Smallest Field is Field

Theorem
The ring $$\left({\left\{{0_R, 1_R}\right\}, +, \circ}\right)$$ is the smallest algebraic structure which is a field.

Proof
The null ring, which contains one element, is not a field as it is trivial.

Therefore any field must contain at least two elements.

For $$\left({\left\{{0_R, 1_R}\right\}, +, \circ}\right)$$ to be a field:


 * $$\left({\left\{{0_R, 1_R}\right\}, +}\right)$$ must be an abelian group. This is fulfilled as this is the Parity Group.


 * $$\left({\left\{{0_R, 1_R}\right\}, \circ}\right)$$ must be a commutative division ring. This is fulfilled, as $$\left({\left\{{0_R, 1_R}\right\}^*, \circ}\right) = \left({\left\{{1_R}\right\}, \circ}\right)$$ is the Trivial Group.


 * $$\circ$$ needs to distribute over $$+$$. This follows directly from Ring Product with Zero and the behaviour of the identity element in a group.