# Every irreducible polynomial of degree $m$ over $\mathbb F_p$ divides $x^{p^m}-x$

We consider $F=\mathbb F_p$ for $p$ prime, $f(x)$ an irreducible polynomial of degree $m$ over $F$ and $g(x)=x^{p^m}-x$. I want to show that $f(x)\mid g(x)$.

From the fact that the field $A=F[x]/(f)$ which has $p^m$ elements how does the $f(x)|g(x)$ follows?

• @JyrkiLahtonen It's irreducible. Thank you. – Test123 Aug 21 '14 at 18:54
• The field $A$ has $p^m-1$ non-zero elements. Assuming $m>1$ we have that $x$ is not a factor of $f$, so $x+(f)$ is a non-zero element of $A$. What does Lagrange's theorem (from group theory) tell us about $(x+(f))^{p^m-1}$? – Jyrki Lahtonen Aug 21 '14 at 18:56
• @JyrkiLahtonen Now I understand. We consider the multiplicative group $A^*$ of non-zero elements so $(x+(f))^{p^m-1}=1_F+(f)$. Right? – Test123 Aug 21 '14 at 19:03
• @JyrkiLahtonen Yes. From the above it follows that $g(x) +(f)$ is the zero element. Thanks for all the help. – Test123 Aug 21 '14 at 19:05
• Correct. Well done! You may consider typing this as an answer. I will upvote it (and your question) tomorrow, for I have used all my votes for today. That's why I starred the question, so I will remember to come back to it. – Jyrki Lahtonen Aug 21 '14 at 19:07

Hint $\$ Generalizing Fermat's little Theorem, in a finite field field $\Bbb F_q\,$ of size $\,q,\,$ the nonzero elements form a group of order $\,q\!-\!1\,$ so, by Lagrange's Theorem $\,a^{q-1} = 1\,$ for all $\,0\ne a\in\Bbb F_q$
In particular, if $\,q=p^m\,$ then $\,x^{q-1}\! = 1\,$ in $\,\Bbb F_q\cong \Bbb F_p[x]/(f)\$ i.e. $\ x^{q-1}\!-1\in (f) = f\,\Bbb F_q[x],\,$ or, said equivalently, $\,f\mid x^{q-1}\!-1\,$ in $\,\Bbb F_q[x].$
The multiplicative group $A^*$ of $A$ has order $p^m-1$. Since $f$ irreducible and assuming $m>1$ we have that $x\not\in (f)$, so $x+(f)$ is a non-zero element of $A^*$ and by Lagrange we have that $x^{p^m-1}+(f)=(x+(f))^{p^m-1}=1_F+(f)$. Then $g(x)+(f)=x(x^{p^m-1}-1_F)+(f)=(f)$ the zero element in $A$ which means that $f(x)|g(x)$.