Suppose $R$ is a Boolean ring ($\forall r \in R, r^2 =r$) which is in fact a field. Show it then must be isomorphic to $\mathbb Z / 2 \mathbb Z$.
I already managed to show that every Boolean ring is commutative and that it has characteristic two. $\forall r \in R$ we have $r+r=0$. By a theorem 6 chapters ahead, we can already say that $R \cong \mathbb Z / 2 \mathbb Z. $ ( classification of finite fields). I am however not really allowed to use this yet. I cannot really construct an isomorphism if I can't describe it. How would I construct this?
To construct an isomorphism I think I need to show that $|R|=2$. So far I am not really allowed to use any characterisation theorems yet.
The only idempotents in a field are $0$ and $1$. Hence, the only field that is also a Boolean ring is the one with two elements, which is $\mathbb{Z}/2\mathbb{Z}$.