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Thread: 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 $\displaystyle \mathbb{K}^m $ be a field containing m elements and $\displaystyle n \in \mathbb{N}$ be positive.

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

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

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


    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)
    $\displaystyle \Rightarrow$
    If $\displaystyle x \in \mathbb{K}^m$. The order of x is a divisor of m ? At least is $\displaystyle x^{m-1}=1$ 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 $\displaystyle \mathbb{K}^m$ doesn't have to be m-1 yeah???
    If so the equation could be multiplied by x and what results is $\displaystyle x^m-x=0$ after subtracting x on both sides and this direction is proved.
    $\displaystyle \Leftarrow$
    Here from $\displaystyle x^m-x=0 $ follows that $\displaystyle x^m=x$ so by dividing by x we have $\displaystyle x^{m-1}=1$ thus x has order m-1 so it is contained in the subset $\displaystyle \mathbb{K}^m$ which consists of elements with order m-1 ????

    Let's have a look at exercise 2)
    So $\displaystyle \mathbb{K}^4$ is isomorphic to $\displaystyle \{ 1, \alpha, \alpha^2 , \alpha^3 \} $ where $\displaystyle \alpha^4=1$?
    $\displaystyle g(1)=2=1+1 \in \mathbb{K}^2$ ?? Why?
    $\displaystyle g(\alpha)=\alpha^3-\alpha$ ...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:

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

    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,

    $\displaystyle K^4=\{0,1,\alpha,\alpha^2\}$ with $\displaystyle \alpha^3=1$. Also $\displaystyle K^2=\{0,1\}$.
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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 $\displaystyle \Leftarrow $ for exercise 1)

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

    Obviously if x=0 then x is in that field. So if $\displaystyle 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 $\displaystyle x \in \mathbb{K}^{m^n} $ but how? $\displaystyle \| \mathbb{K}^m ^* \|=m-1$ which is cyclic regarding multiplication but why can I conclude now that x is in $\displaystyle \mathbb{K}^m $ ? Is is just because the order of x is a divisor of m-1 and the number of elements in $\displaystyle \mathbb{K}^m ^* $ (=the units of $\displaystyle \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 $\displaystyle x\in K^{m^n}$ that satisfy $\displaystyle x^m-x=0$. Now each $\displaystyle x\in K^m$ is a root of this polynomial and there are m such. So the elements of $\displaystyle K^m$ form the complete set of solutions to $\displaystyle x^m-x=0$. So if $\displaystyle x\in K^{m^n}$ satisfies $\displaystyle x^m-x=0$, $\displaystyle x\in K^m$.
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