Let $R$ be a finite Boolean ring with $1 \neq 0$. Show that $R\cong \mathbb{Z}_2\times \mathbb{Z}_2 \times \dots \times \mathbb{Z}_2$.
This is exercise 2 on p267 in the book abstract algebra of Dummit and Foote. The author gives the hint to use that if $e$ is an idempotent of a ring $R$ with $1$, then
$$R \cong Re \times R(1-e)$$
where $Re$ is a ring with identity $e$ and $R(1-e)$ a ring with identity $1-e$.
My attempt:
If $R = \{0,1\}$, then $R \cong \mathbb{Z}_2$ is trivial.
Hence, we may assume that $\{0,1\}$ is properly contained in $R$. Thus,we can pick an element $x \notin\{0,1\}$. We obtain because every element in a Boolean ring is an idempotent:
$$R \cong Rx \times R(1-x)$$
$Rx$ is a ring with identity $x \neq 0$ and $R(1-x)$ is a ring with identity $1-x \neq 0$. Moreover, $Rx$ and $R(1-e)$ have cardinality strictly smaller than the cardinality of $R$. These rings are also Boolean. Induction yields the result. $\quad \square$
Is this correct?