= Solution
Take $\mathcal A$ to be the category of finitely presented <commutative rings> with identity, and put $\mathcal C=\mathcal A^{\mathrm{op}}$. In its presheaf topos, the tautological ring $U(A)=A$ is generic. The additional domain axioms are <coherent sequents>, so they are imposed by a <quotient-theory coverage> on $\mathcal C$.
Concretely, declare the zero ring covered by the empty family. For every finitely presented $A$ and elements $a,b\in A$ with $ab=0$, declare the two opposite quotient arrows associated with
$$
\boxed{A\longrightarrow A/(a),\qquad A\longrightarrow A/(b)}
$$
to be a covering family at $A$. These <quotient rings> are finitely presented. <Pullback> and transitivity generate a Grothendieck topology $J$ from these families. The empty cover forbids $0=1$; the two quotient covers make every zero product locally have a zero factor. Conversely, any internal integral domain satisfies exactly the continuity conditions prescribed by these generating covers. Thus
$$
\boxed{\mathbf{Sh}(\mathcal A^{\mathrm{op}},J)\text{ classifies integral domains},}
$$
and its generic domain is the associated sheaf $K=a_JU$, with the ring operations transported through the left-exact sheaf reflector.
This coverage is \b[not standard], meaning not all <representables> are sheaves; in modern terminology it is not subcanonical. For an explicit obstruction, use $A=\mathbb Z/4\mathbb Z$ and $a=b=2$. The two quotient arrows are the same map $A\to\mathbb Z/2\mathbb Z$, so their generated sieve is a singleton cover. Consider the <representable> on $\mathcal C$ corresponding to $R=\mathbb Z[t]$:
$$
yR(A)=\operatorname{Hom}_{\mathrm{Ring}}(R,A).
$$
Its distinct sections $t\mapsto0$ and $t\mapsto2$ become equal after restriction to $\mathbb Z/2\mathbb Z$. Hence this <representable> is not even separated for $J$. The <nilpotent element> has to disappear in the generic domain, which explains this failure of standardness.
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