= Solution
Under the unital <ring> convention used elsewhere in this wiki, \b[no such pair exists]. In a two-element ring the elements are $0$ and $1$. Additive cancellation forces $1+1=0$, and the identity axiom forces $1\cdot1=1$. The <distributive property> forces products involving zero to vanish. Every operation is therefore determined, giving the <prime field> $\mathbb F_2$ as the unique possibility up to <ring isomorphism>.
If the word ring allows a <nonunital ring>, there are instead \b[two non-isomorphic examples: $\mathbb F_2$ and the <zero-multiplication rng> on the additive group $\mathbb Z/2\mathbb Z$]. In the second example the unique nonzero element $u$ satisfies $u+u=0$ and $u^2=0$; every product is zero, so associativity and both distributive laws hold. An isomorphism would have to send $1$ to $u$ because it preserves zero and is bijective, but then it would send $1^2=1$ to both $u^2=0$ and $u$, which is impossible. These exhaust the <rngs with two elements>, since $u^2$ can only be $0$ or $u$. Specifying the identity convention is essential for this part.
Back to article page
