Let $K$ be a commutative ring, $A$ a commutative $K$-algebra and $S \subseteq K[x_1, \ldots x_n]$ a set of polynomials.
It is well known that the set of simultaneous solutions in $A^n$ of the set of polynomials $S$ is $\operatorname{Hom}(K[x_1, \ldots, x_n]/\langle S \rangle, A)$, where $\langle S \rangle$ is the ideal in $K[x_1, \ldots, x_n]$ generated by $S$, as usual.
Is there a similarly nice way of expressing the set of elements of $A^n$ which satisfy at least one of the polynomial equations in $S$?
If not, what if we restrict to the case where $S \subseteq K[x]$ is the multiplicative submonoid of monic polynomials? In this case the set of solutions in $A$ to at least one equation in $S$ is the integral closure of $K$ in $A$, and I was hoping that there might be a nice way of expressing it.
I'll assume $K$ is an integral domain.
Since it is commutative $A= \prod_l A_l$ is isomorphic to a product of rings whose zero divisors are nilpotent, assume they are integral domains.
Then $Hom(R,A)= \prod_l Hom(R,A_l)$ (homomorphisms of $K$-algebra)
$$S= \{s_1,\ldots,s_M\}, \qquad I= s_1 K[X] \cap \ldots \cap s_M K[X]$$
Let $\phi \in Hom(K[X]/I,A_l)$, $a= \phi(X) \in A_l$.
$\prod_m s_m \in I$ thus $\prod_m s_m(a) = 0$ and $\exists m, s_m(a)=0$.
Conversely if $\exists m, s_m(a) = 0$ then $\forall f \in I, f(a)=0$ so that $\phi: X \mapsto a $ is in $Hom(K[X]/I,A_l)$.
If $A_l$ has nilpotent elements then we can have $\prod_m s_m(a) = 0$ with none of the $s_m(a) = 0$ and I'm not sure if there is a general answer to that problem.