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Field Extensions

  1. Mar 28, 2010 #1
    5vztl5.jpg

    I think these are related:

    Definition. Let F be an extension field of K and let u be in F. If there exists a nonzero polynomial f(x) in K[x] such that f(u)=0, then u is said to be algebraic over K. If there does not exist such a polynomial, then u is said to be transcendental over K.

    Proposition. Let F be an extension field of K and let u in F be an element algebraic over K. If the minimal polynomial of u over K has degree n, then K(u) is an n-dimensional vector space over K.

    Definition. Let F be an extension field of K. If the dimension of F as a vector space over K is finite, then F is said to be a finite extension of K.
    The dimension of F as a vector space over K is called the degree of F over K, and is denoted by [F:K].

    Proposition. Let F be an extension field of K and let u be in F. The following conditions are equivalent:

    (1) u is algebraic over K;

    (2) K(u) is a finite extension of K;

    (3) u belongs to a finite extension of K.

    What I know:

    I've been staring at [K(x):K] = 2n+1 , n>=0, for a while now.

    So this means the degree of the minimal polynomial is odd

    A basis of K(x) over K is 1, x, x2, ..., x2n

    Not sure what to do with all this information though.

    Any help would be appreciated. Thanks
     
  2. jcsd
  3. Mar 28, 2010 #2

    Office_Shredder

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    To show that K(x)=K(x2), you just need to show that x is in K(x2).

    Write down the minimal polynomial for x, and see if you can use that to write x as an element in K(x2)
     
  4. Mar 28, 2010 #3
    So would the min poly look like

    a2n+1x2n+1+a2nx2n+...+a1x+a0?

    for a in K

    Awesome fast reply thanks
     
  5. Mar 28, 2010 #4

    Office_Shredder

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    Minor point of clarification, x is a root of an equation not a variable. It might be better to right down the polynomial as a function of t, say m(t), and then notice m(x)=0. Can you solve for x in terms of the a's and some x2s then?
     
  6. Mar 28, 2010 #5
    So i can write something like

    a2n+1(x2)n.x + a2n(x2)n+...+a1(x2).x-1+a0(x2).x-2

    is that enough, or is it wrong? Seems wrong, I don't really see how it uses the 2n+1 property, would have had the same result with any degree.

    ah i just noticed your edit, hang on a sec for an edit
     
  7. Mar 28, 2010 #6

    Office_Shredder

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    That's not an equation. You need to use that when you plug x into the minimal polynomial you get zero
     
  8. Mar 28, 2010 #7
    So i missed a step?

    I'm not sure how to find the equation though
     
  9. Mar 28, 2010 #8

    Office_Shredder

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    If the minimal polynomial for x over K is [tex]m(t)=\sum_{i=0}^{2n+1}a_i t^i[/tex] then you get the equation:

    [tex]m(x)=\sum_{i=0}^{2n+1}a_i x^i = 0[/tex]

    Notice that xk is in K(x2) if k is even. If k is odd, you can write it as xk-1*x and you have an element of K(x2) times x. Can you use this to solve for x in terms of things from K(x2)?
     
    Last edited: Mar 28, 2010
  10. Mar 28, 2010 #9
    are you asking me to show that every element of the summation is an element of K(x2)?
     
  11. Mar 28, 2010 #10

    Office_Shredder

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    No, because that's not true. Suppose I gave you this formula:
    am + b + cm + d=0

    Can you solve for m?

    m in this case is going to be the x that's factored out of every odd degree monomial.

    Do you understand what the ultimate goal of doing this is? It sounds like you're a bit lost as to why I'm making this suggestion

    Also, I thought of an alternative method if you like this better:

    [tex][K(x):K] = [K(x):K(x^2)][K(x^2):K][/tex]. What can you conclude about [tex][K(x):K(x^2)][/tex] using the fact that [tex][K(x):K][/tex] is odd?
     
  12. Mar 28, 2010 #11
    Ok, so I believe its a summation divided by another summation in the same way as your simple factoring example above.

    Where its the summation of all the even a terms divided by the summation of all the odd a terms.

    edit: actually i just wrote out all the factoring, is it something to do with taking each element of the summation pairwise?

    this is deadly confusing for me

    Edit: saw your edit, im not too concrete with all the theroy yet, it's new stuff so I'm not 100% sure on how and why and where to manuipulate things.
     
  13. Mar 28, 2010 #12
    i like the new method, it seems cleaner, with [K(x):K] being odd then that implies either [K(x):K(x^2)] is 1 or [K(x^2):K] = 1

    So either K(x)=K(x^2) or K(x^2)=K

    so i need to show that K(x^2) is not equal to K

    Any tips?

    I don't knwo any elements of K(x^2) or K I don't think.
     
  14. Mar 28, 2010 #13

    Office_Shredder

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    Umm, they could both just be odd numbers.

    The key is that [K(x):K(x2)] is odd. But also it can only be either one or two. Do you see why?
     
  15. Mar 28, 2010 #14
    not particularly, not sure how it can be 2
     
  16. Mar 28, 2010 #15

    Office_Shredder

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    [tex] Q(\sqrt{2})[/tex] is a degree two extension at most because [tex] \sqrt{2}[/tex] satisfies the polynomial [tex] t^2-2=0[/tex].

    Can you find some element a of [tex]K(x^2)[/tex] so that x satisfies [tex] t^2-a=0[/tex]?
     
  17. Mar 28, 2010 #16
    a part of me wants to say root a but ill go out on a limb here

    b+cx2 is in K(x2), b, c in K


    so (b+cx2)2 - a = 0

    (b2+2bcx2+c2x4) -a = 0

    unless (b2+2bcx2+c2x4) is in K(x2) then I don't see how i can find an a in K(x2) that will cancel it out

    hmm..
     
  18. Mar 28, 2010 #17

    Office_Shredder

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    It's really easier than that.

    You want to find a value of a so that [tex] x^2-a=0[/tex].
     
  19. Mar 28, 2010 #18
    not root(a) then? otherwise im stumped
     
  20. Mar 28, 2010 #19

    Office_Shredder

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    x is fixed, x is what you're extending the field by. You want to find a
     
  21. Mar 28, 2010 #20
    oh so x^2?
     
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