystyle v$ can be represented as a linear combination of elements of the basis for $\displaystyle F(u)$). And $\displaystyle \{1\}$ spans a linear subspace of both spaces. So, $\displaystyle \span(\{1\}) \subseteq \span(\{1,v^1, \ldots, v^{\deg(u)-1}\}) \subseteq F(u)$. Since $\displaystyle F(v)$ can be viewed as the linear space spanned by the basis of powers of $\displaystyle v$, we have $\displaystyle F \subseteq F(v) \subseteq F(u)$.
    Thanks SlipEternal, that is pretty clear, but just one issue ...

    You write:

    " ... ... if we consider $\displaystyle F(u)$ as a vector space over $\displaystyle F$ with basis $\displaystyle \{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace ... ..."

    How do we know for sure that $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ is a subspace ... why is it not possible that $\displaystyle \{u^0,u^1,\ldots, u^{\deg(u)-1}\} \subseteq \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ - that is $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\} $ is a superspace of F(u) - that is $\displaystyle F(u) \subseteq F(v) $

    Can you clarify?

    Peter
    Last edited by Bernhard; Oct 14th 2013 at 06:29 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,337
    Thanks
    1299

    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

    Quote Originally Posted by Bernhard View Post
    Thanks SlipEternal, that is pretty clear, but just one issue ...

    You write:

    " ... ... if we consider $\displaystyle F(u)$ as a vector space over $\displaystyle F$ with basis $\displaystyle \{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace ... ..."

    How do we know for sure that $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ is a subspace ... why is it not possible that $\displaystyle \{u^0,u^1,\ldots, u^{\deg(u)-1}\} \subseteq \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ - that is $\displaystyle \{v^0, v^1, \ldots, v^{\deg(u)-1}\} $ is a superspace of F(u) - that is $\displaystyle F(u) \subseteq F(v) $

    Can you clarify?

    Peter
    You are given that $\displaystyle v \in F(u)$, so since $\displaystyle F(u)$ is a field, $\displaystyle v^i \in F(u) \forall i\in \mathbb{Z}$. Hence, every element of $\displaystyle \{1,v,\ldots, v^{\deg(u)-1}\}$ is expressible as a linear combination of elements of $\displaystyle \{1,u,\ldots, u^{\deg(u)-1}\}$.
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Field Theory - Nicholson - Section 6.2 - Exercise 31
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Oct 8th 2013, 06:04 PM
  2. Replies: 2
    Last Post: Sep 30th 2013, 08:00 PM
  3. Replies: 2
    Last Post: Sep 30th 2013, 07:58 PM
  4. Replies: 2
    Last Post: Sep 28th 2013, 11:02 AM
  5. Replies: 3
    Last Post: Sep 28th 2013, 09:53 AM

Search Tags


/mathhelpforum @mathhelpforum