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Let p be any prime and let $f(x) \in \Bbb{F}_p[x]$ be any polynomial with coefficients in $\Bbb{F}_p$. Let F be a field of characteristic p and let $\alpha \in F$ be an element for which $f(\alpha)=0$. Prove that $f(\alpha^p)=0$ as well.

Fermat's little theorem: for any prime p and any $a \in \Bbb{Z}$,that $a^p \equiv a$(mod p)

$f(x)=c_o+c_1x+...+c_{n-1}x^{n-1}+x^n$. How am I supposed to argue with $(c_o+c_1x+...+c_{n-1}x^{n-1}+x^n)^p$ in $\Bbb{F}_p[x]$?

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    $\begingroup$ Can you prove $f(a)^p=f(a^p)$? $\endgroup$ – Angina Seng May 31 at 19:43
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It is essentially the Frobenius homomorphism.

$F : \mathbb{F}_p \to \mathbb{F}_p \qquad a \mapsto a^p$

Since $\mathbb{F}_P$ is a finite field then the above homomorphism is actually an automorphism. What you need now is the linearity of $F$, namely

$ F(c_0+c_1\alpha+ \cdots + c_{n-1} \alpha^{n-1} + \alpha^{n} ) = (c_0+c_1\alpha+ \cdots + c_{n-1} \alpha^{n-1} + \alpha^{n} )^p = \\=c_0^p + c_1^p (\alpha^p)^1 + \cdots + c_{n-1}^p (\alpha^p)^{n-1} + (\alpha^p)^n$

As you said $a^p \equiv a \bmod p$. In particular $c_i^p = c_i$ in $\mathbb{F}_p$ for every $i = 0, \dots, n-1$.

Finally, by the fact that $F$ is a homomorphism and so $0 = F(0)$:

$0=F(f(\alpha)) = F(c_0+c_1\alpha+ \cdots + c_{n-1} \alpha^{n-1} + \alpha^{n} ) = \\=c_0 + c_1 (\alpha^p)^1 + \cdots + c_{n-1} (\alpha^p)^{n-1} + (\alpha^p)^n =f(\alpha^p)$

i.e.

$0=f(\alpha^p)$

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