# The Frobenius endomorphism

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?

• The deduction that ${\rm im}(\phi) \cong F$ and ${\rm im} (\phi) \subseteq F$ together imply ${\rm im}(\phi) = F$ is flawed. It might be helpful to think of the additive group $\mathbb{Z}$ and the map $\psi: \mathbb{Z} \to \mathbb{Z}$ with $\psi(z) = 2z.$ This is an injective group endomorphism, but is certainly not surjective. – Geoff Robinson Jul 15 '12 at 8:29
• If F is finite, then you can show the Frobenius map is an automorphism. – Holdsworth88 Jul 15 '12 at 8:32
• @GeoffRobinson Is it a problem that your example above is for a group and not a field? – user38268 Jul 15 '12 at 8:36
• @youngtableaux By the way when you say $\im \phi \cong \Bbb{F}$ you surely mean isomorphic as rings yes? – user38268 Jul 15 '12 at 8:44
• @youngtableaux: I was using the $\mathbb{Z}$ example to illustrate the logical flaw, which ultimately comes from the fact that because one infinite object is contained in another and in bijection with it, it is not necessarily equal: indeed the possibility that this fails for infinite sets is the essence of "infiniteness". – Geoff Robinson Jul 15 '12 at 9:15

## 1 Answer

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.

• 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. – Jyrki Lahtonen Jul 15 '12 at 9:08