If \(f\in k[x]^{\irr}\) with \(\ch k = p\), then there is a unique separable \(g\in k[x]^{\irr}\) such that \(f(x) = g(x^{p^e})\) for some unique \(e \ge 0\).
Solution
Let \(f\in k[x]\) be irreducible and \(\operatorname{char}k=p>0\).
For any polynomial \(h(x)=\sum a_i x^i\), \[ h'(x)=0 \quad\Longleftrightarrow\quad a_i=0\text{ whenever }p\nmid i \quad\Longleftrightarrow\quad h(x)=h_1(x^p) \] for some \(h_1\in k[x]\). If \(h\) is irreducible and \(h=h_1(x^p)\), then \(h_1\) is irreducible, since any factorization of \(h_1\) would induce one of \(h\).
Choose \(e\ge0\) maximal such that \[ f(x)=g(x^{p^e}) \] for some \(g\in k[x]\). Such an \(e\) exists because \(e=0\) works and \(p^e\mid\deg f\), so \(e\) is bounded. The polynomial \(g\) is irreducible. By maximality, \(g'(x)\ne0\); since \(g\) is irreducible, this is equivalent to \(g\) being separable. Thus the required representation exists.
For uniqueness, suppose \[ f(x)=g_1(x^{p^{e_1}})=g_2(x^{p^{e_2}}) \] with \(g_1,g_2\) separable. Assume \(e_1<e_2\). Comparing coefficients shows that every exponent occurring in \(g_1\) is divisible by \(p^{e_2-e_1}\), hence in particular by \(p\). Thus \(g_1'(x)=0\), contradicting separability. Therefore \(e_1=e_2\), and then equality of the two substituted polynomials forces \(g_1=g_2\).
Hence both \(e\) and the separable irreducible polynomial \(g\) are unique.