D+F 13.5.8

Prove that $f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]}$.

$\textit{Proof}$ : First note that since we are in a field of charachteristic $p$, then we may use the properties of the Frobenius automorphism: $$(a+b)^p \equiv a^p + b^p \quad and \quad (ab)^p \equiv a^pb^p$$ where we have used "$\equiv$" to remind ourselves that we are working in charachteristic $p$.
Similarly for more terms, we have $$(a+b+c)^p \equiv ((a+b)+c)^p \equiv (a+b)^p+c^p \equiv a^p + b^p + c^p$$ $$(abc)^p \equiv ((ab)c)^p \equiv (ab)^pc^p \equiv a^pb^pc^p$$ Let $$f(x)=a_nx^n+ \cdots +a_0 \quad a_i\in{\mathbb{F}_p}$$ Then by induction, $$f(x)^p \equiv (a_nx^n+ \cdots + a_0)^p \equiv (a_nx^n)^p + \cdots + (a_0)^p \equiv (a_n)^p(x^n)^p + \cdots + a_0^p$$ $$ \equiv a_n(x^n)^p+ \cdots + a_0 \equiv f(x^p)$$ Since $a^p \equiv a (modp)$.
$\therefore \quad f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]} \quad \blacksquare$