Power of a polynomial in Galois field

Question

Let $f(x) \in GF[q](X)$, where $q = p^m$ and $p$ prime. Is the following true?

$$f^{p^m}(X) = f(X^{p^m}).$$

I tried to prove the assertion above and got stuck at the following:

$$ \begin{align} f^{p^m}(X) &= (f_0 + f_1 X + f_2 X^2 + \ldots f_{n-1}X^{n-1})^{p^m}\\ &= f_0^{p^m} + (f_1 X + f_2 X^2 + \ldots f_{n-1}X^{n-1})^{p^m} + \text{cross products}. \end{align} $$

For the assertion to be true, the cross products should go to zero. Once they go to zero, we can keep on applying the binomial theorem to eventually get the proof of our result. If the assertion is not true, we should have some reason for the cross products to be nonzero. Can anyone please guide whether the cross products vanish or not, eventually guiding if the assertion is true or not.

Answer 1

In a field $K$ of characteristic $p$, the identity $$(x+y)^p=x^p+y^p$$ holds for all $x,y\in K$. The reason is simply that in the binomial expansion $$(x+y)^p=\sum_{k=0}^p \binom{p}{k}x^{p-k}y^k,$$ the coefficients $\binom{p}{k} = \frac{p!}{k!(p-k)!}$ are multiples of $p$ if $0<k<p$, so the corresponding term $\binom{p}{k}x^{p-k}y^k$ is zero.

Now if you define $\Phi:K\to K:x\mapsto x^p$, $\Phi$ is an endomorphism (called the Frobenius endomorphism). This implies that for all $m$, $\Phi^m$ is an endomorphisme as well, and thus $$f(X^{p^m})=f(\Phi^m(X))= \Phi^m(f(X))=(f(X))^{p^m}.$$

Answer 2

let $f(X)=X$, you have $X^{p^m}\neq X$. But the functions defined by $f$ and $f^{p^m}\in GF[X]$ coincide on $GF$. To see this remark that for $x\in GF$, $f^{p^m}(x)= (f_0+f_1X+...+f_nx^n)=f_0^{p^m}+f_1^{p^m}x^{p^m}+...+f_n^{p^m}(x^n)^{p^m}=f(x)$ this results from the binomial formula, the facts that the characteristic is $p$ and $u^{p^m}=u,u\in GF$.