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We have $GF(p)$ where $p$ is some prime. The polynomial $f(x)$ is irreducible over $GF(p)$. Show that $f(x)$ divides $g(x) = x^{p^{n}}-x \in GF(p)[x]$ if and only if deg$(f(x))$ divides $n$.

I assume that $f(x)$ divides $g(x)$. Then $g(x) = q(x)f(x)$ for some polynomial $q(x)$, and deg$(g(x))$ = deg$(q(x))$deg$(f(x))$. So the degrees of $q$ and $f$ must be some powers of $p$, say $s$ and $r$ respectively, so we have $p^n = p^{s}p^{r}$. This is where I'm stuck. I would appreciate some hints so I can move forward, but please don't post entire solution.

Auclair
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  • $GF(p)$ is the finite field with $p$ elements? Also, shouldn't $\text{deg}(g) = \text{deg}(f) + \text{deg}(q)$? – John Martin May 16 '16 at 15:17
  • This has been covered (or at least referred to) on our site many times. The key point is that the zeros of $g(x)$ are exactly the elements of $GF(p^n)$. So if $f(x)\mid g(x)$, then $f(x)$ has a zero $\alpha\in GF(p^n)$. So if you adjoin $\alpha$ to $GF(p)$ you get a subfield of $GF(p^n)$, and the law of extension degrees in a tower of fields tells us that $\deg f\mid n$. The other direction is similar. – Jyrki Lahtonen May 16 '16 at 18:30
  • @JohnMartin Yes, of course that should be the relation between the degrees. Thank you. – Auclair May 16 '16 at 19:13
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    @JyrkiLahtonen Thank you. Just before I had to leave I remembered that $GF(p^n)$ is the splitting field of $g(x)$, and now it all seems so obvious. Thanks for the answer though. – Auclair May 16 '16 at 19:16

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