Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$.

share|cite|improve this question
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
up vote 2 down vote accepted

The first theorem of the following paper provides the proof:

Or look at Theorem $7.6$ of the following article

share|cite|improve this answer

@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 $t^{(p^{n})}-t$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.