I'm studying a bit of field theory, and had a quick question about the Frobenius endomorphism, $\phi : K \rightarrow K$. For any prime $p$, is the Frobenius endomorphism on $F_p$ (the finite field with $p$ elements) the identity automorphism?
My reasoning is from working in $\mathbb{Z}/p\mathbb{Z}$, as $F_p \cong \mathbb{Z}/p\mathbb{Z}$. By Fermat's Little Theorem, $\forall x\in \mathbb{Z}$, $x^p\equiv x \mod p$, so $ \phi(x) = x^p \mod p = x \mod p $ and therefore $\phi$ fixes all elements of $\mathbb{Z}/p\mathbb{Z}$ and is the identity on $F_p$.
Is this correct? I'm asking because I haven't been able to find anything to support this claim anywhere-- I'm not sure if I've simply created a false statement and made an error in my justification for it. Thanks!