= Solution
\b[No such ring exists:] every <finite integral domain is a field>. Here is the proof. For $a\ne0$, the map $L_a:R\to R$ defined by $L_a(x)=ax$ is <injective>, since $ax=ay$ gives $a(x-y)=0$ and therefore $x=y$. An <injective> self-map of a <finite set> is <surjective>, so some $x$ satisfies $ax=1$. Since an <integral domain> is commutative, $xa=1$ as well. Thus every nonzero element has a <multiplicative inverse>, making $R$ a <field>.
The conclusion still holds if an identity was not initially required. Surjectivity first gives an element $e$ with $ae=a$. Then $a(ex-x)=0$ implies $ex=x$ for every $x$, and commutativity makes $e$ a two-sided identity. Applying the preceding argument with $e$ supplies all inverses.
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