Let $k$ be a unital commutative ring. Let $(X_\alpha)_{\alpha \in J}$ be a family of indeterminates, let $I$ be an ideal in the ring $K[ (X_\alpha)_{\alpha \in J} ],$ and set $A := K[ (X_\alpha)_{\alpha \in J} ]/ I.$ For each $\alpha \in J,$ set $x_\alpha = X_\alpha + I.$
I know that we can define a functor from (commutative, unital) $k$-algebras to Sets by associating a $k$-algebra $R$ with the set of all points $V(R) \subseteq R^J$ that vanish when evaluated on each element of $I.$ Moreover, this functor is representable because we can identify $(r_\alpha) \in V(R)$ with the $k$-algebra map $x_\alpha \mapsto r_\alpha.$ Thus, $\mathrm{Hom}_k(A, - ) \cong V(-).$
What I was confused about is the following. Suppose that $R$ is a $k$-algebra, and suppose that $(s_\alpha) \in V(R).$ Suppose also that there exists a unital subring $S$ of $R$ such that $(s_\alpha) \in S^J$ but such that $S$ is NOT a $k$-algebra. Then how do we speak of $S$-points algebraically? (For example, we could have $k$ is the rationals, $R$ is the reals, and $S$ is the integers.) So, I was wondering, is the correct way to speak of the $S$-points of $V$ algebraically by identifying the $S$-points of $V$ with $\mathrm{Hom}_k(A, k \otimes_{\mathbb{Z}} S),$ where we identify $(s_\alpha) \in V(R) \cap S^J$ with the map $x_\alpha \mapsto 1_k \otimes_{\mathbb{Z}} s_\alpha$?
The correct way to speak about $S$-points is to not speak about them at all. Notice in particular that the condition that some $(s_\alpha)\in S^J$ is in $V(R)$ depends on how $S$ is embedded in a $k$-algebra $R$, since without such an embedding we cannot evaluate elements of $I$ at the $s_\alpha$. So you could embed $S$ into $k$-algebras in two different ways and get different sets of $S$-points. Moreover, this notion of $S$-points also depends on the presentation chosen for the algebra $A$, since you are requiring only the generators $x_\alpha$ to map to $S$ (this is not an issue if $S$ is a $k$-algebra since then if the generators map to a $k$-algebra so does all of $A$).