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Math Help - Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page 282

  1. #1
    Super Member Bernhard's Avatar
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    Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page 282

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

    Example 14 on page 282 (see attachment) reads as follows:

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

    Example 14. Let  E \supseteq  F be fields and let  u, v \in E .

    If u and u + v are algebraic over F, show that v is algebraic over F.

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

    In the solution, Nicholson writes the following:

    Solution. Write L = F(u + v) so that L(u) = F(u, v). ... ... etc etc



    Can someone please show me (formally and exactly) why  L = F(u + v) \Longrightarrow  L(u) = F(u, v)  .

    Peter
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

    L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. L(u) is the smallest field that contains all members of L and u. Any such number, then, can be written as (c+ d(u+ v))+ jv where c and d are members of F and j is a member of L, not necessarily a member of F. So any such number is of the form (c+ d(u+ v))+ (e+ f(u+ v))v= c+ d(u+ v)+ ev+ fuv+ fv^2. Of course, uv and v^2 are members of F(u,v) as are c, d(u+v), and ev. So it is clearly true that L(u) is a subset of F(u,v). It remains to be shown that F(u,v) is a subset of L(u).

    Any member of F(u,v) is of the form a+ bu+ cv. That is the same as "c+ d(u+ v)+ ev+ fuv+ fv^2= c+ du+ (d+e)v+ fuv+ fv^2" if we take c= a, d= b, e= c- b, and f= 0.
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

    1\cdot (u+v) + (-1)\cdot u = v That's about all there is to it.
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

    Quote Originally Posted by HallsofIvy View Post
    L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. L(u) is the smallest field that contains all members of L and u. Any such number, then, can be written as (c+ d(u+ v))+ jv where c and d are members of F and j is a member of L, not necessarily a member of F. So any such number is of the form (c+ d(u+ v))+ (e+ f(u+ v))v= c+ d(u+ v)+ ev+ fuv+ fv^2. Of course, uv and v^2 are members of F(u,v) as are c, d(u+v), and ev. So it is clearly true that L(u) is a subset of F(u,v). It remains to be shown that F(u,v) is a subset of L(u).

    Any member of F(u,v) is of the form a+ bu+ cv. That is the same as "c+ d(u+ v)+ ev+ fuv+ fv^2= c+ du+ (d+e)v+ fuv+ fv^2" if we take c= a, d= b, e= c- b, and f= 0.
    Thanks for the help, HallsofIvy!

    Just one issue:

    You write:

    " L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. ... ... etc"

    But F(u + v) is a vector space over F with basis  \{ (u+ v)^0 ,   (u+ v)^1,  (u+ v)^2, ... ... , (u+ v)^{deg(u + v) - 1} \}

    So shouldn't we write:

    "L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as  a_0 + a_1 (u + v) + a_2 (u + v)^2 + ... ... + a_{deg(u + v) - 1} (u + v)^{deg(u + v) - 1}

    So writing: "Any such number can be written as a+ b(u+ v) where a and b are members of F." seems to imply that the degree of the minimal polynomial - or the degree of (u + v) is 2?

    Can you clarify?

    Peter
    Last edited by Bernhard; October 15th 2013 at 04:36 PM.
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

    Quote Originally Posted by Bernhard View Post
    Thanks for the help, HallsofIvy!

    Just one issue:

    You write:

    " L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. ... ... etc"

    But F(u + v) is a vector space over F with basis  \{ (u+ v)^0 ,   (u+ v)^1,  (u+ v)^2, ... ... , (u+ v)^{deg(u + v) - 1} \}

    So shouldn't we write:

    "L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as  a_0 + a_1 (u + v) + a_2 (u + v)^2 + ... ... + a_{deg(u + v) - 1} (u + v)^{deg(u + v) - 1}

    So writing: "Any such number can be written as a+ b(u+ v) where a and b are members of F." seems to imply that the degree of the minimal polynomial - or the degree of (u + v) is 2?

    Can you clarify?

    Peter
    You can generate every element of L = F(u+v) by all products of the form a+b(u+v). This is obvious since [0+1(u+v)]^k gives the basis you mentioned. So, to show that L(u) = F(u,v), you only need to show that for any a + bu \in L(u) with a,b \in L, there exists x \in F(u,v) such that x = a+bu. Then, for any c + du + ev \in F(u,v) with c,d,e \in F, show that there exists y \in L(u) such that c+du+ev = y.

    So, since (u+v) \in L, u \in L(u), we know v \in L(u). Since F \subseteq F(u+v) = L \subseteq L(u), we know c,d,e \in L(u). Hence, c+du+ev \in L(u).
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

    V
    Quote Originally Posted by SlipEternal View Post
    You can generate every element of L = F(u+v) by all products of the form a+b(u+v). This is obvious since [0+1(u+v)]^k gives the basis you mentioned. So, to show that L(u) = F(u,v), you only need to show that for any a + bu \in L(u) with a,b \in L, there exists x \in F(u,v) such that x = a+bu. Then, for any c + du + ev \in F(u,v) with c,d,e \in F, show that there exists y \in L(u) such that c+du+ev = y.

    SlipEternal, HallsofIvy, thanks so much for your help on this Example!!

    peter
    So, since (u+v) \in L, u \in L(u), we know v \in L(u). Since F \subseteq F(u+v) = L \subseteq L(u), we know c,d,e \in L(u). Hence, c+du+ev \in L(u).
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

    SlipEternal, HallsofIvy, Thanks so much for your help on this example

    PS sorry for the problem with the formatting of this message on the previous post

    Peter
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