Let \(F\) be a field of characteristic \(p\) and let \(f \in F[x]\) be a polynomial, \(f(x) = \sum_i f_i x^i\). Give necessary and sufficient conditions on the \(\{f_i\}\) for \(f(x^p)\) itself to be a \(p\)-th power, i.e. \(f(x^p) = g(x)^p\) for some \(g \in F[x]\). In particular, prove that your condition is necessary.
Solution
The necessary and sufficient condition is \[ f_i\in F^p:=\{a^p:a\in F\} \qquad\text{for every }i. \] Equivalently, every coefficient of \(f\) must have a \(p\)-th root in \(F\).
1. In characteristic \(p\), taking the \(p\)-th power of a polynomial applies Frobenius to its coefficients and multiplies every exponent by \(p\).
Let \[ g(x)=\sum_j g_jx^j\in F[x]. \] Since the binomial coefficients \[ \binom pk \] are divisible by \(p\) for \(0<k<p\), one has \[ (a+b)^p=a^p+b^p \] in every ring of characteristic \(p\). Applying this repeatedly to the finite sum defining \(g\) gives \[ g(x)^p =\sum_j(g_jx^j)^p =\sum_j g_j^p x^{jp}. \]
2. If every \(f_i\) is a \(p\)-th power in \(F\), then \(f(x^p)\) is a \(p\)-th power in \(F[x]\).
For each \(i\), choose \(g_i\in F\) such that \[ g_i^p=f_i. \] Set \[ g(x):=\sum_i g_i x^i. \] By 1, \[ g(x)^p =\sum_i g_i^p x^{ip} =\sum_i f_i x^{ip} =f(x^p). \] Thus the condition is sufficient.
3. If \(f(x^p)=g(x)^p\) for some \(g\in F[x]\), then every \(f_i\) is a \(p\)-th power in \(F\).
Write \[ g(x)=\sum_j g_jx^j. \] By 1, \[ g(x)^p=\sum_j g_j^p x^{jp}. \] On the other hand, \[ f(x^p)=\sum_i f_i x^{ip}. \] Equality of polynomials implies equality of the coefficient of \(x^{ip}\) for every \(i\), hence \[ f_i=g_i^p. \] Therefore \[ f_i\in F^p \] for every \(i\). This proves necessity and completes the equivalence.