I have the following question,
Let $F$ be a field of characteristic $p\neq 0$. Find an example where the homomorphism $$ \begin{align*} \sigma : &F\rightarrow F\\ & a\mapsto a^p \end{align*} $$ is not surjective.
Trying a lot of things I found that if I take $a\in Ker(\sigma)$ then, $a^p=0 \Rightarrow a^{p-1}a=0$, so $a$ would be a zero divisor and that contradicts $F$ being a field. So $\sigma$ is injective. Then if $F$ is finite, $F$ is an automorphism! That gives me a hint, I must find an infinite field of characteristic $p>0$.
The only infinite field in that fashion that I know is $\mathbb{Z}_p(x)=\{\frac{p(x)}{q(x)}| \ p(x),q(x)\in\mathbb{Z}_p[x], q(x)\neq0\}$, so I was trying to prove $\sigma$ is not surjective.
If I take $a=p(x)\in\mathbb{Z}_2[x]\subset\mathbb{Z}_2(x)$ then, $\sigma(p(x))=p(x)^2$ and therefore, $$deg(\sigma(p(x)))=2deg(p(x))$$
So all the images of $\sigma$ would have even degree (for $a\in\mathbb{Z}_p[x]$), but I'm not sure if I can generalize that for every $a\in\mathbb{Z}_p(x)$. I'm not sure about what's the degree of $\frac{p(x)}{q(x)}$ or if I'm going in the right way.
Any help would be awesome. Or any other example for this case (since I only know one case of infinite fields of possitive characteristic. Thanks in advance.
