I was pondering on affine varieties and ideals. And a curious (and possibly irrelevant) question came to mind, when we talk about for a variety $V\subset \mathbb A^n$, $I(V) =\{ f\in \mathbb C[x_1,...,x_n] \mid f(x) =0 \ \forall x\in V\} $. Now this definition isn't restricted to varieties alone. We can consider $V$ to be an arbitrary set, just like how $V(I)$ can actually be defined on a set of polynomials, other than ideals. So my question is, for an arbitrary set $S\subset \mathbb A^n$, what is the relation between $S$ and $V(I(S))$?
A similar question about coordinate ring can be asked. For an arbitrary set $S$, by considering the restriction of polynomials to $S$, we can see that $\mathbb C[S] \cong \mathbb C[x_1,...,x_n]/I(S)$. Would it make sense to talk about Hilbert's basis theorem /Nullstellensatz on this "coordinate ring"?
I suspect that these questions are irrelevant because we don't really care about arbitrary sets as they aren't as geometric. But I'm curious about it. Thanks in advance.
Fulton's Algebraic Curves lists these properties for general sets, not just algebraic subsets or ideals. In general,
The proof of these facts is just by the definition of the ideal and variety as sets.
As for the second question, the Hilbert basis theorem is just a statement about ideals in a polynomial ring $R[x_1,...,x_n]$, where $R$ is noetherian. Since $I(S)$ is a (radical) ideal for any $S$, you still have the Hilbert basis theorem.
I'm not sure how the Nullstellensatz would play out here. The usefulness of that theorem says that the coordinate ring of an irreducible variety is an integral domain. If $I(S)$ is prime, then we have that $\mathbb C[S]$ is a domain. I don't know of a criterion for this ideal $I(S)$ to be prime, other than the usual definitions.