Elementary proof that all fields of four elements are isomorphic to each other

Solution 1:

Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.

Since $a\ne 0$, it has a multiplicative inverse $a^{-1}$. But then we have $$1 + 1 = a a^{-1} + a a^{-1} = (a+a) a^{-1} = 0 a^{-1} = 0$$

Note that this in turn implies that all elements of the field are their own additive inverse, since $$x+x = 1x + 1x = (1+1)x = 0x = 0$$ Or in short, any finite field with an even number of elements must be of characteristic $2$.

Solution 2:

The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.