$f(x)^p=f(x^p)$ for any polynomial $f(x) \in \mathbb F_p[x].$

Question

Show that $f(x)^p=f(x^p)$ for any polynomial $f(x) \in \mathbb F_p[x].$

I know that the mapping $x \mapsto x^p$ from $\mathbb F_p$ onto $\mathbb F_p$ is an isomorphism. Hence any element $y \in \mathbb F_p$ can be written as $y=z^p$ for some $z \in \mathbb F_p$. Now I really do not understand from where to start. Any hint or solution will be very helpful for me.

Answer 1

Let $R\subseteq\mathbb F_p[x]$ be defined as

$$R:=\{f\in\mathbb F_p[x]\mid f(x^p)=f(x)^p\}$$

So $R$ is the subset of those polynomials which satisfy the equation $f(x)^p=f(x^p)$ and our goal is to prove that, actually, $R$ is the entire $\mathbb F_p[x]$.

Note that there are some obvious elements of $R$:

• Any constant $f(x)=c\in\mathbb F_p$ is in $R$. Namely, if $f=c$, then $f(x)^p=c^p$ and $f(x^p)=c$, but $c^p=c$ per Little Fermat's theorem.
• $f(x)=x$ is in $R$ because $f(x)^p=x^p=f(x^p)$.

$R$ is also closed for multiplication: if $f(x)=g(x)h(x)$ with $g,h\in R$, then:

$$f(x)^p=(g(x)h(x))^p=g(x)^ph(x)^p=g(x^p)h(x^p)=f(x^p)$$

Finally, $R$ is closed for addition. If $f(x)=g(x)+h(x)$, then:

$$\begin{array}{rcl}f(x)^p&=&(g(x)+h(x))^p\\&=&\sum_{k=0}^p\binom{p}{k}g(x)^kh(x)^{p-k}\\&=&g(x)^p+h(x)^p\\&=&g(x^p)+h(x^p)\\&=&f(x^p)\end{array}$$

This is because most of the binomial coefficients $\binom{p}{k}$ in the sum above are divisible by $p$ so they all cancel out. ($\mathbb F_p[x]$ is a ring of characteristic $p$.) The exceptions are $\binom{p}{0}=\binom{p}{p}=1$, i.e. in the sum above, only the first and the last term ($g(x)^p$ and $h(x)^p$ remain.)

(In fact, to elaborate a bit: $\binom{p}{k}=\frac{p!}{k!(p-k)!}$ so if $1\le k\le p-1$, then $1\le p-k\le p-1$ and so the numerator is divisible by $p$ but none of the factors in the denominator are divisible by $p$ as $p$ is prime.)

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The bottom line here is that $R$ is a subring of $\mathbb F_p[x]$, containing all constants and $f(x)=x$ and so it must be the entire $\mathbb F_p[x]$ because you can "reach" any polynomial in $\mathbb F_p[x]$ by applying additions and multiplications finitely many times starting with constant polynomials and the polynomial $f(x)=x$.

Answer 2

The main result you need is this: for $a,b\in\mathbb{F}_p$ $$ (a+b)^p=a^p+b^p\ , $$ which itself hinges on ${p\choose k}=0$ in $\mathbb{F}_p$ for $0<k<p$.

Try proving the result above and generalising to multinomial expressions, i.e. $$ (a_1+a_2+\cdots+a_n)^p=a^p_1+a^p_2+\cdots+a^p_n\ . $$

Can you take it from here?