# Irreducible polynomial over field of order p

Let $p$ be a prime and $F=\mathbb{Z}/p\mathbb{Z}$ and $f(t)\in F[t]$ be an irreducible polynomial of degree $d$.

I need to show that $f(t)$ divides $t^{(p^{n})}-t$ if and only if $d$ divides $n$.

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I have intentionally written $t^{(p^{n})}$ since this is different from $(t^{p})^{n}$. – neelp Aug 26 '12 at 14:41
those parentheses are unnecessary for those who know what the standard meaning of mathematical notation is: $a^{b^c}$ always means $a^{(b^c)}$, because if you want to write $(a^b)^c$, well, that equals $a^{bc}$ and you could just have written that! You won't find $t^{(p^n)}$, written with those parentheses, in any math book. Someone who intends $(t^p)^n$ should just write $t^{np}$. – KCd Aug 26 '12 at 15:33

The first theorem of the following paper provides the proof:

http://www.jstor.org/stable/2316211

Or look at Theorem $7.6$ of the following article

http://www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/ff/ffchap3.pdf

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@pritam: Thanks for the resource!

For the users who might want to know the answer from the link, here it is:

Suppose $f$ divides $t^{(p^{n})}−t$. Then, if $a$ is a root of $f$, then $a^{(p^{n})}=a$, so $a$ is contained in a field of order $p^{n}$ so $F[a]$ is a subfield of $F^{'}=\mathbb{Z}/p^{n}\mathbb{Z}$. But since $f$ is a polynomial of degree $d$, $[F[a]:F]=d$ and $[F^{'}:F]=n$ and so $d$ divides $n$ since $n=[F^{'}:F[a]]d$.

Conversely, suppose $d$ divides $n$. Then $F^{'}=\mathbb{Z}/p^{n}\mathbb{Z}$ contains $F^{''}=\mathbb{Z}/p^{d}\mathbb{Z}$ as a subfield. If $a$ is a root of $f$, then $F[a]=F^{''}=\mathbb{Z}/p^{d}\mathbb{Z}$. Thus, $a\in F^{''}$ so $a\in F^{'}$. So $a^{(p^{n})}=a$ so every root of $f$ is a root of $t^{(p^{n})}-t$. Thus $f$ divides $t^{(p^{n})}-t$.

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