Let \(F\) be a field of characteristic \(p>0\) and let \(f\in F[x]\) be irreducible. Show that there are a unique integer \(e\ge0\) and an irreducible separable polynomial \(g\in F[x]\) such that \[ f(x)=g(x^{p^e}), \] and that every root of \(f\) in an algebraic closure has multiplicity \(p^e\).
Solution
Choose \(e\ge0\) maximal such that every exponent occurring in \(f\) is divisible by \(p^e\). Then there is a unique polynomial \(g\in F[x]\) with \[ f(x)=g(x^{p^e}). \]
The polynomial \(g\) is irreducible. Indeed, if \(g=uv\) with nonconstant \(u,v\in F[x]\), then \[ f(x)=u(x^{p^e})v(x^{p^e}) \] would be a nontrivial factorization of \(f\).
By maximality of \(e\), not every exponent occurring in \(g\) is divisible by \(p\). Hence \(g'\ne0\). Since \(g\) is irreducible, \(\gcd(g,g')=1\), so \(g\) is separable.
Now work in an algebraic closure \(\overline F\). Since \(g\) is separable, \[ g(y)=a\prod_{i=1}^r(y-\beta_i) \] with distinct \(\beta_i\). For each \(i\), choose \(\alpha_i\in\overline F\) with \(\alpha_i^{p^e}=\beta_i\). In characteristic \(p\), \[ x^{p^e}-\beta_i=x^{p^e}-\alpha_i^{p^e}=(x-\alpha_i)^{p^e}. \] Therefore \[ f(x)=g(x^{p^e}) =a\prod_{i=1}^r(x-\alpha_i)^{p^e}. \] Thus every root of \(f\) has multiplicity exactly \(p^e\).
Uniqueness of \(e\) follows from maximality of the common \(p\)-power dividing all occurring exponents; then \(g\) is forced coefficient-by-coefficient.