When is there a one to one correspondence between closed sets in the Zariski topology of the prime spectrum of a ring and radical ideals of that ring? My current understanding: this is always true
Let $A$ be a ring. Let $S$ denote a subset of the ring. Let $V(S)$ be the set of prime ideals of $A$ containing $S$. Define closed sets in $X= \operatorname{Spec} (A)$ to be those of the form $V(S)$.
As I understand, to any radical ideal $I$, we can uniquely associated the closed set $V(I)$. Conversely, to any closed set $V$, we can associate the radical ideal given by the intersection of all elements of $V$.
I can provide proofs I have written to these if this helps, I haven't added them now to avoid drowning my question.
My worry
I haven't been able to find anything on closed sets - radical ideal correspondence, except when we are specifically with algebraic sets in $\mathbb{A}^n_K$ and radical ideals in $K[X_1, ..., X_n]$, such as on this page linked in this question. I understand this is a fundamental result in algebraic geometry (which I am only just beginning to study), but I fail to see why the result may not be more general.
My question
Is my current understanding correct, or have I made a mistake such as forgetting a fringe case ?
 A: You're correct, there is a bijective correspondence between closed subsets of $\operatorname{Spec} A$ and radical ideals $I\subsetneq A$ and this always holds.
This correspondence gets discussed a lot in "classical" algebraic geometry (the typical/original formulation of Hilbert's Nullstellensatz, for instance) because "classical" algebraic geometry can't distinguish between all the different ideals which have the same set of zeroes - $V(x^2)=V(x)$ as affine varieties in $\Bbb A^n$ to any algebraic geometer in the 1800s, for instance. Part of the importance of upgrading to schemes is that now algebraic geometry can distinguish $V(x^2)$ and $V(x)$ because they have different subscheme structures, and that's usually information we keep track of. So it's slightly less important to talk about the underlying closed set.
That doesn't mean we completely forget about that, though - for any closed subset $Y$ in any scheme $X$, there is a unique way to turn this in to a closed subscheme called the reduced induced subscheme structure. Affine-locally this corresponds to exactly what you're talking about here - given a closed subset which can be written as the set of points underlying $V(I)\subset \operatorname{Spec} A$, consider it as the subscheme $V(\sqrt{I})$.
