Let $R$ be a ring. Call a subset $A$ of $R$ diophantine if it is of the form $$ \lbrace x \in R \mid \exists z_1, \ldots, z_n \in R : f(x, z_1, \ldots, z_n) = 0 \rbrace $$ for some $n \in \mathbb{N}$ and some polynomial $f \in \mathbb{Z}[X, Z_1, \ldots, Z_n]$. Call $A$ positive-existential if it is a finite union of finite intersections of diophantine sets.
If $R$ is a domain, then clearly a finite union of diophantine sets is again diophantine. If additionally the fraction field of $R$ does not contain the algebraic closure of its prime field (i.e. $\mathbb{Q}$ if the characteristic is $0$ or $\mathbb{F}_p$ if the characteristic is $p > 0$) then one can use the existence of an anisotropic form in two variables over this field to show that an intersection of diophantine sets is again diophantine. Hence in this case, all positive-existential sets are diophantine.
For a counterexample for the implication positive-existential $\Rightarrow$ diophantine in general, it suffices to note that, for a ring $R$, the diophantine subsets of the product ring $R \times R$ are precisely those of the form $A \times A$ for some diophantine subset $A$ of $R$. This makes it easy to construct positive-existential subsets which are not diophantine by taking the union of two such sets. For example, let $R$ be a field, set $$ A = \lbrace x \in R \mid \exists y \in R : x \cdot y = 1 \rbrace = R\setminus \lbrace 0 \rbrace , $$ then $A \times A \cup \lbrace (0,0) \rbrace$ is positive-existential but not diophantine.
Question: can we find a domain $R$ in which not all positive-existential subsets are already diophantine, i.e. there exists two diophantine subsets of which the intersection is not diophantine? Could $R$ even be a field? I have been trying with $\mathbb{C}[T]$ and $\mathbb{C}(T)$ but have had no luck.