If $g(x)$ is a polynomial in $GF(q)$, then $[g(x)]^q$ = $g(x^q)$.

Question

If $g(x)$ is a polynomial in $GF(q)$, then $[g(x)]^q$ = $g(x^q)$.

I came across this statement as lemma 2.1 in Factoring polynomials over large finite fields, ER Berlekamp.

The author also mentions that this Lemma is discussed among the others in two other publications. However, I could not find adequate proofs there either. The books just state them as fact.

Is there something obvious here that I am missing that can be used to prove this statement?

Answer 1

1. Show that the claim is true when $g$ is constant
2. Expand 1. to monomials
3. Show (e.g., using properties of binomial coefficients) that the claim is true for $g_1+g_2$ if it is true for $g_1$ and $g_2$.

Answer 2

Hint:

A finite field $\mathbf F_q$, where $a=p^n$ is a prime power, is the splitting field of the polynomial $X^q-X$, and the map from $\mathbf F_q$ to itself defined by $x\longmapsto x^q$ is the $n$-iterate of the Frobenius automorphism $x\longmapsto x^p$.