when $a\to a^2$ is a permutation of field $\mathbb F$, what is it's characteristic

Question

So in class my professor posed a question.

Let $\mathbb F$ be a field satisfying the condition $a\to a^2$ is a permutation of $\mathbb F$. What is the characteristic of $\mathbb F.$

I was told it should be $2$, but I am not sure why. I believe the solution involved $-1$ and the something about the inverse but I am not completely sure.

Answer 1

If $\text{char}(\Bbb F) \ne 2$ then $1\ne -1$, and the map $a\mapsto a^2$ cannot be injective as $-1\mapsto 1$. On the other hand if $\text{char} (\Bbb F) = 2$ then

$$|\Bbb F^\times|=2k+1$$

is odd, hence the squaring map is an automorphism of this group, in particular it is injective. As $0\mapsto 0$ this shows the map is injective on all of $\Bbb F$. It is also clearly a ring homomorphism by the binomial theorem. So $a\mapsto a^2$ is a field automorphism iff $\text{char}(\Bbb F)=2$ (as we are assuming, per the tag in the op, that we are dealing with finite fields.)