Let $F$ be a field of characteristic $p$, and define $\phi: F \to F$ by $\phi(a) = a^p$, i.e. let $\phi$ be the Frobenius endomorphism of $F$.
In the text by Dummit and Foote, there is a result that states that $\phi$ is surjective if $F$ is finite. The reason for this is because $\phi$ is injective (the kernel is $\{0\}$) and $F$ is finite, which thus implies surjectivity.
I'm wondering why I can't drop the finiteness assumption by using the first isomorphism theorem for rings. We have $F/ker(\phi) = F/\{0\} \cong \phi(F)$, which gives $\phi(F) \cong F$. This would mean that $\phi$ is surjective, right?
“Right?” No, fortunately, not right. You can see very clearly by noting that even in characteristic zero, if $x$ is an indeterminate, $$ \Phi:\Bbb Q(x)\to\Bbb Q(x), \qquad \frac{P(x)}{Q(x)}\mapsto\frac{P(x^2)}{Q(x^2)} $$ is a field morphism, so with zero kernel, so one-to-one, but the image (range) is $\Bbb Q(x^2)$, a proper subfield.