## Characterization of finite Boolean rings

Let $R$ be a finite Boolean ring with $1 \neq 0$. Prove that $R \cong \mathbb{Z}/(2)^k$ for some $k \geq 1$.

We proceed by induction on the cardinality of $R$. If $|R| = 2$, then $R \cong \mathbb{Z}/(2)$.

Now suppose that for some $k \geq 2$, every Boolean ring with identity of cardinality at most $k$ is isomorphic to $\mathbb{Z}/(2)^n$ for some $n$. Let $R$ be a Boolean ring of cardinality $k+1$. Now $R$ contains an element $e$ not equal to 1 or 0 which is idempotent (since $R$ is Boolean). By a previous theorem, $R \cong Re \times R(1-e)$. Note that if $Re = 0$, then $e = 0$, and likewise if $R(1-e) = 0$, then $e = 1$. Thus $|Re|$ and $|R(1-e)|$ are at most $k$, and we have $R \cong \mathbb{Z}/(2)^n \times \mathbb{Z}/(2)^m$ for some $n$ and $m$. Thus $R \cong \mathbb{Z}/(2)^{n+m}$, and the result holds by induction.