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

  1. #1
    Super Member Bernhard's Avatar
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    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  E \supseteq F  be fields and let  u \in E be algebraic over F .

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

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

    The proof begins as follows:

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


    My problem is as follows:

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

    Would appreciate some help.

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

    Since v \in F(u), if we consider F(u) as a vector space over F with basis \{u^0,u^1,\ldots, u^{\deg(u)-1}\}, then \{v^0, v^1, \ldots, v^{\deg(u)-1}\} must span a linear subspace (since v can be represented as a linear combination of elements of the basis for F(u)). And \{1\} spans a linear subspace of both spaces. So, \span(\{1\}) \subseteq \span(\{1,v^1, \ldots, v^{\deg(u)-1}\}) \subseteq F(u). Since F(v) can be viewed as the linear space spanned by the basis of powers of v, we have F \subseteq F(v) \subseteq F(u).
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

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

    You write:

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

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

    Can you clarify?

    Peter
    Last edited by Bernhard; October 14th 2013 at 07:29 PM.
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    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 F(u) as a vector space over F with basis \{u^0,u^1,\ldots, u^{\deg(u)-1}\}, then \{v^0, v^1, \ldots, v^{\deg(u)-1}\} must span a linear subspace ... ..."

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

    Can you clarify?

    Peter
    You are given that v \in F(u), so since F(u) is a field, v^i \in F(u) \forall i\in \mathbb{Z}. Hence, every element of \{1,v,\ldots, v^{\deg(u)-1}\} is expressible as a linear combination of elements of \{1,u,\ldots, u^{\deg(u)-1}\}.
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