Question

Prove that every nonzero element c of a finite field K of order ${\mathbf{p}}^{\mathbf{n}}$satisfy ${\mathbf{C}}^{{\mathbf{p}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}\mathbf{=}{\mathbf{1}}_{\mathbf{K}}$conclude that C is a root of ${\mathbf{x}}^{{\mathbf{p}}^{\mathbf{n}}\mathbf{-}\mathbf{1}\mathbf{}}\mathbf{-}\mathbf{x}$and use this fact to prove theorem.

Short answer

Cis a root of$x^{p^n-1}-x$.

Step-by-step solution

Definition of field.

A field is a commuatative ring with unity in which every nonzero element has a multiplication inverse.

Step-2: Showing that C is the root of xpn-1-x

Let K be a finite field of order $p^n$. The $p^n$distinct element of K are all the possible roots. Let $C_1,C_2,......C_k$be all the nonzero element of K and u be the product of $C_1,C_2,......C_k$. This is $u=C_1,C_2,......C_k$

The n element $C{C}_1,C{C}_2,......C{C}_k$are all distinct since $C{C}_i=C{C}_j$. This implies:

$C_i=C_j$

So they are just nonzero element of K in some other and their product is the element u. Therefore

$$\begin{gathered} u=\left(c{c}_1\right)\left(c{c}_2\right).......\left(c{c}_k\right) \\ =c^k\left(c_1{c}_2....c_n\right) \\ =u{c}^k \end{gathered}$$

Cancelletion of u shows $c^k=1_k$. Hence $c^{k+1}=c$.

Since $K+1=p^n$. Therefore $c^{p^n-1}{1}_k$

This implies that c is a root of$x^{p^n-1}-x$