Question

If c is a root of $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\in }{\mathit{Z}}_{\mathbf{p}}\mathbf{\left[}\mathit{x}\mathbf{\right]}$prove that ${\mathit{c}}^{\mathbf{p}}$is also a root.

Short answer

It is proved that $c^p$is also a root.

Step-by-step solution

Statement of lemma

To prove that if c is a root of $f\left(x\right)\in Z_p\left[x\right]$then $c^p$is also a root of this. Leema statement is if field F is a finite field with p element then for every c belonging to F satisfy $c^p=c$.

Showing that cp=c is also a root

As the field $Z_p\left[x\right]$ is a finite field with p element for every c belonging to $Z_p\left[x\right]$. Now to show that $c^p=c$satisfy.

For $c=0$. Left hand side and right hand side of this is zero so it satisfy $c^p=c$.

Now if $c\ne 0$. This is a nonzero element of$Z_p\left[x\right]$. Then the nonzero element from a group of order $p-l$under multiplication.

Use the fact that is for any element of a finite group G implies that $a^G=l_G$.

Then for all the non zero element $c\ne 0$of the finite field $Z_p\left[x\right]$satisfy $c^{q-a}=1$.

Now multiply this expression by c then this implies that $c^q=c$.

Therefore this satisfy that $c^q=c$. As is a root of $f\left(x\right)\in Z_p\left[x\right]$and it is proved that role="math" localid="1659168382104" $c^q=c$. Then this implies that $c^q=c$is also a root of $f\left(x\right)$.

Thus for any finite field $Z_p\left[x\right]$that if c is a root of $f\left(x\right)\in Z_p\left[x\right]$then $c^p$is also a root of $f\left(x\right)$.