Suppose that \(K\) is a finite field with \(|K| = q\). Show that if \(f \in K[x]\) is irreducible, then \(f\) divides \(x^{q^n} - x\) in \(K[x]\) if and only if \(\deg f\) divides \(n\).
Solution
Let \(d=\deg f\) and let \(\alpha\) be a root of \(f\) in an algebraic closure \(\overline K\).
1. The field \(K(\alpha)\) has \(q^d\) elements.
Because \(f\) is irreducible of degree \(d\), it is the minimal polynomial of \(\alpha\) over \(K\), so \[ [K(\alpha):K]=d. \] Since \(|K|=q\), the finite extension \(K(\alpha)\) therefore has \(q^d\) elements. Thus \[ K(\alpha)\cong \mathbf F_{q^d}. \]
2. We have\[ f\mid x^{q^n}-x \]if and only if \(\alpha\in \mathbf F_{q^n}\).
The roots of \(x^{q^n}-x\) in \(\overline K\) are exactly the elements of the unique field \(\mathbf F_{q^n}\subset\overline K\). The derivative of \(x^{q^n}-x\) is \(-1\), so all its roots are simple.
If \(f\) divides \(x^{q^n}-x\), then every root of \(f\), in particular \(\alpha\), is a root of \(x^{q^n}-x\), hence lies in \(\mathbf F_{q^n}\).
Conversely, if \(\alpha\in\mathbf F_{q^n}\), then \(x^{q^n}-x\) vanishes at \(\alpha\). Since \(f\) is the minimal polynomial of \(\alpha\) over \(K\), it divides every polynomial in \(K[x]\) vanishing at \(\alpha\). Hence \(f\mid x^{q^n}-x\).
3. We have\[ \mathbf F_{q^d}\subseteq \mathbf F_{q^n} \quad\Longleftrightarrow\quad d\mid n. \]
If \(d\mid n\), write \(n=dr\). Every \(a\in\mathbf F_{q^d}\) satisfies \(a^{q^d}=a\), hence iterating Frobenius gives \(a^{q^n}=a\); therefore \(a\in\mathbf F_{q^n}\).
Conversely, suppose \(\mathbf F_{q^d}\subseteq\mathbf F_{q^n}\). Then by the tower law, \[ n=[\mathbf F_{q^n}:\mathbf F_q] =[\mathbf F_{q^n}:\mathbf F_{q^d}]\,[\mathbf F_{q^d}:\mathbf F_q], \] so \(d\) divides \(n\).
4. Therefore\[ f\mid x^{q^n}-x \quad\Longleftrightarrow\quad \deg f\mid n. \]
By 1 and 2, divisibility is equivalent to \[ K(\alpha)=\mathbf F_{q^d}\subseteq\mathbf F_{q^n}. \] Apply 3.