Question

Let F be a field and $\mathit{f}\left(x\right)$be a monic polynomial in $\mathit{F}\mathbf{\left[}\mathit{x}\mathbf{\right|}$whose roots are all distinct in any splititing field K . Let E be the set root of $\mathit{f}\left(x\right)$in K . If the set E is actually a subfield of K prove that F has characteristic p for some prime and that $\mathit{f}\left(x\right)\mathbf{=}{\mathit{x}}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{-}\mathit{x}$for some $\mathit{n}\mathbf{\ge }\mathbf{1}$.

Short answer

It is proved that F has characteristic p for some prime and that $f\left(x\right)=x^{p^n}-x$for some $n\ge 1$.

Step-by-step solution

Definition of subfield

A subset of a mathematical field that is itself a field. That means a subset of a given field which is a field in itself is called as subfield.

Showing that f(x)=xpn-x 

Let E be the root of $f\left(x\right)$ in K and E is a subfield of K . The root of $f\left(x\right)$are finite so if they form a field R . It is finite and therefore has characteristic p .

The order of field R is $p^n-1$. So every nonzero $u\in R$satisfy $u^{p^{n-1}}=I_R$.

Hence every nonzero root u of $f\left(x\right)$is a root of $x^{p^n-1}-I_R$and therefore all root of $f\left(x\right)$are root of $x^{p^n-1}-x$are equal.

Therefore $f\left(x\right)$and $x^{p^n-1}-x$are monic and have same roots. Thus $f\left(x\right)$and $x^{p^n-1}-x$are equal.

As $1\in R$ and R has characteristic p which mean p is the least such integer such that $1+1+1+1+\dots +1p$ times is equal to 0. 1 is a basic of F so this implies p does same to every element of F.

Thus F has characteristic p.