Question

Show that a field K of order pⁿ contains all $\mathbf{kth}$root of ${\mathbf{1}}_{\mathbf{k}}$where $\mathbf{k}\mathbf{=}{\mathbf{p}}^{\mathbf{n}}\mathbf{-}\mathbf{1}$

Short answer

It is proved that field K of order p contain all kth root.

Step-by-step solution

Definition of splitting field

A splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial split or decomposes into linear factors.

Showing that K of order p contain all kth root

Let K a finite field of order pⁿ . That is$\left|k\right|=p^n$.

Every nonzero $u\in K$satisfies role="math" localid="1659093257824" $u^{p^n-1}=1_k$. So that u is a root of polynomial $x^{p^n-1}-1_k\in K\left[x\right]$. But this implies that u is a root of $x\left(x^{p^n-1}-1\right)=\left({x^p}^n-x\right)\in Z_p\left[x\right]$.

Hence there are pⁿ distinct root of $u\left({x^p}^n-x\right)$including 0, so that $\left({x^p}^n-x\right)$splits over K. Since $u\in K$was arbitrary. Thus K is the splititing field of $\left({x^p}^n-x\right)$.

Therefore K contain all kth roots.