# Let F be a field of characteristic p and let $\alpha \in F$ be an element for which $f(\alpha)=0$. Prove that $f(\alpha^p)=0$

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 $$\Bbb{F}_p[x]$$?

• Can you prove $f(a)^p=f(a^p)$? – Angina Seng May 31 at 19:43

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.

$$0=f(\alpha^p)$$