If $R$ is a finite boolean ring then $R \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \dots \times \mathbb{Z}_2$.
So this makes sense for a lot for some obvious reasons, and I feel like I could exhibit an isomorphism by defining a map by having the $ith$ element of $R$ make the $ith$ tuple a one instead of a zero, etc. It seems strange though that this isomorphism suggests a certain linear independence of the elements of $R$, can somebody give me some insight into this phenomonon? Thanks!
Hint If $R$ is a boleean ring, then $\{ 0,1 \}$ is a subring which is isomorphic to $\mathbb Z_2$. Show that $R$ becomes a vector space over $\mathbb Z_2$.
Simpler approach If $R \neq \{ 0,1 \}$ then for each $x \in R \backslash \{ 0, 1 \}$ you have $$R= Rx \oplus R(1-x)$$
Now each of these are boolean rings (the units are $x$ and $1-x$ respectively) which are smaller than $R$, use strong induction.