# two exercises concerning fields: order of a field

• Sep 18th 2013, 10:37 AM
huberscher
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
• Sep 18th 2013, 01:58 PM
johng
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.

$\displaystyle K^4=\{0,1,\alpha,\alpha^2\}$ with $\displaystyle \alpha^3=1$. Also $\displaystyle K^2=\{0,1\}$.
• Sep 19th 2013, 03:57 AM
huberscher
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
• Sep 19th 2013, 07:12 AM
johng
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 $\displaystyle x^m-x=0$, $\displaystyle x\in K^m$.