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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?

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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 $2=(0,0)=0\in\mathbb F_{2^2}$.

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$2=0$ 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 $1+x$.

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

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