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Math Help - two exercises concerning fields: order of a field

  1. #1
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    two exercises concerning fields: order of a field

    Hi there

    I don't know how to solve these exercises.

    Let \mathbb{K}^m be a field containing m elements and  n \in \mathbb{N} be positive.

    Exercise 1)
    Let x \in \mathbb{K}^{m^n}. Show:

    x \in \mathbb{K}^m \Leftrightarrow x^m-x=0

    Exercise 2)
    Let g: \mathbb{K}^4 \rightarrow \mathbb{K}^4 ,  g(x)=x^3+x. Find all elements of \mathbb{K}^4 that map to \mathbb{K}^2


    Notes: Those fields are finite at least are we doing this topic actually. m does not have to be a prime we have to do the proof for 1) in general.


    And thats my problem for exercise 1)
    \Rightarrow
    If x \in \mathbb{K}^m. The order of x is a divisor of m ? At least is x^{m-1}=1 true? I don't understand why this is true at this point already... If m is not prime (and we can't assume that) the number of units in \mathbb{K}^m doesn't have to be m-1 yeah???
    If so the equation could be multiplied by x and what results is x^m-x=0 after subtracting x on both sides and this direction is proved.
    \Leftarrow
    Here from x^m-x=0 follows that x^m=x so by dividing by x we have x^{m-1}=1 thus x has order m-1 so it is contained in the subset \mathbb{K}^m which consists of elements with order m-1 ????

    Let's have a look at exercise 2)
    So \mathbb{K}^4 is isomorphic to \{ 1, \alpha, \alpha^2 , \alpha^3 \} where \alpha^4=1?
    g(1)=2=1+1 \in \mathbb{K}^2 ?? Why?
    g(\alpha)=\alpha^3-\alpha ...have no idea

    Do you know how this works? Could you please give me a hint?

    Regards
    Huberscher
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  2. #2
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    Re: two exercises concerning fields: order of a field

    Hi,
    How much do you know about finite fields? In particular, up to isomorphism, there is exactly one field of order q for q a power of a prime p. So you can assume:

    K^m\subseteq K^{m^n}

    Now in a finite field F, the multiplicative group of non-zero elements of F is cyclic with generator, say a. For your first problem, then, use this fact and the fact that in any field a polynomial of degree m has at most m zeros.

    For your second question,

    K^4=\{0,1,\alpha,\alpha^2\} with \alpha^3=1. Also K^2=\{0,1\}.
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  3. #3
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    Re: two exercises concerning fields: order of a field

    Hello and thank you for the reply.

    I think now it's all clear but the following:
    direction  \Leftarrow for exercise 1)

    x^m-x=x*(x^{m-1}-1)=0

    Obviously if x=0 then x is in that field. So if  x^{m-1}-1=0 \Leftrightarrow x^{m-1}=1 the order of x is a divisor of m-1.

    Now it comes in that  x \in \mathbb{K}^{m^n} but how?  \| \mathbb{K}^m ^* \|=m-1 which is cyclic regarding multiplication but why can I conclude now that x is in  \mathbb{K}^m ? Is is just because the order of x is a divisor of m-1 and the number of elements in  \mathbb{K}^m ^* (=the units of  \mathbb{K}^m ) is m-1 and that's it?

    Regards
    Huberscher
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  4. #4
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    Re: two exercises concerning fields: order of a field

    There are at most m values x\in K^{m^n} that satisfy x^m-x=0. Now each x\in K^m is a root of this polynomial and there are m such. So the elements of K^m form the complete set of solutions to x^m-x=0. So if x\in K^{m^n} satisfies x^m-x=0, x\in K^m.
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