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Let $\mathbf F$ be a field of prime characteristic $p$. It is known that the Frobenius map $c\phi=c^p~~\forall c\in\mathbf F$ is an endomorphism of $\mathbf F$. Moreover, since the only ideals of $\mathbf F$ are $\{0\}$ and $\mathbf F$, we know that $\ker(\phi)=\{0\}$. This implies that $\phi$ is injective. It is known that $\phi$ is not surjective in general. However, if we consider the following argument below, it seems to me that $\phi$ must be an automorphism (i.e., a bijective endomorphism).

"We claim that $\mathbf F/\{0\}\cong\mathbf F$. Consider the homomorphism $\theta:\mathbf F/\{0\}\rightarrow\mathbf F$ given by $(\{0\}+r)\theta=r$, where $r\in\mathbf F$. Then it is easy to show that it is both injective and surjective; and so our claim holds. However, we also know by the First Ring Isomorphism Theorem that $\mathbf F/\ker(\phi)\cong \text{im}(\phi)$. So we conclude that $\text{im}(\phi)\cong\mathbf F$. But $\text{im}(\phi)\subseteq\mathbf F$ so necessarily $\text{im}(\phi)=\mathbf F$. Therefore $\phi$ is an automorphism of $\mathbf F$."

Could someone tell me where the mistake lies?

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    @BenjaLim: Yes, thanks a million!2012-08-05

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Your mistake comes in assuming that because $\textrm{im} \phi \subseteq \Bbb{F}$ and $\textrm{im} \phi \cong \Bbb{F}$ then $\textrm{im} \phi = \Bbb{F}$. Here's a good example to consider. I will produce for you an example of two fields $E,H$ one of which is contained in the other with $E \cong H$ (as rings) but $E \neq H$.

Consider $F = \Bbb{Z}/2\Bbb{Z}$ and $t$ an indeterminate. Let $E = F(t)$, $H = F(t^2)$. Then you can see that $H \subseteq E$ and $H \cong E$ (as rings) by the map $f$ that is constant on $F$ and sends $t \mapsto t^{2}$. However clearly $H \neq E$. To see this begin by noticing that $F(t^2)$ is the fraction field of $F[t^2]$. If $t \in F(t^2)$, we have that $\frac{p(t^2)}{q(t^2)} = t$ for some polynomials $p$ and $q$. Then you have $tq(t^2) = p(t^2)$. But then the guy on the left has all odd powers while the guy on the left only even powers, a contradiction.

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    It might be better, if you specified $F=\mathbb{F}_2$, because then the mapping $t\mapsto t^2$ **would be the Frobenius endomorphism**. +1 all the same.2012-07-15