This map $f(x)=x^{-1}$ is an automorphism if and only if $F$ has at most four elements

Question

I'm trying to solve this question:

If $K$ is a field and $f:K\to K$ is defined by $f(0)=0$ and $f(x)=x^{-1}$ for $x\neq 0$, show $f$ is an automorphism of $K$, if and only if, $K$ has at most four elements.

The converse seems easy, but I'm really stuck in the first implication.

Anyone has an idea?

Thanks

Answer 1

Even easier than my original solution, if $x\in F$ and $x\neq 0$ and $x+1\neq 0$, then $$(x+1)^{-1}=f(x+1)=f(x)+f(1)=x^{-1}+1$$ Multiply both sides by $x(x+1)$ and re-arrange, and you get: $$x^2+x+1=0$$

So $x(x+1)(x^2+x+1)$ has all the elements of your field as roots. So your field cannot have more than four elements.

Answer 2

$f(x+y) = f(x) + f(y)$ implies $x^2 + y^2 + xy = 0$ and thus $x^3 = y^3$ for all non zero $x,y \in \mathbb{F}$ and in particular for $y=1$. Thus $x^3 = 1$ for all non zero $x \in \mathbb{F}$

What can you conclude now?

EDIT: This approach is wrong. Since we have to take care of the case $x+y \neq 0$ as Thomas Andrews points out. I apologize for the erroneous solution.