Vtyjkyuk

Prove that $f(x)$ divides $x^p^n - x$ if and only if $d := deg f(x)$ divides $n$.

I believe that I have the backward direction covered: Let $d mid n$ say $n = dq$ for some $q$ in $mathbbF_p[x]$. Consider the field $mathbbF_p[x]/(f(x))$ which has $p^d$ elements. Take an element $x+I$ from the field (here $I = (f(x))$) so we have: $(x+I)^p^n = (x+I)^p^dq$. As long as you keep factoring out $(x+I)$ with the $p^d$ power you will get $(x+I)$ so $x^p^n - x in (f(x))$.

I am having trouble getting to the other direction.

Hints:

(i) Show that the splitting field of the polynomial $,p(x):=x^p^n-xinBbb F_p[x],$ over the prime field $,Bbb F_p,$ is the field $,Bbb F_p^n,$

(ii) One way to go: show that the set of roots of the above polynomial $,p(x),$ in some algebraic closure of $,Bbb F_p,$ is a field...

(iii) Prove that $,Bbb F_p^d,$ is a subfield of $,Bbb F_p^n,$ iff $,dmid n,$

Of course, take into account that $,f(x)mid p(x)Longrightarrow,$ all the roots of $,f(x),$ are also roots of $,p(x),$

This question is very old, but there is a more direct solution worth noting.

Proof.

Suppose $f(x)$ divides $h(x):=x^p^n -x$. Then since $h(x)$ splits over $mathbbF_p^n$, so does $f(x)$. Let $alpha in mathbbF_p^n$ be a root of $f(x)$. Then $mathbbF_p(alpha)subset mathbbF_p^n$, and $[mathbbF_p(alpha): mathbbF_p] = d$. Finally, we have that $n = [mathbbF_p^n: mathbbF_p]= [mathbbF_p^n: mathbbF_p(alpha)][mathbbF_p(alpha): mathbbF_p]$, completing the proof that $d | n$.