Frobenius injective for finite fields - what about $\mathbb{F_{p^n}}$

Question

Quick question about the Frobenius endomorphism. My lecture notes and wikipedia say that the Frobenius is injective for finite fields. However, if we look at $\mathbb{F_4}$, we have $$\text{Frob}(2) = 2^2 = 0 = \text{Frob}(0)$$ Am I missing something?

Answer 1

Perhaps you are missing the fact that $\mathbb F_{2^2}\simeq\mathbb F_2\times\mathbb F_2$ is a $2$-dimensional vector field over $\mathbb F_2$. So in fact $2=(0,0)=0\in\mathbb F_{2^2}$.

Answer 2

$2=0$ in $\mathbb F_4$, so there's no contradiction. One way of constructing $\mathbb F_4$ is as $\mathbb F_2 [x] / (x^2+x+1)$. Then Frob fixes $\mathbb F_2$ and swaps $x$ and $1+x$.

Answer 3

A ring homomorphism from any field to a ring is necessarily injective since its kernel is an ideal of the field, which is trivial.