The trivial ring

Detailed Solution

In $R=\{0\}$, the single element $0$ serves as both the additive identity and the multiplicative identity, so $0=1$ in this ring. In any ring with unity where $0=1$, every element $x$ satisfies $x=x\cdot1=x\cdot0=0$, forcing the ring to be trivial; so $\{0\}$ is the *only* ring where the two identities coincide, proving (A). $R$ does satisfy all ring axioms (associativity, distributivity, etc.) trivially with just one element, ruling out (B) and (C). By convention, a field is required to have $0\neq1$ (a nontrivial multiplicative identity distinct from $0$), so the trivial ring is explicitly excluded from being a field, ruling out (D). Thus the correct option is $\boxed{\mathrm{A}}$.