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 forexercise 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 atexercise 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