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(Linear Algebra) Vector Space and Fields

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\omega = \frac{1}{2} + \frac{\sqrt7}{2}i[/tex]


    (a) Verify that [tex]\omega^2 = \omega - 2[/tex]

    (b) Prove that [tex]F = \{a + b \omega : a, b \in \mathbb{Q} \} [/tex]is a field, using the usual operations of
    addition and multiplication for complex numbers.

    (c) Recall that we can think of [tex]F[/tex] as a vector space over the subfield [tex]\mathbb{Q}[/tex] (i.e. using
    the rational numbers [tex]\mathbb{Q}[/tex] as the scalars). Find a basis for [tex]F[/tex] over [tex]\mathbb{Q}[/tex], justifying
    your answer. Hence write down the dimension of [tex]F[/tex] over [tex]\mathbb{Q}[/tex].

    (d) Is the field [tex]F[/tex] algebraically closed? Explain your answer.

    (e) Explain briefly why [tex]F[/tex] is the smallest subfield of [tex] \mathbb{C}[/tex] containing [tex]\omega[/tex], i.e. if [tex]K \subseteq \mathbb{C} [/tex]
    is any field containing ! then [tex]F \subseteq K[/tex].


    2. Relevant equations



    3. The attempt at a solution

    (a)
    I've done this with no worries.

    (b)
    Am I right in thinking that since [tex]F[/tex] is a subfield of [tex]\mathbb{C}[/tex], I only need to show that it is closed under addition, multiplication, it contains the identity and inverse of the operations? I.e The 'algebraic' properties of commutativity, associativity and distribution are 'inherited'?

    (c)
    I am not too sure how to approach this question. Intuitivlely, I would have proposed a basis of [tex] (1,\omega)[/tex] but I'm not sure, perhaps it seems too simple?
    And I know to find the dimension, I just need to find out how many elements of the basis there are.

    (d)
    I know the definition of algebraically closed; if every polynomial has solutions within the field, then it is.

    But I am not sure how to check this for the field [tex]F[/tex]. I am thinking that since it is in [tex]\mathbb{C}[/tex] then it is... but I really am not sure.

    (e)
    I was really not sure how to approach this question.

    Thanks in advance.
     
  2. jcsd
  3. Aug 25, 2009 #2
    (b) OK but say this about 'inherited' in your write-up

    (c) Yes, that is a basis, use (b) to show it.

    (d) Wait! R is in C, but R is not algebraically closed. So that argument is no good. Think how you show R is not algebraically closed, see if you can do something similar in F

    (e) Think about it. Use (b).
     
  4. Aug 26, 2009 #3
    Ok So I have shown that [tex]F[/tex] meets all the field axioms, except I am a little stuck on the multiplicative inverse.

    Let [tex]x= a+b\omega[/tex]

    So I want some number [tex]x^{-1}\times x=1[/tex]

    So I propose that [tex] x^{-1} =\frac{1}{a+b \omega}[/tex]

    Let [tex]\bar{x} =a - b \omega[/tex]

    I then manipulate it, multiplying by [tex]\frac{\bar{x}}{\bar{x}}[/tex]

    and I get [tex]a^{2}+2ab+b^{2}[/tex] on the denominator. Here is where I get stuck, I know that the multiplicative inverse exists for [tex]x \in F, x\neq0[/tex], provided [tex]a^{2}+2ab+b^{2}\neq0[/tex].

    How do I show that [tex]a^{2}+2ab+b^{2}\neq0[/tex]?
     
  5. Aug 26, 2009 #4

    HallsofIvy

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    Don't let [itex]\overline{x}= a- b\omega[/itex]. Let [itex]\overline{x}= a- b\overline{\omega}[/itex] where
    [tex]\overline{\omega}= \frac{1}{2}-\frac{\sqrt{7}}{2}i[/tex]
    of course.
     
  6. Aug 27, 2009 #5
    oops, I see.. but I still get a denominator of [tex]
    a^{2}+2ab+2*b^{2}
    [/tex]

    I am really new to this stuff and I'm also having a hard time trying to think of an example which would demonstrate that the field is not algebraically closed. Any hints would be greatly appreciated.

    edit: I'm thinking [tex]x+4\omega[/tex] demonstrates that the field is not closed, since it has rational coefficients, yet no solutions in [tex]\mathbb{Q}[/tex] is that the correct way of thinking?
     
    Last edited: Aug 27, 2009
  7. Aug 27, 2009 #6

    Dick

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    Homework Helper

    I get a denominator of a^2+ab+2*b^2. You don't actually need to show that is nonzero if a and b are nonzero. You know it is. It's the product of a+bw and its complex conjugate. And a+bw is zero only if a=0 and b=0. x+4w=0 doesn't work. The coefficients of x (1) and 1 (4w) ARE in F. But the root x=(-4w) is ALSO in F. You want to find an equation whose coefficients are in F and which has a root that is NOT in F.
     
  8. Aug 29, 2009 #7

    SFB

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    Hi I am kinda new to this topic two . I was wondering how can I prove that the following expressions define scalar product. All I can guess that I need to show that they follow the properties of the scalar product.

    But how? If possible, help me with an example .


    1. (f,g)=[tex]\int f(x)g(x)w(x)dx[/tex] where w(x)>0 where x=[0,1]
    2. (f,g)=[tex]\int f'(x)g'(x)dx[/tex] +f(0)g(0)
     
    Last edited: Aug 29, 2009
  9. Aug 29, 2009 #8

    Dick

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    This thread isn't even about inner products. You should really start your own thread on the subject. While you are doing that write down what the properties of an inner product are and tell us which property is giving you trouble.
     
  10. Aug 29, 2009 #9

    SFB

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    Sorry for the inconvenience , I am kinda new to this forum.............already started a new thread
     
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