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My question: If $f(x)$ is irreducible of degree $d$ in $GF(q)$ then $f(x)\mid x^{q^n}-x$ if and only if $d\mid n$.

My try: Consider $f(x)\mid x^{q^n}-x$. Assume that $\alpha$ be a root of $f$ then $\alpha^{q^i}$, $1\leq i \leq d-1$ are other roots of $f$. The roots of $f$ are elements of the finite field $GF(q^d)$. The elements of $GF(q^d)$ are solutions of $x^{q^d}-x=0$ which results in $x^{q^d}-x \mid x^{q^n}-x$. The last relation is true iff $d\mid n$.

Is my proof correct?

Thanks for any suggestions.

Amin235
  • 1,867

1 Answers1

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Even more is true: in fact, $x^{p^n} - x$ is the product of all monic irreducibles in $\Bbb Z_p$ of degree $d \,|\, n.$

Proof. Consider the set $\Bbb F_{p^n}$ of zeros of $x^{p^n} - x$ in $\overline{\Bbb Z}_{p}.$ We note that $\Bbb F_{p^n}$ is the unique field extension $\Bbb Z_{p^n}$ of $\Bbb Z_p.$ Each element $\alpha \in \Bbb F_{p^n}$ is algebraic over $\Bbb Z_p$ of degree $d \,|\, n,$ hence each element $\alpha \in \Bbb F_{p^n}$ is the root of some monic irreducible $p_d(x)$ of degree $d \,|\, n$ in $\Bbb Z_p[x].$ Furthermore, every root of an irreducible $p_d(x)$ of degree $d \,|\, n$ lies in some subfield $\Bbb F_{p^d}$ of $\Bbb F_{p^n}.$ Combining these facts, it follows that $x^{p^n} - x = \prod_{\alpha \in \Bbb F_{p^n}} (x - \alpha) = \prod_{d \,|\, n} p_d(x),$ as desired.

Observe that this proof makes use of each of the facts that you used in your proof.

Dylan C. Beck
  • 4,811