Question

Let E be the set of roots of $\left({x^p}^n-x\right)\mathbf{\in }{\mathbf{Z}}_{\mathbf{p}}\left[x\right]$in some splititing field. If $\mathbf{a}\mathbf{\in }\mathbf{E}$prove that $\mathbf{-}\mathbf{a}\mathbf{\in }\mathbf{E}$.

Short answer

It is proved that $-a\in E$.

Step-by-step solution

Definition of splititing field

A splititing 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 -a∈E 

Let E be the set of root of $\left({{\mathrm{x}}^{\mathrm{p}}}^{\mathrm{n}}-\mathrm{x}\right)\in {\mathrm{Z}}_{\mathrm{p}}\left[\mathrm{x}\right]$in some splititing field and $\mathrm{a}\in \mathrm{E}$.

Since E is the set of roots of $\left({{\mathrm{x}}^{\mathrm{p}}}^{\mathrm{n}}-\mathrm{x}\right)\in {\mathrm{Z}}_{\mathrm{p}}\left[\mathrm{x}\right]$. Therefore if $a\in E$then ${a^p}^n-a=0$.

Simplify further:

$$\begin{gathered} a\left({a^p}^n-1\right)=0 \\ a=0,a^{p^n}-1=0 \end{gathered}$$

If a=0 then it is true for -a also, so $-a\in E$.

If ${a^p}^n-1=0$first if p is an odd integer again and then $\left(p^n-1\right)$will be an even integer. Therefore ${a^p}^n-1=0={\left(-a\right)}^{p^n}-1$. Thus $-a\in E$.

Second if p is an even number then pⁿ is an even integer again and then pⁿ will be an even integer. Thus $-a\in E$.

In both case -a is root of the equation $\left({{\mathrm{x}}^{\mathrm{p}}}^{\mathrm{n}}-\mathrm{x}\right)\in {\mathrm{Z}}_{\mathrm{p}}\left[\mathrm{x}\right]$. Hence $-a\in E$.

It is proved that $-a\in E$.